(1/2) a a (1/2) 6. Area of ABC will be 127cm because two congruent (1/2) 8. Given, the an gles of a tri an gle are 5( y 1) 180 x x (1/2) (1/2)

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Download "(1/2) a a (1/2) 6. Area of ABC will be 127cm because two congruent (1/2) 8. Given, the an gles of a tri an gle are 5( y 1) 180 x x (1/2) (1/2)"

Erin Conley
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1 Sample Question Paper (etailed Solutions) Mathematics lass th. Given, a and b b a ( a b ) ( ) (/) ( 8 ) ( ). In the given figure, AB E EBA EBA 0 a a (/) [alternate interior angles] In ABE, EBA EAB AEB 80 (/) [by angle sum property of a triangle] 0 0 AEB 80 0 AEB 80 AEB 80 0 AEB 0 (/) /. We have, ( ) n m m/ n ( ) [ a ( a) ] ( ) ( ) a m. Let the angles of triangle be, and. (/) (/) m a Then, 80 (/) 80 0 [by angle sum property of a triangle] Hence, the three angles of a triangle are 0, 0 and 0. (/). Let the angle Then, its supplement angle 80 According to the question, 80 (/) ( 80 ) (/). Area of AB will be cm because two congruent triangles always have equal areas. (). Let f( ) p Then, the given equation can be written as f( ) ( ) q( ) As, ( ) is a factor of f( ). f( ) 0 () ( ) ( ) p( ) 0 80 p 0 p 8 p () 8. Given, the an gles of a tri an gle are ( y ), ( y ) and y. We know that in a triangle, sum of all the angles is equal to 80. ( y ) ( y ) y 80 y y y 80 0 y 80 0 y () So, angles are ( 0 ), ( 0 ) and 0, i.e., and 0. So, the given triangle is a right angled triangle. (). (i) Here, num ber of ob ser va tions, n 0 which is an even num ber. n + n Median (/)

2 0 + 0 [ median, given] () (ii) The value of will be remain same as we add on minimum and maimum value. (/) 0. Given observations are,,,,,,,, and mean of these observations Sum of observations Mean Number of observations () On arranging observation in ascending order, we get,,,,,,,, Here, occurs most frequently, i.e. times. So, the required mode is. (). Given dimensions for herbarium l 0 cm, b cm and h 0 cm (i) Area of the glass panes Total surface area of glass panes which is cuboidal in shape. ( lb bh hl) ( ) ( ) ( 0) 00cm () (ii) Length of the tape ( l b h ) ( 0 0) 0 cm (). In the given figure, AB EF and E is a transversal. A 0 Therefore, sum of two interior angles is 80. i.e. FE E 80 0 E 80 [ FE 0, given] E 0 () Since, AB and B is a transversal. AB B [alternate angles] B E 0 F BE E. Equation of the sides are 0 0 [ BE 0, given] 0 () AB : y 0 (/) B : (/) : y (/) A : (/) Required area sq units (). (i) A terminated line can be produced indefinitely. () (ii) A circle can be drawn with any centre and any radius. () (iii) All right angles are equal to one another. (). Let radius of the outer surface of the roller be R cm. Given, width of roller, h cm Area of base cm R R R R cm [taking positive square root] () Given, thickness of roller cm Then, inner radius of roller, r ( ) cm (/) Volume of the roller Volume of the hollow cylinder. Let the lin ear equa tion be ( R r ) h [( ) ( ) ] (/) [ ] 0 00 cm () a by c 0 (i) Since, (, ), ( 0, 0 ) and (, ) are the solutions of linear equation, therefore it satisfies the Eq. (i). At point (, ), a b c 0 (ii) At point (0, 0), 0 0 c 0 c 0 (iii) At point (, ), a b c 0 (iv) () From Eqs. (ii) and (iii), we get a b 0 0 a b b a a b On putting a b and c 0 in Eq. (i), we get b by 0 0 b by 0 ()

3 b( y) 0 y 0, b 0 Hence, y 0 is required form of the linear equation. (). We have, We have, a( b c ) b( a c ) b a abac abbc c bc m n mn [ ( a ) a ] ac () ( ) a m mn a n a (/) () ab ac ab bc bc ac ab ac ab bc bc ac ab ac bc bc ac ab (/) / / / ( ) / ( ) / / / / ( ) ( ) ( ) / / ( ) ( ) ( ) ( a ) a and a m n mn m / / () m a () [ ( a m ) n a mn ] 0 0 () 8. onstruction raw RP, QT and RT. P Q 8 Given PQ QR RS and PQR () Now, PTQ QPR O PTS 8 () and ROS RTS () Given, A, B, and are four points on a circle and A, B intersect at a point E, such that E 0 and BE 0 T R S B A From the figure, BE E 80 [linear pair aiom] 0 E 80 [ BE 0, given] E 80 0 E 0 (i) () In E E E E 80 [ sum of all angles of a triangle is 80 ] 0 0 E 80 [from Eq. (i) and E 0, given] E 80 0 E 0 (ii) () Now, BA B [ angles in the same segment are equal] BA E [ B E] BA 0 [from Eq. (ii)] (). We know that sum of the two sides of a triangle is greater than the third side. () In PQS, PQ QS PS (i) (/) and in PRS, PR SR PS (ii) (/) On adding Eqs. (i) and (ii), we get ( PQ QS ) ( PR SR) PS (/) PQ ( QS SR) PR PS PQ QR RP PS In AB and B, E 0 AB Hence proved. (/) AB B (Given) B B (ommon) () AB B (SAS congruence rule) A B. (By c.p.c.t.) () 0. (i) From the given data, minimum class size is. The modified lengths of the rectangles are proportionate to the class size. Adjusted frequency Minimum class size lass size of given data Now, we get the following table: 0 Frequency (/)

4 Number of children 8 0 Age (in years) Number of children (frequency) Width of the class Adjusted frequency 8 8 () We take age of the children along the -ais with suitable scale i.e. cm unit and number of children along the -ais with suitable scale cm unit. So, the correct histogram with varying width is given below: (ii) With the help of histogram, we cannot determine the mean. (/) 8 0 Age (in years) () Given data is 0, 8,, 0,,, 0, 8,,. Sum of observation (i) Mean, Number of observation () 0 (ii) Write the data in ascending order,, 8, 8, 0, 0, 0,,, Here, n 0 (even) n n + Median () (iii) Write the data in ascending order,, 8, 8, 0, 0, 0,,, Since, the maimum frequency of 0 is. So, mode 0 (). (i) Since, AB and AF are on the same base A and between the same parallel lines A and BF. ar ( AB) ar ( AF ) (ii) We have, ar ( AB) ar ( AF) ar ( AB) ar (quadrilateral AE) ar ( AEF) ar ( ABE ) (½) ar ( AF) ar (quadrilateral AE). Let A be the event of getting atleast one tail. (½) It is given that three coins are tossed simultaneously 00 times. Total numbers of trials 00 () (i) Number of trials in which event A happened Number of trials in which A happened P( A) Total number of trials () (ii) Let B be the event of getting two heads and one tail. Number of trials in which B happened 0 Number of trials in which B happened P( B) Total number of trials (). (i) Given, AB is a right angled isosceles triangle in which B 0 and AB B. A () [ sides opposite to equal angles are equal] Now, in AB, we have A B 80 [by angle sum property of a triangle]

5 A 0 A 80 [ B 0 and A ] () A 0 A Hence, A () (ii) are and concern towards animals is depicted in the paragraph. (). By factor theorem, ( ) will be a factor of p( ), if p( ) 0. Here, p( ) 0 On putting in p( ), we get p( ) ( ) 0( ) ( ) 0 0 ( ) is a factor of p( ). () By long division method, we have p( ) ( )( ) Hence, ( ) is a factor of P( ) and p( ) ( )( ). Let sides of a triangle be a, b and c. Then, a b c s () Hence proved. () (i) (/) Area of triangle, s ( s a)( s b)( s c) (/) Let the sides of new triangle be a, b and c. ( a b c) Then, s a b c s [from Eq. (i)] (/) Area of new triangle, s( sa)( sb)( s c) (/) s ( s a)( s b)( s c ) s ( s a) ( s b) ( s c) (/) Now, increase in area (/) Percentage increase 00% (/) 00% (/). Let h be the height and r be the radius of given cylinder. Then, h m and radius r m. iameter of the base m Radius ( r ) m (/) learly, curved surface area of the cylindrical portion rh m () and the area of the top of a cylinder r Area of the canvas required to make a tent ( ) m (/) urved surface area of cylindrical portion + Area of the top of a cylinder ( ) m () learly, maimum length of the rod put in the cylinder, B ( AB) ( A) ( ) ( ). 0m () Given, outer diameter of cylindrical tube, d cm r Outer radius of cylindrical tube 8 cm r Inner radius of cylindrical tube (r thickness of cylindrical tube) ( 8 ) cm [d cm] and h 00 cm () Volume of metal used in cylindrical tube Outer volume of cylindrical tube A Inner volume of cylindrical tube m m r h r h h( r r ) () 00 ( 8 ) 00( 8 )( 8 ) [ a b ( a b)( a b)] cm Hence, 8800 cm of iron has been used in making the cylindrical tube. () m B

6 . Given, work done by a body on ap pli ca tion of a con stant force is di rectly pro por tional to the dis tance travelled by the body, i.e. Work done ( W ) istance ( s ) W F s ' where, F is an arbitrary constant. Now, let be the distance and y be the work done. Then, the equation will be y F. Which is the required linear equation in two variables and y. If the constant force is units, i.e. F, then equation is y (i) When 0, then y 0 When, then y () Thus, we have the following table y Points (, y) 0 0 O( 0, 0) A (, ) () Now, plot the points O ( 0, 0) and A (, ) on graph paper and join them to get a line OA which represents the required graph. (0, 0) 0 B (,0) 8 y= A (, ) (, 0) (0, 0) O 8 ' istance (s) () Work done (w) (i) From point (, 0 ), draw a line parallel to O to cut y at B and then from B, draw a line parallel to -ais which intersect the -ais at ( 0, 0 ). (/) learly, B represents (,0). So, work done, when the distance travelled by the body is units, is 0 units. (ii) learly, y 0, when 0, so the work done when the distance travelled by the body is 0 unit, is zero unit. (/) Let, age of Amit yr and age of Akhil y yr (i) According to the question the linear equation for the above situation is y y () 0 0 y 0 () (ii) From the graph, when Amit s age yrs, then Akhil s age yr. () 8. Steps of con struc tion (i) First, draw a horizontal ray OA. (ii) Taking O as centre and some radius, draw an arc of a circle which intersects OA, say at a point B. (iii) Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point. () (iv) Now, draw the ray OE passing through. Then, EOA 0. 0 O B A (½) (v) Taking B and as centres and with the radius more than B, draw arcs to intersect each other, say at. 0 0 (0, ) (0, ) (, 0) E 0 (vi) raw the ray O then O is the bisector of the EOA, i.e. EO AO EOA ( 0 ) 0 (½). The graph obtained by plotting the given points P, Q and S is given below. To find R, draw a line parallel to PS and passing through Q. Also, draw a another line parallel to PQ and passing through S. learly, the point of intersection of above two lines will be R. ()

7 () For the coordinate of point R, Abscissa perpendicular distance from -ais dinate perpendicular distance from -ais Hence, the coordinates of point R are (, ). () 0. We know, (/) ' ( ) S (,) R (,) P (,0) Q (,0) O ' [ ( a b) a b ab], given...(i) (/) [taking square root on both sides] Now, cubing both sides of Eq. (i), we get ( ) (/) (/) [ ( a b) a b ab( a b)] [using Eq. (i)] (/) (/) Let ( ) be a rational number equal to. Then, ( ) ( ) ( ) [ ( a b) a b ab] () () Now, is rational is rational is rational. () But, is irrational. So, it is a contradiction that is rational. Hence, ( ) is irrational. ()

Mathematics. Sample Question Paper. Class 9th. (Detailed Solutions) 2. From the figure, ADB ACB We have [( 16) ] [( 2 ) ] 3.

Mathematics. Sample Question Paper. Class 9th. (Detailed Solutions) 2. From the figure, ADB ACB We have [( 16) ] [( 2 ) ] 3. 6 Sample Question Paper (etailed Solutions) Mathematics lass 9th. Given equation is ( k ) ( k ) y 0. t and y, ( k ) ( k ) 0 k 6k 9 0 4k 8 0 4k 8 k. From the figure, 40 [ angles in the same segment are