MTHEMTICS MODEL QUESTION PPERS WITH NSWERS SET 1 Finish Line & Beyond CLSS X Time llowed: 3 Hrs Max. Marks : 80 General Instructions: (1) ll questions are compulsory. (2) The question paper consists of 30 questions divided into 4 sections:, B, C and D. Section comprises of 10 questions of 1 marks each. Section B comprises of 5 questions of 2 marks each. Section C comprises of 10 questions of 3 marks each and section D comprises of 5 questions of 6 marks each. (3) ll questions in section are to be answered in one word, one sentence or as per the exact requirement of the question. (4) There is no overall choice. However, internal choice has been provided in one question of 2 marks each, three questions of 3 marks each and two questions of 6 marks each. You have to attempt only one of the alternatives in all scuh questions. (5) In questions on constructions, the drawing should be neat and clean and exactly as per the given measurements. (6) Use of calculator is not permitted. Section Question numbers 1 to 10 carry 1 mark each. p 1. If is a rational number ( q 0 ), what is the condition on q so that q the decimal representation of q p is terminating. If the prime factorization of q is of the form 2 n.5 m, where, n, and m are nonnegative integers, then the decimal representation of q p is terminating. 2. Write the zeroes of the polynomial x²+ 2x + 1. x² + 2x + 1 (x+1) (x+1) So, the value of x² + 2x + 1 will be zero When, x+10 i.e., x -1 Or, x+10 i.e., x -1 So, the zeroes of the given equation are -1 and -1
3. Find the value of k so that the following system of equations has no solution: 3x y 5 0; 6x 2y k 0 a 1 3 1 Here, a2 6 2 b 1 1 1 b2 2 2 c1 5 5 c2 k k a 1 b1 c1 If, a2 b2 c2 Then the system of equations will have no solution. If, k 10, then k 5 2 1 So, if k 10, then the equations will have no solution. 4. The nth term of an P is 7 4n. Find its common difference. Let t n denote the nth term of the P. Then t n 7 4n Common Difference d t n - t n -1 Or, d 7 4n [ 7 4 (n 1) ] Or, d 7 4n 7 + 4n 4 Or, d -4 lternate Method, t 1 7-4 3 t 2 7-8 -1 t3 7-12 -5 It is clear that common difference is -4
5. In the given figure, D 4 cm, BD 3cm and CB 12 cm, find the value of cotθ. 4 D 3 B 12 θ C In BD, using Pythagoras theorem value of B can be calculated as follows: B² D² + BD² 4² + 3² 16 + 9 25 Or, B 5 BC Now, In BC, cotθ B 12 5 6. In the given figure, P and Q are points on the sides B and C respectively of BC such that P 3.5 cm, PB 7 cm Q 3 cm and QC 6 cm. If PQ 4.5 cm, find BC. P Q B C
P 3.5 1 B 10. 5 3 Q 3 1 C 9 3 So, PQ BC PQ 1 So, BC 3 BC PQ 3 4.5 3 13.5 7. In figure, PQ 24 cm, QR 26 cm, PR 90, pa 6 CM ND ar 8 CM. Find QPR. Q P R Solution. In right PR, we have PR² P² + R² (6)² + (8)² 36 + 64 100 (10)² PR 10 cm QPR, PQ² + PR² (24)² + (10)² 576 + 100 676 PQ² + PR² (26)² QR² QPR 90 8. In figure, O is the centre of a circle. The area of sector OPB is 18 5 of the area of the circle. Find x.
O P B Solution. rea of a sector OPB 18 5 x rea of a circle x 360 X Πr² 18 5 Πr², where r O OB x 18 5 x 360 100 Thus, x 100. 9. Which measure of central tendency is given by the x-coordinate of the point of intersection of the more than ogive and less than ogive? Solution. The median of a grouped data of central tendency is given by the x- coordinate of the point of intersection of the more than ogive and less than ogive. 10.From a well-shuffled pack of card is drawn at random. Find the probability of getting a black queen. Solution. Well-shuffling ensures equally likely outcomes. Since there are 52 cards in a pack, therefore, the total numbers of possible outcomes 52. There are Two black queens in a pack of 52 cards. Let E be the event getting a black queen, then the number of outcomes favourable to E 2. 2 1 So, P(E) 52 26 Section B Question numbers 11 to 15 carry 2 marks each. 11.Find the zeroes of the quadratic polynomial 6x² - 3 7x and verify the relationship between the zeroes and the coefficients of the polynomial.
Solution. We have 6x² - 3 7x 6x² - 7x 3 6x² - 3 7x 6x² - 9x + 2x 3 6x² - 3 7x 3x(2x 3) + (2x 3) 6x² - 3 7x (2x 3)(3x + 1) So, the value of 6x² - 3 7x is zero when 2x 3 0 or 3x + 1 0, i.e, when x 2 3 1 or x 3 3 1. Therefore, the zeroes of 6x² - 3 7x are and. 2 3 3 1 Now, Sum of zeroes + ( ) 2 3 9 2 6 6 7 ( 7) (Coeff of - 6 Coeff of x² 3 1 1 Product of zeroes x ( ) - 2 3 2 x) ( 3) 6 Const.Term Coeff of x² Without using the trigonometric tables, evaluate the following : 11 sin 70 4 cos53 cosec37. -. 7 cos 20 7 tan15 tan 35 tan 55 tan 75 Solution. We have 11 sin 70. 7 cos 20 4 cos53 cosec37 -. 7 tan15 tan 35 tan 55 tan 75 11 sin(90 20 ). 7 cos 20 4 cos(90 37 ).cosec37 -. 7 tan15 tan 35.tan(90 35 ).tan(90 15 ) 11 cos 20. 7 cos 20 11 cos 20. 7 cos 20 4 sin 37.cosec37 -. 7 tan15 tan 35.cot 35.cot15 [ sin( 90 θ ) cosθ,cos(90 θ ) sin θ, tan(90 θ ) cot θ ] 4 sin 37.cosec37 -. 7 (tan15.cot15 )(tan35.cot 35 ) 11 4 1. (1) -. 7 7 (1)(1 ) 11 4-7 7
7 7 1. 12.For what value of p, are the points (2,1), (p, -1) and (-1, 3) collinear? Solution. Since the given points are collinear, therefore, the area of the triangle formed by them must be zero, i.e. 1 [x1 (y2 y 3) + x 2(y 3 y 1) + x 3(y 1 y 2)] 0, where 2 x 1 2, y 1 1, x 2 p, y 2-1, x 3-1, y 3 3 1 [ 2(-1-3) + p(3 1) + (-1)(1 + 1)] 0 2 1 [- 8 + 2p 2] 0 2 1 [- 10 + 2p] 0 2-5 + p 0 p 5 Verification : rea of V 1 [2(-1-3) + 5(3 1) + (-1)(1 +1)] 2 1 [- 8 + 10 2] 0 2 13.BC is an isosceles triangle, in which B C, circumscribed about a circle. Show that BC is bisected at the point of contact. Solution. BC is an isosceles triangle, in which B C, circumscribed about a circle with centre O. Since tangents drawn from an external point to a circle are equal in length. F E..(1) [Tangents from ] BF BD..(2) [Tangents from B] CD CE..(3) [Tangents from C] dding (1), (2) and (3), we get F + BF + CD E + BD + CE B + CD C + BD But B C (given) CD BD BC is bisected at the point of contact D. F O E B D C Or
In figure, a circle is inscribed in quadrilateral BCD in which B 90. If D 23 cm, B 29 cm and DS 5 cm, find the radius of the circle. Q R B O D P S C Solution. Since tangents to a circle is perpendicular to the radius through the point. OPB OQB 90 It is given that B 90. lso, OP OQ. Therefore, OPBQ is a square. Since tangents drawn from an external point to a circle are equal in length. DR DS [Tangents from D] R Q [Tangents from ] nd BP BQ [Tangents from B] Now, DR DS DR 5 [Q DS 5 cm (given)] D R 5 23 R 5 R 23 5 18 Q 18 [R Q] B BQ 18 29 BQ 18 [Q B 29 cm (given)] BQ 29-18 11
BQ 11 cm But OPBQ is a square, therefore, OP OQ BP BQ Hence, OP 11 cm, i.e., r 11 cm. 14. die is thrown once. Find the probability of getting (i) a prime number (ii) a number divisible by 2. Solution. s we know that a die has six faces with 1, 2, 3, 4, 5 and 6 written on them. Thus, the total number of outcomes when a die is thrown once are 6, i.e., 1, 2, 3, 4, 5, 6. (i) Since there are 6 equally likely outcomes : 1, 2, 3, 4, 5 or 6 in a single throw of a die and there are 3 ways of getting a prime number, namely, 2, 3 or 5. P(getting a prime number) 3 6 1 2 (ii) Since there are 6 equally likely outcomes : 1, 2, 3, 4, 5 or 6 in a single throw of a die and there are 3 ways of getting a number divisible by 2, namely, 2, 4 or 6. P(getting a number divisible by 2) 3 6 1 2 Section C Question numbers 16 to 25 carry 3 marks each. 15. Show that 5-2 3 is an irrational number. Solution. Let us assume, to contrary, that 5-2 3 is rational. That is, we can find coprime a and b (b 0) such that 5-2 3 a b Therefore, 2 3 5 - a b 2 3 5b a b 3 5 b a 2b Since a and b are integers, 5 b a 2b is rational, and so 3 is rational. But this contradicts the fact that 3 is irrational. This contradiction has arisen because of our incorrect assumption that 5-2 rational. So, we conclude that 5-2 3 is irrational. 3 is 16. Find the roots of the following equation :
1 x + 4-1 x 7 11, x -4, 7. 30 Solution. We have 1 x + 4-1 x 7 11 30 s x -4, 7, multiplying the equation by (x + 4)(x 7), we get (x 7) (x + 4) 11 (x + 4)(x 7) 30 x 7 x -4 11 (x + 4)(x 7) 30-11 11 (x + 4)(x 7) 30-30 (x + 4)(x 7) - 30 x² - 7x + 4x -28 x² - 3x + 2 0 So, the given equation reduces to x² - 3x + 2 0, which is a quadratic equation. Here a 1, b -3, c 2. So, b² - 4ac (-3)² - 4(1)(2) 9 8 1 > 0 Therefore, x 3 ± 1 3 ± 1 i.e., x 2 or x 1 2 2 So, the roots are 1 and 2. 18. Represent the following system of linear equations graphically. From the graph, find the points where the lines intersect y-axis. 3x + y 5 0; 2x y - 5 0 Solution. Given equations are : 3x + y 5 0 y 5 3x..(1) 2x y - 5 0 y 2x 5..(2) Let us draw the graphs of the equations (1) and (2) by finding two solutions for each of these equations. They are given in tables : Y 5 3x y 2x 5 X 0 2 Y 5-1 B X 0 3 y -5 1 C D Y
6 5 4 (0,5) 2x y 5 0 3 X 2 1-6 -5-4 -3-2 -1 O -1-2 -3 D(3,1) 1 2 3 4 B(2,-1) 5 6 X -4-5 -6 C(0, -5) -3x + y 5 0 Y In figure, we observe that the two lines representing the two equations are intersecting at the point B (2, -1). Hence, x 2 and y -1. The line B cuts the y-axis at the point (0,5) and the line CD cuts the y-axis at the point C(0,-5). 19. The sum of n terms of an.p. is 5n² - 3n. Tind the.p. Hence, find its 10 th term. Solution. Let S n denote the sum of first n terms of an.p., then S n 5n² - 3n nd S n-1 5(n 1)² - 3(n 1) t n S n - S n-1 5n² - 3n [5(n 1)² - 3(n 1)] 5[n² - (n 1)²] 3[n (n 1)] 5[n² - n² + 2n 1] 3(1) 5(2n 1) -3 10n 8 Putting n 1, 2, 3,, we get
t 1 10 x 1 8 2, t 2 10 x 2 8 12, t 3 10 x 3 8 22, Clearly, d t 2 t 1 12 2 10 and d t 3 t 2 22-12 10 Thus, the.p. is 2, 12, 22, 32,.. Now, 10 th term of the.p. t 10 a + (10 1 )d 2 + 9(10) 2 + 90 92 Hence, the 10 th term of the.p. is 92. 20.Prove that : cot cos cos ec 1 cot + cos cos ec + 1 Solution. We have cot cos L.H.S cot + cos - cos cos sin + cos 1 cos 1 sin 1 cos + 1 sin 1 1 sin 1 + 1 sin cos ec 1 cos ec + 1 R.H.S Prove that: Or (1 + cot cosec )(1 + tan + sec ) 2 Solution. We have
L.H.S (1 + cot cosec )(1 + tan + sec ) cos 1 1 sin 1 + sin sin 1 + + cos cos sin + cos 1 cos + sin + 1 sin cos [(sin + cos ) 1][(sin + cos ) + 1] sin cos (sin + cos )(sin + cos )² - (1)² sin cos sin ² + cos ² + 2sin cos ) 1 sin cos 1 + 2sin cos ) 1 sin cos 2sin cos sin cos 2 RHS 21.Determine the ratio in which the line 3x+4y-90 divides the line segment joining the points (1, 3) and (2, 7). Let the required ratio be k:1 in which the line segment joining the points (1, 3) and (2, 7) be divided by the point R. Then the coordinates of R are 2k + 1 7k, k + 1 k + + 3 1 * * * k:1 * (1, 3) R (2, 7) Since the line 3x+4y-9 0 Divides the line segment joining the points (1,3 ) and (2, 7), therefore R lies on the line 3x+4y-90
2k + 1 7k + 3 3 + 4 9 0 k + 1 k + 1 (6k+3)+(28K+12)-(9k-9) 0 (6k+28K-9k)+(3+12-9) 0 25k+6 0 6 k - 25 6 Hence the required ratio is - :1, 25 i.e., -6:25 internally Or, 6:25 externally 22. Construct a BC in which B6.5 cm, B 60 and BC 5.5 cm. lso construct a triangle B C similar to BC, whose each side is 2 3 times the corresponding side of the BC. C
X C 5.5 cm 6.5 cm B B 1 2 3 Y Steps of Construction: 1. Draw a line segment B6.5 cm 2. t B construct BX 60 3. With B as centre and radius BC5.5 cm draw an arc intersecting BX at C. 4. Join C. Triangle BC so formed is the required triangle. 5. Construct an acute angle BY at on opposite side of vertex C of BC. 6. Locate three points (the greater of 3 and 2 in 2 3 ) 1, 2, 3 on Y such that 1 1 2 2 3. 7. Join 2 (the 2 nd point, 2 being smaller of 2 and 3 in 2 3 ) to B and draw a line through 3 parallel to 2B, intersecting the extended line segment B at B. 8. Draw a line through B parallel to BC intersecting the extended line segment C at C. Triangle B C so obtained is the required triangle.
23. If the diagonals of a quadrilateral divide each other proportionally, prove that it is a trapezium. Given a quadrilateral BCD whose diagonals C and BD intersect each other at O such that O BO OC OD To Prove: Quadrilateral BCD is a trapezium, i.e., B DC D C E O B Construction: Draw OE ll B, meeting D in E. Proof: In BD, we have OE ll B E BO So, ED OD O BO But, OC OD From above equations it is clear that E O ED OC So, from Parallel line s theorem it can be said that EO ll DC EO ll B So, DC ll B or, B ll CD Hence, BCD is a trapezium. lternate Question: Two s BC and DBC are on the same base BC and on the same side of BC in which D 90. If C and BD meet each other at E, show that E. EC BE.ED
E D B C In BC and BCD BC BDC (Right ngle) EB DEC (Opposite Cone) So, By theorem of similar triangles, BC DBC So, E BE ED EC Or, E.EC BE.EC proved 24. If the distances of P(x,y) from the points (3,6) and B(-3, 4) are equal, prove that 3x+y 5. Here P(x, y), (3. 6) and B(-3, 4) are given points. It is given that distances of P(x, y) from (3, 6) and B(-3, 4) is equal. So, P BP Or, P 2 BP 2 Or, (x-3)²+(y-6)² (x+3)²+(y-4)² Or, (x²-6x+9)+(y²-12y+36) (x²+6x+9)+(y²-8y+16) Or, -6x-12y+45 6x-8y+25 Or, 6x+6x-8y+12y45-25 Or, 12x+4y 20 Or, 3x+y 5 proved 25. In the given figure, find the perimeter of the shaded region where DC, EB and BFC are semi-circles on diameters C, B and BC respectively. D E 2.8 cm B 1.4 cm F C
The sum total of perimeters of semicircles EB, DC and BFC will give the perimeter of the shaded figure. d Perimeter of Semicircle EB 22 1. 4 4.4 2 7 22 4.2 Perimeter of Semicircle DC 6.6 7 2 Perimeter of Semicircle BFC 22 0. 7 2.2 7 So, the required perimeter 4.4 + 6.6 + 2.2 13.2 cms lternate Question: Find the area of the shaded region in the figure, where BCD is a square of side 14 cm. B D C
reas of four circles subtracted from the area of square will give the area of the shaded region. Radius of one circle will be equal to 1/4 th of the side of the square. rea of Square Side² 14² 196 22 7 7 rea of one circle Πr² 7 2 2 22 7 7 So, rea of four circles 4 22 7 154 7 2 2 So, rea of shaded region 196 154 42 cm² SECTION D Question numbers 26 to 30 carry 6 marks each. 26. In a class test, the sum of the marks obtained by P in the Mathematics and Science is 28. Had he got 3 more marks in Mathematics and 4 less marks in Science, the product of marks obtained by him would have been 180. Find the marks obtained in two subjects separately. Let us assume that marks obtained by P in Math M and in Science S So, as per question, M+S 28 ----------------------------- (1) (M+3)(S-4) 180 --------------------------- (2) From equation (1) M 28-S Putting value of M in equation (2) we get (28-S+3)(S-4) 180 (31-S)(S-4) 180 35S -S²-124 180 Or, -S²+35S 304 Or, S²-35S+304 0 Or, S²-19S-16S+304 0 Or, S(S-19)-16(S-19) 0 Or, S16 or S19 If, S 16 then M28-16 12 Then (12+3)(16-4) 15 12 180 If, S19 then M 28-19 9 Then (9+3)(19-4) 12 15 180 lternate Question: The sum of reas of two squares is 640 m². If the difference in their perimeters be 64 m, find the sides of the two squares. Let us assume that the side of one square is S 1 and that of another square is S 2
Then, S 1²+S 2² 640 nd, 4S 1-4S 2 64 Or, S 1-S 2 16 ----------------------------- (1) Or, S 1 16-S 2 Then, (16-S 2)²+( S 2)² 640 Or, 256+S 2²-32S 2+ S 2² 640 Or, S 2²+16S 2+128 320 Or, S 2²+16S 2-192 0 Or, S 2²+24 S 2-8 S 2-192 0 Or, S 2(S 2+24)-8(S 2+24) 0 Or, S 2-24 or +8 s side of a square can t be negative so lets take 8 as the side of one of the squares. From equation (1) it is clear that the side of another square is 24 m. Sum of reas 8²+24² 64+576 640 Difference in Perimeters 96-32 64 Note: lways cross check your answers by testing if they fulfill the conditions given in the question. 27. statue 1.46 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60 and from the same point, the angle of elevation of the top of the pedestal is 45. Find the height of the pedestal. (use 3 1.73) 14.6 m B 60 45 C D
BC In BCD tan 45 CD Or, BC 1 CD Or, BC CD C 1.46 + In CD tan 60 CD CDBC Or, 1.46 + BC 3 BC Or, 1.73 BC 1.46 + BC Or, 1.73 BC BC 1.46 Or, 0.73 BC1.46 Or, 1.46 BC 2 m 0.73 28. Prove that the ratio of areas of two triangles is equal to the ratio of the squares of their corresponding sides. Using the above results, prove the following:
In BC, XY is parallel to BC and it divides BC into two equal parts areawise. BX 2 1 Prove that D B 2 B G C E H F Given : BC DEF. rea BC BC² B² C² To Prove: rea DEF EF² DE² F² Construction: Draw G BC and DH EF Now, ar( BC) ar( DEF) 1 2 BC G 1 2 BC EF DH Or, ar( BC) G ar( DEF) EF DH Now, In BG & DEH B E GB DHE Hence, BG DEH So, B G DE DH But, B BC DE EF So, G BC DH EF So, ar( BC) BC BC BC² ar( DEF) EF EF EF² Similarly it can be proved that
ar( BC) B² C² ar( DEF) DE² DF² Second Part of the Question: In XY and BC, we have XY B (Corresponding ngles on Parallel Lines) So XY BC ar( XY ) X ² So, ar( BC) B² ar( XY ) X ² Or, 2ar( XY ) B² X ² 1 Or, B² 2 X 1 Or, B 2 B BX 1 Or, B 2 BX 1 Or, 1- B 2 BX 1 Or, 1 B 2 Or, BX B 2 1 Proved 2 29. gulab jamun, when ready for eating, contains sugar syrup of about 30% of its volume. Find approximately how much syrup would be found in 45 such gulab jamuns, each shaped like a cylinder with two hemispherical ends if the complete length of each of them is 5 cm and its diameter is 2.8 cms. The volume of gulab jamun can be calculated by adding volumes of two hemispheres and one cylinder. Here the radius of cylinder and that of hemisphere is 1.4 cms and height of cylinder is 5 cms. Volume of syrup in one gulab jamun will be 30% of its volume. 5 cms 2.8 cms
4 Volume of 2 hemispheres r 3 4 22 1.4 1.4 1. 4 3 7 Volume of cylinder r²h 2.2 22 5.6 So, volume of gulabjamun 1.4 1. 4 + 2. 2 7 3 12.2 22 0.2 1.4 3 Hence, volume of syrup in 45 gulabjamuns 12.2 22 0.2 1.4 30 % 45 338.184 cm 3 30. container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice-cream. This ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with icecream. Radius of Cylinder 6 cm Height of Cylinder 15 cm Volume of Cylinder r²h 6² 15 540 Volume of Cones 3 1 Π r²h 3 1 Π3² 12 36 Π Volume of Hemispherical top 3 2 Πr³ 3 2 Π3³ 18Π So, volume of ice-cream Π(36+18) 54Π
Number of cones required 540 54 10 cones 31. survey regarding the heights (in cms) of 50 girls of class X of a school was conducted and the following data was obtained. Height in cms 120-130 130-140 140-150 150-160 160-170 Total Number of Girls 2 8 12 20 8 50 Find the mean, median and mode of the above data The cumulative frequency distribution with the given frequency becomes: Height (In cms) Number Of Girls (f 1) Cumulative Frequency (cf) Class Mark (x 1) d 1x 1-145 x1-145 U1 10 120-130 2 2 125-20 -2-4 130-140 8 10 135-10 -1-8 140-150 12 f 0 22 cf 145a 0 0 0 150-160 20 f 1 42 155 10 1 20 160-170 8 f 2 50 165 20 2 16 Total N f i50 f iu i f 1u1 From the table, n f i 50 2 n 25 a 145 h 10 Using the formula for calculating the mean: Mean a+ fiui fi h 24 145 + 10 50 149.8 Now, 150-160 is the class whose frequency 42 is greater than 2 n 25 Therefore, 150-160 is the median class. Thus, the lower limit (l) of the median class is 150.
n cf Median l + 2 h f 25 22 150+ 10 20 151.5 Since the maximum number of girls is 20, therefore, the modal class is 150-160. Thus, the lower limit (l) of the modal class is 150. Using the formula for calculating the mode: fi fo Mode l + 2 fi fo f 2 20 12 150 + 10 2 20 12 8 8 150+ 10 20 154 h