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 equal] Now, in P, P P P 80 0 P 40 80 P 80 0 40 0 [angle sum property] (/) P 0 (/). We know that, rea of parallelogram ase ltitude Given, E 9. m, F. m and m ar (parallelogram ) ase ltitude E 9. [ m and E 9. m] (i) and ar (parallelogram ) F. [ F. m] (ii) (/) 9.. 9..4 (/). 4. Since, the an gle sub tended by the centre is dou ble the an gle sub tended by cir cum fer ence. Refle PR PQR 0 40 P 0 Q Q R In PR, PR 60 40 0 gain, in PR, P R [radii of the circle] RP PR [by property of an isosceles triangle] In PR, PR RP PR 80 [ sum of all angles of a triangle is 80 ] (/) PR PR 0 80 [ RP PR] PR 80 0 60 60 PR 0 (/). We have [( 6) ] [( ) ] / / 4 / / 4 [( ) ] / m n mn [ ( a ) a ] (/) / [( ) ] ( ) (/) 6. Pos tu lates are the as sump tions used spe cially for ge om e try and a i oms are the as sump tions used through out Mathematics and not specifically linked to geometry. 7. Given lin ear equa tion can be writ ten as y 7 7 y [y in terms of ] (i) (/) n putting in Eq.(i), we get ( ) 7 y 6 7 So, the point (, ) is not on the given line. (/) n putting 0 in Eq. (i), we get y 0 7 0 7 7 (/) n putting in Eq. (i), we get y 7 7 Hence, the required points are 7 0, and (, ). (/)
7 8. To tal num ber of times a coin tosses 0 Number of times tail appears 0 Then, number of times head appears 0 0 0 (/) Required probability of getting a head Number of times head appears Number of times a coin tosses 0 0 No, because the coin is biased, then the above probability will be different. (/) 9. Given, total number of students 7 6 9 0 Total number of trials Let E be the event of getting the number of students whose weights are more than 46. Number of trials in which E happened 9 0 4 Number of trials in which E happened Now, P( E) Total number of trials 4 0.76 Hence, the required probability is 0.76. 0. Let be a rhombus whose diagonals and intersect at E. Let be centre of the circle with diameter. We know that the diagonals of a rhombus intersect each other at a right angle. E 90 Thus, circle with as diameter passes through E.. Given, E is a regular pentagon. 08 (/) [ in a pentagon, each interior angle is 08 ] In, we have 80 [by angle sum property of a triangle] 08 80 80 08 E 7 [ ] 6 (/) Now, E 08 [ each angle of a pentagon is 08] c 08 c 08 6 7...(i) gain, E 08 In E, E a 80 (/) [by angle sum property of a triangle] 08 a 80 a 8008 a 7 [ E a ] a 6 Hence, a 6 and c 7 (/). Given, perimeter of an equilateral triangle 7 m Let side of a triangle be a. Perimeter of equilateral triangle a 7 a 7 a 4 m (/) Now, 7 s 6 m and let side of triangle be a 4 m, b 4 m, c 4 m rea of triangle s( s a)( s b)( s c ) (/) 6 ( 6 4)( 6 4)( 6 4) [by Heron s formula] 6 44 m. Given n equilateral triangle in which, E and F are the mid-points of sides, and respectively. To prove The centroid and circumcentre are coincident. onstruction raw medians, E and F. (/) Proof Let G be the centroid of i.e. point of intersection of, E and F. In E and F, we have 60, and F E E ~ F (/) E F F G E (i)
8 Similarly, F ~ F (ii) E F (/) E F [multiply by ] [ centroid of a triangle divide of the median in : ] G G G G is equidistant from the vertices G is the circumcentre of Hence, the centroid and circumcentre of equilateral triangle are coincident. Hence proved. (/) Given right angled in which is right angle and is double of. To prove onstruction Produce upto such that and join. Proof In and, we have [by construction] (i) [each 90 ] and [common side] [by SS congruence rule] [by PT] (i) and (say) [by PT] Now, [ ] [ sides opposite to equal angles are equal] [ and from Eq. (i)] Hence proved. 4. Given, and. To prove Proof Given, n subtracting from both sides, we get (i) In and, [given] [given] and [from Eq. (i)] [by SS congruence rule] [by PT] Hence proved. 9 9. We have, y ( ) ( y ) (/) ( y )[( ) y ( y ) ] 6 6 ( y )( y y ) [ a b ( a b)( a ab b )] 6 6 ( y)( y y )( y y ) (/) [ a b ( a b)( a ab b )] We have, ( y z)( 4 y 9z y yz 6z) ( y z)[( ) ( y) ( z) ( y) ( y) z z ] ( ) ( y) ( z) ( y) z [ a b c abc ( a b c ) ( a b c ab bc ca)] 8 y 7 z 8yz 6. Rationalising the denominator of each term of LHS, we have LHS 4 6 7 8 6 6 4 4 6 6 7 8 7 8 4 4 6 7 6 7 4 6 7 8 9 4 4 7 8 7 8 4 4 6 7 6 7 8 9 8 9 8 9 8 9 4 4 6 6 7 7 8 8 9 9 RHS 7. Let number of boys be and number of girls be y. ccording to the question, y 400... (i) and y 7... (ii) lso, y 0...(iii)(/) From Eq. (i) we have y 400
9 ( y)( y) 400 [ a b ( a b)( a b)] 0( y) 400 [ from Eq.(iii)] y 40 (iv) (/) n adding Eqs. (iii) and (iv), n putting in Eq. (iii), we get y 0 y Hence, the total number of students 40 Required ratio : y : : Total cost Value of pen Number of pens y 6 Table for Eq. (i) is 0 4 6 y 0 6 48 64 80 96 Now, the graph for the above values is 8 96 80 64 48 6 Y 0 4 6 7 8 From the graph, cost of 6 pens ` 96. 8. Given, and GE is a transversal line, then (i) GE GE [alternate interior angles] GE 6 (/) Now, GEF GE FE GEF 6 90 [ FE 90 ] (/) GEF 6 lso, FGE GE 80 [linear pair aiom](/) FGE 6 80 FGE 806 4 (/) Hence, GE 6, GEF 6 y=6 and FGE 4. 9. In the given figure, two same triangle in the blades of magnetic compass. lso, the sides of triangle cm, cm and cm. X Let a cm, b cm, c cm a b c s rea of triangle s( s a)( s b)( s c ) 9 (/) [Heron s formula] 4 9. 96.. 49 cm () 4 rea of blades of magnetic compass rea of triangle. 49 4. 98 cm (/) Let the three sides of a triangular plot be m, m and 7 m. (/) Given, perimeter 00 m Sum of all sides of a triangle Perimeter 7 00 00 0 (/) The sides of triangle are 0, 0, 7 0, i.e. 60 m, 00 m, 40 m. Let a 60 m, b 00 m, c 40 m a b c 60 00 40 Then, s 0 m rea of triangle 00 (/) s( s a)( s b)( s c ) [by Heron s formula] 0( 0 60)( 0 00)( 0 40) (/) 0 90 0 0 0 0 0 0 0 0 00 cm 0. Write in ascending order of given scores,,,,, (/) 6 4 Here, n (odd) n Median Value of th scores (/) 6 Value of th scores 6 Value of rd scores
0 6 4 4 Now, value of 4 4 6 6. Total number of workers 8 7 86 46 00 Total number of trials 00 (i) P (person is 40 yr or more) P (person having age 40-49 yr) P (person having age 0-9 yr) P (person having age 60 yr and above) 86 46 0. 67 0. 68 00 00 00 00 (ii) P (person is under 40 yr) P (person having age 0-9 yr ) P (person having age 0-9 yr) 8 7 6 0. 0. 00 00 00 (iii) P (person having age under 60 but over 9 yr) P (person having age 40-49 yr) P (person having age 0-9 yr) 86 46 00 00 0 00. 66. Given, To prove and Proof In and, we have [ is the mid-point of ] [vertical opposite angles] and [ is the mid-point of ] ~ [by SS congruence rule](/) Then, [by PT] and [by PT] (/) (i) ut and are alternate interior angles formed when transversal intersects at and at. (/) Then, and. Hence proved. (/). Let the sides of triangular park a 0 m and b 80 m, c 0 m. (let) 0 m 80 m a b c s 0 80 0 0 rea of park s( s a)( s b)( s c ) m [by Heron s formula] ( 0)( 80)( 0) 4 7 7 m () Length of the wire needed for fencing Perimeter of the park Width of the gate 0 47 m ost of fencing ` ( 0 47) ` 4940 Given, radius of cylindrical flower base, r 4. cm and height, h. cm Surface area of one cylindrical flower base urved surface area rea of a base rh r 4.. ( 4. ) 4. (. 4 4. ) 4. 6. 6 7. cm () Surface area of 0 cylindrical flower base 0. 76 cm Now, amount spent for purchasing 00 cm mount spent for purchasing 76 cm ` 0 0 76 `.0 00 4. Given, chords 0 cm and 4 cm Let Q and P. We know that, the perpendicular drawn from the centre of a circle to a chord bisects the chord. 0 Therefore, P P cm [ 0 cm] and Q Q 4 cm [ 4 cm] gain, let P be and radius of the circle be r, then Q 7. 4 cm cm Q r cm (7 ) cm m 0 m r cm cm P 0 cm cm
In right angled P and Q, we have P P [by Pythagoras theorem] r (i) and Q Q r ( 7 ) ( ) (ii) (½) ( 7) 7 ( ) [ ( a b) a b ab] 89 4 44 4 408 408 4 cm n putting cm in Eq. (i), we get r ( ) ( ) 44 69 r cm [taking positive square root] (½). (i) Let the to tal num ber of stu dents be. Then, number of students planned to visit historical monuments (/) and number of students planned to visit old age homes 7 The number of students decided to teach poor children 0 y using all the above informations, we get the 7 polynomial, p( ) 0 (/) (ii) The degree of the polynomial is. (/) (iii) Given, total number of students are 96. i.e. 96 The number of students planned to visit the historical monuments ( 96 ) 96 768 (iv) (a) First group of students shows deep interest in pre history and historical background of country. (b) Second group of students shows high standard of morality. (c) Third group of students shows their duties for the society or deprived society. (½) 6. Given, / / / / / / ( ) n cubing both sides, we get / / ( ) { ( )} (/) / ( ) ( ) ( ) ( ) / / / {( ) ( ) ( ) } / / ( 8) ( 6 ) { } / / 8 6 { } / / 6 8 6{ } 6 8 6( ) 6 8 6 6 Therefore, 6 6 0 (½) 7. Steps of con struc tion (i) First, draw a line segment 6. cm. (ii) raw an arc with as centre and 7. cm as radius and draw another arc with as centre and 6 cm as radius to intersect each other at. (iii) Join and. Thus, we get the required, on measuring all the three angles, we get 0, 7 and (iv) Now, draw the angle bisector of and, which intersect each other at, since is the smallest angle and is the largest angle in. (v) n measuring the acute Y formed by the bisecting rays X and Y at the point of intersection, we get Y 6. a( bc ) 8. LHS b b( ac ) a 7. cm y 6 cm X 6. 0 7 6. cm abac babc c ba c ( ) abac ba bc ac bc m n mn a ( a ) a and a m n a mn ( ba) c bc ac ac bc bc ac ac bc bc ac 0 a m a m RHS Firstly, we draw a line 4.7 units. Now, from point, mark a distance of unit. Let this point be. Let be the mid-point of. (/)
Now, draw a semi-circle with centre and radius. Let us draw a line perpendicular to passing through point and intersecting the semi-circle at point. The distance 4.7 (/) E 4.7 () raw an arc with centre and radius, which intersects the number line at point E, then the point E represents 4.7. 9. First, plot the points ( 4, 4), ( 6, 0 ), ( 4, 4 ) and (, 0 ) on a graph paper and join all these points. ( 4,4) ( 6,0) (,0) X X 6 4 0 4 6 ( 4, 4) 6 4 4 4.7 6 Y () We obtained a quadrilateral which is a rhombus because its all sides are equal. i.e. and diagonals are not equal. Now, area of a rhombus d d 4 8 6 sq units Y [ d 4 and d 8] (/) Hence, area of the quadrilateral formed by joining is 6 sq units. (/) 0. Let be the field and be the part of the field where a pit is to be dug out. Volume of the earth dug out ( 6. ) m 4 m (i) [ volume of cuboid lbh] rea of remaining part of the field rea of the field rea of pit ( 0 4) ( 6 ) 80 8 6 m Let h m be the raised over the field uniformly. 6 m Volume of earth dugout 6 h 4 6 h [from Eq. (i)] 4 h 0.78 m 6 7.8 cm learly, the roller is a right circular cylinder of height, h 40 cm and radius of its base, r 84 cm rea of covered by the roller in one revolutions urved surface area of the roller (/) 84 40 7 670 cm So, area covered by the roller is 000 revolutions 670 000 670 00 000 00 67 m Hence, cost of levelling the playground 0 ` 67 00 ` 80.6 m 0 m 4 m (½)