Solutions to PULLOUT WORKSHEETS FOR CLASS VIII SARASWATI HOUSE PVT. LTD.

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1 Solutions to Mathematics PULLOUT WORKSHEETS FOR CLASS VIII Since 90 SARASWATI HOUSE PVT. LTD. (An ISO 900:008 Company) EDUCATIONAL PUBLISHERS 9, Daryaganj, Near Telephone Office, New Delhi-000 Ph: 6600 (00 lines), 80 Fax: Website: Branches Ahmedabad: 988 Bengaluru: (080) Chandigarh: (0) 688 Chennai: (0) 0 Jaipur: (0) 0060 Kochi: (08) 988 Lucknow: (0) 06 Mumbai: (0) 80 Patna: (06) 00

2 CONTENTS. Rational Numbers Worksheets ( to ).... Linear Equations in One Variable Worksheets (8 to ) Understanding Quadrilaterals Worksheets ( to 0) Practical Geometry Worksheets ( to 6) Data Handling Worksheets ( to ) Squares and Square Roots Worksheets ( to 0) Cubes and Cube Roots Worksheets ( to ) Comparing Quantities Worksheets (8 to ) Algebraic Expressions and Identities Worksheets ( to 60)... 06

3 0. Visualising Solid Shapes Worksheets (6 to 6).... Mensuration Worksheets (68 to ).... Exponents and Powers Worksheets (6 to 8).... Direct and Inverse Proportion Worksheets (8 to 88) Factorization Worksheets (89 to 9) Introduction to Graphs Worksheets (96 to 0) Playing with Numbers Worksheets (0 to 0)... PRACTICE PAPERS ( to )... 8

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5 Solution to PULLOUT WORKSHEETS AND PRACTICE PAPERS [Summative Assessments]

6 Chapter RATIONAL NUMBERS WORKSHEET. (D) Negative of ( ) [ ( a) a]. (C)Multiplicative identity for any rational number.. (B) Reciprocal of ( ).. (D) Reciprocal of.. (A) The given property is commutativity under multiplication. 6. (B) 9 6 ( ) (B)Since addition is associative for rational numbers. Therefore, for rational numbers a, b and c, a + (b + c) (a + b) + c. 8. (A) Since is a negative number, so it is on the left of 0 on the number line. 9. (D) Reciprocal of zero is not defined. 0. (D) Rational numbers between and R A T I O N A L N U M B E R S. (C) 0 must be negative. between and 0. and 0.8, lie 00 0., , is between and 6.. (B) The sum, subtraction and multiplication of two rational numbers is always a rational number.. (D) is not defined and so it is not a 0 rational number.. (A)Area of rectangle Length Breadth. (C) Length Area 8 6 m. 6. (A) Additive inverse of 9 cm. Breadth ( ) 6 ( )

7 WORKSHEET. (i) (ii) 0. (i) 0 < 0 (ii) > <..... (i) Let us first find the LCM of 6,, and LCM of 6,, and 0 6,,,,,, Now,,,,,,, < 6 < < < 0 > 0 0 < 6 0 < 0 < 0 < 0 0 or < < 6 < <. (ii) LCM of,, 0 and 0 Now, < < 9 < or 6 0 < 0 < 9 0 < 0 8 < < 0 <..(i) LHS RHS Clearly, LHS RHS Hence verified. (ii) LHS ( 8) RHS ( 8) Clearly, LHS RHS Hence verified. 6 M A T H E M A T I C S VIII

8 . Yes. + + Multiplicative inverse of 0 ( ) Let us is represented on the number line by divisioins Therefore, 8 is represented by 8 i.e., 8 divisions. Thus, the point A represents 8 on the number line.. There are infinitely many rational numbers less than. Five of them are:,, 0,,. 8. (i) (ii) ( ) R A T I O N A L N U M B E R S 8 0 (iii) 9. (i) + (ii) 9 9 ( ) [ ( a) a] [ LCM (,, ) ] [ (a) a] + 9 [ LCM (, 9) 9] WORKSHEET. (i) LHS ( 8 ) ( 8 ) ( ) ( ) ( )

9 RHS ( ) () { } () { } () () As, LHS RHS, the given rational numbers satisfy the property of multiplication. (ii) LHS ( ) 8 ( ) RHS ( ) { } ( ) ( ) 8 ( ) 8 8 As LHS RHS, the given rational numbers satisfy the property of multiplication.. Substitute a, b and c in LHS and RHS separately. L.H.S a (b c) ( ) ( ) 8. RHS (a b) c ( ) 8. Hence, a (b c) (a b) c verified.. (i) Reciprocal of Reciprocal of (ii) Reciprocal of (iii) Reciprocal of (iv) Reciprocal of. Reciprocal of....(i) (ii) ( ). ( ) () M A T H E M A T I C S VIII

10 . (iii) And ( ) ! 9 or! ! (i) Substituting x and y 9 in x + y, we get x + y [ LCM (, 9) 6] 6 () Substituting x and y 9 in y + x, we get y + x ( ) () From equations () and (), we have x + y y + x R A T I O N A L N U M B E R S (ii) Substituting x 8 and y in x + y, we get x + y ( ) 8 0 [ LCM (, 8) 0] 6 0 Substituting x 8 y + x, we get y + x ( ) () and y in 0 () From equations () and (), we have x + y y + x.. Total number of students 6. Number of students liking cricket of 6 6 Number of students liking football 6 of Total number of books 80 Number of books on literature 8 of

11 Number of books on fiction of A 6 m 6 + m 6 m. 8 + l m m m Area of a rectangle is given by A l b b A l m. 89. (i) So, the width of the rectangle is WORKSHEET Point A represents line. (ii) 89 m. on the number Point B represents on the number line.. (i) ( 6). 8 6 (ii) [ LCM (,, 6) 6]. Perimeter [ LCM (,, 8) 8] m As we know that additive inverse of a a (i) Additive inverse of And (ii) Additive inverse of Substituting x 6 and y 9 in x + y, we get x + y () [ LCM (6, 9) ] Substituting x 6 and y 9 in y + x, we get 0 M A T H E M A T I C S VIII

12 y + x () From equations () and (), we obtain x + y y + x. 6. Let the other rational number be x. Then, x + ( ) x + or x (Transposing + 9 to RHS) Thus, the other rational number is 9.. Let the other rational number be y. 8. Then, y + ( ) y or y 8 (Transposing 6 +. to RHS) Thus, the other rational number is. 88. () ( ) R A T I O N A L N U M B E R S 9. LHS RHS Since LHS RHS. Therefore, verified ( ) ( ) LHS ( ) ( ) + ( ) RHS ( ) 8 9 ( 8) ( 8) Clearly, LHS RHS. 8 Hence, ( ) verified.. Additive inverse of 9 9 ( ) ( ) 8. is

13 WORKSHEET. Substituting x x + y, we get and y M A T H E M A T I C S in x + y () [ LCM (, ) 60] Substituting x and y in y + x, we get y + x () From equations () and (), x + y y + x is verified.. Other rational number. 9 ( ) ( ) ( ) ( ) ().. LHS 8 0 RHS 8 0 ( ) ( ) Since, LHS RHS. Therefore, is verified ( ) LHS ( ) RHS ( ) 8 9 ( 8) ( 8) Since LHS RHS, so the given statement is proved. ` m ` Cost of Cost of m cloth ` m 9 m 66 + ` 6 m cloth ` VIII

14 ` 6 9 ` ` ` 8. Since any rational numbers between 8 and are:, 6,, and Therefore, rational numbers between 8 and are:, 6,, 9. Sum of and Product of and Now, the required quotient 6 0. LCM of, 0 and 0. and R A T I O N A L N U M B E R S Now, and 0 < 8 < < 8 0 < 0 0 or 0 < <. 0 0 ( 0 ) ( 0 ) ( ). ( 0 0 ) ( 0 ) 0 ( 0 ) ( ) ( 0) ( ) Total expenditure Expenditure on shopping + Expenditure on groceries Thus, Reema spent ` in all.

15 WORKSHEET 6. Let the other number be x. Then, x + 9 x 9 (Transposing 9 9 So, the required number is 9.. Let y should be added. Then, to RHS) 9 y or y or y So, should be added. [ LCM (8, 9) ] 0 or. Let p should be subtracted. Then, or So, 8 p p or 8 8 p or 8 should be subtracted.. Let A should be added. Then, p p A 0 9 So, should be added.. Let B should be added. Then, B + ( + + ) or B + ( ) 0 or B + 0 B 0 So, 9 0 should be added Let x should be subtracted. Then, or or or ( ) x x x x or x. 9 0 x A + + So, should be subtracted. M A T H E M A T I C S VIII

16 . Let y should be added. Then, y y So, should be added. 8. Let M should be added. Then, M + ( ) or M or M + 60 M 60 or () ( 8 ) ( ) + ( ) ( 6 ) ( ) + ( 8 ) ( ) ( ) ( ) ( ) ( ) +( ) ( ) So, 60 should be added. 9. Let x should be subtracted. Then, or or So, ( ) x x x x or should be subtracted. 0 x. (i) WORKSHEET R A T I O N A L N U M B E R S

17 (ii) x or ( ) x.. Let you should be multiplied by x. Then, x x 8 8. ( ) 6. Let the other number be y. Then, y 8 y So, the other number is.. Let the other number be M. Then, M M So, the other number is.. Let the required number be x. Then, 0 or 0 x x or x 6 or So, 0 should be divided by 6. LCM of and 8 Sum of 6 Difference of 6 and and 6 8 Now, the required result Amount of iron filings for cartons kilos Amount of iron filings for carton kilos 9 or kilos So, the required amount of iron filings is 9 kilos. 6 M A T H E M A T I C S VIII

18 8. The point A represents 8 B represents 8 and the point on the number line. 9. The required three rational numbers are:,, ( 6 ) ( ) 9 ( ) ( ) ( ) + ( ). (i) (ii) [ LCM (, ) 86] R A T I O N A L N U M B E R S

19 Chapter LINEAR EQUATIONS IN ONE VARIABLE WORKSHEET8. (C) x x +. y. (A). y... y.6.. (D) 8x + 6 (x ) + or 8x + 6 x 0 + or 8x x or x x.. (C) m m 8 or 8 m m or 8 m m. z. (B) z ( + ) or z z + 0 z z 0 + or z or z z. x 6. (B) + x x x or (Transposing) or x x 6 0 x 9 or or x i.e., x 0.. (C) 0.(0t 9) 0.(t ) or.t. t. or.. t.t (Transposing) or 0.9. t or t i.e., t 0.6. y 8. (A) y 9 or 9y y 0 or y 0 y 0. x 9. (D) x or or or x x + x x+ x + x x + x or x 0 or x (B) Substituting m m, we get or ( ) ( ) or Which is true. + in m or So, the given equation is satisfied by m. 8 M A T H E M A T I C S VIII

20 . (B) y or y. y. (A) Let x should be added. Then, x + x (B) Let the required number be y. Product of y and y y. Sum of this product and y + According to the given condition, or y + 60 y 60 y 60 y (C) Let present age of Anand x years Then, present age of his father x years According to the given condition, (x + ) + (x + ) 8 x 8 8 or x 0 0 years. And x 0 0 years.. (B) Let the number of boys be x and the number of girls be x. So, x x 0 x 0 0 x (D) Let the numerator of the original number be x. Then its denominator x + 6. So x + x + 6+ or x + 0 x + 0 x And, x Hence, the original number is.. (D) Let ten s digit x. Then units s digit x + So [0x + x + ] + [0(x + ) + x] 99 or x + 99 x And, x + + The original number 0x + x +.. (i) x 6 or WORKSHEET9 x 6 x. (ii) or x 60 x 60 (Dividing throughout by ) (Multiplying throughout by ) x 0. L I N E A R E Q U A T I O N S I N O N E V A R I... 9

21 (iii) or x x (Multiplying throughout by ) x. (iv) x or x + + (Adding to both sides) x. OR (i) x + or x + (Subtracting from both sides) or x or x ( ) ( ) (Multiplying throughout by ) x. (ii) x + or x + (Subtracting from both sides) or x or x. (Dividing throughout by ) (iii) x or + x (Multiplying throughout by ) or + x (Subtracting from both sides) x 0. (iv) (x + ) or x + or x + (Dividing throughout by ) or x + (Subtracting from both sides) x.. (i) x 0 x x+ 0 x or 9 0 Multiplying both sides by 0, we get x + 0 x 90 or x 90 0 (Transposing 0 to RHS) or x 60 Dividing both sides by, we get x 0. (ii) 8(x + 0).(x + 8) or 8x + 0 x + or 8x x 0 (On transposing) or x 08 Dividing both sides by, we get x 6.6. OR (i) x x Multiplying both sides by, we get 9x + x On transposing, we get 9x x + or x 6 Dividing both sides by, we get x. (ii) y + y y + y + 0 M A T H E M A T I C S VIII

22 Cross-multiplying, we have (y + ) (y + ) (y ) (y + ) or y + y + y + y + y y or y y + y + y y + y or 6y 8 Dividing both sides by 6, we get. x + x 9 y. Cross-multiplying, we have 9(x + ) (x ) or 8x + 9 x 0 or 8x x 0 9 or x 9 Dividing both sides by, we get Verification: LHS x + x 8 + x 9. 9 ( ) 9 ( ) + (Substituting x 9 ) RHS Hence verified. 6. Let Mr. Sharma s son s age now x years Then Mr. Sharma s age now x years After years, Mr. Sharma s age (x + ) years 9 years ago, The son s age (x 9) years According to the given condition, x + (x 9) or x + x 6 or 0 x or 0 x (Dividing both sides by ) x 0 0 years. Thus, Mr. Sharma s age is of 0 years and his son is of 0 years now. OR Let breadth b of the rectangle be x, i.e., b x Then, length l + b + x Area A l b ( + x) x New length l l 0 + x 0 x New breadth b b x New area A l b (x ) (x ) But, A A 08 (x ) (x ) ( + x)x 08 or x 6x + 9 x + x 08 or x 6x x x 08 9 or x Dividing both sides by, we get x 9 b 9 m L I N E A R E Q U A T I O N S I N O N E V A R I... i.e.,

23 Further, l + x m Thus, length 6 m and breadth 9 m.. Let weight of box B x kg Then weight of box A x + x + And weight of box C x + x + x + kg x + kg Total weight of the three boxes 9 or x + + x + x + 9 or x + + x + x + 6 or x + 6 or x 6 or x Weight of box B x kg Weight of box A x + kg. And weight of box C x + kg (i) x + ( x ) 0 or x + x 0 or x+ x+ 0 or x + x + 0 or x + 0 or x x. (ii) or a + + a + ( a+ ) + ( a+ ) ( a+ )( a+ ) a + 0 a + 0 or (a + ) (a + 0) + (a + ) (a + 0) (a + ) (a + ) or a + 0a + a a + 0a + a + 0 a + a + a + or a + a a 6a 0 or a 6 or a (iii) ( ) x 0. x x Multiplying both sides by, we get 0.x 0 ( 0. x ) 6x Again multiplying both sides by, we get.x 0 ( x ) 0. 0x or.x 0x + 0 0x or 0 0x.x + 0x or 0 9.x x M A T H E M A T I C S VIII

24 WORKSHEET0. (i) x + x + or x + (x + ) or x + x + or x x or x x. (ii) m m (m ) m or m m or 0m m 0. OR p + (i) p + 6p or or or p+ p+ 6p p + 8p 8p p or p p. (ii) ( y ) y 0 or (y ) 0y or y 0y or 0y y or y y Thus, y.. (i) Let the number be x. Twice this number x x and times this number x x According to given condition, we obtain x + x 0 This is the required equation. Let us solve it. x + x 0 or 6x 0 x 0 6 Thus, the number is. (ii) Let the cost of a chair be ` y Then the cost of a table ` (y + 0) According to given condition, we obtain (y + 0) + y 0 This is the required equation. Let us solve it. (y + 0) + y 0 y y 0 or y or y y Cost of chair is `60 and cost of table is ` 80. OR (i) Let three consecutive number be x, x + and x +. Sum of these numbers x + (x + ) + (x + ) This is the required equation. Let us solve it. x + (x + ) + (x + ) or x + or x or x L I N E A R E Q U A T I O N S I N O N E V A R I... 9

25 x and x Hence, the required numbers are 9, 8 and. (ii) Let the number be y. Its one-third y y According to given condition, we obtain y This is the required equation. Let us solve it. y y or + or y Thus, the required number is.. (i) Let an odd number be x Then the next odd number x + And again the next odd number x + + x + Sum of these three numbers 6 or x + x + + x + 6 or x or x 6 6 or x 9 x And x Therefore, the required numbers are 9, and. (ii) Let the numbers be y and 8y. Their sum i.e., y + 8y or y or y y.(i) and 8y 8 Therefore, the required numbers are and. x + x + x 0 xx ( + ) + ( x) x ( x)( x+ ) or 0 ( x )( x+ ) x or x + x + x x x + 0 or x x x + 0 or x + 0 or x i.e., x. (ii) 6a + a + or (6a + ) (a + ) or 8a + a + 0 or 8a a 0 or a a. (iii) x Dividing both sides by, we get. (i) x x 6. or or ( a ) a a a6a 6 a + a a + 9 Multiplying both sides by, we get a a + 8 or 8 a + a M A T H E M A T I C S VIII

26 or a or a a. (ii) or or (x+ ) ( x+ ) x x+ x x x + x By cross multiplying, we have (x + ) x or 6x + 69 x or x x. 6. Let side of the square be x metres. Then length of rectangles, l (x + ) m And breadth of rectangle, b (x ) m Perimeter of the rectangle (l + b) (x + + x ) x m. According to given perimeter, we have x 6 or x 9. Therefore, the side of the square is 9 m. OR Let the number of total children in the group be y. Then number of children playing in the y park Number of remaining children y y y Number of children busy in studies y y 8 Number of children doing yoga 9 Consequently, we obtain y y y or y y+ y or y y y + 9 or y y or 9 or y Thus, the number of total children in the group is.. (i) x WORKSHEET (x + ) 8(x + ) or x (x + ) 8x + Multiplying both sides by, we get or or 6x x x + x x 8x or 8 x or 6 x i.e., x 6. (ii) x + 8 x x Multiplying both sides by 8, we get x + 8x x or + x + x or 6 9x 6 or 9 x or x x. L I N E A R E Q U A T I O N S I N O N E V A R I...

27 (iii) ( x 6) or x (x + 6) x + Multiplying both sides by, we get or or x + 6x x + x or 8x 8 x i.e., x 8. (iv) or y + 0 y + 0 y + 8 or y 8 y y 8 0( y 8) Multiplying both sides by, we get (y + 0) (y 8) or 8y + 60 y 0 or y 8y or 00 y or 00 y i.e., y 00. (v) x ( x+ )( x+ ) 6 x + Multiplying both sides by (x + ), we get x (x + ) (x + ) 6(x + ) or x (x + x + x + ) 6(x + ) or x x x 0x + 6 or 6 0x + x or 8 x or 8 x i.e., x (vi) x + ( x ) 8. x x + Multiplying both sides by, we get x + 9 (x 8x + ) x + or x + 9 x + 8x x + or or or 9 x 8x x x i.e., x.. (i) Let the number be x. Thrice x x According to given condition, we have x 60 This is the required equation. Let us solve this equation. x 60 x 60 or (Dividing both sides by ) or x 0. Therefore, 0 is the required number. (ii) Let the number be y. Subtracting 60 from y, we get y 60. According to given condition, we have y 60 This is the required equation. Let us solve this equation. y 60 or y Therefore, is the required number. 6 M A T H E M A T I C S VIII

28 (iii) Let the numbers be z and 8z. According to given condition, we have z + 8z 0 This is the required equation. Let us solve this equation. z + 8z 0 or z 0 or z 0 0 z 0 0 and 8z Therefore, 0 and 80 are the required numbers. OR (i) Let present age of Sumi s brother x years. Then present age of Sumi (x + 9) years. After 0 years, age of Sumi (x ) years (x + 9) years. 0 years ago, age of Sumi s brother (x 0) years. According to given condition, we have x + 9 (x 0) or x + 9 x 0 or x x or 9 x x Therefore, present age of Sumi is 8 years and present age of her brother is 9 years. (ii) Let Mintu s present age be x years and Shanu s present age be x years. Four years later, Mintu s age (x + ) years. and, Shanu s age (x + ) years. According to given condition, we have x + x + Cross-multiplying, we get or x + 0x + 6 or x 0x 6 or x x 0 and x 8 Therefore, the age of Mintu is 0 years and the age of Shanu is 8 years.. (i) ( x) 8 or 6 x 8 or 6 8 x or x x. (ii) x + (x + ) + (x + ) 0 or x + x + + x + 0 or x 0 or x 00 or x 00 x. (iii) x + or x + (Multiplying both sides by ) or x 8 x 8.. Let number of ten rupee notes be x. Then number of five rupee notes x +. Amount by ten rupee notes ` (x 0) ` 0x L I N E A R E Q U A T I O N S I N O N E V A R I...

29 Amount by five rupee notes Sum of these amounts ` {(x + ) } ` (x + ) ` 0x + `( x + ) ` (x + ) This is given to be ` 9. x + 9 or x 9 80 or x 80 x + +. Thus, Rohan has notes of ten rupees and notes of five rupees. WORKSHEET. (i) x 0 or x 0 x. (ii) y 6 or y 6 y. (iii).x. or x.. x. or x x (iv) or x ( ) x 0. (v) p or p or p p.. (i) x 0 0 or x x 0. (ii) p 0 or p 0 + p. (iii) a 0 0 or a a 0. (iv) x 6 or 6 + x x. (v) x or x + x. or x 6 +.(i) Let the number be x. Then x (ii) Let the number by y. Then y 0. (iii) Let the number be p. Then p 6. (iv) Let Romi s age be q years. Then q 00. (v) Let the number be r. Then 8r (i) x + x + Cross-multiplying, we get x + 8x + 8 or 8 8x x or x or x i.e., x. Check: Numerator of LHS of given equation ( ) M A T H E M A T I C S VIII

30 69 + And its denominator ( ) LHS (ii) 6 + RHS. y + y Cross-multiplying, we get y + (y + ) or y + y + or y y y. Check: Numerator of LHS ( ) + Denominator of LHS + LHS RHS.. (i) Let the number be x. Seven times of x x x It is given to be 9. x 9 x 9 or (Dividing both sides by ) x. Thus, is the required number. (ii) Let the number be y. One and half + + times y y y This is given to be 00. or y 00 y 00 (Multiplying both sides by ) or y Thus, 00 is the required number. OR Let the larger part be ` x. Then the smaller part ` (00 x) 0% of x 0 00 x x 0 8 8% of (00 x) (00 x) 00 0 x According to given condition, we have x 0 ( 0 x ) 60 x or x 60 or x 0 + x or x+ x 0 80 or 9x or x 9 x 000 L I N E A R E Q U A T I O N S I N O N E V A R I... 9

31 00 x So, the larger part is ` 000 and the smaller part is ` Let length of the rectangle be x m. Then its breadth (x 0) m. So the perimeter (length + breadth) (x + x 0) (x 00) m. But the perimeter is given to be 80 m x or x or x 80 9 m Therefore, length 9 m. And breadth x m. OR Let one multiple of be x. Then the next one + x. Sum of these two multiples x + + x x + But this is given to be x + or x 0 or x 0 + x + 0. Therefore, and 0 are the two required multiples. x x. (i) WORKSHEET + 0 Multiplying both sides by LCM (,, ), we get x (x ) or x x or + 0 x x or 6x 0 M A T H E M A T I C S or 6 x i.e. x. (ii) ( ) (x ) x or ( x ) x 8 x + x x x Multiplying both sides by LCM(, ) 6, we get (x ) (x ) (x ) or 6x x + x or x x or + x + x 0 or 0 0x or 0 x or x i.e., x. (iii) m or or or m m( m) m ( m ) m + m m + 8 m or m (iv) or or Thus, m. x + x + x ( ) x x + x x+ x + x + x + x + VIII

32 Multiplying both sides by LCM (, ), we get x + 9 x x + or x or x (v) Thus, x p p + + p Multiplying both sides by LCM (,, ), we get p 6 p p 8 or 8 p or p or p i.e., p x 9 (vi) x + 9 Putting x y, we get y 9 y + 9. Cross-multiplying, we get 9y 8 y or 9y + y 8 or y 6 or y 6 x ( y x ) or x ± or x ±. (vii) or or x + x + x + x x x+ 8 x x 6 x or x + or x 60 9 or x 9 Thus, x.. (i) Let the number be x. 0 less than x is x 0 This is given to be 80 x 0 80 This is the required equation. Let us solve it x 0 80 or x or x 0 Therefore, the required number is 0. (ii) Let Gaurav s age be y years Then Preeti s age (y ) years But this is given to be 8 years. y 8 This is the required equation. Let us solve it. y 8 or y 8 + Therefore, Gaurav s age is years. (iii) Let the number be z. 0 0% of z 00 z z This is given to be 0. z 0 This is the required equation. Let us solve it. z 0 or z 0 00 Therefore, the required number is 00. L I N E A R E Q U A T I O N S I N O N E V A R I...

33 . Let one of the two numbers be x. Then the other one x + 6. Sum of these two numbers x + x + 6 x + 6 This is given to be 60. x or x 60 6 or x x Hence, the two numbers are and 8. OR Let larger part be y toffees. Then smaller part (y ) toffees. Sum of these two parts y + y (y ) toffees Since the total number of toffees is y or y + 8 or y 8 y 0 Therefore, the larger part is toffees and the smaller part is 0 toffees.. (i) Let the number be x. Adding to 8 times x, we get 8x +. So, the required equation is 8x (ii) Let the number be y. Multiplying y by 9, we get 9y So, the required equation is 9y. (iii) Let the number be z. Subtracting 0 from z, we get z 0 So, the required equation is z OR (i) Let a number be x. 0 of x 0 x x. 0 So, the required equation is x. 0 (ii) Let the sum be y. 0 0% of y 00 y y y. So, the required equation is y 00. (iii) Let the number be z. and half + + of z z So, the required equation is z 0. WORKSHEET. (i) x or x + or x or x. (ii) x (Adding to both sides) (Dividing both sides by ) M A T H E M A T I C S VIII

34 or x (Multiplying both sides by ) or x + x. (Dividing both sides by ) (iii) x + (x ) or 0 0 x or x i.e., x.. Let the number be x. One-fourth x x. Since x is 8 more than. or x + x or x + or x x 0 0. (iv) x x + x 8 Multiplying both sides by LCM (,, ), we get 6(x ) (x + ) (x 8) or x 8 x 9x or 8 + 9x x + x or x i.e., x. (v) ( ) (x ) x x x or 8 x + x x (x ) Multiplying both sides by LCM (, ) 6, we get (x ) (x ) (x ) or 6x x + x or + + x 6x + x or 0 0x x + 8 or x Thus, the required number is.. Let digit in units s place of the given number be x. Then digit in ten s place 9 x. Given number 0(9 x) + x. A number obtained by interchaning its digits 0x + (9 x). This obtained number Given number [0x + (9 x)] [0(9 x) + x] 0x + 9 x (90 0x + x) 0x + 9 x x x 8x 8 This is given to be. 8x 8 or 8x or x i.e., Digit in unit s place 6 And digit in ten s place 9 x 9 6. Now, given number Thus, the required number is 6. L I N E A R E Q U A T I O N S I N O N E V A R I...

35 OR Let the numerator of the original rational number be x. Then denominator will be x + 6. x So, the original rational number. x + 6 Numerator of new rational number x + 9. And its denominator (x + 6) x +. So, the new rational number x + 9. x + This is given to be. x + 9 x + Cross-multiplying, we get x + x + 8 Transposing x to LHS and to RHS, we get x x 8 or x or x x Therefore, the rational number is.. Let Vedant s salary before the increase ` x. Increase in the salary 0% of x 0 00 x ` 0 x. Salary after the increase x x ` x + ` ` 0 ( x + ) ` x 0 This is given to ` 800. or x 0 0 x (Multiplying both sides by 0 ) or x So, Vedant s salary before the increase was ` m +. 66m Cross-multiplying, we get 9m + m or 9m + m or m 0 or 0 m 0 i.e., m (i) or or or or b ( b)( b+ ) b ( b + b b ) b ( b + b ) b b b+ b + M A T H E M A T I C S VIII

36 Cross-multiplying, we get b + 0 or b 0 or b or b. y (ii) + y y + y Cross-multiplying, we get (y )( + y) ( + y)(y ) or 8y + y 8 0y y + y y or 8y + y 8 8y + y or 8y + y 8y y + 8 or 0y or y 0 or y. L I N E A R E Q U A T I O N S I N O N E V A R I...

37 Chapter UNDERSTANDING QUADRILATERALS WORKSHEET. (A) As we know that the sum of angles of a polygon (n ) 80, n number of sides. So, the sum of the angles of a quadrilateral ( ) 80 [ n ] (D) By the definition, a concave polygon has any angle greater than 80 (i.e., reflex angle).. (C) A quadrilateral has sides, angles (or vertices) and diagonals.. (C) Sum of all exterior angles of a polygon 60. It is constant.. (D) Sum of all interior angles of a polygon (n ) 80. (It is a formula) 6. (A) As we know that each exterior angle of a polygon 60, n n number of sides. 60 n Measureof each exteriorangle By putting one by one obtains we find that with angle measure, number of sides is in whole number otherwise it is in decimals which is not possible. A regular polygon is possible with each exterior angle of.. (B) Sum of exterior angles 60 x x (C) The diagonals of a rhombus (or a square) bisect each other at (C) AB ydc and AC is a transversal. DCA BAC 0. In ODC, OCD + ODC + COD (From figure, COD 90 ) (A) Sum of two adjacent angles of a parallelogram 80 x + x 80 ( Given ratio : ) x 80 x 80 6 Required angle (B) Diagonals of a rhombus intersect each other at right angle, i.e., 90.. (B) Consider QPS + RSP PQ y SR.. (D) Diagonals of a rectangle are equal to each other. 6 M A T H E M A T I C S VIII

38 . (A) A rectangle is a kind of a parallelogram.. (A) A rhombus has all sides of equal length. 6. (A) A rectangle has four sides and four right angles so it is a convex quadrilateral.. (B) As we know that diagonals of a square are equal and bisect each other at right angle. So, from figure HJ IK or HO IO x + x + x x x. 8. (D) Number of side of a polygon 60 Measureof an exteriorangle Now, in quadrilateral ABCD, ABC + BCD + CDA + DA 60 (Angle sum property) 0 + x x 60 0 x 0. (ii) Using angle sum property in a quadrilateral, x 60 x x 80.. Let ABCD be a parallelogram in which A 0. Two adjacent angles of a parallelogram are supplementary. WORKSHEET 6. (i) Concave polygon (ii) Concave polygon (iii) Concave polygon (iv) Convex polygon. Note: A convex polygon has the whole parts of each diagonal in its interior region.. (i) Octagon (ii) Decagon. (i) From figure, EDC + ADC 80 (Linear pair) 90 + ADC 80 ADC (i) A + B B 80 B Also, we know that opposite angles of a parallelogram are equal. A C and B D C 0 and D 0 Thus, other angles of the parallelogram are 0, 0 and 0.. Let one angle of the parallelogram be x. According to question, Two adjacent angles are equal so another angle also be x. U N D E R S T A N D I N G Q U A D R I L A T E R A L S

39 Since, the sum of two adjacent angles are supplementary. x + x 80 x 80º x As we know that a parallelogram whose each angle is of measure 90 is a rectangle. 6. As we know that a quadrilateral whose all sides are equal is called rhombus. But the quadrilateral has one of the angle is 90 so adjacent angle of the rhombus is also 90. Thus, the quadrilateral is a square. In other words, a rhombus with one of the angle of 90 is called a square.. The given three angles of a quadrilateral are, and 0. Let fourth angle be x. Using Angle sum property, x 60 x Let ABCD be a rhombus in which AC 6 cm and BD 8 cm. AO AC 6 cm BO BD 8 cm In right triangle AOB, AB AO + OB AB cm Thus, required side of the rhombus is cm. 9. Let the length of a rectangle x and breadth x. Given perimeter 90 cm. We know that perimeter of the rectangle (l + b) (x + x) 90 9x 90 x 90 8 l x cm b x 0 cm. 0. Suppose two adjacent angles of a parallelogram are x and x respectively. Since adjacent angles are supplementary. x + x 80 9x 80 x x 0 00 and x 0 80 Thus, the four angles of the parallelogram are 00, 80, 00 and 80.. Let one adjacent angle of an angle of measure 0 be x in the given parallelogram. So x [ Two adjacent angles are supplementary] x Also opposite angles of these angles are 0 and 60. Thus remaining angles are 60, 0 and M A T H E M A T I C S VIII

40 WORKSHEET. (i) Convex quadrilateral (ii) Convex quadrilateral. (iii) Concave quadrilateral. [A convex quadrilateral has all angles less than 80 but a concave quadrilateral has any angle more than 80.]. (i) Pentagon (ii) Heptagon. No, because sum of the four angles Let, fourth angle of the quadrilateral be x. Three given angles are, and 0. Using Angle sum property, we have x 60 + x 60 x 60.. Let the common factor of the angles be x. So, the four angles of the quadrilateral are x, x, x and 9x. Using Angle sum property, x + x + x + 9x 60 x 60 x 60 Thus, four angles are:,, and 9 i.e.,,, 0 and. 6. Let each equal angle of the quadrilateral be x. Measure of one given-angle 9 We know that sum of the angles 60 x + x + x x 60 9 x 6 89 So, the three equal angles are 89, 89 and 89.. Suppose the common factor be x. So remaining three angles are x, x and x. Mean of these angles 6 x+ x+ x 6 x 6 x 6 6 Therefore, the three angles are 6, 6 and 6 i.e.,, 8 and So, fourth angle 60 sum of three angles 60 ( ) Thus, required angles are, 8, and (i) From figure, (Linear pair) and (Linear pair) We know that sum of interior angles of a pentagon ( ) 80 U N D E R S T A N D I N G Q U A D R I L A T E R A L S 9

41 + + x x x 0 x 0 0 x 0. (ii) Each interior angle of a regular hexagon (6 ) 80 6 x (iii) Sum of all exterior angles of a polygon x 60 x Using linear pair axiom, x x and z z Using Exterior angle property in a triangle, y Thus, x + y + z Alternative Method: We know that sum of all exterior angles in a polygon is 60. So, x + y + z 60 [From given figure] 0. Convex polygon Concave polygon WORKSHEET 8. As the sum of the angles of a quadrilateral 60. So, x 60 + x 60 x 60.. Let the measure of the fourth angle be x. Three acute angles are given as 0 each. So, x (Using Angle sum property) x x Let each equal angle of the quadrilateral be x. So, x + x + x + x 60 x 60 x Let each of the three equal angle be x. Fourth Angle 0 (Given) Using Angle sum property, we have x + x + x x x M A T H E M A T I C S VIII

42 . Let the common factor of the angles be x. So the four angles are x, x, x and 6x. Using Angle sum property, x + x + x + 6x 60 8x 60 x Therefore, four angles are: 0, 0, 0 and 6 0 i.e., 80, 60, 00, and Each interior angle of a regular hexagon ( 6 ) All the angles are 0, 0, 0, 0, 0 and 0.. (i) Since sum of all exterior angles of a polygon x x 60 x (ii) As the sum of all exterior angles of a quadrilateral x x 60 x Let each equal adjacent angle be x. As the sum of two adjacent angles of a parallelogram 80 So, x + x 80 x 80 x Let common factor of adjacent angles be x. So the two angles are x and x. As we know that adjacent angles of a parallelogram are supplementary. So, x + x 80 x 80 x 80 6 x 6 and x 6 08 Thus, four angles are, 08, and In a parallelogram, one given angle 0. Let one adjacent angle of the given angle be x. As the two adjacent angles are supplementary. So, x x Thus, required angles are 0, 60, 0 and 60.. Two adjacent sides of a parallelogram are 0 cm and cm. As the perimeter of a parallelogram (sum of two adjacent sides) (0 cm + cm) cm cm.. Given: Shorter side of a parallelogram 0 cm According to question, longer side shorter side 0 cm 0 cm. So the perimeter of the parallelogram (sum of two adjacent sides) (0 cm + 0 cm) 0 cm 60 cm. U N D E R S T A N D I N G Q U A D R I L A T E R A L S

43 WORKSHEET 9. (i) A + B 80 (Since two adjacent angles are supplementary) x 80 ( A 60 (given)) x Also, A C and B D [In a parallelogram, opposite angles are equal) y 60 and z 0. (ii) A + D 80 ( Two adjacent angles are supplementary) x (Given D 0 ) x A C (Opposite angles are equal) y 0 and z x (Corresponding angles are equal) z 0. (iii) B D (Opposite angles are equal) y 0. In ACD, ACD + CDA + DAC 08 (Using angle sum property) x x (iv) A + B 80 (Adjacent angles are supplementary) x (Given B 0 ) x Since opposite angles of a parallelogram are equal. So, B D y 0. Also, y z (Alternate interior angles are equal; AD BC) z 0.. Suppose the two adjacent angles of a parallelogram are x and x. As the adjacent angles are supplementary. So, x + x 80 x 80 x x 60 0 Then, four angles are 60, 0, 60 and 0.. Suppose ABCD is a rectangle with AB 6 cm and BC cm. We have to find diagonal AC. In right triangle ABC. Using Pythagoras theorem,. AC AB + BC AC cm.. Two adjacent sides of a parallelogram are given as cm and cm. Perimeter of a parallelogram (sum of two adjacent sides) ( cm + cm) 9 cm 8 cm.. In figure, PQRS is a rectangle with sides PQ 0 cm and QR 8 cm. Diagonal PR? M A T H E M A T I C S VIII

44 In right PQR, PR PQ + QR (By Pythagoras theorem) (0) + (8) PR 6 cm. 6. Let the number of sides of a regular polygon be n. Given: Each exterior angle But we know each exterior angle of a 60 regular polygon Number of sides So, 60 n n 60 n 60.. Let each of the three equal angles be x. Given: One angle of the quadrilateral. Using Angle sum property, x + x + x + 60 x + 60 x x Let the number of sides of a regular polygon be n. Given: Each interior angle 6. But each interior angle of a regular ( ) 80 polygon with n sides n n 6 ( n ) 80 n 6 n (n ) 80 6 n n n 80 n 6 n (80 6 ) 60 n 60 n Each interior angle of a regular hexagon (6 ) 80 6 ( n ) 80 Using n Using Linear pair axiom at vertex A, BAP + BAF 80 x x Similarly, y z p q r 60 So, x + y + z + p + q + r Alternative Method: We know that the sum of exterior angles of a polygon 60. Here, x, y, z, p, q and r are the exterior angles of a regular hexagon. So, x + y + z + p + q + r 60. WORKSHEET 0. Given: one angle of a parallelogram is 00. Let one of the adjacent angles to the given angle be x. x ( Two adjacent angles are supplementary). x Other three angles are 80, 00 and 80. U N D E R S T A N D I N G Q U A D R I L A T E R A L S

45 . Given: Perimeter of a parallelogram 80 cm Let shorter side of the parallelogram be x. So longer side of it is x + 0. Now, perimeter (sum of two adjacent sides) 80 (x + x + 0) 80 (x + 0) 80 x x x 0 x 0 0 x Thus, the adjacent sides of the parallelogram are 0 m and 0 m.. (i) Sum of all the four angles of a quadrilateral x x 60 x (ii) Using Angle sum property, x x x (iii) Using Angle sum property, x x 60 x (iv) Using Angle sum property, x x 60 x 60 x.. In gm ABCD, A 0. We know that the sum of two adjacent angles is 80. So, A + B B 80 B 80 0 B 0 Also, A Cand B D ( Opposite angles are equal) C 0 and D 0.. In figure, ABCD is a rhombus with diagonals AC cm and CD 6 cm. We know that diagonals of a rhombus bisect each other at 90. M A T H E M A T I C S AO OC AC 6 cm and BO OD BD 8 cm. Now, in right-angled AOB, AB AO + OB (Using Pythagoras theorem) AB 00 0 cm. Thus, required side of the rhombus is 0 cm. VIII

46 6. In trapezium ABCD, ABy DC and A B 0. Since AByDC and AD is a transversal. So A + D 80 (Co-interior angles are supplementary) 0 + D 80 D 80 0 D 0 So, x 8 or y x 8 or y + x 9 or y. (ii) In the given figure, ABCD is y gm. ADyBC and DE is a transversal. Similarly, C 80 B 0.. As given two diagonals of a rectangle are x + and x + But we know that diagonals of a rectangle are equal to each other. So, x + x + x x x Now, putting the value of x in given expressions, we get each diagonal + or + cm. 8. (i) As we know that opposite sides of parallelogram are equal. ADC BCE 0 + z 80 z Again, ADB CBD 0 y Further, ADC + DAB 80 (Sum of adjacent angles) 80 + x 80 x Thus, x 00, y 0 and z 0. U N D E R S T A N D I N G Q U A D R I L A T E R A L S

47 Chapter PRACTICAL GEOMETRY WORKSHEET. (C) A quadrilateral has sides, angles and diagonals in which any parts we need to construct it uniquely.. (A) If the measures of four sides and one of the diagonals are given to construct a quadrilateral then we firstly draw a triangle containing the given diagonal of it.. (B) According to given procedure in question, we required to complete a ABC so the next step is to mark for AC.. (C) Measures of PR and S can t help in the construction of quadrilateral PQRS.. (D) According to given measures and steps, when we draw firstly OR and R then next step would be to draw O. 6. (D) Only square can be constructed using a side because its all sides are equal and each angle be 90 are known.. (B) To construct a rectangle, angles are already known so we need either two adjacent sides or one side and one diagonal. In the case diagonal is not given so we choose one adjacent side among the given choices. 8. (C) After drawing a side AB and an arc with radius AD and centre as A we would draw a diagonal BD to determine another vertex D. 9. (C) To construct a quadrilateral we need total parts out of 0. Here two diagonals are given so we need sides. 0. (D) To construct a unique parallelogram, we need two more parts out of a adjacent side, a diagonal and an angle or only two diagonals.. (D) For constructing a quadrilateral, if we first draw a triangle using available data then we try to determine fourth vertex.. (C). (C) Two diagonals are sufficient to construct a rhombus because the diagonals are perpendicular bisector of each other.. (C) Diagonals of a rhombus bisect each other perpendicularly.. (D) MO, OR, RE, EM 6. (C) Length of the side of a rhombus d d WORKSHEET. Steps of construction:. cm.. Draw a line segment EA 6 cm. Using ruler and compass, draw an angle of at E and another angle of 90 at A. 6 M A T H E M A T I C S VIII

48 . Taking the radius of cm and centre as E, mark the point D at other arm of E.. Similarly, taking radius of. cm and centre as A. Mark the point R at other arm of A.. Steps of construction:. Draw a line segment PQ cm.. Using protractor, make an angle of measure 00 at the end P and another angle of measure 80 at Q.. Taking radius of cm and centre as Q, draw an arc to mark the point R at other arm of Q,. Further, make an angle of measure 00 at R. Thus, we observe that another arms of P and R meet each other at a point say S.. Now, join DR. Thus, the required quadrilateral DEAR is formed.. Hence, the required quadrilateral PQRS is obtained.. Steps of construction:-. Draw line segment A 8. cm.. At the end A, make an angle of using ruler and compass then draw ray AX.. Further, make an angle of 60 at L with the help of protractor and draw a ray LY.. Taking radius of cm and centre as L draw an arc to mark the point P on ray LY.. Now, make an angle of 8 at P with the help of protractor and draw a ray P R A C T I C A L G E O M E T R Y

49 PZ which intersects the ray AX at a point say N. Thus, the required quadrilateral PLAN is formed.. Steps of construction:. Draw a line segment of measure 6 cm. Name it as PL. 6 cm and. cm and centres as U and P respectively. These arcs cut each other at a point say S.. Now, join PS and VS. Thus, PLUS is the required rectangle is so obtained.. Using ruler and compass, make an angle of 90 ( because each angle of a rectangle is right angle) at the end L and draw a ray LX.. Taking radius of. cm and centre as L, cut the line segment of measure. cm (say LU) from ray LX.. Further, draw the two arcs of radii 8 M A T H E M A T I C S VIII

50 6. ruler and compass and draw ray LR.. Taking radius of. cm and centre as L, draw an arc to mark the point M at ray LR.. Further, draw two arcs of radii. cm and 8 cm with centres as K and M respectively. Then cut each other at N. Thus, PQRS is the required rhombus.. Thus, ABCD is the required parallelogram. WORKSHEET. Steps of construction:. Draw a line segment of length 8 cm and name it KL.. As we know that a rectangle has all four angles of measure 90. So, we make a right angle at L using. Join KN and MN. Thus, the required rectangle KLMN is so formed.. Steps of construction:. Take a line segment MN. cm and then make an angle of at M using protractor.. Taking radius of. cm and centre as M, draw an arc which cuts the other arm MR of M at P.. Further, draw two arcs of the same radii as. cm with the centres as N and P. Thus, they cut each other at O.. Now join the intersecting point O to P and N. P R A C T I C A L G E O M E T R Y 9

51 Thus, the rhombus MNOP is so formed.. Steps of construction:. Draw a line segment of length 6. cm and name it as PQ.. Since each angle of a square has of measure 90. So we draw a right angle at Q using ruler and compass.. Taking the radius of 6. cm and centre as Q, draw an arc to mark the point R at ray QY.. Draw two arcs taking the same radii 6. cm and centres as P and R respectively. They cut each other at a point say S.. Now join PS and RS. Thus, the required square PQRS is so obtained.. The length of diagonal BD 0. cm. 0 M A T H E M A T I C S VIII

52 . 6. Steps of construction:. Draw AB. cm.. Using ruler and compass, make an angle of 60 at A and draw another arm ray AX.. Taking radius of cm and centre as A draw an arc to cut the segment AD cm from ray AX.. Further, taking radii as cm and. cm with centres as B and D respectively, draw two arcs that cut each other at C.. Join BC and DC to obtain the required parallelogram ABCD.. WORKSHEET. Steps of construction:. Draw a line segment AB 6 cm.. We know that all angles of a rectangle has of measure 90. So, we make right angles at B as well as at A using ruler and compass. Then draw rays BX and AY.. Also we know that a rectangle has equal diagonals. So, we take equal radii as cm and centres as A as well as B to mark the point C and D on the rays BX and AY respectively. P R A C T I C A L G E O M E T R Y

53 . Taking the same radius of cm but centres as D again draw two arcs on either side of BD. Thus, they cut the previous arcs at A and C respectively.. Now, join AB, AD, CB and CD. Hence, the required rhombus ABCD is so obtained.. Now, join CD to get the required rectangle ABCD.. Steps of construction:. Draw a line segment HO cm.. Using ruler and compass, make an angle of 60 at H and draw a ray HX.. Taking a radius of cm and centre as H, cut the segment HS cm.. Further, draw two arcs of the same radii cm with centres as O and S. Thus, they cut each other at T.. Now, join ST and OT. Thus, the required rhombus HOTS is so obtained. Steps of construction:. Draw a line segment (diagonal) BD 6 cm.. Taking a radius of cm and centre as B, draw two arcs on either side of BD.. Steps of construction:. Draw a line segment of length cm and say it diagonal AC.. Draw perpendicular bisector (say PQ) of the segment AC. Let it intersect at O.. As we know that diagonals of a rhombus bisect each other at 90. So we take the radius as half of other diagonal (i.e., cm) and centre as O then mark the points B and D at PQ on either side of AC.. Now, join BA, BC and DA, DC. M A T H E M A T I C S VIII

54 Thus, we obtain the required rhombus ABCD. 6. Consider P + Q + R As we know that sum of all the four angles of a quadrilateral is equal to 60. But in this case, the sum of only three angles is equal to 60. So the construction of PQRS is not possible... P R A C T I C A L G E O M E T R Y WORKSHEET. Steps of construction:. Draw a line segment QR 6 cm.. Using ruler and compass, make an angle of 90 at R and draw a ray RY.. Also, using ruler and compass, make an angle of 0 at Q and draw a ray QX.. Further, take a radius of. cm and centre as Q then mark P on the ray QX.. Now, using ruler and compass, make an angle of at P and draw a ray PZ.

55 6. Thus, the two rays RY and PS cut each other at S. Hence, the required quadrilateral PQRS is so obtained.. Steps of construction:. Draw a line segment DE cm.. Using ruler and compass, make an angle of at D and draw a ray DX.. Taking a radius of cm and centre as D, draw an arc to mark the point G on DX.. Further, taking the same radii of cm with centres as G as well as E, draw two arcs that cut each other at a point F.. Now, join the segments FG and FE. Thus, the required rhombus DEFG is so formed.. Steps of construction:. Draw a line segment of length cm, name it as KM.. Draw perpendicular bisector of the segment KM and name it as XY.. Further, taking half of KM (i.e., KO OM) as radius and centre as O, cut ON as well as OL from XY on either side of KM. [Note: Diagonals of a square are bisect each other at 90.]. Now, join LK, LM and NK, NM. Thus, the required squared KLMN is so formed. M A T H E M A T I C S VIII

56 . Steps of construction:. Draw a line segment of length 6. cm and name it as DU.. Using protractor make an angle of 80 at the point D and draw a ray DX.. Taking radius of cm and centre as D, draw an arc that cut the ray DX at K.. Further, take two radii as 6. cm and cm and centres as K and U, then draw two arcs that cut each other at C.. Now, join CK and CU. Thus, the required parallelogram DUCK is so formed.. The measurement of PS.6 cm. P R A C T I C A L G E O M E T R Y

57 6... Further, taking the same radii of. cm with centres as A and C, draw two arcs on other side of AC. Thus, they cut at point D. WORKSHEET 6. Yes, since AB + BC > AC and CD + AD > AC. Steps of construction:. Draw a line segment AC 8 cm.. Taking radius of cm and centre as A draw an arc on one side of AC.. Taking radius of 6 cm and centre as C draw another arc on the same side of AC in which previous arc is drawn. Thus, they cut each other at B.. Join B to A and C. Thus, a triangle ABC is formed. 6. Now, join DA and DC. Hence, quadrilateral ABCD is so formed.. Steps of construction:. Draw a line segment AB 6 cm. At the ends A and B, make angles of 6 M A T H E M A T I C S VIII

58 measures 90 and 60 respectively using ruler and compass. Also draw the rays AX and BY.. Taking radius of cm and centre as B, draw an arc to mark C on ray BY.. Further, using protractor, make an angle of measure 0 at C and draw a ray CZ.. Thus, the two rays AX and CZ cut each other at a point D.. Steps of construction:. Draw a line segment of length. cm and name it as PR. Hence, the required quadrilateral ABCD is so formed.. Steps of construction:. Draw a line segment GO cm.. Using ruler and compass, make and angle of 90 at G and draw a ray GX.. Using protractor, make an angle of 80 at O and draw a ray OY.. Further, taking radius of 6 cm and centre as O, draw an arc which cut OY at L.. Now, using ruler and compass, make an angle of 0 at L and draw a ray LZ that intersect the ray GX at D. Thus, the required quadrilateral GOLD is so formed. P R A C T I C A L G E O M E T R Y

59 . Draw perpendicular bisector LM of PR and name the intersecting point as O.. Now take the radius as half of other diagonal (i.e.,. cm) and centre as O and hence cut OS and OQ on LM.. Now join QP, QR, SP and SR. Thus, the required rhombus PQRS is formed M A T H E M A T I C S VIII

60 Chapter DATA HANDLING WORKSHEET. (A) Class size Upper class limit Lower class limit.. (A) For the class 0-0, lower class limit 0 and upper class limit 0.. (A) Class size Upper class limit Lower class limit 0.. (C) Upper limit 60 and lower limit 0.. (C) The lowest frequency is which corresponds to the class (B) The height of a bar in a histogram shows the corresponding frequency.. (D) There is no gap between any two consecutive bars in a histogram. 8. (D) The bar corresponding to is the highest. So the number of students is maximum in (B) Central angle (C) Required per cent 0 00% 0%. + 8 Number of all balls + + Required probability 8.. (C) Draw a horizontal line. The spinner is divided into 8 equal parts. There are parts in the unshaded portion. Required probability 8.. (C) Numbers less than or equal to are:,,,,. Required probability 0. WORKSHEET 8. (i) Let us draw pictograph.. (C)Probability.. (C)Favourable outcomes: (, ), (, ), (, ), (, ). Number of all possible outcomes 6 Probability (A) Number of non-black balls Fig.: Pictograph D A T A H A N D L I N G 9

61 (ii) Let us draw bar graph. (iii) Let us draw bar graph. Fig.:bar graph 60 M A T H E M A T I C S VIII

62 (iv). Frequency Table: Hobbies of Tally No. of Students Marks Students Art Book reading Dance Instrumental music Music 8 Total. Frequency Table: Weight (in kg) Tally Marks No. of Students Total 0 D A T A H A N D L I N G 6

63 .. WORKSHEET T9 9 (ii) Marks Obtained Tally Marks Frequency Total 0. (i) Since pink has the largest central angle, so it is the favourite colour. (ii) Lemon and blue were equally liked by the family, as they have equal central angles. (iii) Since green has the smallest central angle, so it is liked the least.. (i) Expenditure Modes Expenditure ( in %) House rent 0 Household items 0 Daughter s fees 0 Savings Petrol (ii) Expendi- Expenditure Central ture In Angle Modes (in %) fraction. (i) Lowest observation 8 marks Highest observation 6 marks. House rent Household items M A T H E M A T I C S VIII

64 Daughter s fees 0 Savings Percentage chance of occurrence of head 6 (H) 00% 60% 0 Percentage chance of occurrence of tail (T) 00% 0% 0 Petrol WORKSHEET 0. (i) Required probability P( on st die and 6 on nd die) + P(6 on st die and on nd die) Fig.: Pie-chart 6. For the given information, we have to draw a double bar graph (ii) Let E represents the event that sum of two numbers is 0. E {(, 8), (, ), (, 6), (, ), (6, ), (, ), (8, )} n(e). n(s) P(E) n(e) n(s) 6. (iii) P(both the numbers are even) P(an even number on st die) P(an even number on nd die) Number of heads (H) 6; Number of tails (T) Total number of outcomes 0 D A T A H A N D L I N G (iv) Required probability P(an odd number on st die and an even number on nd die) + P(an even number of st die and an odd number on st die)

65 (v) Required probability P(6 is not on st die) P(6 is not on nd die) (i) Total number of marbles (ii)(a) P(a red marble) + +. Number of red marbles Total number of marbles. (b)p(a blue marble) Number of blue marbles Total number of marbles. (c)p(a green marble) Number of green marbles Total number of marbles.. On tossing a coin three times, the sample space S is given by S {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} n(s) 8. (i) Let E Event of showing up three heads E {HHH} n(e ) n(e ) P(E ) n(s) 8. (ii) Let E Event of showing up three tails {TTT} n(e ) n(e ) P(E ) n(s) 8. (iii) Let E Event of showing same side all the three times {HHH, TTT} n(e ) n(e ) P(E ) n(e ) 8. 6 M A T H E M A T I C S (iv) Let E Event of tails showing up times and heads once {HTT, THT, TTH} n(e ) n(e ) P(E ) n(s) 8. (v) Let E Event of heads showing up times and tails once. {HHT, HTH, THH} n(e ) n(e ) P(E ) n(s) 8.. Percentage of no heads Number of outcomes having no head Total number of outcomes 00% 6 00% 6 %. Percentage of head Number of outcomes having head Total number of outcomes 00% VIII

66 00 00% 6 % %. Percentage of heads Number of outcomes having heads Total number of outcomes 00% 6 00% % Percentage of heads Number of outcomes having heads Total number of outcomes 00% 6 00% 6 %. No. of Heads Frequency D A T A H A N D L I N G Percentage 0 6 % % % 6 % Total 6 00%. Ishita s marks in English Ishita s marks in Hindi Ishita s marks in Maths Ishita s marks in Science Ishita s marks in S.St Total number of items (i) Probability of removing a rubber 0. (ii) Probability of removing a pencil WORKSHEET The most common outcome is as its frequency is the highest, i.e., Probability of.. Number of H s 0 Chance (Probability) of occurrence of H 0 0 Number of T s 0 Chance (Probability) of occurrence of T (i) The sale of T.V. was maximum in July. (ii) The sale of T.V. was minimum in March and May. 6

67 (iii) 0 T.V. were sold in each of January, April, June, August, September and December. (iv) T.V. were sold in October. (v) 0 T.V. were sold in March and May each. (vi) Required number of T.V. Total number of T.V. sold in January, February and March (vii) Required number of T.V. Total number of T.V. sold in April, May and June (viii) Required number of T.V. Total number of T.V. sold in July, August and September (ix) Required number of T.V. Total number of T.V. sold in October, November and December (x) The sale was maximum in the third quarter and it was minimum in the first quarter of the year.. The appropriate graph of the given data is a double bar graph. In order to draw a bar graph, you have to follow the steps: Step I. Draw a pair of perpendicular lines say OX and OY on a graph paper. OX is called x-axis or horizontal axis and OY is y-axis or vertical axis. Step II. Write names of students on the x-axis and percentage of marks on the y-axis taking an appropriate scale as No. of students unit 0% on the y-axis Step III: Draw the bars on the x-axis such that any two bars of Maths and Science touch each other with equal width for a student. The gap between any two consecutive pairs of bars should be equal. Step IV: The heights of these bars correspond to the percentage of marks.. In order to draw a histogram, you have to follow the steps given below: Step I: Take a graph paper and draw a pair of perpendicular lines OX and OY on it. The horizontal line OX is called x-axis and the vertical axis OY is called y-axis. 66 M A T H E M A T I C S VIII

68 Step II: Mark heights of the students on the x-axis and number of students on the y-axis by taking a scale as unit cm height on x-axis and unit student on y-axis. Step III: Draw the bars on the x-axis such that the width of each bar is same and there is no gap between any two consecutive bars. Step IV: The heights of these bars are proportional to the number of students. 6. Favourite Tally Frequency Dish Marks (No. of Children) French fries Macroni 8 Pizza 0 Sandwich Total (ii) People of - 0, -0 and 0- age groups spent equal number of hours at the Gym. (iii) People of - 0 age group spent hour at the Gym.. (i) Required number of members + 8. (ii) 0 members are in the age group of -0. (iii) Age group of -6 has the maximum number of members.. Total number of students 80 (i) Number of students liking Basketball (ii)number of students liking Badminton WORKSHEET T Pictograph: ` 00 Coin represent ` 00. Expenses Amount 0 s Fig: Pictograph. (i) People of 0- age group spent maximum time for working out at the Gym. D A T A H A N D L I N G Number of students liking Cricket Required number of students (iii) Number of students liking Tennis Required ratio [Using part (i)] :. 6

69 . Frequency Distribution Table: Score Obtained Tally Marks Frequency 6 6 Total 6. In order to draw a histogram, you have to follow the steps given below: Step I: Take a graph paper and draw a pair of perpendicular lines OX and OY on it. The horizontal line OX is called x-axis and the vertical axis OY is called y-axis. Step II: Mark daily earnings (in ` ) of given drug stores on the x-axis and number of stores on the y-axis by taking a scale as. unit storeson the y-axis. Step III: Draw the bars on the x-axis by taking it as base, such that the width of each bar is same and there is no gap between any two consecutive bars. Step IV: The heights of these bars are proportional to the number of stores.. (i) Range Maximum wage Minimum wage ` 00 ` 00 ` 00. (ii) workers are getting ` 0 each. (iii) workers are getting minimum wages of ` 00 each. (iv) The highest amount of wages earned by workers is ` 00 each. (v) workers get ` 00 each as weekly wages. WORKSHEET. (i) Mark Tally Marks No.of Students Total 0 (ii) 00 marks is the highest score (iii) marks is the lowest score. 68 M A T H E M A T I C S VIII

70 (iv) Range marks. (v) Number of failures Sum of number of students having less than 0 marks. (vi) Number of students having or more marks (vii) 00 is beyond the class (viii) Number of students having less than 0 marks +.. (i) Lower limit of class 0-60 is (ii) Class marks of class Class marks of class (iii) Size of each given class is 0.. (i) Since the classes are 0-0, 0-0, 0-0,.... Therefore, the class size is 0. (ii) Number of students in the class interval 0-0 are. So, students obtained less than 0 marks. (iii) Number of students obtaining 0 or more marks but less than 0 Number of students in the class interval (iv) Class 0-80 is of highest marks and students are there in this interval. (v) Number of failures (i) In the age group of -0, the number of literate females is the highest. In the age group of 0-, the number of literate females is the lowest. (ii) 00 is the lowest frequency. (iii) Class-mark of class D A T A H A N D L I N G Class-mark of class Class-mark of class Class-mark of class Class-mark of class Class-mark of class (iv) Width of each class 0.. Frequency Distribution Table: Marks Tally Marks Frequency Total 0 (i) Range 0 8 marks (ii) Highest marks 0 and lowest marks 8. (iii) 0 marks are occurring most frequently. 69

71 Chapter 6 SQUARES AND SQUARE ROOTS WORKSHEET T. (B) 6 6 or (B) (A).. (B) The square ends in.. (D) (C) (B) (B) So, the given number should be divided by. or 9 < 88 < 0 Since 88 8 is smaller than Therefore, 88 is approximately 9.. (C) Area Side 89 m.. (C) 6.. (C) Required number. 6. (D) Square root is the inverse operation of square and vice-versa.. (C) 9, 9. (D) So, the given number should be multiplied by. 0. (B) So, must be added to 00.. (A) (C) 8 < 88 < 00 Here 8 < 88 < 00,. 8. (D) 6 8, 8 9. (A) LCM Required number (C) Using the given pattern, we get ( + ) Missing number (D) 9 9. (A) M A T H E M A T I C S. VIII

72 WORKSHEET OR. (i). (ii) 6. (i). (i) 9 (ii) (iii) 0. (iv) 9.. (ii) OR First find the LCM of 8, 9 and 0. S LCM Required number 60 Q U A R E S & Two perfect square numbers are: (a) and (b) 9.. (i) So, the Pythagorean triplet is,,. (ii) So, the Pythagorean triplet is 6, 8, 0. S Q U A R E R The given numbers must be multiplied by The given number must be divided by.. Perimeter of a square Side O O T S

73 or 8 Side Side m. or 8 Area Side 69 m. 8. Area Side 6 Side or Side or Side Side 68 m. Side 9. Let one of the required numbers be x. Then the other number 6x Their product x 6x 6x This is given to be 96 6x 96 or x (i) 69. (ii) (a) 6. (i) (ii) 6. (b) (c) 6 m m m 6 m m. WORKSHEET 6 Thus,.6.6. or x x 6 9 Hence, the numbers are and 9. OR Let the two consecutive natural numbers be x and x +. Then, (x + ) x 9 or x + x + x 9 or x 8 or x 9 x + 0 Now we can write and the required numbers are 0 and 9. Thus, (approximately). OR 0.86 (approx.) Thus, 0.9 (approx.). M A T H E M A T I C S VIII

74 . Let the number be x. Then x x.88 or x.88 (ii) x (i) (i) (approx.). Now, (ii) (i) (approx.). OR..6. (approx.). (ii) Now, S Q U A R E S & S Q U A R E R O O T S

75 (approx.).. No. Reasons: Given equality is Square of LHS 0. Square of RHS (0.) Since 0. is not equal to 0.0 i.e., or So, 0 should be divided by to make it a perfect square. OR 6.. Required number So, should be divided by to make it a perfect square. 0. Since the square of is an even number, so the square of is also an even number WORKSHEET T 8. Required number ( ) ( ) Clearly, 880 is not perfect square.. 8 ( ) 6 So, 8 must be multiplied by 6 to make it a perfect square. M A T H E M A T I C S VIII

76 ( ) So, 90 must be divided by 0.. We know that A B (A + B)(A B) Substituting A and B 0, we get 0 ( + 0)( 0) Let the two consecutive natural numbers be x and x + such that (x + ) x or (x + + x) (x + x) or (x + ) or x or x 0 or x x So, is written as We know that a number may be a perfect square if its unit s digit is either 0,,,, 6 or 9. So, 0668 is not a perfect square number 8. Square of 8 is an even number, as square of 8 is 6 which is an even number ( ) + ( ) So,, and 9 do not form a Pythagorean triplet. 0. [ (a) (a) a ] 69.. First find the LCM of, 6 and 0. LCM (, 6, 0) 60. The prime factorization of 60 is: Now, the required number OR ( ) ( ) (i) (ii) (approx.).. Area of square Side Side Side or Side 9 m. S Q U A R E S & S Q U A R E R O O T S

77 . We know that for m >, (m + ) (m) + (m ) Let m +. Then m m So, m and m Hence, the Pythagorean triplet is:,,. WORKSHEET T8 8. Let the number of students in each row be x. Then the number of rows is also x. Total number of students x x x But this is given to be 8. x 8 x 8 9 So, 9 students stand in each row Area 6 m Side 6 Required number Side 6 9 m.. The least number of digits is 000. We have to make 000 as a perfect square. For which, we have to add a least number to it. 000 Therefore, the required number is 0. Side of square Area 600. Perimeter Side 980 m. 6 M A T H E M A T I C S VIII

78 9.. WORKSHEET T Now, Clearly, is not a perfect square.. ( ) ( ) 6.. Area Side Area m m ( ) ( ) ( ) ( ) Here does not make its pair. Therefore, the required number is. Clearly, 89 is a perfect square digits perfect square number are 6,, 6, 9, 6, and 8. Therefore, the required number is ( ) ( ) ( ) Consequently, we get that we should divide 88 by to make it a perfect square.. 8 ( ) ( ) does not make its pair. Therefore, 8 must be multiplied by to make it a perfect square.. (i) Since, unit s digit of is 9. Therefore, unit s digit of 8 is also 9. (ii) Since, unit s digit of is. Therefore, unit s digit of 0 is also ( + 09)( 09) 0 0. OR 9 9. S Q U A R E S & S Q U A R E R O O T S

79 . 0 ( ) ( ) ( ) Let the number of students in the school was x. So each student paid ` x. Collection x x ` x This is given to be ` 000 x 000 x 000 Let us find 000 x 0. Thus, the number of students in the school was 0. OR Area of square Side 6 Side Side 6 m. Now, distance covered by the man Perimeter of the square Side 00 m. Speed of the man 0 km/hour 0 km hour 0 8 Time taken 0 000m 600 s m/s 0 9 m/s. Distance covered speed s (80 + 8) s min 8 s. Thus, the man returns after minutes and 8 seconds. 9. First find the LCM of 8,, and 0. So, LCM ( ) ( ) 0 We have to multiply this LCM by 0 to make it a perfect square. So, required number Thus, 600 is the least square number which is exactly divisible by 8,, and Largest -digit number 999 Smallest -digit number Also Required number M A T H E M A T I C S VIII

80 . (i) (a) 0000 ( ) ( ) ( ) ( ) (b) Hence, 0000 is the perfect square number. WORKSHEET 0. (i) ( ) ( ) ( ) ( ) 8. (ii) 8 0 9a b c ( a ) ( b ) ( c ) a b c a b c.. (i) ( ) ( ) ( ) ( ) Hence, 96 is the perfect square number. (ii) 0 ( ) ( ) ( ) ( ) ( ) 0. (ii) 6 6. (i) Clearly, (ii). Clearly does not make its pair. Therefore, 0 should be multiplied by to make it as a perfect square. Clearly, 9. S Q U A R E S & S Q U A R E R O O T S 9

81 AB i.e., According to the Pythagoras property, we have. Hypotenuse Sum of squares of other two sides Hypotenuse m. Thus, the length of the hypotenuse is m. 6. AC is string, BC is wall, the flyer is at A and kite is at C (see fig.). AC 0 m, BC 0 m m. Therefore, the flyer is at a distance of 0 m from the wall.. Let each side of the wall be x metres. Area of the square wall Side x m Expenditure on paving at m ` Expenditure on paving at x m ` x ` x This is given to be ` 600. x 600 Dividing both sides by, we get x ( ) ( ) ( ) ( ) x 8 So, the length of each side of the wall is 8 metres. 8. Using Pythagoras property, we have AB + BC AC AB or AB ( ) ( ) ( ) ( ) M A T H E M A T I C S VIII

82 9. First find LCM of, and (ii) LCM (,, ) 60 Now, make 60 as a perfect square by multiplying it by a least number. so, first find the least number. 60 (obtained above) ( ). Since and do not occur in pairs, so, is the least number Multiplying 60 by, we get Hence, 900 is the required smallest number OR Let us use long division method to obtain square roots. (i) Thus, Thus, ( ) The prime factor does not occur in pair. Therefore, must be multiplied by to make it as a perfect square. Hence, the required number is.. Represent as its prime factors ( ) ( ) ( ) ( ) ( ). Here, a does not occur in pair. Therefore the required least number is.. Smallest -digit number 0000 Greatest -digit number Their sum S Q U A R E S & S Q U A R E R O O T S 8

83 Chapter CUBES AND CUBE ROOTS WORKSHEET. (C) (C).. (C) i.e., 9 is equal to 9.. (A) which is not a perfect cube.. (B) 6 is not in triplet, so the required multiplier is. 6. (A) ( ) ( ) A prime is not a group of three. So is the required divisor.. (D)The symbol denotes square root The symbol denotes cube root. 8. (C) and So, estimated value of 688 is (B). 0. (A) (D) (C) Comparing corresponding terms between the equations + + x + ( y) and ( ), we obtain x and y.. (B) According to the given pattern, the number of consecutive odd numbers whose sum provides n is n. Therefore, the required number is 9.. (A) One s digit of 00 One s digit of One s digit of.. (A) Unit digit of unit digit of. 6. (C) One s digit of cube of a number ending with 6 One s digit of (D) ( ) (A) If a perfect cube number ends with 0, then its cube root also ends with 0. WORKSHEET. (i) Cube of x x x x x (ii) 9. (iii) ( ) (iv) ( 8) ( 8) ( 8) ( 8) 8.. (i) ends in. Therefore, is odd. (ii) 6 ends in 6. Therefore, 6 is even. (iii) 00 ends in 0. Therefore, 00 is even. (iv) ( 98) ends in. Therefore ( 98) is even.. Volume of cube (Edge) (.) cm. 8 M A T H E M A T I C S VIII

84 . (i) 9 9 ( ) ( ) 8 9 is a perfect cube number as prime 9 factor is in the group of three. (ii) Here, ( ) ( ) is a perfect cube number as each prime factor appears in group of three. (iii) Here, ( ) ( ) is a perfect cube number as each prime factor occurs in the group of three. (iv) Here, 6000 ( ) ( ) ( ) 6000 is not a perfect cube number as a does not occur in the group of three Here, 08 ( ) The prime factor does not occur in the group of three. To make 08 a C U B E S A N D C U B E R O O T S perfect cube, you will have to complete this group. For this, multiply 08 by. Hence, the required number is. 6. (i) Here, 000 ( ) ( ) ( ) ( ) (ii) Here, 8 ( ) 8. (iii) Here, 068 ( )

85 (iv) 9 9. ( ). Volume of a cube (Edge) Edge Volume Thus, edge of the cube is cm. 8. (i) ( ) ( ) Therefore, cube root of (ii) Here, ( ) Therefore, cube root of. 9. Let us take 8 and as two even natural numbers , which is even. 8, which is even. 0. Here, 66 ( ) ( ) Clearly all the s do not appear in the groups of 66 three. To complete such 8 groups, we should multiply 9 by. Product Now, 968 ( ) ( ) 9 ( ) ( ) Therefore, cube root of WORKSHEET. (i) (ii) ( ).6.. (iii) ( ) ( ) ( ) ( ) 9.. (i) Unit digit of Unit digit of. (ii) Unit digit of Unit digit of. (iii) Unit digit of 8 Unit digit of. (iv) Unit digit of 8888 Unit digit of 8.. (i) 08 ( ) 08 is not a perfect cube. (ii) 6 ( ) ( ) 6 is a perfect cube. 8 M A T H E M A T I C S VIII

86 (iii) ( ) ( ) is a perfect cube. ( ). First represent 600 as its prime factors. 600 ( ) ( ) The prime factor does not 0 appear in a group of three. If we divide the number by, then the prime factorisation of the quotient will not contain. So, the required smallest number is.. (i). (ii) 8 ( ) 8 8. (iii) (i) 6 ( ) ( ) ( ) ( ). (ii) 8 ( ) ( ). C U B E S A N D C U B E R O O T S. (i) 8 ( ) ( ) ( ) ( ) Now, 8 (ii) ( ) ( ) 0 (0.) Now, 0.00 (0.) (i) ( ) 00 ( ) (0 0) ( ) 0 Therefore, (ii)... (.) (. ) (.). Therefore,

87 9. Volume of a cube Edge Edge Volume 8 ( ). Thus, edge of the metallic cube is cm. WORKSHEET. Volume 6 ( ) m We know that: Volume of a cube Side Side Volume ( ) 6 6 m.. Volume ( ) 9 ( ) 0 (.8) m. We know that: Volume of a cubical box Side Side Volume (.8).8 metres.. (i) (ii) Therefore, ( ) 0 (0.) (0.) Therefore, ( ) 9 ( ) (iii) (0.) (0) (0. 0). Therefore, First represent 600 as its prime factorisation. 600 ( ) Here, one s, two s and two s do not occur in the groups of three each. For happening this, we should multiply 600 by M A T H E M A T I C S VIII

88 So, the required smallest number is 60. Therefore, product Hence, 6000 ( ) ( ) ( ) ( ) ( ) (60) Thus, the cube root of the product is 60.. First represent 89 as its prime factorisation. 89 ( ) ( ) ( ) ( ) 89 A prime factor does not appear in its group of three So, we should divide 89 by to make it a perfect cube. 0 Thus, is the required smallest number. 6 8 Quotient Further,096 ( ) Thus, the cube root of the quotient 096 is (i) is not a perfect cube number. C U B E S A N D C U B E R O O T S. (ii) is a perfect cube number. or ( ) Therefore, 8. ( ) ( ) is the cube of The five natural numbers are, 6, 9, and Now, obtain the cubes of these numbers. Cube of. Cube of Cube of Cube of 8. Cube of. WORKSHEET. (i) 6 ( ) ( ) 6 is not a perfect cube number. (ii) 6 ( ) ( ) 6 is a perfect cube number (iii) 6000 ( ) (0 0 0) 6000 is a perfect cube number First, represent 9 as its prime factors. 9 ( ) The prime factor does not occur in a 8

89 group of three. To make this group, we need one. And then 9 will make a perfect cube. In this case, 9 ( ) ( ) which is a perfect cube. Hence the required smallest number is.. Let the given number be a Its cube a a a a (i) New number Double of given number a Cube of new number a a a 8a (ii) From equations (i) and (ii), we have cube of new number 8 Cube of given number For example, if a then a 8, a and (a) 6 Here, i.e., (a) 8 a Hence if a given number is doubled, then its cube becomes eight times the cube of the given number.. (i) ( ) ( ) ( ) ( ) ( ) 8. (ii) ( ). (i) ( ) ( ) ( ) () (ii) 9 ( ) 9 ( ) 99 ( ) (iii) Here, 68 ( ) ( ) ( ) ( ) ( ) ( ) And ( ) 0 6 ( ) Taking cube root both sides, we get Here, And M A T H E M A T I C S VIII

90 ( ) C U B E S A N D C U B E R O O T S 0 (.) 0 Taking cube root both the sides, we get.. 8. (i) i.e., (ii) 9 9 i.e., (i) and ( 6) (ii) 0. (0.) 0.. (iii) WORKSHEET 6. Side of a cube. cm Volume of a cube Side (.) ( ) cm. (i) ( 0) ( 0) ( 0) ( 0) (ii) ( ) (iii) (.) ( ) 0. ( ) 8... (i) 096 ( ) ( ) ( ) ( ) ( ) Clearly, 096 is a perfect cube number. (ii) 9 Clearly, 9 is a perfect cube number. (iii) Clearly, 689 is a perfect cube number is a cube of an even number as this ends in 0.. Volume 8688 mm 8688 ( ) ( ) ( ) 96 ( )

91 Volume Edge 9 Edge Volume 9 9 mm. 6. (i) Unit digit of cube of 0 is same as unit digit of cube of. 6 Clearly, unit digit of is. Hence, unit digit of 0 is. (ii) Unit digit of cube of is same as unit digit of cube of. 9 Clearly, unit digit of is. Hence, unit digit of is.. (i) 9 ( ) 9 0 ( ) 0 ( ) ( ) Cube root of 9 9. (ii) 68 ( ) Cube root of Volume 8000 cm ( ) ( ) ( ) ( 0) Side Volume 0 0 Thus, the measure of side is 0 cm (i) (ii). ( ) 0. 0 ( ) ( ) ( ) ( ).. ( ) 0. (i) And (ii) And M A T H E M A T I C S VIII

92 WORKSHEET. Volume of the box 6 cm Side of the cubical store. m. 00 cm 0 cm Volume of the store Side cm. (i) Number of boxes Volume of the store Volume of box Thus, 000 boxes can be put in the store. (ii) Length, breadth and height of the box are of equal measurement as it is a cube. Edge Volume 6 Thus, dimensions of the box are cm, cm, cm.. ( ) The prime factor does not appear in the groups of three absolutely. If we divide by 9, this will happen. So, the required smallest number is 9.. (i) The only prime factor of 689 is 9 which appears in triplet. So, 689 is a perfect cube number. (ii) 088 ( ) ( ) ( ) The prime factors of 088 are, and. Each of them appears in triplet. So, 088 is a perfect cube number.. Volume of cube Side (.) ( ) cm.. (i) (ii) (iii) (iv) (v) 6 ( 6) 6. C U B E S A N D C U B E R O O T S 9

93 (vi) (i) Unit digit of cube root of 698 Unit digit of cube root of. (ii) Unit digit of cube root of 66 Unit digit of cube root of Side 0.8 cm 8 0 cm 8. Volume Side 8 ( ) cm First, represent 0 in its prime factors. 0 ( ) ( ) ( ) 60 The prime factor does not occur in the triplet. If we divide 0 by, all the prime factors occur in the triplets. 9 Quotient 0 00 In this case, quotient ( ) 6 Cube root of the quotient M A T H E M A T I C S VIII

94 Chapter 8 COMPARING QUANTITIES. (D) WORKSHEET 8 m 6 km m 6000 m 00 : 00.. (A) : 00% 80%. (C) Required number 8% of (A) Bill amount ` 0 + ` 0 ` 0 + `.0 ` (A) Price before VAT ` ` 00. ` 6. (C) CP for each article ` 8 SP for each article CP + Profit ` + ` 00 ` 8.. (B) CP ` 0 + ` 0 ` 00 SP ` 60 Since SP > CP, therefore there is a gain % 00 0 % 0%. 8. (A) Marked price `.0 9. (C) Let CP x, ` ` 0. Then SP x x SP > CP Gain%.08 x x x 0. (D) Let a single discount be x%, Then x 0 CP CP or 00x x 8.. (A) SI ` CP ` 00.. (C) P ` 000, A ` 0, n years, R? R A ( ) n P + gives R ( ) Gain% SP CP CP 00% C O M P A R I N G Q U A N T I T I E S or R +.0 or R %. 00 9

95 R. (C) P ( ) P + gives 00 ( ) R + 00 R Further, 8P ( ) n n 8 P + gives 00 n or or n years.. (A) P ` 600,. Loss percentage % CP CP 00 0 or CP CP ` 960. The man wants to a gain of %. SP CP + CP CP R 0 % per half annum, n half years. A 600 ( ) ` (D) ( ) P + gives P.0 or P l 860. WORKSHEET 9. SP ` 6 Loss percentage Loss percentage CP 6 CP 8 % % CP SP CP or CP 00 CP 6800 or CP Thus, the cost price of the chair is ` ` 00. Thus, he must sell the furniture for ` 00.. Let single discount be x%. x Single discount CP. 00 st out of two successive discounts CP 0 00 CP. And nd out of two successive discounts CP ( CP ) 9 M A T H E M A T I C S 0 00 CP. According to the given condition, x CP 00 CP + CP or x Thus, the required discount is 8%.. Amount paid by a customer Marked Price Discount ` 6 Thus, the amount paid by a customer is ` 6. VIII

96 . Marked Price CP + CP 0 00 CP + CP 0 CP. 0 Discount 0% of marked price SP < CP as So, there is a loss CP CP 00 SP MP Discount CP CP. CP CP < CP 00 Loss CP SP CP 99 CP CP Loss percentage Loss CP 00 CP 00 CP 00 %. Thus, the shopkeeper loses by %. 6. Let the constant of ratio be x. Then cost of calculator ` x and cost of typewriter ` 9x. 9x 60 or x 0 Therefore, the cost of the calculator is ` 0.. Let the constant of ratio be y. Then Miti has y stamps and Gunjan has y stamps. After taking 0 stamps, Miti has (y + 0) stamps. After giving 0 stamps, Gunjan has (y 0) stamps. Since, finally both have same number of stamps. C O M P A R I N G Q U A N T I T I E S. y + 0 y 0 or y y or 60 y or 0 y y 0 0 Therefore, Miti has 0 stamps. R 8. A P( ) n + gives 00 8 A 000 ( ) ` 8 CI A P ` 8. Thus, the compound intersect is ` CP ` 000 Sales Tax % of CP ` 0 Cost for a buyer SP for the seller CP + Sales tax ` ` 0 ` CP ` 80 8 Sales tax 8% of CP ` 6.0. Actual cost price CP + Sales tax ` 80 + ` 6.0 ` Let original cost price be ` x. VAT 8% of x ` 0.08x Now, x x 6 or.08x 6 x 6.08 ` x 9

97 OR Let Kishore s savings be ` x. Expenditure on a car of x ` x Now, required percentage x WORKSHEET 0. (i) ` to ` ` ` (ii) hours to 80 minutes. (i) : x 00% 00% 0%. hours 80 minutes :. 60 minutes 80 minutes :. 00% 6.%. (ii) 0 : % (i) %. OR 6 8 :. (ii) :.. Let the other number be x. Then 0 x 6 Crossmultiplying, we have 6 x 0 x Thus, the other number is 00.. Let the constant of ratio be x. Then Rushil s amount ` x and Timmy s amount ` x Also Timmy s amount ` 6 more than Rushil s amount ` (6 + x) There are two amounts of Timmy here, compare them, we get x 6 + x x x 6 or x 6 Rushil s amount ` 6 ` 8 And Timmy s amount ` 6 ` Now, Total amount ` 8 + ` `.. For each goat CP ` 00 For one goat, loss % of CP 00 ` SP for this goat SP Loss ` 00 ` 60 ` 0. For second goat, profit 0% of CP ` 0. SP for this goat CP + Profit ` 00 + ` 0 ` 0. Thus, selling price of one goat is ` 0 and that of other one is ` M A T H E M A T I C S VIII

98 6. Increase in the population Final population Initial population,00,000,,000,,000. Increase in percentage Increase Initial population %.. A man sells a cow for ` 00 at a loss of %. This means if CP ` 00, then SP `. or if SP `, Then CP ` 00 Therefore, if SP ` 00, CP ` ` Now, the selling price to gain % CP + Gain CP + % of CP CP + 00 CP CP 9600 ` 00 Thus, the must sell the cow for ` 000. OR Let CP x rupees. Loss 9 Now, SP CP Loss x 00 x 9 or x CP of CP 9 x 9 C O M P A R I N G Q U A N T I T I E S x ` 800 Therefore, the cost price of the article is ` CP for Bebo ` 000 SP for Bebo ` 000 Loss ` 000 % of ` 000 ` ( ) ` ( ) ` 00 CP for Monika SP for Bebo ` 00 SP for Monika ` 0 Profit for Monika SP CP ` 0 ` 00 ` 0 Profit % for Monika %. Thus, cost price for Monika is ` 00 and profit is 0%. 9. Marked Price ` 00, SP ` 00 Discount Marked price SP ` 00 ` 00 ` 00. Rate of discount Discount Marked Price %. 0. P ` 00, R 0 % %, n years n R A P( 00 ) ( ) 9

99 00 ( ) `.0 CI A P.0 00 `.0 l.0 Thus, the compound interest is `.0. WORKSHEET. Number of boys 0% of Number of girls 0 Number of boys So, the boys are 0 and girls are 0.. Let the store contains x vegetables in altogether. Then x 00 or x 00 Therefore, there are vegetables in the store.. Let ratio of constant be x. Then, weight of Sanya 8x kg and weight of Guddu x kg. But it is given that weight of Sanya is 0 kg 8x 0 or x 0 8. Hence, weight of Guddu x kg.. Let Deepali s age x And Anuj s age y x Half of Deepali s age and One-third of Anuj s age y These last two ages are given to be equal x y gives x y x or y or x : y :. Thus, the required ratio is :.. Let original salary be ` x. Then New salary x + 0% of x 0x x + ` 6 00 x. This is given to be ` x 0000 or x or x 000 Thus, Mr. Verma s original salary is `,, Let marked price be ` x. Then Discount % of x x 00 0 x Now, SP Marked price Discount x 9x x 0 0 This is given to be ` x or x Thus, marked price of the T.V. set is `, CP ` 80 Tax charges % of CP `.0 98 M A T H E M A T I C S VIII

100 Now, actual cost CP + Tax charged ` Thus, actual cost of the item is ` CP ` 9900 and SP ` 9000 Here, it is clear that CP > SP. So, Billu made a loss. Loss CP SP ` 9900 ` 9000 ` 900. Loss per cent Loss CP Thus, Billu s loss per cent is 9. CP ` 00, VAT 0%. Price before VAT CP 9 % ` 000. Thus, price of a sofa set before VAT added was ` P ` 000, R 8%, n or T years Let us first find simple interest. SI PRT ` 800 i.e., Simple interest ` 800. Now, find compound interest. n R 8 A P( + 00 ) 000( ) ( ) ` 8 CI A P ` 8. i.e.,compound interest ` 8 Required difference ` 8 ` 800 `.. P ` 0, R 8.% (i) At the end of second year, there are years elapsed n years A P ( R + ) 00 n 8. 0 ( ) ` l ` Therefore, amount at the end of second year received by Ritu is ` (ii) The amount A obtained in part (i) will be the principal for the third year. P ` R Now, A ( ) P ( + ) Required interest Thus, interest for the third year is ` 9.8. C O M P A R I N G Q U A N T I T I E S 99

101 WORKSHEET. Let constant of the ratio be x. Then Mr. Lal s wife had ` x. and four sons had ` x in all. Since all the sons has an equal share. Each son had ` x. Now, x 000 x x Thus, each son got ` 60.. P ` 000, R 8% per annuam % half yealy. n years half-years. R A P( ) n + gives 00 A 000 ( ) ( ) `.86 l `.86 CI A P `.86. Thus, the compound interest is `.86.. P ` 0000, n years, R 0% Using formula. R A P( ) n +, we get 00 0 A 0000 ( ) ( ) ( ) ` 0 CI A P ` 0. Thus, the compound interest is ` 0.. CP ` 00 0 ` 000 Since 0 milk bars had to be thrown away due to be rotten. Number of remaining bars SP ` 0 ` 0 Since SP > CP Therefore, Suman made a profit. Profit SP CP ` 0 ` 000 ` 0. Profit% Profit CP %. Thus, Suman s profit was.%.. CP of each almirah ` 800. SP of one almirah CP Loss CP 0% of CP 00 M A T H E M A T I C S VIII

102 SP of other almirah CP 0 00 CP CP CP 0 9 CP ` 60. CP Loss CP % of CP CP 00 CP CP CP CP ` 6. Thus, Rinku sold one almirah for ` 60. and other one for ` 6. OR CP of unit ` 8 ` SP of unit ` 0 0 ` Clearly, SP is greater than CP. Therefore, there is a gain Gain on unit ` ` ` Gain% Gain CP Hence gain is %. 6. CP of microwave oven after adding VAT % of ` 696. C O M P A R I N G Q U A N T I T I E S. Decrease in number of people Per cent decreased Thus, the decrease in number of people was 8.%. 8. Total number of parts Let the percentage of milk be x. Then, x % of 0 8 or x x %. Thus, the percentage of milk in the can is 80%. OR Profit SP CP 8 0 ` 6. Profit % Profit CP %. Thus, the profit is ` 6 and profit per cent is Let n games were played in all. According to given condition, we have 0% of n 0 or n n 0 Thus, 0 games were played in all. 0

103 WORKSHEET. (i) : (ii) : % % 00% 8%. 00%.%. % of students wear glasses. And (00 )% or % of students do not wear glasses.. Let Babita s income be ` 00. Then Anita s income ` (00 0) ` 80. So, Babita s income is ` (00 80) ` 0 more than Anita s income. Require percentage Thus, Babita s income is % more than Anita s income.. CP ` 00 Total CP CP + Sales tax CP + 6% of CP 6 CP + 00 CP CP `. Therefore, Poonam paid ` to the shopkeeper.. CP ` 000. Additional expenditure ` Total CP CP + Additional expenditure ` SP ` 0000 Profit SP Total CP ` Profit per cent Profit Total CP or l M A T H E M A T I C S % Thus, Mr. William's profit was % or 89.6%. 6. CP ` 0000 Profit % of ` 00 SP CP + Profit ` 000. Thus, selling price is ` P ` 800, n, R 0%. R 0 A P( + ) 800 ( ) n 800 ( ) (.). 800 (.) CI A P ` 6.0. Thus, amount is ` 06.0 and compound interest is ` 6.0. VIII

104 8. Let constant of ratio be x. Then, speed of car x km/hr and speed of bus x km/hr Now, x 6 gives x x. Therefore, the speed of the bus is km/hr. 9. Total number of students 00. Number of present students 8% of Number of absentees Total number fo students Number of present students students Thus, 6 students were absent on Monday. OR (i) Let the original price of a soap be ` x. 8 Then, VAT 8% of x 00 x 8 00 x 8 CP x + 00 x x. But this is given to be ` x 00 This gives x (approx.) Thus, the original price of the soap was `.. (ii) Let the original price of a shampoo be `y. Then VAT 8% of y 8 00 y 8 00 y. 8 CP y + 00 y y. But this is given to be ` y 80 This gives y (approx.) Thus, the original price of the shampoo was ` CP of each television ` 0000 Loss on one television 0% of ` SP of this television CP Loss ` 0000 Profit on other television % of ` SP of this television CP + Profit ` 600. Total SP ` 000 Total CP ` SP > CP Therefore, the shopkeeper made a profit. Profit SP CP Thus, the shopkeeper made a profit of ` 00 on the whole transaction. C O M P A R I N G Q U A N T I T I E S 0

105 WORKSHEET. Let Divya s salary before the increase be ` x. Then, the increase in her salary 0% of x x x. 0 So, her salary after increase x x x. But this is given to be ` x 6600 x ` Thus, Divya s salary before increase was ` 6, 0, Number of good students 6% of Number of students which are not good SP CP + 0% of CP or 9 CP + 0 CP 00 or 9 CP + CP 0 CP CP ` 0. Therefore, the cost price of the almirah was ` 0.. Let Rohan s income be ` 00 Then Amit s income ` If Amit s income is `, then Rohan s income is less by ` When Amit s income is ` 00, Rohan s income is less by ` 00 i.e., ` 0. Therefore, Rohan s income is 0% less than Amit s income.. Let CP of mango be ` x Then CP of 8 mangoes ` 8x SP of 6 mangoes ` 8x SP of mango ` 8 x 9x ` 6 8 Now, profit on mango 9 x ` ( ) 8 x ` 8 x x Profit % % x 8 Therefore, the gain is.%. 6. Marked price ` 80 Discount 0% of marked price 0 80 ` 8 00 SP 80 8 ` Profit SP CP CP Now, using the formula, Profit % Profit CP 00, we get CP 6 CP 00 or 6 CP CP or 6 CP 00 or CP 00 6 or CP 00 Therefore, the cost price of the article is ` M A T H E M A T I C S VIII

106 OR Increase in the price 0% of ` 8000 New price Price last year + Increase in the price ` 8000 Thus, the new price of the scooter is ` P ` 6000, R n years Using the formula, R A P( ) n +, we get 00 A 6000 ( ) ( ) + 00 % %, A 6000 ( ) ( ) CI A P ` 68. Thus, Roma paid ` 68. as compound interest. 8. P 6, n 9 months quarters, R 6% per annum 6 C O M P A R I N G Q U A N T I T I E S 8 i.e., % quarterly. Using the formula, R A P( ) n +, we get 00 A 6 ( ) ( ) ( ) ` 6 CI A P 6 6 ` 9. Thus, the compound interest is ` Let the rate of VAT be x %. Then x % of 0 x or 0 or 00 0 x x 0 0%. Thus, the rate of VAT is 0%. 0. Total CP + ` 0. SP ` 00 Profit SP Total CP 00 0 ` 60. Profit per cent %. 0 OR Discount Marked price Selling price 0 00 ` 0 Discount per cent Discount Marked price %. 0

107 Chapter 9 ALGEBRAIC EXPRESSIONS AND IDENTITIES WORKSHEET. (C) The expression a b is a binomial because it has terms.. (A) x and x are like terms because they are formed from same variable and the powers of the variable are the same.. (D) Adding, pq pq + pq + pq + pq + pq +.. (A) Subtracting, xy yz zx + 0xyz xy + yz zx + xy yz + zx + 0 xyz.. (A) x y xy x y xy. 6. (B) ( a) ( a ) a a a a 6.. (C) (a + b) (a + b) a (a + b) + b(a + b) 6a + 8ab + 9ab + b 6a + ab + b. 8. (D) (a + b) (a b + c) (a b)c a(a b + c) + b(a b + c) ac + bc a ab + ac + ab b + bc ac + bc a b ab + bc ac. 9. (B) a + b! ab x + y! xy. 0. (D) (x 6) x 6 x 0.. (B) 8ab ab 8 aa b b 9a b.. (B) We have x + x + x x(x + x + ) x(x +) Therefore, x + x + x xx ( + ) xx ( + ) xx ( + ) x +.. (C) (a + b) (a b) a(a b) + b (a b) a ab + ab b a b.. (B) (a + b) a + ab + b.. (A) The factorization of x 6y is a binomial as it is a binomial. 6. (B) ( + c) + c + (c) 9 + 0c + c. 6 x x+ 0. (D) x 8 x x 8 x 6 x x+ 0 6 x x + 6x + 0 6x (C) (x y z + x y z + x y z ) x y z (x + y + z) ( x y z + x y z + x y z ) xyz (x + y + z). 06 M A T H E M A T I C S VIII

108 9. (B) (x + )(x ) x(x ) + (x ) x x + x 6 x + x (B) (a b) a ab + b. WORKSHEET 6 6. (i) x + 9x x + x. (ii)8p p + 9p.. (i)ab (ii) x + xy 6x +. (i) Adding, The required sum is 6x + x. (ii) Adding, pq qr + qr rp pq + rp The required sum is 0.. (i) Subtracting, x x + x x + x x + x x 6x + x y + x + x y xy y + x + x y xy + + y + x + 0 xy The required subtraction is y + x xy. (ii) Subtracting, a a + a + a a a a a + a + 6 The required subtraction is a a + a + 6. (iii) Subtracting, b ab b + ab + b 6ab The required subtraction is b 6ab.. (i) ( x)(x ) ( ) x x (ii) ( )( ) 0x. y y ( ) y y y. 6. (i) Area of a rectangle Length(l) Breadth (b) (i) l x, b 8y Area l b ( x) ( 8y) ( ) ( 8)x y 6xy. (ii) l ab, b a b Area l b ab ( a b) ( )ab a b 8a b. OR (i) Remember the identity: (a b) a ab + b Put a 00 and b. (00 ) or (ii) Remember the identity: (a + b) a + ab + b Put a 90 and b. (90 + ) or A L G E B R A I C E X P R E S S I O N S A N D I D E N... 0

109 . (i) Volume of a cuboid Length Breadth Height ax by cz a b c x y z 0abcxyz cubic units. (ii) Volume of a cuboid Length Breadth Height (xy) ( y) ( x) ( ) ( ) xy y x 8x y cubic units. 8. Let y x(8x ) x 8x x x x. Substituting, x, we get y ( ) ( ) 6 OR Remember the identity: (a b)(a + b) a b Putting (i) a 0 and b, we have (0 )(0 + ) 0 or 6 (0 0) ( ) (ii) a 00 and b, we get (00 ) (00 + ) 00 or 99 0 (00 00) ( ) or Required value WORKSHEET (x xy + y ) (x + y ) x xy + y x y (x x ) xy + (y y ) x xy + y. 08 M A T H E M A T I C S OR Remember the identity: Substituting (a + b) a + ab + b a x and b y, we get (x + y) (x) + x y + (y) or 9x y or 9x + y + or 9x + y or 9x + y.. (i)(a + b)(a + b) a(a + b) + b(a + b) a a + a b + b a + b b 0a + ab + 8ab + b 0a + ab + b. (ii) ( x)( + x + x ) ( + x + x ) x ( + x + x ) + x + x x x x x x x.. (i) Substituting a and b in (a + )(b ), we get (a + )(b ) ( + )( ) 0 0 ( m 0 0) (ii) Substituting x 0 and y in (x y )(x + y ), we get (x y )(x + y ) (0 )(0 + ) VIII

110 . (i) We have (a )(a + ) (0 )(0 + ) ( 0 0) ( ) (). a (a + ) (a + ) a + a a (a ) (a + ) + a + a a + a + a a 0. (ii) (t + s )(t s) t(t s) + s (t s) OR (i)(x y )(x y ) t ts + s t s. x (x y ) y (x y ) 9x 6x y 6x y + y 9x x y + y. (ii) ( a + b ) ( a b ) a( a ) + ( a ) a 6a b b b + 6a b 9 b a 9 b.. (i) (a + 6)(a + 6) (a + 6) Remember the identity: Put (A + B) A + AB + B A a and B 6 to get (a + 6) a + a or (a + 6)(a + 6) a + a + 6. (ii)(a )(a ) (a ) Remember the identity: (A B) A AB + B b A L G E B R A I C E X P R E S S I O N S A N D I D E N... Put A a and B to get (a ) (a) a + or (a )(a ) 9a 66a +. (iii) Remember the identity: Put (A B)(A + B) A B A x and B to get (x )(x + ) (x) 6. (i) Remember an identity: x 9. (A B) A AB + B Substituting A a and B, we get (a ) a a + a a +. (ii) Remember an identity: (A + B) A + AB + B Substituting A a and B, we get ( a ) + ( a ) 9a 6 OR (i)(x + x + )(x x + ) (ii) We have a + ( ) + + 6a + 6. x (x x + ) + x(x x + ) + (x x + ) x x + x + x x + x + x x + x +. (x + y)(x y) x + x(x y) + y(x y) x 6xy + 6xy 9y x 9y 09

111 (x + y)(x y)(x + 9y ) (x 9y )(x + 9y ) x (x + 9y ) 9y (x + 9y ) 6x + 6x y 6x y 8y 6x 8y.. (i) Remember the identity: (a b)(a + b) a b Putting a 00 and b, we get (00 )(00 + ) 00 or 9 0 (00 00) ( ) (ii) Remember the identity: a b (a + b)(a b) Putting a 6 and b, we get 6 (6 + )(6 ) WORKSHEET8. Adding, x + x x + x + x x + 8 x. (a + ab + b )(a b) a (a b) + ab(a b) + b (a b) a a b + a b ab + ab b a b.. (x x + ) (x + x + ) x x + 9 x x 8 x 9x +.. Area of a rectangle Product of two consecutive sides 6x x + x 6x + 6 6(x + ) square units.. Use the identity: (a + b) a + ab + b (i) Substituting a x and b, we have ( x ) + ( x ) 0 M A T H E M A T I C S x 9 + x + + x +. (ii) Substituting a x y and b xy, we have (x y + xy ) (x y) + x y xy + (xy ) x y + x y + x y. OR (i) We have (x + )(x ) x(x ) + (x ) 6x x + x 0 6x + x 0. And (x + )(x ) x(x ) + (x ) x x + x 6 x + x 6 Therefore, (x + )(x ) + (x + )(x ) 6x + x 0 + x + x 6 8x + x 6. VIII

112 (ii) We have (6x + y )(6x y ) (6x ) (y ) [Using the identity: (a + b)(a b) a b ] (6x 6x ) (y y ) 6x y Therefore, (6x + y )(6x y ) 6. Take the identity: 6 x y x y. (a b) a ab + b (i) Substituting a x and b y, we get (x y) (x) x y + (y) x 0xy + y. x y (ii) Substituting a and b, we get x y ( ) x ( ) x y y + ( ) x xy 6y (i) Remember the identity: (a b) a ab + b Substituting a 000 and b, we get (000 ) or (ii) Remember the identity: (a + b) a + ab + b Substituting a and b 0., we get ( + 0.) (0.) or (.) Perimeter of a square Side (x + y ) 6x + y. 9. Perimeter of a rectangle (length + breadth) (x + x + + x x ) (x x + ) 8x x x + x (x x) + x + x x + + x ( + ) + (x + x) + (x x ) + x + x. WORKSHEET 9. Side x + 8y 8 Perimeter of a square Side (x + 8y 8) 6x + y.. Perimeter of a rectangle (one side + other side) (8x + x + + x x ) (x + x ) x + 8x 8.. Substituting p, q, s in (p q s ), we get (p q s ) { ( ) ( ) } A L G E B R A I C E X P R E S S I O N S A N D I D E N...

113 ( + ) ( ). Substituting r, s in (r s ), we get (r s ) { ( ) } ( ) ( ) Therefore, (p q s ) (r s ) ( ) +.. (i)(x + y)(x + y) x(x + y) + y(x + y) x + xy + x y + y. (ii)(l + lp + p )(l p) l (l p) + lp(l p) + p (l p) l l p + l p lp + p l p l p. OR (i) Substituting m, n in (m n)(m n), we get (m n)(m n) { ( )} { ( )} ( + )( + ). (ii) Substituting a, b in (a + b), we get (a + b) () + () (a + b)(a + b) (i) Take the identity: (x + a)(x + b) x + (a + b)x + ab Substituting a 8 and b, we get (x 8)(x ) x + ( 8 )x + ( 8) ( ) x 0x + 6. (ii) Take the identity: (A B) A AB + B Substituting A a and B, we get a ( ) a a a ( ) a a + a. (iii) Take the identity: a + ( ) a (a + b)(a b) a b Substituting a x and b y, we get (x + y )(x y )(x ) (y ) (x x ) (y y ) x y. 6. (i) Take the identity: (a b)(a + b) a b Put a 00 and b 0 to get (00 0)(00 + 0) 00 0 or 90 0 (00 00) (0 0) (ii) Take the identity: (a b) a ab + b Put a 000 and b to get (000 ) or Adding, x y + x y x y + x y + 9 Thus, the required sum is x y + 9. M A T H E M A T I C S VIII

114 8. (i)xy y (xy 8y) xy y xy + 8y (xy xy) + ( y + 8y) xy + y. (ii) a b 8b (a b + ab b ) a b 8b a b ab + b ( a b a b) + ( 8b + b ) ab a b b ab. WORKSHEET 60. (i) Adding the three expressions, xy x y xy + x y + xy x y 0 + x y. Required sum is x y. (ii) Adding the three expressions, x x + x + x x x x x + ( + + ) x ( + ) x+ 6 + Required sum x + ( ) x x x x + x + 0 x x +. + x. (i) Subtracting, x + x + x + 6 x x + x + + x + x x + (ii) Subtracting, bc + ab 6 bc + ac + ab bc ac + Required subtraction ab 6 + 6bc ac +.. (i) ( x ) ( xy ) ( yz ) ( ) ( ) ( ) x xy yz ( ) x y z x y z. (ii) ( 9 ) ( ) abc 9 ab ( b c) 9 ( ) 9 ( ) abc a b b c ( ) ( ) a b 6 c a b 6 c.. (i) 6a(a ) 6a a and a( + a) a + a a 6a(a ) + a( + a) a (6a a) + a + a a 6a + a + a + a 6a + a a(6a + ). (ii) st(s t) s t st s (t t ) s t s t A L G E B R A I C E X P R E S S I O N S A N D I D E N....

115 t (s s) s t st and st(s t) s t st st(s t) s (t t ) t (s s) s t st (s t s t ) + st(s t) (s t st ) + s t st s t st s t + s t s t (s t s t + s t) + st + s t st + ( st + st st ) + (s t s t ) s t st + s t st(s t + st).. (i) Use the following identity: (a + b) a + ab + b Put a ( ) + x y x and b y to get ( ) ( ) x + or ( ) ( ) + + x y x (y ) + (y ) x y 9 x + 8 x y + 9y ( a a a) (ii) Use the following identity: (a b)(a + b) a b Put a 6x and b y to get (6x y )(6x + y ) (6x ) (y ) (iii) Use the following identity: (a b) a ab + b 6x 9y. M A T H E M A T I C S Put a x and b y to get ( ) x y ( ) ( ) x ( ) x y or ( x y ) ( x y ) 6. (i) We have (a b)(a + b) a b Substituting get + ( ) y y. x xy + a 0 and b, we (0 )(0 + ) 0 or (ii) We have Substituting get a b (a + b)(a b) a 8 and b, we 8 (8 + )(8 ) 0 0. OR Product xy(xy + y ) xy xy xy y x y xy Adding it to x y xy, we get xy(xy + y ) + x y xy x y xy + x y xy ( x y + x y ) + ( xy xy ) x y xy. VIII

116 0 Chapter VISUALISING SOLID SHAPES WORKSHEET 6. (B) The relation among the numbers of faces F, vertices V and edges E of a plyhedron is given by F + V E.. (C) Using F + V E, we get E F + V (B) Using Euler s formula: F + V E, x V E + F y F E + V z E F + V (C) A cuboid has 6 faces and edges.. (A) A tetrahedron has vertices. 6. (B) The top view is shown in the part B.. (C) The side view of the given figure is 8. (B) A cube has 6 congruent faces. 9. (A) Each vertex of a cuboid is formed by meeting of faces. 0. (A) The lateral faces of a pyramid are triangles with a common vertex.. (D) A prism having a square base has edges. Fig.: Prism with square base.. (C) A pyramid with rectangular base has vertices.. (B) E 0, F 6. Using F + V E, we get V (C) The match box is a cuboid.. (B) The solid shown in option (B) is made up of a cylinder and a cone, so it is a nested solid. 6. (C) A sphere has neither vertex nor flat face.. (B) A prism has the given properties. WORKSHEET 6. Name Example (i) Cylinder Drum (ii) Cone Tent (iii) Sphere Ball.. (i) Tetrahedron Example -Tent No. of triangular faces - No. of vertices - No. of edges - 6. Fig.: Tetrahedron (ii) Hexahedron (cube) Example - Die Number of square faces-6 Number of vertices - 8 Number of edges -. Fig.: Hexahedron V I S U A L I S I N G S O L I D S H A P E S

117 . We know that a cube has 6 faces, i.e., f 6 and 8 vertices, i.e., v 8 Using Euler s formula: f + v e, we have e f + v We know that an icosahedron has 0 faces, i.e., f 0 and vertices, i.e., v. Using Euler s formula: f + v e, we have e f + v So, f + v 0 + and e Solid f v e f + v e + (i) Cube 6 8 (ii) Icosahedron (i) (ii) WORKSHEET 6. (i) (ii).. (i) (ii). (i) A tetrahedron has vertices. (ii) A cuboid has 8 vertices. (iii) A pentagonal pyramid has 6 vertices. (iv) An octagonal pyramid has 9 vertices.. (i) A cuboid opened at the top. (ii) A cone surmounted on a cylinder. (iii) A cylinder. (iv) A hollow cylinder.. (i). (i) (ii). Line of symmetry (ii) The given figure is a rhombus which has lines of symmetry. 6 M A T H E M A T I C S VIII

118 . (i). (ii). 6. A cube has planes of symmetry.. 8. V I S U A L I S I N G S O L I D S H A P E S

119 9. (ii) A cone has edge. (iii) An octahedron has edges. (iv) A rectangular prism has edges.. (i) WORKSHEET 6. (i) A triangular prism has faces. (ii) A hexagonal pyramid has faces. (iii) A pentagonal prism has faces. (iv) An octagonal prism has 0 faces.. (i) A square prism has edges. (ii) Fig.: Cube Fig.: Cuboid 8 M A T H E M A T I C S VIII

120 . F 6, V 8, E? Using Euler s formula, F + V E or E F + V Thus, the given polyhedron has edges.. (ii) Fig.: Net a cube 6. (i) The given net is of a cone. (ii) The given net is of a tetrahedron. (iii) The given net is of a cylinder opened at the top.. (i) Tetrahedron. (ii) Cuboid opened at the top. 8. (i) (iii) Fig.: Cuboid 9. A sphere has infinitely many planes of symmetry. V I S U A L I S I N G S O L I D S H A P E S 9

121 . WORKSHEET 6 Using Euler s formula, F + V E Here, F + V E 0 + 0! Therefore, no polyhedron is possible.. Fig.: Tessellations of an isosceles triangle. Yes, we can use right angled triangle, 8, square etc.. The required net is given below.. 8. (i) (ii) (iii) Fig.: Net..... (i) A decagonal prism has faces. (ii) A pentagonal pyramid has 0 edges. 9. (i) Cylinder. (ii) Square pyramid. (iii) Triangular prism. WORKSHEET 66. (i) (iii) (iii) Each face of a tetrahedron is in the shape of a triangle. (iv) Hexagonal prism.. In the given figure: f 9, v 9, e 6 Now, f + v And e (ii) Therefore, f + v e F 0, E 0, V 0 M A T H E M A T I C S VIII

122 .. (i) Isosceles trapezium, kite. (ii) Rhombus, rectangle, square. (iii) Rhombus, rectangle, square.. (i) (ii) x + 6 or x nd Column: F, V y, E 9 Substituting these values in the Euler s formula, we get + y 9 or y rd Column: F 0, V, E z Substituting these values in the Euler s formula, we get 0 + z or z Therefore, the complete table will be: Number of Faces 8 0 Number of Vertices 6 6 Number of Edges Solid f v e f + v e + Octahedron 8 6 Dodecahedron Let unknown numbers in st column be x, in nd column be y and in rd column be z. Euler s formula is F + V E st column: F x, V 6, E Substituting these values in the Euler s formula, we get 8. Figure (i): The given figure is of a tetrahedron. Number of faces Number of edges 6 Number of vertices Figure (ii) The given figure is of a triangular prism. Number of faces Number of edges 9 V I S U A L I S I N G S O L I D S H A P E S

123 Number of vertices 6 Now, we can make a table as given below: (ii) Fig. No. of No. of No. of faces edges vertices (i) 6 (ii) 9 6 WORKSHEET 6. (i). (i) Yes, tessellation is possible by using equilateral triangle. The figure is given below: (ii) (ii) Yes, tessellatiion is possible by using regular pentagon. The figure is given below:. (i). Fig.: Base design M A T H E M A T I C S VIII

124 . Diamond is the example of octahedron. 8. Fig.: Diamond The octahedron has 8 faces, 6 vertices and edges i.e., f 8, v 6, e. 6. (i) The given tessellation is made up of regular hexagon and rhombus. (ii) The given tessellation is made up of rectangles.. 9. Fig.: Net of a cone V I S U A L I S I N G S O L I D S H A P E S

125 Chapter MENSURATION WORKSHEET (A) Area d d 8 0 cm.. (B) Area Sum of parallel sides. (A) Area Distance between them ( + ) 8 0 cm. 6 Sum of parallel sides Altitude Altitude.. cm.. (D) Ar (ABCDE) Ar(ABC) + Ar(ACD) + Ar(AED) ( 8.8. ) ( ) cm.. (C) 6a 9 a 9 6 a cm. 6. (A) Required number + ( ) 9 Volume of cuboid Volume of cube (D) Length of rod Length of diagonal ( ) cm. 8. (C) Curved surface πr h πrh 9. (B) πr h 9 r 0 9 r 9. 0 r.. cm d r. cm. 0. (B) Volume...86 m Capacity l 860 l.. (C) Capacity Volume cm l.. (D) Volume πr h 0 60 cm Capacity 60 l.6 l (A) Surface area of the roller πrh cm. Area of the road cm m.. (C) m 000 l. M A T H E M A T I C S m VIII

126 . (A) 000 cm l cm 000 l 0000 cm (A) Area Base Height M E N S U R A T I O N l 0 l cm. WORKSHEET 69. (i) l cm, b cm Area of rectangle l b 6 cm Perimeter of the rectangle (l + b) ( + ) 6 cm. (ii) l.9 m, b 0.66 m Area of rectangle l b m. Perimeter of the rectangle (l + b) ( ).8.6 m. OR (i) The given solid is a cuboid. Its measurements are given below: Length l mm, Breadth b 9 mm, Height h 9 mm. Volume V l b h mm Surface area (lb + bh + hl) ( ) ( ) 9 9 mm Lateral surface area (l h + b h) ( ) (08 + 8) 89 8 mm Diagonal l + b + h mm. (ii) The given solid is a cube. Edge a Volume a Surface area 6a Lateral surface area a Diagonal a.. (i) Side of square a 0. cm Area a cm and perimeter a 0..0 cm. (ii) Side of square b. cm. Area b... cm and perimeter b.. cm.. l.6 cm, A.8 cm, b? A l b or.8.6 b b cm.

127 OR Let constant of ratio be x. Then, l x, b x, h x Total surface area (lb + bh + hl) (x x + x x + x x) (x + x + 6x ) x This is given to be 88 m. x 88 x 88 x. l x, b x, c x 6. Therefore, the dimensions are m, m, 6 m.. Length of a rectangle Area Width 00 m 90 m 80 m. Thus, length of the rectangular field is 80 m.. Side of a square Area Thus, side of the square is 0 m. OR Let length of side a and height h The given triangle is ABC. Draw perpendicular AD on BC. In right triangle ADC, a h a + ( ) or h a a ( CD a ) a h a Now, area a h 6 or a a 6 or a 6 or a 6 or a Thus, length of side is m. 6. Let l x and b x Then, perimeter (l + b) (x + x) 0x But this is given to be 00 cm 0x 00 cm This gives, x 0 l x 0 0 cm and b x 0 00 cm. OR Base Hypotenuse Side Now, 69 cm area Base Height 0 cm.. Side of square m Perimeter of the square Side 8 m 6 M A T H E M A T I C S VIII

128 For rectangle, l 8 m and b m Perimeter of the rectangle (l + b) (8 + ) m. Clearly, the square has larger perimeter than that of the rectangle. 8. The floor is in the shape of a rectangle. For the floor, l 0 m and b 8 m In right AMF, AM + MF AF (Pythagoras property) or AM + AM 96 or AM 6 cm Area of the floor l b m Now, area of ABF AM BF Side of a tile 0. m 0 m m Area of a tile Side m Now, the required number of tiles Area of the floor Area of a tile 60 ( ) OR Area of rectangle BCEF BC BF 0 0 cm ABF is an isosceles triangle with AB AF cm cm Since ABF and CDE are congruent Area of CDE Area of ABF 0 6 cm So, area of the given polynomial Area of ABF + Area of rectangle BCEF + Area of CDE (6 + 6 ) cm. 9. Let the side of a square be a. Then its area a But the area is given to be 00 m a 00 or a or a ( 0) BM MF 0 cm cm a 0 0 m. Now, perimeter a 0 80 m. Thus, the perimeter of the square is 80 m. M E N S U R A T I O N

129 WORKSHEET 0. Side a 8. cm 8 0 cm Area a 8 ( ) cm Perimeter a cm. OR Let edge of a cube be b. Surface area 6b 06 b b 6 m Now, volume b m.. Area of floor Length Breadth 8 Area of a tile Length Breadth Required number of tiles Area of floor Area of a tile Thus, 6 tiles are required to pave the floor. OR (i) Area of the parallelogram Base Height...68 cm. (ii) Area of the parallelogram Base Height cm.. l 6 m, b 8 m Perimeter (l + b) (6 + 8) 8 6 m. Required length of wire Perimeter 6 m.. Perimeter of the field Side..60 m. Distance covered by Chulbul Perimeter m.. Side of square a cm Area of the square a 6 m. Cost of cultivating on 00 m ` 0 Cost of cultivating on m ` 0 00 Cost of cultivating on 6 m ` ` 00 0 ` 6.0 Thus, the cost of cultivating the field is ` 6.0. OR (i) Area of trapezium Sum of parallel sides Distance between them ( + 9 ) 8 cm. 8 M A T H E M A T I C S VIII

130 (ii) Area of trapezium Sum of parallel sides Distance between them ( + ) 6 cm. 6. Let other side of the rectangle be x. Then, area x But this is given to be 86 sq.m. x 86 x 86 8 m. Perimeter of the rectangle ( + 8) 60 0 Thus, the perimeter of the field is 0 m.. Area of a trapezium Sum of parallel sides Distance between them ( +.) m 8. d 8 m, d m d d m, m Area dd 8 m d d Side ( ) ( ) + 8. M E N S U R A T I O N. m. 9. Base. cm, height cm Area of a parallelogram Base Height. 0. cm. 0. Base 8. cm, area.6 cm Area of a parallelogram Base Height Height Area Base cm.. Area of the triangle AB AC cm. WORKSHEET. Base. cm, height. cm Area of a parallelogram Base Height cm.. Area Base Height s h sh square unit.. Area 0 m, h m Area Base Height Base Area 0 Height 9 8 m. 9

131 . Sum of parallel sides.8 m +.6 m 9. m and distance between them 0.6 m Area of a trapezium Sum of parallel sides Distance between them m.. d 8.6 cm, d. cm Area of rhombus dd cm. 6. Area 9 m,d 6 m, d? Area dd 9 6 d d 9. 6 Thus, the length of the other diagonal is m.. Let constant of given ratio be x. Then length l x and breadth b x. So, Area l b x x x This is given to be 60 m. x 60 x 60 or x l x 60 and b x 6 Now, perimeter (l + b) (60 + 6) 96 9 Cost of fencing Perimeter Cost of fencing perimeter Thus, The cost of fencing around the field is ` 600. OR Area of floor 6000 m cm Base of parallelogram b cm and corresponding height h 0 cm. Area of a tile b h 0 0 cm Required number of tiles Area of floor Area of a tile Thus,,00,000 tiles are required to cover the floor. 8. (i) Side a cm 9 cm Area Side cm 6 Perimeter Side 9 9 cm. Side b. cm Area Side.. 6. cm Perimeter Side. 0 cm. 0 M A T H E M A T I C S (ii) VIII

132 OR The given figure is a right circular cylinder. r.. cm, h 8. cm Volume πr h (.) cm Curved surface area πrh cm. 9. ABC is the given triangle. Draw AD BC In right triangle ACD, AC AD + CD AD 0 00 AD cm Now, Area of ABC 0. Area h b M E N S U R A T I O N BC AD 0 m. h Area b m.. (i) Area of ABC 8 cm Area of small square cm Area of shaded portion Area of ABC Area of small square 8 80 cm. (ii) The shaded portion represents six equilateral triangles each of side cm. Area of one such triangle Side cm. Area of shaded portion 6 cm cm. OR Join GD. Let A Area of trapezium ABCH and A Area of rectangle CDGH. Areas of trapeziums ABCH and GDEF are equal. Sum of parallel sides Distance between them A ( ) (8 + 8 ) 8 ( + )

133 6( + ) cm BD cm Now area of ABC AC BD A Length Breadth HC CD (8 + ) ( + ) ( + ) cm Now, area of ABCDEFGH A + A 6( + ) + ( + ) 6( + ) cm. WORKSHEET. Area of a trapezium Sum of parallel sides Distance between them (8 + 8) cm.. (i) Draw BD AC So, BD DC cm cm. (ii) Area of trapezium Sum of parallel sides Distance between them (BC + AD) BF (0 + 0) 8 0 cm. OR (i) Area of ABC 0 AC BD cm Area of small rectangle 0 cm Area of shaded portion cm. (ii) Area of parallelogram ABCD AB EF 0 60 cm. Area of ABE AB EF In right BDC, BD + DC BC BD BC DC ( ) 0 0 cm. Area of shaded portion cm. M A T H E M A T I C S VIII

134 . Let length of required side be x. Area of trapezium Sum of parallel sides Distance between them ( + x) 80 or 00 or + x 00 or x 0 0 Thus, the length of the required side is m. OR ABCD is the given quadrilateral. We have to find AC. Area of ABCD Area of ABC + Area of ADC AC + or AC M E N S U R A T I O N AC or AC 8. Thus, length of the other diagonal is 8. m.. Area of floor Length Breadth 8 6 m Area of a tile m Area of floor Required number of tiles Area of a tile 6 08 Thus, 08 tiles are required to pave the floor. OR l 0 m, b 0 m, h 6 m Area of floor of the pool l b m Area of walls of the pool (l + b) h (0 + 0) m Total area of the floor and walls m Cost of cementing ` 80 ` Area 60. sq.m, b 98 m, h? Area bh gives h Area b h Thus, height is.0 metres. 6. Let initially, base b and height h. Then finally, base b and height h So initially, area A b h bh and finally, area A b h bh Dividing A by A, A A bh bh bh bh or A A Hence, area of the triangle will be times.

135 OR Area of a square Side Side Side Thus, length of the side of the square is m.. Area 80 cm, d cm, d? (ii) Area of polygon ABCD Area of ABC + Area of ACD AC BN + AC DM Area d d Area d 80 0 d Therefore, the measure of other diagonal is 0 cm. OR l , b 8 cm, h 8 cm Volume l b h cm Surface area (lb + bh + hl) ( ) 6 cm. 8. (i) The given polygon is made up of 6 equilateral triangles. Height of one such triangle ( ) Area of one triangle Area of the polygon cm cm. cm AC(BN + DM) 8 ( + ) 0 cm. (iii) Area of trapezia ABCD and HEFG are same as they are congruent. Area of ABCDEFGH Area of trapezium HEFG + Area of rectangle ADEH (6 + ) cm. 9. (i) Let old side a Then new side a So, Old area a + 6 New area (a) a New area Old area a a or new area Old area Thus, the new area will be four times. M A T H E M A T I C S VIII

136 (ii) Let old length l and old breadth b Then new length l and new breadth b Old area l b and new area l b l b So, New area Old area l b l b or New area Old area Thus, the new area of the rectangle will be four times. OR (i) Let old base b and old altitude h Then new base b and new altitude h Old area bh and new area b h bh So, New area Old area bh bh or new area Old area Thus, the new area will be four times. (ii) Let old base b and old height h Then new base b and new height h Old area bh and new area b h bh New area So, Old area bh bh or new area Old area. Thus, the new area will be four times. M E N S U R A T I O N WORKSHEET. l m, b m, h m Surface area (lb + bh + hl) ( + + ) ( + + ) 9 8 m.. l m, b 0.8 m, h 0. Total surface area Outer surface area + Inner surface area Outer surface area (lb + bh + hl) ( ).88.6 m.. l m, b 8 m, h m Area of walls (l + b) h ( + 8) 0 00 m Area of ceiling l b 8 96 m Total area of the walls and ceiling 00 m + 96 m 96 m 8 m can be painted by can m can be painted by 8 can 96 m can be painted by 8 96 cans or cans Thus, cans of paint will be required. OR Area of region A B C D E F Area of A E F + Area of A B D E + Area of B C D π π

137 m Area of region A B C D E F Area of A E F + Area of A B D E + Area of B C D π ( + ) + (90 + 8) 90 + π ( + ) m Now, area of the shaded portion m.. Total surface area of the room (lb + bh + hl) ( + + ) m Cost of white washing ` ` 666. OR Area of the garden Base Height m Cost of levelling ` 960 ` 90.. Area of walls. m (l + b) h. or ( +.) h.. h. m Let side a 6a 0 a 0 or a m. 6. r cm, h 0 cm Curved surface area πrh 0 60 cm. 8. h 0 cm, r 8 cm cm Area covered to level the road Curved surface area of the roller Number of revolutions πrh cm m m. 9. Edge of the cube a mm. cm Space covered by the cube Volume a cm. 6 M A T H E M A T I C S VIII

138 0. r cm 6 cm, h 0 cm Volume of cylinder πr h M E N S U R A T I O N cm.. Let height of the raised platform be hm. Volume of platform Volume of earth dug out or h π( ) 6 or h 6 or h or h or h. Thus, the height of the raised platform is. m. OR Required area Area(rectangle ABCD) + Area(semicircle) AB BC + ( ) π BC m. WORKSHEET. Volume of a cuboid l b h cm.. Volume of a cube a cm.. Volume of resulting cuboid l b h ( ) cm.. Let l x, b x and h x. Then, l b h x x x 0000 or 0000 x 000 or x 0 l x 0, b x 0 and h x 0. Thus, dimensions are: 0 cm, 0 cm, 0 cm. OR Area of a trapezium Sum of parallel sides Height 6 Height + 6 cm.. l 00 cm, b 0 cm, h 8 cm Volume l b h cm Weight grams 00 kg. OR Volume l b h 0 cm 0 8 h 0 This gives, 0 h cm. 6. Volume πr h or ( ) 0 h 0000 This gives, h cm.. Since the rainwater falls 0 cm, Therefore the height of water level on the roof is 0 cm.

139 (i) Volume of rainwater (0 00) ( 00) cm 08 m. 08 (ii) Rise in water level 0. m 0 cm. 8. r m m, h 0 m. A well is in the form of right circular cylinder. Volume of earth taken out πr h 0 00 m. 9. πr h 80 h 80 h cm. OR Volume 8000 l m 8 m Let depth of the water h. Now, 6. h 8 8 h 6. m. 0. (i) The given solid is a cube with edge a. m Volume a....8 m Total surface area 6a m Lateral surface area a.. m. (ii) The given solid is a cuboid with measures: l 6 cm, b cm, h cm Volume l b h 6 0 cm. Total surface area (lb + bh + hl) ( ) 8 cm Lateral surface area (l + b) h 0 00 cm.. Let edge of the cube formed be a. Volume of the cube formed Sum of volumes of three given cuboids or a or a. cm (approx.). WORKSHEET. (i) Area of shaded portion π π. (9.).0 cm. (ii) Area of shaded portion π 0 π. Volume πr h 08 (00 ) 0. cm. 8 M A T H E M A T I C S or r 08 r 08 or r 8 cm Lateral surface area πrh 8 cm. VIII

140 . Let raise in height of the plot be h. Volume of plot Volume of earth dug out or 0 8 h 8 or 80h h. m. 80 OR Let the edge of the cube be a. Volume of cube a a cm Surface area of the cube 6a cm.. Total surface area πrh + πr πr(h + r) 9 (60 + 9) cm. OR h m, (l + b) 0 m or l + b m Area of four walls (l + b) h 90 m.. Total surface area 6a cm Volume a cm. New volume 6. Old volume ( a ) 8 a Thus, the volume will be 8 times. M E N S U R A T I O N. Volume πr h cm. Curved surface area πrh.8 0 cm. OR a 0 cm 0. m Volume of ice a (0.) 0. m Weight of the ice kg. kg. 8. a a a cm Surface area 6a 6 9 cm. OR In the pool, length of water 0 m, breadth of water 0 m. Let height of water level h So, 0 0 h 0 0 h 0. m 0 cm (i) The given solid is a cylinder Volume πr h mm Curved surface area πrh mm Total surface area πrh + πr mm. 9

141 (ii) The given solid is a hollow cylinder. r cm, r 9 cm Volume π(r r ) h (9 ) 8 86 cm Curved surface area π(r + r )h ( + 9) 8 86 cm Total surface area π(r + r )h + π(r r ) 86 + (9 ) 0. cm. 0. h m, (l + b) 0 m or l + b m Area of four walls (l + b) h 90 m. 0 M A T H E M A T I C S VIII

142 Chapter EXPONENTS AND POWERS WORKSHEET6 6. (B) In a b, exponent is b In 0, exponent is.. (D)...five times.. (B) a a (D)( ) ( ) () ( ) 6.. (D) a 0 for a 0 ( + + ) 0 ( + + 0) 6. (D)( ) 6 ( ) ( ) 6 ( ) ( ) 6.. (C) ( 8 ) (C) p 0 p 0 +p 0 + p 0 p. 9. (B) At x, x(x x ) x ( ) 8 6. { } 0. (A) ( ) ( ) ( ) E X P O N E N T S A N D P O W E R S ( ) (A) a m b m (a b) m (ab) m.. (B) (C) 0 km km km.. (A) + + ( + + ).. (B) kg.0 0 kg. 6. (C) (D) ( ), ( ) and ( ) ( ) ( ) ( ) is false. 8. (B) (A) (C) nanometre m. (i) () 9 0 WORKSHEET 00 m 0 9 m.. (ii) (i) (ii)

143 . (i) ( ) (ii) ( ). (i) ( ) 6 (ii) ( ) ( ) + + ( ) ( ) ( ) 0 ( ) 0 0. ( ) ( ) ( ) ( ). + ( ) 0 ( ) () ( ) ( ) ( ) ( ). (i) (ii)( ) ( ) 6. (i) 8 (ii) ( ) 6 ( ) ( + ) ( + ). 8 ( ) 6 9 (iii) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 9. M A T H E M A T I C S ( ) ( ). ( ) ( ) + ( ). (i)( ) ( )(+) (+). (ii) (i)() ( ) (ii) ( ) ( ) 6 6 OR (i)( ) ( ) 0 0. n n a 0 (ii)( ) ( ) a (i) (ii) OR Let the required number be x. Then x ( ) ( ) 60. VIII

144 or x ( ) or 9x or x 9 or x Thus, ( ) E X P O N E N T S A N D P O W E R S should be multiplied by. WORKSHEET8 8. (i) ( ) ( ) ( ) (ii) ( ) ( ) ( ) ( 8 6 ) 6 ( ) (i) ( ) ( ) ( ) (ii) ( ) ( ) ( ) ( ) ( ) ( ) ( ) () ( ) ( ) (i) (ii) (i) (ii) OR Thickness 0. mm. 0 mm. (i) ( ). 0 mm. ( ) 6 ( ) x or ( ) 6 ( ) x Comparing the exponents as the bases are same, we get 6 x or x or x. (ii) x () x or 6x x or 6 x Cross-multiplying, x 6 or x. 6. (i) ( ) ( ) ( ) (ii) ( ) ( ) + ( ) ( ) 6. ( ) ( ) ( ) 6 ( ) ( ) ( ) 0 ( ) 6 ( ).. ( ) 6

145 . a ( ) b a b ( ) ( ) 9 6 ( ) 6 ( ) 0 OR ( ) ( ) 6 ( ) 6.6 (i) (ii) ( ) 6 ( ) 6 or ( ) m + m ( ) 6 ( ) 6 6 Comparing exponents as the bases are same, we have + m or m + or m. 9. Let the required number be x. ( ) Then, x ( or ) x Cross-multiplying, x 9 or x 9 Thus, ( ) should be divided by. 0. (i) ( ) ( ) ( ) Reciprocal of ( ) (ii) ( ) M A T H E M A T I C S { ( ) Reciprocal of ( ). (i) ( ) { 6. ( ) } 8 ( ) } WORKSHEET9 9 ( ) ( ) Reciprocal of ( ) ( ) { } (ii) ( ) ( ) Reciprocal of ( ) ( ) (i) ( ) (ii) ( ) ( ) ( ). { } OR ( ) ( ) ( ) ( ) ( ) ( ) ( ) 6 ( ) 8. ( ) ( ). ( ) 6. ( ) ( ). VIII

146 . (i)( 8 ) ( ) ( ) (ii) ( ) 0. ( ) 6 8 ( ) ( ). ( ) 6 9 or ( ) 6 or ( ) 6 9 Comparing, x y x 9 y [ a m b m (ab) m ] x y x y 9. 6 OR (i) (ii) (i) (ii) (i) (ii). 0 (iii) (i) Thickness of a paper. 0. mm 0 mm. 0 mm. (ii) Size of bacteria mm mm. (iii) Size of a plant cell m. ( ) 6 ( ) 6 or ( ) or ( ) m. m 0. 0 m. 6 x y x y x y Comparing the bases of both sides as exponents are same, we get ( ) x y or x y. 8. (i)(6 8 ) + ( ) ( ) + ( ) 6 8 ( ) ( ) (ii)( + 8 ) ( ) ( ) + 6 ( ) ( ) E X P O N E N T S A N D P O W E R S

147 9. Let (8) be multiplied by x. Then x (8) 0 x or 8 0 or x 8 0 or x Thus, the required number is. 0. Size of a blue tablet m 00 0 m Size of a red tablet m 0 m Clearly, size of the blue tablet is larger by ( )m, i.e.,. 0 m m Ratio of their sizes. (i) ( ) 0 m 0 WORKSHEET 80 ( ) ( ) 0 :. 6. (ii)( ) 6 ( ) Let () be divided by x. Then, or () x x () or x Cross-multiplying, x or x Thus, the required number is.. (i) ( ) (ii) ( ) 8. (i) (ii) 8 ( ) ( ) ( ) ( ) 6 a ( ) 6. ( ) 6 ( ) a a 0 0 a8 a a 8 a 6 a.. Let () should be divided by x. Then, () x ( ) or Cross-multiplying, 6x x 6 M A T H E M A T I C S VIII

148 or x 6 8 So, the required number is Let ( ) ( ) Then, x or x should be divided by x. ( ) ( ) 8 or x Cross-multiplying, x 8 x Thus, the required number is ( ) or ( ) ( ) 0 + ( 0) or ( ) ( ) x ( ) x ( ) x Comparing the exponents as bases are same, we get x or x Thus, x. 8. x ( ) ( ) ( ) ( ) ( ) 6 E X P O N E N T S A N D P O W E R S So x ( ) 6 ( ) 6 x x ( ) 6 ( ) 6 or x ( ) OR (i) ( 6 ) ( ) or x ( ) (ii)( + ) ( ) + ( ) (i) (ii) ( ) ( ) (i) ( 6 ) ( ) 6 6 ( ) 6. ( ) 6 0 ( ) (ii)( ) ( ) ( ).

149 (i) 6 6 OR 6.. (i) (ii) () ( ) 6. (ii) (iii) ( ) ( ) 0. WORKSHEET 8. (i) x x or or x+. Comparing the exponents as bases are same, we get x + or x or x.. x ( ) or x ( ) (x) ( ). (i) micron ( ) ( ) 6 ( ) 6 or x ( ) metre metre metre.. (ii) Size of bacteria metre metre metre. metre. (i) 6 (ii) (i)() ( ) (ii) ( ) ( ) ( ) ( ) ( ) M A T H E M A T I C S ( ) 6. ( ).. (i) ( ) ( ) ( ). (ii)( + ) ( ) ( ) + + ( ) 6 6 ( ) 6 6. VIII

150 { } 8. ( ) ( ) ( ) x ( ) or x ( ) (x) ( ) 9. (i) OR ( ) ( ) 6 0 (ii)() () (iii) ( ) 8. ( ) + ( ) 6 or x ( ) ( ). 6 () () () 6 ( ) 0. ( ) ( ) (iv) (i) ( ) ( ) ( ) or ( ) or ( ) 0 OR ( ) x ( ) x ( ) x 0 Comparing the exponents as the bases are same, we get 0 x or x 0. (ii) ( ) or ( ) ( ) x + ( ) x or ( ) 0. ( ) x Comparing the exponents as the bases are same, we get x or x. 0. x ( ) or x ( ) +6 ( ) 6 ( ) Taking cube root on both the sides, we get [ ] x ( ) or x. E X P O N E N T S A N D P O W E R S 9

151 Chapter DIRECT AND INVERSE PROPORTION WORKSHEET 8. (C) Time Distance Speed 0 90 hours.. (B) x my At x 0, y 6; 0 6m m At x 0; 0 y y.. (D) Cost ` 0. (A) x 8.8 x x 8 ` (A) Required number of tools (B) A pole, its shadow, another pole (say h), that s shadow must be in proportion. 0 0 h or h 60 cm 0 6 m 0 cm.. (D) Distance covered in the map 0 cm cm (B) Cost of articles increases as the number of purchasing articles increases. 9. (B) x y 8(xy ) or x y 8xy or (xy ) 0 or xy or x y So, x and y are in inverse proportion. 0. (C) Required weight kg 9. kg.. (B) The working power and the time taken to complete a work are in inverse variation. Ratio of numbers of days :.. (B) The general equation representing x and y are in inverse proportion is xy k, k being a constant.. (A) Let x > 0. x decreases as x increases and x increases as x decreases. So, x and are in inverse proportion. x. (C)Number of men and number of days are in inverse proportion. Number of days (D) a 0.b Speed WORKSHEET 8 km hour 000 m s 0 m/s.. 0 m/s 0 m 0 km 000 s hr km/hr 08 km/hr. 0 M A T H E M A T I C S VIII

152 . Time 8 min 8 60 hour Distance 800 m km 8 0 km Speed Distance Time 6 km/hr. 8/0 8/60. Let k be the constant of proportionality. Then x ky At x and y ; k At x 9, y a and k ; x ky gives a At, x b, y 6 and k ; x ky gives b Thus, a and b.. Geeta s day s work 8 day s work of both Geeta and Meeta 6 So, Meeta s day s work 6 8. Therefore, Meeta alone can finish the whole work in days. 6. Mr. Menon s day s work 6 Mr. Kumar s day s work day s work of both of them Therefore, both of them together can write the chapter in days.. In 6 hours Ritu knits a whole sweater So, in hour she would knit of the 6 sweater So, in hours she would knit 6 i.e., of the sweater. Thus, Ritu will knit part of the sweater. 8. (i) x and y very inversely, if the product xy is constant. (ii) If x y is constant for each pair of values of x and y, then x and y vary directly. (iii)if x ky, where k is a constant, then x and y vary directly. WORKSHEET 8. hour s work of Radha and Medha together 0 hour s work of Radha alone hour s work of Medha alone Therefore, Medha will take 0 hours to do the whole work.. Distance travelled in 0 minutes 60 km. Distance travelled in 60 minutes 60 km 0 km. Therefore, the speed of the car is 0 km/hr. D I R E C T A N D I N V E R S E P R O P O R T I O N

153 . day s work of A alone 0 day s work of both A and B 6 day s work of B alone So, B alone can do the work in days.. Distance covered in 0 minutes 000 m km Distance covered in 60 minutes 6 km 6 km. Therefore, Lily s speed is 6 km/hr.. After joining 6 more people, the family has 8 people. Numbers of people and days are in inverse proportion. So, number of days decreases as number of people increases. Let the required number of days be x. Then, 8 x 60 x Thus, the gas cylinder lasts after 0 days. 6. Let the height of the tree be x metres. The heights of an object and its shadow are in direct proportion. 0 x x 0 Thus, the tree is m high.. Let the required number of sheets be x. The number of sheets and their weights are in direct proportion. : 0 x : ( 000 ) or 0 x x 0 0 Thus, 0 sheets weigh kg. 8. When parts of red pigments, parts of base 8 When parts of red pigments, parts of base 8 When parts of red pigments, parts of base 8 6 When parts of red pigments, parts of base 8 96 When parts of red pigments 0, parts of base Thus the complete table is: Parts of red pigments 0 Parts of base Let the number of machines required be x. Numbers of machines and days are in inverse proportion. x 6 6 x School time in a day 8 minutes 9 periods are of 8 minutes period is of 8 minutes, i.e., 0 9 minutes Thus, each period will be of 0 minutes. M A T H E M A T I C S VIII

154 . 8 shops require m shop requires 8 m 0 shops will require 0 8 m WORKSHEET 8 80 m.. Let the required number of balls be x. Then, 8 6 x x Thus, Kanwar Singh should sell 6 cosco balls.. Required number of dollars Since number of packets and their cost vary directly. Required number of packets Let required number of packets be x. Numbers of packets and cartons are in direct proportion 0 : 0 x : 0 This gives, x Let required number of men be x. Since, numbers of men and days are in inverse proportion. 0 x This gives, x Distance Length of train 0 m Speed 0 km/h m/s 600. Let 9 m/s. Time Distance Speed ( ) D I R E C T A N D I N V E R S E P R O P O R T I O N seconds. x k, k is a constant y At x 9 and y., k 9. At y 8 and k, x 8 6 At y. and k, x. 6.0 Therefore, the complete table is: x y (i) Directly (ii) Directly (iii) Directly (iv) Directly. 9. Workdone by Rohan in day Workdone by Rohan in days 0. In kg of sugar, number of crystals In kg of sugar, number of crystals (i) In kg of sugar, number of crystals (ii) In. kg of sugar, number of crystals

155 . Let required number of time be x days. After joining 0 girls, number of girls Since number of girls and food provision vary inversely 0 80 x This gives, x. 80 Thus, the provision will last after days. WORKSHEET 86. One month and ten days 0 days In 0 days, kg is consumed by persons In day, kg is consumed by 0 persons In day, kg is consumed by persons 0 0 In 0 days, kg is consumed by 0 persons In 0 days, 0 kg is consumed by 0 0 persons i.e., persons. 0 Therefore, the required number of persons is.. Numbers of buses and tourists vary directly. Number of tourists Height of wall 0 m 8 m 8 00 cm 8 cm Height of wall Number of bricks Height of a brick 8.. At g per child, 0 children take days At g per child, child takes 0 days At g per child, child takes 0 days At g per child, children take days At 00g per child, children take 0 days i.e.,. days. 00 Therefore, the rice is enough for. days.. Fare ` per km `.0 per km. Required distance km. km km Number of steps 6. Gunjan s speed Time 0 0 steps/min [ hour 0 min.] 8 steps/min. Required number of steps Speed Time M A T H E M A T I C S VIII

156 . 8 days wage ` 00 day s wage ` 00 ` 8 0 days wage ` 0 ` In 00 days, 0 men can eat the food. In days, 0 men can eat of the 0 food. So, 9 of the food remains. 0 Further, 0 men can eat the food in 00 days So, 90 men can eat the food in days So, 90 men can eat of the food in days, i.e., 60 days. Thus, the remaining food lasts after 60 days. 9. Let required number of men be x. Number of days Number of men 8 x Note that less the number of days, more the number of men. Therefore, this is a case of inverse proportion. So, 8 x or 8 x or x Thus, men should be require to repair the machine. 0. xy k At x 6 and y 6, k So xy 96 Hence, x y WORKSHEET 8. day s work of both Rita and Mita D I R E C T A N D I N V E R S E P R O P O R T I O N and day s work of Rita 6 day s work of Mita 6 Consequently, we obtain that Mita alone can do the work in days.. Let Roma can do x of the work in days. Number of days 0 Quantity of work x Number of days and quantity of work are in direct proportion 0 or x x Thus, Roma can do work in days.. day s work of both A and B 0 A alone can do the work in, i.e., 60 days day s work of A alone 60 So, day s work of B alone Consequently, we obtain that B alone can do the work in 0 days.. day s work of both X and Y 0 day s work of both X and Y 0 0

157 Remaining work day s work of Y alone 0 So, Y alone can finish the whole work in 0 days. So, Y alone will finish 9 of the work 0 9 in 0, i.e., days. 0. days s income of girls ` 80 days income of girl ` 80 ` 0 day s income of girl ` 0 ` 0 day s income of girls ` 0 ` 0 6 day s income of girls ` 0 6 ` 900. Thus, girls will earn ` 900 in 6 days. 6. Let the length of the bridge be x m Speed 60 km/hr m/s m/s Time 90 s Distance Length of the train + Length of the bridge (600 + x)m Now, distance Speed Time x 0 90 or x m Thus, length of the bridge is 900 metres.. Money on Raghu Number of machines Price of machine ` 00 After discount, CP of a machine ` 00 ` 0 ` 0 Number of required machines ` 00 CP of machine ` ` 0 Thus, Raghu can buy 00 machines. 8. Let m cows will graze the field in 0 days. Number of cows m Number of days 6 0 Numbers of cows and days are in inverse proportion 6 m 0 Which gives, m 6 0 Thus, cows will graze the same field in 0 days. 9. Let the required number of days be x. Income increases as number of days of work increases. So, income and days of work vary directly x x Thus, the man works for 6 days. 6 M A T H E M A T I C S VIII

158 WORKSHEET 88. Let x ky as x and y vary directly. At x and y 6, k 6 So, x y At x 9, y 9 6 At y 8, x 8 At y 6, x 6 9 At x, y At y, x At x, y Therefore, the complete table is: x 9 9 y Cost of mango ` 6 ` Cost of 9 mangoes 9 Cost of mango 9 ` `.. Speed 0 km/hr 0 60 km/min Time min. Distance Speed Time km.. Let x men will dig the trench in days. Number of men increases as number of days decreases. So, numbers of men and days are in inverse proportion. or 6 x or x x 68 men. 6. Cost of books Cost of book 00 ` 00 New cost of book ` 00 + ` ` 6 Required number of books Thus, Veena will be able to buy 0 books. 6. Let carpets can be woven in x days Numbers of carpets and days vary directly or x or x x Thus, Jojo can weave carpets in days.. Let required number of hours per day be x. Since, numbers of hours per day and days vary inversely. 8 x 8 8 or x or x 8 Thus, Kamla should work hours per day. 8. Let required number of words be x. Number of words and time vary directly x ( hour 60 minutes) 6 or x Thus, Geeta can type 6 words in 6 minutes 9. (i) directly D I R E C T A N D I N V E R S E P R O P O R T I O N

159 (ii) direct (iii) x 8 y 0 0 ( xy constant) 0. day s work of both X and Y 0 day s work of both Y and Z day s work of both X and Z day s work of X s, Y s and Z s day s work of all the X, Y, and Z 8 Now, day s work of X alone 8 day s work of Y alone and day s work of Z alone Consequently, we obtain that X alone, Y alone, and Z alone can finish the work in days, 0, i.e., days and 0 days respectively. 8 M A T H E M A T I C S VIII

160 Chapter FACTORIZATION WORKSHEET 89. (C) Factors of are and itself.. (C) xy + y + x + y(x + ) + (x + ) (y + )(x + ).. (B) is a common factor of abc and pqr as these are divisible by.. (A) 6m n 6(m n).. (D) (x + a)(x + b) x + (a + b)x + ab is an identity. 6. (B) x + x + x +.. (B) x 6x x x x 6x x (A)Let us take identity (a + b)(a b) a b Put a x and b y (x + y)(x y) (x) (y) or 0xy(x + y)(x y) 0xy(6x 9y ). 9. (D) z 80 (z 6) (z + )(z ). 0. (D) (p q 6 p 6 q ) p q p q (q p ) p q q p.. (C) is a factor of 6.. (D) a b (a + b)(a b) At a 9 and b 8, 9 8. F A C T O R I Z A T I O N. (B) y y.. (A) 6(x yz + xy z + xyz )6xyz(x + y + z) So, 6(x yz + xy z + xyz ) is divisible by xyz.. (A) 0x x x 96xy x y 08xy x y y HCF x x. 6. (B) a + bc + ab + ac a + ab + ac + bc a(a + b) + c(a + b) (a + b)(a + c). Thus, factors are (a + b) and (a + c).. (C) x + x + x x(x + x + ) x(x + )(x + ) Clearly, x + is not a factor. 8. (D) Factors of are, and Factors of x are, x and x So, all the factors of x are,,, x, x, x, x, x and x. 9. (C) x + (x )x x (x ) + (x ) (x )(x + ). 0. (A) a b (a + b)(a b) Substituting a z and b, we get z (z + )(z ).. (B) 66 So, factors of 66 are,,,, 6,,, and 66 So, number of factors of 66 is 8. 9

161 WORKSHEET 90. (i) The further factors of y(xy + ) are not possible, So it is in the factor form. (ii) x + 8x + 6 (x + ) (x + )(x + ) So, x + 8x + 6 is in the expanded form. (iii) Factors of (x + ) + or x + 0 are possible, so it is in the expanded form. (iv) Factors of x are possible, so it is in the expanded form.. (i) a a a a and a a So, HCF (a, a) a. (ii) x y x x y and xy x y So, HCF (x y, xy) x y xy. (iii) 6x y x x y y and x y x x y So, HCF (6x y, x y) x x y x y. (iv) a b a a a b and a a a So, HCF (a b, a ) a a a.. (i) x x and 8y y HCF (x, 8y) Therefore, x + 8y (x + y). (ii) x x and 9y y HCF (x, 9y) Therefore, x + 9y (x + y). (iii)x x and HCF (x, ) Therefore, x (x ). (iv) 6x x x, x x x x and 6x x x x x HCF (6x, x, 6x ) x x 6x Therefore, 6x x + 6x 6x ( x + 6x ).. (i) x + xy + 8x + 8y (x + xy) + (8x + 8y) x(x + y) + 8(x + y) (x + y)(x + 8). (ii)xy 6x + y (xy 6x) + (y ) x(y ) + (y ) (y )(x + ). (iii) ax ay + bx by (ax ay) + (bx by) a(x y) + b(x y) (x y)(a + b). (iv) z 6 6xy + xyz (z 6) + (xyz 6xy) (z 6) + xy(z 6) (z 6) ( + xy). (v)0mn + m + n + (0mn + m) + (n + ) m(n + ) + (n + ) (n + )(m + ).. (i) x + 0x + x + x + x + x + x + (x + ). (ii) m + 8m + 6 m + m + m + 6 m + m + 6 (m + ). 60 M A T H E M A T I C S VIII

162 (iii) x + x + 60 x + x + x + 60 x(x + ) + (x + ) (x + )(x + ). (iv) x + xy y x + 8xy xy y x(x + 8y) y(x + 8y) (x + 8y)(x y). OR (i) x + x 0 x + 8x x 0 x(x + 8) (x + 8) (x + 8)(x ). (ii) x x + 90 x 0x x + 90 x(x 0) (x 0) (x 0)(x ). (iii) n + n 60 n + 0n n 60 n(n + 0) (n + 0) (n + 0)(n ). (iv) z + z 90 z + 8z z 90 z(z + 8) (z + 8) (z + 8)(z ). 6. (i)x xy + 0y x xy 8xy + 0y x(x y) y(x y) (x y)(x y). (ii)x + xy 0y x + xy 8xy 0y x(x + y) y(x + y) (x + y)(x y). (iii)6x + xy 6y 6x + 6xy xy 6y 6x(x + 6y) y(x + 6y) (x + 6y)(6x y). F A C T O R I Z A T I O N WORKSHEET 9. (i) p 6p + 99 p 8p + 99 (p 8) (p 8) (p 8) (p 8 + )(p 8 ) (p )(p ). (ii) x + x x + x (x + ) (x + ) (x + + )(x + ) (x + 9)(x ). (iii) p + p (p + ) (p + ) 9 (p + + 9)(p + 9) (p + )(p ). (iv) a a (a ) (a ) (a + )(a ) (a + )(a ). (v) y y + ( ) y ( ) ( ) y ( ) + ( ) y (y )(y 8). y + ( ) (vi) z z 6 ( ) z ( ) (z + )(z 6). 6 ( ) z ( ) 6 ( z + ) ( z )

163 . (i)x x + (x x + ) (x x x + ) {x(x ) (x )} (x )(x ). (ii)x 6x 9 x 8x + x 9 x(x 9) + (x 9) (x 9)(x + ). (iii)8a a + 8a a 0a + a(a ) (a ) (a )(a ). (iv)0a 8a 0a 8a + a a(a ) + (a ) (a )(a + ). (v)x x 8 x 6x + x 8 x(x 8) + (x 8) (x 8)(x + ).. (i)(x + )(x + ) x(x + ) + (x + ) x + x + x + x + 8x +. (ii)(x 0)(x ) x(x ) 0(x ) x x 0x + 0 x x (i)y 0y + 8z yz y(y ) z( + y) (y )(y z). (ii) ab bx + ay xy b(a x) + y(a x) (a x)(b + y).. (i) q 0q + (q ) + (q ) (q + )(q ) (q )(q ). (ii) p + 6p 6 (p + ) 6 (p + ) (p + + )(p + ) (p + 8)(p ). WORKSHEET 9. (i) We are given the identity: a b (a + b)(a b) (i) 9q p (q) (p) (ii) 9x y 6 (xy) () (q + p)(q p). (xy + )(xy ). (iii) x 9y (x) (y) (iv) x (x ) () (x + y)(x y). (x + )(x ) (x + ){x ( ) } (x + )(x + )(x ). (v)x 08x x (x 9) x (x ) x (x + )(x ).. (i) x + x + (x + x + ) (x + ) (x + )(x + ). (ii) y + y 6 y + 8y y 6 (iii) x + x y(y + 8) (y + 8) (y + 8)(y ). (x + x ) (x + 6x x ) {x(x + 6) (x + 6)} (x + 6)(x ). 6 M A T H E M A T I C S VIII

164 (iv)x 8x + (x x + ) (x ) (x )(x ). (v)9p + q 9r pq (9p pq + q ) 9r (9p pq pq + q ) 9r {p(p q) q(p q)} 9r (p q)(p q) 9r (p q) (r) (p q + r)(p q r).. (i) x + x 6 x + x x 6 x(x + ) (x + ) (x + )(x ). (ii) m + m + 90 m + 8m + m + 90 m(m + 8) + (m + 8) (m + 8)(m + ). (iii) b b b 8b + b b(b 8) + (b 8) (b 8)(b + ). (iv) a ab + 0b a ab 0ab + 0b a(a b) 0b(a b) (a b)(a 0b).. 8(p 8q) 6(p 8q) (p 8q){(p 8q) } (p 8q)(p q ).. (i)y + y + ( y + y+ ) ( y ) ( ) + + y + ( ) F A C T O R I Z A T I O N ( ) ( ) y + + y + ( ) ( ) y + y + (y + )(y + ). (ii) x + x + 0 ( ) ( ) x + ( ) ( ) x ( x + + )( x + ) (x + )(x + ). WORKSHEET 9. (i)ab 8b 6 + 9a (ab + 9a) (8b + 6) a(b + ) (b + ) (b + )(a ). (ii)8x y + 8x 6xy (8x y) + (8x 6xy) (x y) + x(x y) (x y)( + x). (iii)ab a + 0b 6 (ab + 0b) (a + 6) b(a + ) (a + ) (a + ) (b ). (iv)x 6xy x + y (x 6xy) (x y) x(x y) (x y) (x y)(x ). 6

165 (v)6l 8l lm + m (6l 8l) (lm m) 8l(l ) m(l ) (l )(8l m) (l )(l m).. (i) 6m m m m m m m HCF ( 6m, m ) m m 8m So, 6m + m 8m ( + m). (ii) 0l l l l 0alm a l m HCF (0l, 0alm) l 0l So, 0l + 0alm 0l(l + am). (iii) 6x y x x x y 8xy x y y y HCF (6x y, 8xy ) x y 6xy So, 6x y 8xy 6xy(x y ) 6xy(x + y )(x y ). (iv) 6a a a 6ab a b 6ca c a HCF ( 6a, 6ab, 6ca) a 6a So, 6a + 6ab 6ca 6a( a + b c). (v) p qr p p q r pq r p q q r pqr p q r r HCF (p qr, pq r, pqr ) p q r pqr So, p qr + pq r + pqr pqr(p + q + r).. (i) x x x 6y y y y HCF (x, 6y ). (ii) 8a b a a b b ab a b HCF (8a b, ab) a b 6ab. (iii) 90a bc a a b c 8bc b c HCF (90a bc, 8bc) b c 9bc.. (i)p q 6r (p q 9r ) [(pq) (r) ] (pq + r)(pq r). (ii) s(r + q) + (r + q) (r + q)(s + ). (iii)9x 6 (x) 6 (x + 6)(x 6). (iv)6x 9y (x) (y) (x + y)(x y). OR (i) a ab + b c (a ab + b ) c (a b) c (a b + c)(a b c). (ii)8x y xy 8xy(x y ) (iii)xyz x y 8xy[x (y) ] 8xy(x + y)(x y). xy(z 9x y ) xy[(z) (xy) ] xy(z + xy)(z xy). (iv)0a b 98c (a b 9c ) [(ab) (c) ] (ab + c)(ab c). 6 M A T H E M A T I C S VIII

166 . (i) 9 x y + xy 9 (x + y xy) (x y) ( x + y)( + x y). (ii) x y + xz + z x + xz + z y (x + z) y (x + y + z)(x y + z). (iii) a + ab + b c (a + b) c (a + b + c)(a + b c). WORKSHEET 9. (i) m 0m + m 6m m + m(m 6) (m 6) (m 6)(m ). (ii) p + p p + 9p 8p p(p + 9) 8(p + 9) (p + 9)(p 8). (iii) a + a a + a a a(a + ) (a + ) (a + )(a ). (iv) x x + 0 x x x + 0 x(x ) (x ) (x )(x ). (v)9a + ab + b (a + b) (a + b)(a + b). (vi) 6x x 6x 0x + 9x x(x ) + (x ) (x )(x + ). (vii)x 8x + (x x + ) (x ) (x )(x ). F A C T O R I Z A T I O N (viii)x + x + 0 x + 8x + x + 0 x(x + ) + (x + ) (x + )(x + ). (ix)9a b 6c (ab) (8c) (ab + 8c)(ab 8c). (x) m(x + a) + (x + a) (x + a)(m + ). (xi) x y (x ) (y ) (x + y ) (x y ) (x + y )(x + y)(x y). (xii)9x +6y xy (x y) (x y)(x y).. (i) x + x (x + ) (ii) 6x + 6x (x + ) (x + + )(x + ) (x + )(x ). ( 8x ) [ ( ) x ] ( + x) ( x) ( + x)( + x) ( x ) ( x ).. (i) p q r p p p p q q q r r p q r p p q q q q r r r p q r p p p q q r r r r HCF (p q r, p q r, p q r ) p p q q r r p q r So, p q r + p q r +p q r p q r (p q + q r + pr ). (ii) 0a b a a a b 0ab a b b b 6

167 0a b a a a b b b HCF (0a b, 0ab, 0a b ) a b 0ab So, 0a b 0ab 0a b (iii) 0ab(a + b + a b ). x x x x x x x x x HCF (x, x ) x x x and HCF (x, ) So, x + x + x + x (x + ) + (x + ) OR (x + )(x + ). (i) x 6x + 9 (x ) x 6x+ 9 x (x )(x ) ( x )( x ) x x. (ii) x 6 x (x + )(x ) x 6 x + ( x+ )( x ) x + (x ).. 8a b c abc + ab c Now, 6abc(ab + bc) 8 abc abc+ abc 6abc 6 abc(ab + bc ) 6abc (ab + bc ). WORKSHEET 9. (i)0x 0x 0x x 0x (ii)xy xy xy. xy (iii) a 6 a. (i) 66 M A T H E M A T I C S x y xy a a a a. xy x y x y. xy (ii) a b ( a b) (iii) 6 y + 8y 8y a b. 8 y (9 y + y) 8y 9y + y.. Area a + a a(a + ) b a, l? Area l b a(a + ) l a ( aa+ ) or l a a + x. a Thus, length of the rectangle is +.. x + x + 0 x + 8x + x + 0 Now, x + x+ 0 x + x(x + ) + (x + ) (x + )(x + ) ( x+ )(x+ ) x + x +.. (i) x + x + 0 x + x + x + 0 x(x + ) + (x + ) (x + )(x + ) VIII

168 x + x+ 0 ( x+ )( x+ ) x + x + x +. (ii) x + x + 6 x + x + x + 6 x(x + ) + (x + ) (x + )(x + ) x + x + 6 ( x+ )( x+ ) x + x + x +. (iii) x + 0x + x + 6x + x + x(x + 6) + (x + 6) (x + 6)(x + ) x + 0x+ ( x+ 6)( x+ ) x + x + x + 6. (iv) x + x 6 x + 8x x 6 x(x + 8) (x + 8) (x + 8)(x ) x + x 6 ( x+ 8)( x) x + 8 x + 8 x. (v) x + x + 6 x + 6x + x + 6 x(x + 6) + (x + 6) (x + 6) (x + ) x + x+ 6 ( x+ 6)( x+ ) x + x + x + 6. (vi) x 6 (x ) (x + )(x ) (x + )(x ) (x + )(x + )(x ) x 6 ( x + )( x+ )( x ) x + x + (x + )(x ). 6. (i)x + xy + 6y x + 9xy + xy + 6y x(x + y) + y(x + y) (x + y)(x + y). (ii)6x x + 6 6x 9x x + 6 x(x ) (x ) (x )(x ). (iii) x 8 (x ) 9 (x + 9)(x 9) (x + 9)(x ) (x + 9)(x + )(x ).. Cost of z metres cloth Cost of metre cloth ` (z + z ) ` z ( + z) ` z ( + z) z ` z( + z) ` (z + z ). F A C T O R I Z A T I O N 6

169 Chapter INTRODUCTION TO GRAPHS WORKSHEET 96. (B) A point (x, 0) lies on the x-axis.. (A) (, 0) lies on the x-axis as (x, 0) lies on the x-axis.. (D) x-coordinate of E y-coordinate of E So, the coordinates of E are (, ).. (A) The coordinates of A are (, ).. (B) The coordinates of O are (0, 0) as O is the origin. 6. (C) The coordinates of C are (8, ).. (B) Distance travelled in first hour y-coordinate of the graph at p.m. 8 km. 8. (B) The traveller is the fastest between p.m. and p.m. 9. (B) The traveller stops twice from : 0 p.m. to : 00 p.m. and from p.m. to 6 p.m. 0. (D) Distance km 8 km km.. (C) y Area of square Side x 6 square units.. (D) y Perimeter of square x.. (B) Cartesian plane has axes, namely, x-axis and y-axis.. (A) Coordinates of any points on the y- axis are of the form (0, y) x-coordinate 0.. (B) Equation of a straight line is of the form ax + by + c 0, whose degree is. 6. (D) The x-coordinate of a point is its perpendicular distance from the y- axis. The y-coordinate of a point is its perpendicular distance from the x- axis.. (B) The graph is a straight line as simple interest is directly proportional to the number of years. 8. (C) A line graph changes over time. WORKSHEET 9. (i) The graph represents the measures of temperature of a city from 8 a.m. to p.m. of a day. (ii) The temperature was highest from 9 a.m. to 0 a.m. (iii) The temperature was least at p.m. (iv) Increase in temperature 0 C C C. (v) The temperature at 8 a.m. was C which is less than 0 C.. (i) To draw the required graph, let us take days of the week on the x-axis and temperature on the y-axis. 68 M A T H E M A T I C S VIII

170 (iii) Distance covered in hour 0 km Distance covered in hours km Required distance 0 km. (iv) Yes, we can tell. Distance covered between hours and hours km.. (i) (ii) (ii) To draw the required graph, let us take distance travelled on the x-axis and the cost on the y-axis.. (i) The required set is {, 6, }. (ii) The required set is {,, }. WORKSHEET 98.. (i) 0 km big division on the vertical line. hour big division on the horizontal line. (ii) Distance covered after hours 00 km.. Point corresponding to x, y is (, ) I N T R O D U C T I O N T O G R A P H S 69

171 Point corresponding to x, y 6 is (, 6) Point corresponding to x, y 8 is (, 8) Point corresponding to x, y 0 is (, 0) Point corresponding to x, y is (, ).. (iii) A point with coordinates (0, 0) lies at the point of intersection of the x- axis and y-axis. Joining these points, we obtain a straight line (see graph), Therefore the points lie on a straight line.. (i) If x-coordinate of a point is 0, then the point will lie on the y-axis. (ii) If y-coordinate of a point is 0, then the point will lie on the x-axis. (i) The points lie in the first and third quadrants and at the origin. All the points lie on a straight line (see figure). (ii) x-coordinate and y-coordinate of each point are equal, so x y.. We know that area of a square is the square of its side. For example, if side is a than area a. Hence, S. No. Side of square Area. cm cm. cm 6 cm. cm cm. 6 cm 6 cm. 8 cm 6 cm Let us draw graph using this table and taking side of the square at x-coordinate and its area as y-coordinate. 0 M A T H E M A T I C S VIII

172 . Joining the points, we obtain that the graph is a curve not a line segment. 6. (i) Point (, ) is represented by the letter F. (ii) Point (, 8) is represented by the letter M. (iii) Point (, ) is represented by the letter Q. (iv) Point (, ) is represented by the letter O. WORKSHEET 99. (i) Given point is A(, ) Its x-coordinate is and y- coordinate is. (ii) Given point is B(, ) Its x-coordinate is and y- coordinate is. (iii) Given point is C(0, ) Its x-coordinate is 0 and y- coordinate is. I N T R O D U C T I O N T O G R A P H S. Let us take side of square as x-coordinate and its perimeter as y-coordinate to draw a graph.

173 . From the graph, we conclude that When x 0, y When x, y / When x, y / When x, y When x 6, y So, when x, y and when x, y / The complete table will be as follows x 0 6 y / / / Consequently, we obtain that the relationship between x and y will be y x +.. The given relation between volume (v) and temperature t is Volume (v) v.t. At t 0 C, v. 0 cubic units At t C, v. 0 cubic units At t 8 C, v. 8 cubic units At t 0 C, v. 0 0 cubic units At t 0 C, v. 0 cubic units Let us draw graph, taking t as x-coordinate and v as y-coordinate. Temperature (t) M A T H E M A T I C S 6. VIII

174 . (i) The graph shows the temperature of a city at the time from 8 a.m. to 6 p.m. of a day. (ii) The temperature was highest at 0 a.m. (iii)the temperature was least at 6 p.m. WORKSHEET 00. The coordinates of letter A are (, ) The coordinates of letter B are (, ) The coordinates of letter C are (, ) The coordinates of letter D are (, ). (i) After ploting the points from the given table and joining them, we get a straight line.. Taking years as x-coordinate and number of T.V.s. sold in hundreds as y-coordinate, we get the following graph. (ii). (i) y-coordinate of the point A is 6 (ii) y-coordinate of the point B is 0 (iii) y-coordinate of the point C is (iv) y-coordinate of the point D is.. 6. (i) The meeting was gone from hours to hours. So, the duration was hours. (ii) Distance travelled after hours Distance travelled after hours 0 km. I N T R O D U C T I O N T O G R A P H S

175 (iii) Time taken to travel the first 0 km WORKSHEET 0 was hours.. (iv) Time spent to travel a distance between 0 km and 0 km was hours hour 0 minutes.. (i) x-coordinate of A(0, ) 0 y-coordinate of A(0, ) (ii) x-coordinate of B(6, ) 6 y-coordinate of B(6, ) (iii) x-coordinate of C(, ) y-coordinate of C(, ). 8. Let us make a table of three pairs of points of x and y such that x y. x 6 y 6 Let us plot the points and join them. Yes, the given points lie on a line. This line is parallel to the y-axis.. Coordinates of vertices of the triangle are A(, ), B(/, ) and C(, ). Coordinates of the parallelogram are P(, ), Q(6, ), R(, ) and S(6, ). From the graph, it is clear that the points lie on the line passing through the origin O(0, 0). M A T H E M A T I C S VIII

176 . (i) (ii) 6.. Coordinates of A are (, ) Coordinates of B are (, ) Coordinates of C are (, ) Coordinates of D are (6, ).. y-coordinate of each point is zero. They lie on the x-axis. Yes the given points lie on a line which is parallel to x-axis. y-coordinate of each point is 6. I N T R O D U C T I O N T O G R A P H S

177 . (i) (ii) The vertical line passing through kg intersects the graph at a point which corresponds to ` 600 on the y-axis. So, the cost of kg rice is ` 600. (iii) The horizontal line passing through ` 00 intersects the graph at a point which corresponds to 0 kg on the x-axis. So, 0 kg of rice can be purchased for ` M A T H E M A T I C S VIII

178 Chapter 6 PLAYING WITH NUMBERS WORKSHEET 0 Then,. (A) (D) (D) Reversing the order of the digits of 08, we get 80.. (A). (C) Since 8 is divisible by, therefore 8 is divisible by. 6. (B)Let ten s digit x Then unit s digit 9 x So, 0x + (9 x) 9 0(9 x) + x or 9x 9x + 90 or x 9 x Now, required number (D) + a + + (8 + a) is divisible by 9 if a. 8. (B) + y + (6 + y) is divisible by if y 0,, 6 or (B) 0 is divisible by as it ends with zero. 0. (B) 6 6. (C) (A) Let the missing number be x. P L A Y I N G W I T H N U M B E R S x x or 8x + 6 x or x 8x 6 0 or (x + )(x ) 0 i.e., x or x.. (B) (C). (A) Sum of the numbers in any horizontal or vertical strip is. So, m. WORKSHEET 0. (i) Reversing the digits of 9, we get (ii) Reversing the digits of 6, we get (i) 6 (ii) x x, y and y 8 and z. (i) 8 9 (ii) p 8, q 9, q r 6. r 8

179 . Let unit s digit of required number be x, then ten s digit would be (8 x) So, the required number 0 (8 x) + x On reversing the digits, the new number 0 x + (8 x) According to given condition, 0 x + (8 x) 0 (8 x) + x + 8. or 0x + 8 x 80 0x + x + 8 or 8x 90 or x 8 x 8 So, required number (i) A and B (ii) A, B and C Answer may vary. 6. Yes, such fractions are possible as x y x y for x > 0, y > 0. Example: (i) On dividing 9 by 9, the quotient is. (ii) On dividing 9 by, the quotient is Difference (i) (ii) OR (i) is divisible by but not by 9. Answer may vary. (ii) is divisible by but not by 0. Answer may vary is the closest to 000 such that it is a multiple of 9. WORKSHEET Required quotient 9.. Since ( ) ( ) i.e., is not divisible by. So, 60 is not completely divisible by.. a must be either 0 or.. Let ten s digit be x. Then the required number will be 0 x +, i.e., 0x + Further, 0x + 6 (x + ) or 0x 6x or x 0 or x 0x Hence, the required number is It will be sunday after days, days, days, 8 days,.... So, it will be Tuesday after 0 days.. x and y. 8 M A T H E M A T I C S VIII

180 8. (i). (ii). 9. (i) 8 8. (ii) OR (i) A 8. (ii) A, B 9 and D 0., ( + ), ( + 0), ( + ), ( + 0), ( + ),... or, 8,, 8,, 8,.... Let * m Now, ( ) ( + m) is divisible by or ( m) is divisible by m 6 Thus the number is 666. OR p would be a multiple of 6 if it is multiple of both and. So, p can take values,, or. WORKSHEET0. 00 is divisible by not by 9 as is divisible by not by and 9 00 Thus 99 is the closest to 00 and 00 is the closest to If a number ends with 0,,, 6 or 8, then it is divisible by. So, 60 is divisible by.. Calculation : Calculation : M O R E + S E N D M O N E Y Comparing both calculations, we obtain M, O 0, R 8, E, S 9, N 6, D, Y.. (i) 8 (ii) y y. 6. ( ) ( ) 0 Since 0 is not divisible by, so 98 is not divisible by. OR 800 is divisible by as it ends with 0. Sum of digits is not divisible by as is not divisible by.. Let us think a number A. Double A A A Adding 8 to A, we get A + 8 Taking away 0 from (A + 8), we get A + 8. Half (A + 8) A + Taking away from (A + ), we get A Now, we get the number A itself OR ( ) ( ) 6 Remainder will be 0 when 986 is divided by as is divisible by. P L A Y I N G W I T H N U M B E R S 9

181 8. Let Reema s age be x years Seema s age (8 x)years. Since Reema is years younger than Seema. x + 8 x or x or x 0. year 8 x years. Therefore, Reema s age is 6 months and Seema s age is. years. 9. (i) + 8 Thus, A and B. (ii) A, B 9 and D WORKSHEET is divisible by as the number ends with an even number.. (i) ( ) ( + * + ) (9 + * ) * Put * 0. So, *. (ii) ( + + ) ( * + + 8) (9 + * ) * Put * 0. So, *.. Let us start with a number. Then the first domino may be filled as given below 6 Then the second domino may be filled as given below 6 Further, the last domino must be filled as given below Hence the result is: 6 6 Answer may vary.. Sum of digits of is neither divisible by 9 nor by because is neither divisible by 9 nor by.. The given number 8600 ends with 0, so it is divisible by. Also, the number ends with two zeroes, so it is divisible by. Hence, the given number is divisible by both and. 6. (i) Numbers of eggs and crates vary directly. So the required number of crates Thus, 0 crates will be filled by 000 eggs. (ii) + p + p To make ( + p) as a multiple of, we must put p 9. So, p 9.. Let perimeter of the equilateral triangle as well as the square be a units. Then, area of the triangle ( ) a a sq. units 9 a Area of the square ( ) a 9a sq. units 9 Therefore, the square occupies more area. 80 M A T H E M A T I C S VIII

182 OR Number of books in 8 crates 8 0 Cost of 8 crates of books Number of books Cost of a book Thus, the required cost is ` Calculation : pq r st + uv wx Calculation : Comparing both the calculations, we obtain p, q, r, s 6, t 8, u, v, w 9 and x. 9. (i) (ii) WORKSHEET0. Let the value of * be x. According to the given conditions, we have ( x + ) 6 x + x But * x. So, the result is the number ( * ) itself. P L A Y I N G W I T H N U M B E R S. (i) (ii) Thus, x 6 Thus, x OR (i) is divisible by So, *. (ii) is divisible by So, * is not divisible by. 868 is not divisible by because sum of the digits is not divisible by. 8 is divisible by. 868 is divisible by because number obtained by last two digits is divisible by.. Reversing the order of digits of 986, we get number x8 is a multiple of is a multiple of Also, + and + 8 are multiples of Therefore, x, or. 8

183 Sum of the digits is divisible by 9. So, the required remainder is. OR years months 6 months Incoming of month ` 90 Incoming of 6 months 6 ` 90 ` 000 Thus, the man earns ` 000 in years. 9. ( ) ( ) is divisible by. So, the required remainder is x x To make ( x) as a multiple of, we must substitute x So, x.. The three-digit least number whose digits are in ascending order is. To make as a multiple of, we should replace by. Therefore, the required number is. 8 M A T H E M A T I C S VIII

184 PRACTICE PAPERS Practice Paper- SECTION-A. (A) Since denominator of any rational number cannot be zero.. (D) Consider x x + + x.. (C) We know that diagonals of a rhombus (or a square) bisect each other at 90.. (C) We know that if a dice is rolled then all possible outcomes are:,,,,, 6. Out of them getting a set of prime numbers is,,.. (D) A set a, b, c is said to be Pythagorean triplet if a + b c. Here, ,, is the Pythagorean triplet. 6. (B) Since the cube root of (B) Sum (xy 6z + ) + (0xy + 6z ) (xy + 0xy) + ( 6z + 6z) + ( ) (Regrouping like terms) xy xy. P R A C T I C E P A P E R S 8. (A) Number of edges 6. Edge 9. (D) We have ( ) 8 ( ) ( ) 8 [ a m a n a m n ] ( ) ( ) ( ) ( ) (C) Given: 6 x x 6 x. SECTION-B. In kg of sugar, there are In kg of sugar, there are In kg of sugar, there are crystals crystals crystals crystals. Alternative Method: Weight of sugar kg kg Number of crystals x Here, weight and number of crystals of sugar are in direct variation x x x

185 . We have x Cross-multiplying, x + (x ) (x + ) 9x 8x + 0 Transposing, 9x 8x 0 + x.. Given: B A A Changing into complete system, Putting A [ is the digit that gives product (i.e., ones digit A) when multiplied by ]. Now, ( + 0) 6 and 6. So putting B, A, B.. Given: Sum of two numbers 8 One number Other number 8 L.C.M. of and is. 6.. Given expression x + x + and value of x Value of expression at x is ( ) + ( ) [ ( ) ( )! ]. 6. Given dimensions of cuboid Volume lbh cm 0 cm cm cm 0 cm cm Changing cm to m,. From given figure, (Using linear pair axiom) m Now, using angle sum property of a quadrilateral, we have, + x x x x M A T H E M A T I C S VIII

186 8. For a polyhedron we have F 0, E 0 and V. Putting these values in Euler s formula F + V E. i.e., !. Thus, a polyhedron cannot have the given values. SECTION-C 9. Given: P ` 000, r 8% per annum t years, C.I.? Using formula, t r A P P R A C T I C E P A P E R ` 8 C.I. A P ` 8 ` 000 ` L.H.S. R.H.S Thus, L.H.S. R.H.S. Proved. S. Given number Taking prime factorization, In above factorization, is not in pair, so to make perfect square we should multiply it by.. Consider 8 Taking prime factorization, 8 Thus, we observe that every factors have exponent. Therefore, 8 is a perfect cube.. Let cost price of an article be ` 00. So, from question marked price ` % of ` ` 0 Also discount % Selling price ` 0 % of ` ` 0.60 Now gain S.P. C.P `.60 Gain % %. 8

187 . (i)x x Transposing, x x 8 x 8 Dividing both sides, by x 8 (ii) x + (x ) x + x x Transposing, x 6. x + + x Dividing by both sides, x x Distance covered.6 km m ( km 000 m) 600 m Time taken minutes 0 seconds seconds Distance covered Speed Time taken 600 m 0 s m/s To convert m/s into km/h, we have to 6. Let ABCD be a rhombus whose diagonals AC and BD bisect each other at O. Also let AC 6 cm, BD 8 cm We know that area of a rhombus product of diagonals AC BD 6 8 cm. Again, in right triangle AOB, AB AO + BO [By Pythagoras theorem] (i) AB cm. t 0 t 8 (t! 0) t t 8 () (8) t + +8 t 6 t. (ii) ( 0 + ) m a [ n a m n ] a t multiply the speed obtained by 8. Therefore, speed 8 km/h 8 km/h. + [ a0, a a ] M A T H E M A T I C S VIII

188 8. We have circumference of base of a cylinder 88 cm. pr 88 r 88 p 88 cm Volume of the cylinder pr h () ( Given h cm) cm. SECTION-D 9. Given: Capacity of a cylinder. l Volume of the cylinder. 000 m pr h r r [ 000 l m ] ( h m given) r 00 Now total surface area of closed cylinder pr(r + h) m. P R A C T I C E P A P E R S 0. Steps of construction:. Take a line segment PQ cm.. Make an angle of measure 80 at Q with the help of protractor. Then draw a ray QX.. Taking P as centre and 6 cm as radius, draw an arc which cuts ray QX at R.. Further, taking cm as radii and with centres P as well as R, draw two arcs which cut each other at S.. Now join PR, PS and RS. Thus, the quadrilateral PQRS is formed.. (i) Let B s income be ` 00. So A s income ` 00 0% of ` 00 ` 00 ` 0 ` 60 So the difference between their incomes ` 00 ` 60 ` 0. Since A s income is ` 60 then B s income is ` 0 more than that of A s income A s income is ` then B s income 0 is ` more than that of A s 60 income. A s income is ` 00 then B s income is ` %. 8

189 (ii) Given 0% of x x x From graph plotted to the below, it is clear that the points lie on the same line ABCD.. (i) In the given graph, maximum number of students is and they watched the show for - hours. (iii) The students spent more than hours in watching the show means the students that spent to 6 or 6 to hours. So, the total number of such students (i) Given monomials are 9x y, yz, 8 yz and 6x y z. Product of the monomials (9x y) yz 8 y z (6x y z ) (x x ) (y y y y ) (z z z ) 8. x +. y z + + (ii) The students watched the show for less than hours means that students watched the show for to, to or to hours. So, the total number of such students xyz. (ii) Given number is 8. Using prime factorization, 8 In factorization, we observe that a is not in exponent of. So we need eliminate to 8 a perfect cube. Thus, the least required number is. 88 M A T H E M A T I C S VIII

190 Practice Paper- SECTION-A. (C) L.H.S. x x x x [Using prime factorization] R.H.S.. (D) Starting from 0, firstly we move units to the right on x-axis and then units upward along y-axis. Thus, we reach at the point D which represents (, ).. (A) By the divisibility test of 0, we know that number divisible by 0 has always ones digit as 0.. (B) Subtraction is not commutative e.g.,!.. (C) Putting m in L.H.S, + 6m ( ) 8 9 R.H.S. 6. (A) By theorem, sum of all exterior angles of a quadrilateral (or any polygon) 60.. (B) Class width of a class interval Upper limit Lower limit (D) Consider ( + )( ) [ a b (a + b)(a b)]. 9. (C) (a + b) (a + b)(a + b) a(a + b) + b(a + b) (Distributive property) a + ab + ba + b a + ab + ab + b (Commutative property) P R A C T I C E P A P E R S a + ab + b (Closure property). 0. (B) We have V, F, E? Using Euler s formula, F + V E + E 0 E E 8. SECTION-B. Cube of. (.) (.) (.) (.).9.. Let the number of side of a regular polygon be n. We have measure of each exterior angle n n n 8.. One rational number between and Second rational number between and i.e., the rational number between and + Thus, two rationals between and are,. [Note. There are infinitely many rational numbers between two rational numbers.] 89 +6

191 . Alternative Method: Given rational numbers are and. Taking equivalent rationals with the same denominator (i.e., L.C.M.), and [ L.C.M. of and 6] 6 and 6 Here difference between the numerators and is and we have to find two rational numbers. Therefore, we multiply both rationals (Numerator and Denominator) by. 6 and and < 8 < 8 8 < 9 8 < 8 < 8 8 < Thus, the rationals are 8 and 8 8. (Answer may vary) ( ) Given: a 8, b To verify: a + b b + a L.H.S. a + b R.H.S. b + a + 8 +( 8) 0 Thus, L.H.S. R.H.S. 6. Let a number be x. So eight times of the number is 8x. According to question, now to simplify the equation divide both sides by 8. 8x x Given expression: x + 0x + Find the product x and i.e., x Factorize the product, x Regroup these factors into x two groups such that their sum is equal to the middle 6x term. x So we find such groups as x x x and x i.e., x x and 6x. Now split up the middle term as the sum x + 6x. Expression x + x + 6x + (x + x) + (6x + )(Regrouping ) x(x + ) + 6(x + ) (Taking common) (x + )(x + 6). 8. In a triangle, base (b) 0 cm altitude (h) 6 cm } Given Area of the triangle b h (Formula) cm. 90 M A T H E M A T I C S VIII

192 SECTION-C 9. Side of a small cube cm Volume of the cube (Side) ( cm) cm cm cm Dimensions of a big cuboidal box. m 90 cm cm 0 cm 90 cm cm The number of cubes that can be filled in bigger box Volume of cuboidal box Volume of a cube Let ABCD be a rhombus in which all sides are of m and diagonal AC 0 m. Also let diagonals AC and BD bisect each other perpendicularly at O. AO OC 0 m In right-triangle AOB, using Pythagoras theorem, AO + OB AB + OB OB 69 OB m BD OB m Therefore, area of the rhombus AC BD 0 0 m.. We have base radii of two right circular cones are in the ratio : r r Also their height are same say h. V Therefore, ratio of their volumes V prh prh ( Volume of a cone π rh ) r r r r 9 V : V 9 :.. Given: Curved surface area of a cylinder 0 cm and base diameter cm Radius r cm. We know that curved surface area of the cylinder prh 0 h h 0 Now, volume of the cylinder 0 cm pr h cm. P R A C T I C E P A P E R S 9

193 . Square root of 8 8 Using long division method, 8.9 l.9.. (i) Out of 0 to 9, is the only digit which when added odd number of times the sum also has the ones digit as. So we take A and add three times. Thus, we get B. (ii) Q Rewrite Q Q 8 0 Q Subtract the right column and transfer the digit so obtained Q 8 Length of a train m Speed of the train km/h + 8 m/s Since, the train has to pass a single post, that means train has to cover its length i.e., m. Time taken Distance Speed 6. (i) { } m m/s 0 seconds. (Taking reciprocals). (ii)( + + ) (Taking reciprocals) [ a 0 ]. Cost price of a sofa ` 800 Selling price of the same sofa ` 00 Here, S.P. > C.P. Profit S.P. C.P. ` 00 ` 800 ` 0 Profit % Profit 00 C.P % Cube root of 8 8 Let us factorize M A T H E M A T I C S VIII

194 9. SECTION-D Steps of construction:. Take a line segment AB cm.. Using protractor, make an angle of measure 00 at A and another angle of measure 0 at B.. Draw two rays AX and BY. BAX 00 and ABY 0.. Taking. cm radius with centre B, draw an arc that intersects the ray BY at C.. Again, taking. cm radius with centre A, draw another arc that intersects the ray AX at D. 6. Now Join CD. Thus, we obtain the quadrilateral ABCD. x 0. (i) + (x ) x + x x + ( x ) 6 x( x+) x + 8 x 6 x x x + 8 x( x) Multiplying both sides by, x + 8x (x ) x + 8x x + x + x x x (Transposing) x x. (ii) + a + a + a++ a+ ( a+)( a+) a + a + a+ a +0 a +0 a +0 Cross-multiplying, (a + )(a + 0) (a + a + ) a + a + 0 a + 6a + a 6a 0 (Transposing) a 6 a 6.. (i) The horizontal (x) axis shows the time. The vertical (y) axis shows the distance of the car from City A. (ii) The car started from City A at 8 a.m. (iii) The speed of the car was not the same all the time. (iv) We find that the car was 00 km away from City A when the time was a.m. and also at noon. This shows that the car did not travel during the interval a.m. to noon. The horizontal line segment representing "travel" during this period is illustrative of this fact. (v) The car reached to City B at p.m.. When two dice are thrown together then total possible outcomes are follows: (, ), (, ), (, ), (, ), (, ), (, 6) (, ), (, ), (, ), (, ), (, ), (, 6) (, ), (, ), (, ), (, ), (, ), (, 6) (, ), (, ), (, ), (, ), (, ), (, 6) (, ), (, ), (, ), (, ), (, ), (, 6) (6, ), (6, ), (6, ), (6, ), (6, ), (6, 6) The total number of outcomes is 6. (i) Getting the sum as an even number, the favourable outcomes are: P R A C T I C E P A P E R S 9

195 . (i) (, ), (, ), (, ), (, ), (, ), (, 6), (, ), (, ), (, ), (, ), (, ), (, 6), (, ), (, ), (, ), (6, ), (6, ), (6, 6). The total number of favourable outcomes is 8. P (getting an even number as the sum) Number of favourable outcomes Total numberof outcomes 8 6. (ii) Getting a total of at least 6 means the sum 6. Favourable outcomes are: (, ), (, 6), (, ), (, ), (, 6), (, ), (, ), (, ), (, 6), (, ), (, ), (, ), (, ), (, 6), (, ),... (, 6), (6, ),...,(6, 6) Total number of favourable outcomes is 6. P(getting total of at least 6) (ii). (i) (a + ) (a + a + ) 0a a + a(a + a + ) + (a + a + ) 0a a 0a [Distributive property] a + a + a + a + a + 0a 6a [Distributive property] a + (a + a 0a ) + (a + a 6a) + a a +. (ii)(a + b) (a b + c) (a b)c a(a b + c) + b(a b + c) ac + bc [Distributive property] a ab + ac + ab b + bc ac + bc [Distributive property] a b ab + ab + bc + bc + ac ac [Rearranging] a b ab + bc ac. Practice Paper- SECTION-A. (B) Curved surface area base perimeter height circumference of circular base height pr h prh. 9 M A T H E M A T I C S VIII

196 . (B) ( ) ( a a ) ( a0 ). (A) By the definition of direct variation, two variables are always in constant ratio. So, true equation is x y.. (D)Observing all the pairs, we conclude that one such pair whose both members having or its multiple as a coefficient is xy,.. (C) Origin is the intersecting point of axes where x and y-coordinate are zero. So the coordinates of origin are (0, 0). 6. (B) Expanded form of 80 8 hundreds + 0 tens + ones i.e., (B) Multiplicative inverse of a P R A C T I C E P A P E R 0 (a ) a a 8. (C) By the definition, linear equation in one variable must has a variable of degree. But we observe that + x has its variable of degree x. 9. (C) A convex quadrilateral has two of its diagonals in interior region. 0. (A) In the given data, is appeared frequently four times. So its frequency is. SECTION-B. Let A, B, C, D be the vertices of given quadrilateral. We know that opposite angles of a parallelogram are equal. S B D y Using angle sum property in ADC, 0 + x + y x + 80 x Now DCyAB and Ac is transversal. BAC ACD (Alternate interior angle) z x 8.. Square root of (i) (a + b) (a + b)(a + b) a(a + b) + b(a + b) a + ab + ab + b a + ab + b. (ii) (x + a )(x + b) x(x + b) + a(x + b) x + bx + ax + ab x + (b + a)x + ab x + (a + b)x + ab.. Let a man s original salary be ` x. After 0% increament his new salary is ` 000. x + 0% of x 000 x + x x+0x x x `,0,000. 9

197 . Let breadth of a rectangle be x. So length of the rectangle be x. From question, Area 88 cm x x 88 [ Area lb] x 88 x 88 x cm. Length x cm, breadth x cm. 6. Given number is 69. Sum of the digits in the given number which is not divisible by as well as 9. Thus, 69 is not divisible by and 9.. Cube of. (.) ( ) + 0 (0) SECTION-C 9. Additive inverse of [ a m a n a m+n ] a a m n a 8 m n 8 8 ( ). 0. Let Seema s age x years. According to question, Reema s age (x ) years Given that sum of their ages is 8 years x + x 8 x 8 x 8 + x years and years + months years and 6 months x years months 6 months Hence, Reema s age 6 months Seema s age years and 6 months or. years.. We know that a number is divisible by when the difference between the sums of digits at odd places and even places is 0 or multiple of. So, 6 * is divisible by if ( + 6) ( + * ) 0 or or... 8 * 0 or or... * 0 or (Rest neglecting) * 0 or ± * 0 or * ± * or * 0 or [Not possible] *.. Length of the room. m Breadth of the room.8 m Height of the room. m We know that area of four walls Perimeter of floor height (l + b) h (. +.8) m. 96 M A T H E M A T I C S VIII

198 . Rate of painting the walls ` per m Total cost of painting the four walls Rate Area ` 8.8 `. From graph, we find that the given points lie on the same vertical line. This line is named as y-axis.. On the basis of given information, the spinning wheel may be as shown to the right. It has total number of sectors. Number of green sectors is The probability of getting a green Number of green sector sector Total number of sector. Further, number of blue sector So total number of non-blue sectors The probability of getting a non-blue Number of non-blue sectors sector Total number of sectors.. We know that the least number which exactly divisible by given number is their LCM. Let us find L.C.M. of, 8 and. P R A C T I C E P A P E R S Using division method,,8, LCM, 8 and,,6,, Since we have to find least square number which is exactly divisible by, 8 and. But observing factors contained in LCM, we see that is not a perfect square. So multiplying by i.e. 6, we get 6 which is the required number. 6. Principal `,000, Rate of interest % yearly, Time years, Simple Interest?, Amount? Using formula, S.I. P R T ` 0 Amount P + S.I. ` ` 0 ` 8,0. Thus, Geeta will have to pay her friend `,0 as simple interest and ` 8,0 as amount.. (x ) x x x ( x ) 8 x ( x ) x 8 x 8 x + x x 8 x x x Multiplying both sides by the LCM of denominators, i.e., by 6 (L.C.M. of, ) (8x ) (x ) (x ) 6x x + x Transposing variables and constants, 6x x x + 6x 6x 0x 0 x

199 8. Suppose Kishore s wife s salary be ` 00. So Kishore s salary ` % of ` 00 ` 00 + ` 0 ` 0. So the difference between their salaries ` 0 ` 00 ` 0. Since Kishore s salary is ` 0 then his wife s salary is ` 0 less So Kishore s salary is ` then his wife s salary is 0 0 less So Kishore s salary is ` 00 then his wife s salary is less 00 9 %. SECTION-D 9. (i) The greatest three-digit number 999 Let us try to find the greatest threedigit perfect square number. At first, we try to find square root of 999 using division method. Subtracting remainder 8 from 999, which is the required number. (ii) Given number,66 66 Applying prime factorization method To plot the given data on a graph, firstly, we choose the suitable axes for the quantity of petrol and its cost. Since cost of petrol depend on the quantity of petrol, we take quantity of petrol in litre on x-axis and cost in ` on y-axis. Also we take the scale unit ` 00 on y-axis and unit litres on x- axis. Then plot the points (0, 00), (, 0), (0, 000), (, 0) as shown on the graph.. (i) Steps of construction:. Take a line segment of measure. cm and name it AB.. Using ruler and compass, make an angle of 90 at A and B. Then draw two rays AX and BY.. Taking radius as. cm and centres as A and B, draw two arcs that cut the rays AX at D and BY at C. 98 M A T H E M A T I C S VIII

200 . Join CD. Thus, the square ABCD is obtained. (ii) Steps of construction:. Take a line segment of measure cm and name it AB.. Using ruler and compass, make an angle of 60 at A and draw a ray AX.. Further, taking cm as radius and A as centre, draw an arc that cuts ray AX at D.. Now taking radii of lengths cm and cm with respectively centres B and D, draw two arcs that intersect each other at C.. Join BC and DC. Thus, a parallelogram ABCD is obtained.. (i) Diameter of a rod roller. m Radius (r) of a rod roller. m 0. m P R A C T I C E P A P E R S Length (i.e., height) of the rod roller m Curved surface area prh m Since in revolution the rod roller covers the area 8.8 m So in 0 revolution the rod roller covers the area m. (ii) Lengths of the parallel sides are cm and 6 cm. Distance between the parallel sides i.e., altitude. cm Area of the trapezium (Sum of parallel sides) Altitude ( + 6). 8.. cm.. (i) Consider 6 xx x + yyyz This puzzle has three letters x, y and z whose values are to be found. We study the sum in the ones column. The sum of two letters x and z is 9 so it is clear that the carry cannot be forwarded to the tens column. When we study the tens column, the sum of and y is 6 i.e., + y 6 [ The sum 6 is not possible 99

201 if is added to a digit out of 0 to 9] y Therefore, studying in hundreds and thousands columns, we find that x + y 8 ( Carry is not forwarded to ten thousands column) i.e., x + 8 x. Now putting x in ones column, we get x + z 9 i.e., + z 9 z Thus, x, y, z. (ii) x x x 8 x 8 ( am a n a m+n ) Since bases are equal, exponents also be equal. That means + x x x x (Transposing) x.. From figure, we observe that ABCD is a trapezium whose parallel sides AB 0 cm, DC 0 cm and distance between them is cm ( AD). Area of trapezium ABCD (AB + DC) AD (0 + 0) cm The region unshaded in interior of trapezium ABCD is a rectangle with dimensions cm cm. Area of the rectangle cm cm ( A lb) cm. The area of shaded region 00 cm cm 88 cm. Practice Paper- SECTION-A. (A) The given rationals are 6 and. Let us find their equivalent fractions with the denominator. 6 6 and 6 i.e., and 6 < < 6 Now < < 6 6 < <.. (B) We have 0. 0.x 0. x (C) It is given that sum of any two angles of a quadrilateral is 0. We know that the sum of all four angles of a quadrilateral is 60. The sum of remaining two angles (B) By the property of perfect square numbers ending with different digits, we know that a perfect square number ending with ends with itself i.e.,.. (D) Cube of ( ) 9 [ (a m ) n a m n ] 00 M A T H E M A T I C S VIII

202 6. (C) Before VAT charged, the price of a double bed ` 0,000 Rate of VAT 0% After VAT charged, the price of the double bed % of `,000.. (D) An algebraic expression having only two terms joined with sign + or is called binomial. Here, we find that the expression 6xy y contained two terms 6xy and y joined with ve sign. Therefore, it is a binomial. 8. (A) Counting the vertices of a hexagonal prism drawn to the right, we get that its total number is. 9 (D) We know that area of a parallelogram base height r h rh. 0. (B) [ a m a n... a m + n +... ] + 0. [ a 0 ] SECTION-B. We know that ratio is a comparission between two quantities with same units. So, firstly, we convert the given quantities in same units then find their ratios. (i) minutes to 0 seconds minutes 60 seconds 0 seconds 0 seconds [ minute 60 seconds] 60 0 :. P R A C T I C E P A P E R S (ii) paise to ` paise ` paise 00 paise [ ` 00 paise] 00 8 : 8.. P ` 000, r 8% per annum, t years, C.I.? C.I. A P t r P + P 00 t r P t r A P (08 +00)(08 00) ` 8.. Addition (l + n ) + (m + n ) + (l + m ) + (mn + lm + nl) l + n + m + n + l + m + mn + lm + nl (l + l ) + (m + m ) + (n + n ) + mn + lm + nl l + m + n + mn + lm + nl (l + m + n + lm + mn + nl). 0

203 . x(8x ) x 8x x Putting x, [Distributive property] x x. x x () () It is given that side of a square x y So, area of the square (side) (x y) (x) (x)(y) + (y) [Using identity (a b) a ab + b ] x xy + y. 6. When a train passes a platform (bridge, etc) then it covers the distance equals to the sum of lengths of train and platform. So the total distance is to be covered 00 m + 0 m 60 m Speed of the train km/h 8 m/s [Changing into m/s] m/s The time required Distance Speed. 8. For a polyhedron, given that Number of edges 0, Number of vertices 0 We have to find the number of faces the polyhedron has. Using Euler s formula, F + V E F F 0 F SECTION-C or 60 m m/s 60 6 sec. sec. 0. In 0 days, Rinku can make dress. In day, Rinku can make 0 part of a dress. 0 M A T H E M A T I C S VIII

204 In days, Rinku and Teena together can make dress. In day, Rinku and Teena together can make part of the dress. So, in day, Teena alone can make 0 part of the dress 0 0 part Now, Teena alone can make 0 part of a dress in day So, Teena alone can make full dress in 0 0 days. 0.(i)0x yz 0xy + x x y z x y + x x( x y z y + x ) x(xyz y x ) (ii) x + x + x + (x + x ) + (x + ) x (x + ) + (x + ) (x + )(x + ). x + x 0 (x ) Let us perform long division, x +0 x x + x0 x x + 0 x 0 0 x 0 Quotient x We have to draw a graph to show the reading related to the distance travelled by a car and its average cost. For this, we choose suitable axes on which the given data can be taken and also P R A C T I C E P A P E R S choose their scales. Let us take distance travelled (in km) on x-axis (since it is independent) and average cost (in ` ) on y-axis (since it depends on the distance travelled). On x-axis, unit 0 km and on y- axis, unit ` 00. Now plot the points (0, 00), (0, 80), (0, 0), (0, 00), (0, 680) as shown in the graph. Then join them with free hand curve.. From the situation given in question, we will have to perform the operation of subtraction for algebraic expression. Required expression (x + x + ) (x x + ) x + x + x + x (x x ) + (x + x) + ( ) x + 8x.. (i) 999 (000 ) (000) (000)() + () [Using (a b) a ab + b ] (ii) (.) ( + 0.) () + ()(0.) + (0.) [Using (a + b) a + ab + b ] Given:Area of a square 600 m (Side) 600 0

205 Side 600 m Perimeter of the square side m 980 m. Let us find square root.. The given number is 0,. Applying prime factorization method Observing the above factors, we find that appears only once. Therefore, we have to eliminate it to become 0 a perfect square. Thus, is the least number by which the given number must be divided (i) x 9 y 6 6 x 9 y [ a m b m (a b) m ] 6 6 x 9 y Since exponents are same of both sides, base also be same. x i.e., y 9. (ii) (a) Standard form of (Transfer decimal point from end to 6 places to the left) (b) Standard form of (Transfer decimal point from right to places to the left) SECTION-D 9. (i) Total number of tossing a coin 0 Percentage chance of occurring a Tail 60% Percentage chance of occurring a Head % So the number of times Head has occurred 0% of (ii) We know that the outcomes when an unbiased die is tossed two times is equal to the outcomes when two dice together are tossed one time. Therefore, the total 6 outcomes are: (, ), (, ), (, ), (, ), (, ), (, 6), (, ), (, ), (, ), (, ), (, ), (, 6), (, ), (, ), (, ), (, ), (, ), (, 6), (, ), (, ), (, ), (, ), (, ), (, 6), (, ), (, ), (, ), (, ), (, ), (, 6), (6, ), (6, ), (6, ), (6, ), (6, ), (6, 6). 0. (i) No. Since, ! 6 i.e., 9 (ii) The smallest four-digit number Taking prime factorization, Thus, we observe that 0 M A T H E M A T I C S VIII

206 . each factor has an exponent so we conclude that 000 is a perfect cube. Therefore, The least -digit perfect cube (i) Since ABCD is a parallelogram, diagonals AC and BD bisect each other at O. i.e., AO OC AC and BO OD BD. Given: OB cm, AC BD + cm So AC OB + Steps of construction:. Take a line segment AB. cm.. Using ruler and compass, make an angle of measure at A and draw a ray AX.. Again, make another angle of measure 0 at B and draw a ray BY.. Taking 6. cm as radius with centre B, draw an arc that cut BY at point C.. Further, make a right angle at C using ruler and compass and draw a third ray CZ. 6. Extend rays AX and CZ till they cross each other at a point D. Thus, a quadrilateral ABCD is formed. P R A C T I C E P A P E R S ( OB BD ) AC cm. OA AC 6. cm. (ii) Since given figure is a parallelogram, opposite angles are equal. i.e., x z and y 00 Also adjacent (consecutive) angles are supplementary. x x Thus, x 80, y 00, z 80.. (i) Let us find the sum of digits contained in number 68. Sum of the digits which is divisible by 9 The given number also divisible by 9. (ii) By the divisibility test of, a number is divisible by if the difference between the sum of 0

207 digits at odd places and even places is either 0 or a multiple of. Now, consider the number 9 * 6 ( ) ( + * + ) 0 or or... 6 * 0 or or or... 9 * 0 or or or... 9 * 0 or ± or ± or... 9 * 0 or 9 * ± or 9 * ± or... Neglecting * 9 or 9 * ± or soon... 9 * ± 9 * (Neglect ve sign) * Since a rectangular paper of width cm is rolled along its width to form a cylinder, height of the cylinder is equal to cm. Given radius 0 cm Volume of the cylinder pr h (0) cm. Practice Paper- SECTION-A. (C) Since x and y are in direct proportion, the ratio of x to y is always constant. x i.e., k where k is constant y x ky.. (A) (a b) c (a b + c)(a b c) [Using x y (x + y)(x y)]. (C) The perpendicular distance of a point is denoted by x-coordinate. In this case, that is a.. (C) Consider the number Here, ones digit 9 is not divisible by, and 0 (also). So only option 9 may be its factor. Let us check the sum of digits i.e., which is divisible by 9. Hence the given number also divisible by 9. x y. (D) (x ) y [Multiplying both sides by ] x 6 y x y (C) In the given figure, producing a side, we get [Exterior angle property] 0 Again, x (C) The spinning wheel has total five sectors but the letters inserted are P, Q, R (only three). So the number of all possible outcomes is. 8. (B) [ ] (D) 9% of x 9 x x M A T H E M A T I C S VIII

208 0. (A) We know that variables are represented by small letters of English Alphabet. In the given expression, we observe that those are three namely x, y and z. SECTION-B. (i) Consider 0 We see that the number containing 0 raised to.that means decimal point will move places from right to left (ii) Here,. 0 has 0 raised to (+ve). So decimal point will move places from left to right In 6 hours, Reema completes knitting full sweater. In hour, Reema completes knitting part of the sweater. 6 In hours, Reema completes knitting 6 part of the sweater Let us find prime factors of Dividend Quotient, Divisor?, Using Algorithm Theorem, Dividend Divisor Quotient [ Remainder 0] Divisor Divisor Divisor. Consider 0 A + A This puzzle has two letters A and B whose values are to be found. We observe the sum of ones column and find that B + 8 B. Now puzzle seems to be A B + A B B 8 We study the addition in tens column and find that sum A + a two digit number may be, A + A. Putting A in hundreds column and carry forwarded to this column satisfies the addition. Therefore, A and B. 6. ( 8 ) 8 [Taking reciprocals] P R A C T I C E P A P E R S 0

209 . Side of a cubic wooden block 0 cm Volume of one block (0) cm Dimensions of a cuboidal wooden block are m, 0 cm and 0 cm Volume of the block l b h cm [ m 00 cm] The number of small cubic blocks that can be cut from the big block Volume of cuboidal block Volume of one cubic block In a square, diagonal (d) 90 m (Given) From figure, it is clear that d a a 90 a 90 m Area of the square a m. SECTION-C 9. We observe the given figure and find that it is a regular hexagon that contains six equilateral triangles. So, area of the regular hexagon 6 Area of an equilateral triangle 6 (side) 6 () 6 m. 0. Cost price of a second-hand refrigerator ` 00 Additional cost on its repairing ` 00 Total pure cost ` 000 Selling price of the refrigerator ` 00 Since, Selling price > Total pure cost Rajesh gained his transection.. Yes. Gain ` 00 ` 000 ` 00 Gain % Gain 00 Total cost % M A T H E M A T I C S Since the product of 0. and Therefore, we conclude that 0. and are the multiplicative inverse of each other.. (i) x y (x ) (y ) (x + y )(x y ) [ a b (a + b)(a b)] (x + y )(x + y)(x y). (ii)8p + q 8(p + q ) [ Taking H.C.F. of both terms in common]. Side of a cube cm. Total surface area of the cube 6a 6 () 6 0 cm Volume of the cube a ( cm) cm. VIII

210 . We have given two adjacent sides of a rectangle. We know that, in a rectangle adjacent sides are length and breadth.. The figure drawn shows a rectangle with sides. cm and 6 cm. So, perimeter (length + breadth) (i) Let l 8x + 0, b x Perimeter (l + b) (8x x ) (x + ) 6x +. (ii) Let l m + n, b m n Perimeter (l + b) (m + n + m n ) (m n ) m n (i) We know that the sum of all exterior angles of a polygon is 60.. ( ) n + ( ) ( ) ( ) n + + ( ) ( ) n + 6 ( ) [ a m a n a m + n ] Since bases are equal, exponents also be equal. n + 6 n Expression 8(p q s ) (r s ) So, x x x (ii) Since the given figure is a parallelogram, opposite angles are equal. 8p 8q 8s r + s 8p 8q r 6s. Putting the values p, q, r, s ; we get Value of expression 8p 8q r 6s 8( ) 8( ) () 6( ) () x z and y 0 Also, adjacent angles of a parallelogram are supplementary. x x Thus, x 60, y 0 and z 60. P R A C T I C E P A P E R S 09

211 SECTION-D 9. (i) m + 88m Cross-multiplying, (m + ) (8 8m) m + 6 6m Transposing, m + 6m 6 m m. Verification: L.H.S. m + 88m R.H.S. (ii) x + 8 x 9 Cross-multiplying, 9(x + ) 8(x ) 8x + 9 x 6 Transposing, 8x x 6 9 6x x 6 Verification: L.H.S x + x R.H.S Steps of construction:. Take a line segment AB cm.. Taking radii 6 cm and cm with centres B and A respectively, draw two arcs which cut each other at C.. Again, taking radii cm and. cm with centres C and A respectively, draw two arcs which cut each other at D.. Join AC, AD, BC and CD. Thus, a quadrilateral ABCD is formed.. We know that central angle Particular item Sum of total items 60 In this case, central angle No. of students in a single game No. of total students in all games 60 Number of students in a single game Central angle 60 Total number of students. 0 M A T H E M A T I C S VIII

212 Therefore, number of students in different games as: In Cricket, students. In Badminton, students In Basket ball students In Tennis students (i) The number of students playing cricket 0. (ii) The sum of number of students playing tennis and badminton (iii) The difference between the number of students who play badminton to cricket (iv) The ratio of students playing badminton to tennis 60 : 0 :. (v) The number of students who play neither basket ball nor cricket The number of students who play either badminton or tennis (i) We know that the general form of Pythagorean triplet be m, m, m + Let us take m m + So m is not an integer. Now let us take, m m 6 m 6 6 and m Again let us try, m + m Here also m has not integral value. Thus, the Pythagorean triplet is,,. [Note: All Pythagorean triplets may not be obtained using this general form. e.g., another triplet,, also has as a member.] (ii) Square root of 8 8 Let us find its square root up to two decimal places using long division method. 8.9l.9.. (i) In the graph, time is taken on x-axis. Scale: unit on x-axis hour. P R A C T I C E P A P E R S

213 (ii) The person started the journey at 8 am and reached at.0 am to the place of merchant. So taken time hours 0 minutes or hour. (iv) Yes. During the period of 0 am to 0.0 am, line graph is parallel to x-axis. That means he was stoped. (v) From graph, we observe that the person rode the cycle with different speeds but he covered most distance in his first hour between 8 am to 9 am.. Front view Side view Top view (iii) The place of the merchant is km apart from town. M A T H E M A T I C S VIII