10 th CBSE (SESSION : ) SUBJECT : MATHS SUMMATIVE ASSESSMENT-II SOLUTION _SET-1_CODE NO. 30/1

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2 Pre-foundation areer are Programmes (PP) Division 0 th BSE (SESSION : 05-6) SUBJET : MTHS SUMMTIVE SSESSMENT-II SOLUTION _SET-_ODE NO. 0/. Given : B is diameter B 0 To find P construction : Join O sol : In O as O O O O 0 OP 90 [Radius make an angle of 90 with tangent at point of contact] P + O 90 P P 60. k + 9, k and k + 7 are in.p. a a a - a [where a, a and a are the st, nd and rd term of the.p.] k k 9 k + 7 k + k 0 8 k 8. In B B cos m length of the ladder is 5 m B We have to draw a card from 5 playing cards so the total event of drawing a card is 5 and the event of getting red card and queen is cc to question The probability of getting neither red card nor a queen P( ) P() 5. Let 5, be the roots of + p 5 0 so sum of roots 5 + P and product of roots 5 P( ) If / then P 7 and P( + ) + k 0 have equal roots so D 0 P 4Pk 0 P(P 4k) 0 P 0 & P 4k 0 so 4k p k 4 P 4 7 SOLUTION_0 th BSE(BORD) _SUMMTIVE SSESSMENT-II_MTHS_PGE #

3 6. ( 7) ( ) 4 4 ( 4) ( ) 8 6 y (,y) ( 4, ) coordinates of Q. coordinates of point P( y ) mid of Q is P ( 4) So y 0, y 0 D R 7. S O Q P s we know that tangent from same eternal points are equal SD DR...() Q R...() QB BP...() S P...(4) dding equation (), (), () & (4) SD + Q + QB + S DR + R + BP + P D + B B + D Hence proved 8. To proove : B is a triangle isosceles triangle Proof : B ( 6 ) (0 4 ) (By using distance formula) B B 5 ( ) (0 ) B ( 6 ) (4 ) B 5 Now as B B is isosceles and (B) + () (B) By converse of pythagoras theorem B is a right angle isosceles triangle. 9. Let the first term and common difference of the.p. be a and d respectively. Then, a n a + (n )d a 4 a + (4 ) d 0 a 4 a + d 0 a + d 0 a d... () SOLUTION_0 th BSE(BORD) _SUMMTIVE SSESSMENT-II_MTHS_PGE #

4 a 5 a + (5 ) d a 5 a + 4 d By equation...() a 5 d + 4d a 5 d a a + ( ) d a a + 0d By equation...() a d + 0d a 7d multiply both sides by a d a a 5 Hence proved 0. In OTP OT r, OP r [Given] OTP 90 [radius is perpendicular to tangent at the pair of contact] Let TPO T OT sin OP r r 0 In TOP TOP 60 [By angle sum property] TOP SOP [s s are congruent] SOP is also 60 TOS 0 In OTS as OT OS [OST OTS] OTS + OST + SOT 80 OST OTS + OST 0 O Q S P. B B + 69 B B B 5 rea of shaded region rea of semicircle rea of B r B [.4 ] (5 ) ( ) 6.5 cm. Total S of tent rh + rl [(.) + (.4)] 7 cm O B 5 cm.8 m 0.5 m 7 Total S oftent m m cost Rs. 500 m cost Rs Rs So total cost of canvas needed to make the tet is Rs 6500 m. m SOLUTION_0 th BSE(BORD) _SUMMTIVE SSESSMENT-II_MTHS_PGE #

5 . Given : oordinates of P(,y) P(,y) (a + b, b a) B (a b, a + b) To prove b ay ccording to question P PB (P) (PB) (a + b, b - a) B(a - b, a + b) so accoding to distance formula [ (a + b)] + [y (b a)] [( (a b)] + [y (a + b)] (a + b) (a + b) + (b a) (b a)y (a b) (a b) + (a + b) (a + b)y [(a + b) + (b a)y] [(a b) + (a + b)y] (a + b) + (b a)y (a b) + (a + b)y (a + b) (a b) (a + b)y (b a)y (a + b a + b) (a + b b + a)y b ay b ay hence prove 4. Shaded area rea of larger major sector area of smaller major sector (4) (7) (4 7 ) 60 (47) 5. cm n [a n [a (n )d ] (n )d ] 7n 4n 7 a a (n )d (n )d 7n 4n 7 (n ) a d (n ) a d 7n 4n 7..() n Put m n m n m + m a a (m )d (m )d 7(m ) 4(m ) 7 4m 7 8m 4 7 4m 6 8m SOLUTION_0 th BSE(BORD) _SUMMTIVE SSESSMENT-II_MTHS_PGE # 4

6 6. Let t + t(t ) t(t ) t t t(t )(t ) t t (t + ) (t ) t t(t ) t t t t 4t 0 t (t 4) 0 t 0 t 4 0 t 4 t 0 & 0, 4 7. Volume of cone r h Volume of cone volume of cylinder h 7 7 h cm 8. The rise in the level of water will be due to the volume of sphere 4 (6) diameter 8 cm 9. Let be distance of cliff from man and h + 0 be height of hill which is required. In right triangle B, h B 60º 0º 0 E tan 60º h h h B In right triangle BD, tan 0º D...(i) D 0 0 B 0...(ii) SOLUTION_0 th BSE(BORD) _SUMMTIVE SSESSMENT-II_MTHS_PGE # 5

7 From (i) & (ii) h 0 h 0 m Height of cliff h m. Distance of ship from cliff 0 m 0 (.7) 7. m 0. Sample space while tossing coins S {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} (i) Favourable cases {HHT, HTH, THH} P(eactly heads) Number of favourable outcomes Number of total outcomes 8 (ii) Favourable cases {HHH, HHT, HTH, THH} 4 P(at least heads) 8 (iii) favourable cases {HTT, THT, TTH, TTT} P (at least tails) 8 4. Given r.8 ; h.5 m (ht. of cone) h. m l r (h ).5 m rea of convas required per tent [S of cone + S of cylinder] rl + rh r [.5 + 7] m cost of canvas per tent Rs. 0 Rs Total cost of 500 tents Rs mount shared by each schoo; Rs Rs Given : P and Q are two tangents drawn from a point to a circle (O, r). To prove : P Q. onstruction : Join OP, OQ and O. Proof : In OQ and PO OQ OP SOLUTION_0 th BSE(BORD) _SUMMTIVE SSESSMENT-II_MTHS_PGE # 6

8 [Tangent at any point of a circle is perp. to radius through the point of contact] O O [ommon] OQ OP [Radius] So, by R.H.S. criterion of congruency OQ OP Q P [By PT] Hence Proved.. O P B Steps of constructions are as follows () Draw a circle of radius 4 cm () Let O be its centre and P be any eternal point such that OP 8 cm () Join OP and then taking OP as diameter draw a circle intersecting the given circle at two points and B. Join P and BP. (4) Hence, P and BP are the required tangents D 4. O' X O Let O OX XO r Radius is always perpendicular to tangent, O 90 In DO and O DO O [ommon] DO O [each 90 ] By similarity criteria DO ~ O DO ' O 5. We have + O ' O r r 4 4 ( ) 4 4 ( + 4) ( + 4) 4( + + ) Using quadratic formula b b a 4ac or SOLUTION_0 th BSE(BORD) _SUMMTIVE SSESSMENT-II_MTHS_PGE # 7

9 Q h - 40 M 45 Y m P m X Let PQ be the tower Let PQ h learly XY PM 40 m QM (h 40) Let PX MY QM h In MQY, tan 45 MY 40 h 40...(i) QP h In QPX, tan60 PX h...(ii) From (i) and (ii) we get - 40 ( ) 40 PX ( 40 ) ( ( ) ) 0 ( ) PQ h 0 ( ) 0 ( ) 7.,,,...-,, +,..., 49 S S' S ( ) [ + ] () S ( + ) + ( + ) ( ) 49 S S ( + 50) 49 ( + 50) SOLUTION_0 th BSE(BORD) _SUMMTIVE SSESSMENT-II_MTHS_PGE # 8

10 8. oordinates of D (4) () 8 7 oordinates of D (, ) 9 (6) ()(5) oordinates of E (6) () (4) ()(7) area (DE) [4 + ( 6) + 5 (6 )] (4) [ ] area B [4(5 ) + ( 6) + 7 (6 5)] [ ] 5 [ 4 + 7] area B area DE area B 9 area (DE) 9. Total no. of events 6 {, 4, 4 9, 6, 4, 9, 6, 4, 9, 6 4, 4 4, 4 9, 4 6} Events when product is less than 6 8 {, 4, 9,, 4,, 4, 4 } Probability that sproduct of & y is less than 6 events when product is less than 6 Total no. of events 6 8 B º O r 0. SOLUTION_0 th BSE(BORD) _SUMMTIVE SSESSMENT-II_MTHS_PGE # 9

11 (i) length of sector r 80 In OB B tan O B r tan BO Now, sec so, BO r sec r length of O r So length of B OB O r sec r So perimeter r 80 + B + B + r tan +r sec r r [tan + sec + ] 80. Speed of boat in still water 4 km/hr Let the speed of stream be Upstream Speed of boat 4 Tupstream Distance speed Downstream Speed of boat 4 + T downstream dis tance speed TP T upstream T downstream (4 ) (4 )(4 ) [ ] (4 ) (4 + ) 64 (4) ( + 7) 8( + 7) 0 ( 8) ( + 7) 0 8, 7 speed of stream 8 km/hr SOLUTION_0 th BSE(BORD) _SUMMTIVE SSESSMENT-II_MTHS_PGE # 0

General Instructions:

CBSE Board Class X Summative Assessment II Mathematics Board Question Paper 016 Time: 3 hrs General Instructions: Max. Marks:90 (i) All questions are compulsory. (ii) The question paper consists of 31