L.K. Gupta (Mathematic Classes) www.poineermathematics.com. MOBILE: 98155771, 461771 Solution Of Class 10 th CBSE SA-II Board (Set-1)Mathematics 1. (k 1) k = (k + 1) (k 1) k 1 k = k + 1 k + 1 k 1 = k. Ans. (B) 0 APB 90. Ans. (D) AB = 5 cm PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, 1
L.K. Gupta (Mathematic Classes) www.poineermathematics.com. MOBILE: 98155771, 461771 BC = 1 cm AC =? Pythagoras theorem = AC = AB + BC = 5 + 1 = 5 + 144 AC = 169 AC = 1 cm BC = x + 1 x AB = x + 5 x BP = BQ..(1) (Tangents drawn from an external point to a circle are equal) AP = AR..() CO = CR..() AC = 1 5 x + (1 x) = 1 5 x + (1 x) = 1 x + 17 = 1 x = 4 x = 14 = x = Radius = cm Ans. (C) 4. No of children = Outcomes = BBB GGG BBG BGB GBB BGG PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0,
L.K. Gupta (Mathematic Classes) www.poineermathematics.com. MOBILE: 98155771, 461771 Probability = GBG GGB Total cases = 8 No. of favourablecases Total no. of cases = 7 8 5. In ABC Ans. (A) AB tan0 o BC 150 1 BC BC 150 m Ans.(B) 6. Total numbers= 15 No. divisible by 4 are 4, 8, 1 Outcomes = 1 15 5 Ans. (C) 7. Distance of BD PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0,
L.K. Gupta (Mathematic Classes) www.poineermathematics.com. MOBILE: 98155771, 461771 BD= (0 4) (0 ) = 16 9 5 5cm 8. Ans. (A) AB =? Pythagoras theorem AB = AO + BO = 10 + 10 = 100 + 100 AB = 00 AB = 10 cm Ans.(b) 9. As given equal roots D=0 b 4ac =0 a = 4, b = p, c = p 48 =0 p = 48 p 48 p= 4 10. The numbers divisible by both and 5 are PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, 4
L.K. Gupta (Mathematic Classes) www.poineermathematics.com. MOBILE: 98155771, 461771 110, 10 ------------------------990 d = 10 110 =10 a 110 d 10 an=110 an = a + (n 1)d. 990 = 110 + (n 1)10 990 110 = (n 1)10 880 + 10n 10. 890 n 10 n 89 11. As tangents drawn from an external point to the circle are equal in length 1. So EA = EC..(1) EB = ED.() Adding (1) and () EA + EB =EC + ED AB = CD. Hence. Proved. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, 5
L.K. Gupta (Mathematic Classes) www.poineermathematics.com. MOBILE: 98155771, 461771 AB = AC (Given) BD = BF (length of the tangent drawn from an external point to a circle are equal)..(1) AF = AE.() DC = CE.() Now, AB = AC AF + BF = AE + CE AE + BD = AE + DC (From equation (1), () & ()) BD DC 1. (1, 1) (, 1) (, 1) (4, 1) (5, 1) (6, 1) (1, ) (, ) (, ) (4, ) (5, ) (6, ) (1, ) (, ) (, ) (4, ) (5, ) (6, ) (1, 4) (, 4) (, 4) (4, 4) (5, 4) (6, 4) (1, 5) (, 5) (, 5) (4, 5) (5, 5) (6, 5) (1, 6) (, 6) (, 6) (4, 6) (5, 6) (6, 6) (i) p( No. of each dice is even)= 9 1 6 4 favourable cases (,),(,4),(,6),(4,),(4,4)(4,6),(6,),(6,4),(6,6) (ii) P( sum appearing on both dice is 5)= 4 1 6 9 Favourable (1,4),(,),(,),(4,1) 14. r 46 7 r 46 r 46 7 1 7 r r 49 r 7 Vol. of hemisphere= r PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, 6
L.K. Gupta (Mathematic Classes) www.poineermathematics.com. MOBILE: 98155771, 461771 15. = 7 7 7 7 44 156 = 49 718.6 cm. 16 15 1 x x 1 16 x 15 x x 1 (16 x) (x +1) = 15x 16x + 16 x x = 15x x + 15x + 16-15x=0 x = 16 x +15x + 16 = 15x x 4 16. a5 + a9 = 0 (a + 4d) + (a + 8d) = 0 an a (n 1)d a + 1d = 0 a + 6d = 15 (1) a5 = a8. a + 4d = (a + 7d) a + 4d = a + 1d 4d 1d = a a. d = a. d = a.() Put d in the equation (1) a + a 6 15 a + 4a = 15 5a =15 a PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, 7
L.K. Gupta (Mathematic Classes) www.poineermathematics.com. MOBILE: 98155771, 461771 d = () 6 d = A.P. =, 5, 7,------------ 17. Step of construction (i) Draw a (ii) Then draw acute ABCwith the sides given such that AC = 5.5 cm, AB = 5 cm, BC=6.5cm BAX (iii) Make 5 arcs on line AX such that AA1=A1A=.=A4A5 (iv) Join BA5 (v) Draw B A parallel to BA5 (vi) Similarly drawn C B parallel to CB. Hence C'AB' is the required triangle PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, 8
L.K. Gupta (Mathematic Classes) www.poineermathematics.com. MOBILE: 98155771, 461771 18. In PAB o 000 tan60 y y 000 y 000m In A' B'P o 000 tan0 x y 1 000 x y x y 000 x+000=9000 x 6000m From B to B time taken by aeroplane is 0 second S dis tance 6000 time 0 19. As given PA = PB 00 m / sec (K 1 ) ( K) (K 1 K) ( 5) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, 9
L.K. Gupta (Mathematic Classes) www.poineermathematics.com. MOBILE: 98155771, 461771 Squaring both sides (K 4) + ( K) = 1 + 9 K + 16 8K + 4 + K 4K =10 K 1K + 10=0 K 6K + 5 =0 K 5K K+ 5=0 K(K 5) 1(K 5)=0 K= 1, 5. 0. Let the ratio be K:1 Let P divides AB in the ratio K : 1 K 7K P, K 1 K 1 But 7K 0 K 1 7K =. K = 7 Ratio : 7. Co-ordinates of pt. P 7 7 7,, 0 1 1 7 7 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, 10
L.K. Gupta (Mathematic Classes) www.poineermathematics.com. MOBILE: 98155771, 461771 1. Area of major sector OAYB = = 5 π (4) 6 = 460 cm Area of major sector OCZD = = 5 π 441 6 = 1155 cm Area of shaded region 00 π (4) 60 00 π (1) 60 = Area of major section OAYB -Area of major sector OCZD = 460 1155. = 465 cm PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, 11
L.K. Gupta (Mathematic Classes) www.poineermathematics.com. MOBILE: 98155771, 461771 a = 7 cm Let R = radius of the sphere R = a R = 7 R = 7 cm Volume of the wood left = Volume of cube Volume of sphere = a 4 πr = 4 7 (7) π 4 7 π 7 π = 7 π 7 7 1 8 6 6 =16.cm. Speed = 4 km/hr = 4 1000 m 60 min = 100 = 00 m/min S D T 00 D 10 00 D 10 D m/min 000 m Volume of the water flowing in a canal = Volume of water used for irrigation PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, 1
L.K. Gupta (Mathematic Classes) www.poineermathematics.com. MOBILE: 98155771, 461771 000 8 6 1.5 A 100 6 15 00 100 8 A A 0 00 100 8 15 00 100 4 A 15 50 100 A A 75000 m 4. Area of trapezium = 1 (sum of parallel sides) Height 1 4.5 (AD BC) AB 1 4.5 (10 4) AB 49 = 14 AB 49 AB 14 AB 7 cm PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, 1
L.K. Gupta (Mathematic Classes) www.poineermathematics.com. MOBILE: 98155771, 461771 Area of quadrant = = = = 1 49 4 7 4 4 4 7 11 77 7 cm 8 8 Area of shaded region 1 π (AB) 4 = Area of trapezium Area of quadrant = 4.5 77 8 = 4.5 9.65 = 14.875 cm 5. x x 4 10 x x 5 x,5 (x )(x 5) (x 4)(x ) 10 (x )(x 5) x 5x x 10 x x 4x 1 10 (x )(x 5) x 5x x x 4x 10 1 10 x 5x x 15 x 14x 10 x 8x 15 (x 14x + ) = 10(x 8x + 15) 6x 4x + 66 = 10x 80x + 150 10x 6x 80x + 4x + 150 66 = 0 4x 8x + 84 = 0 x 19x + 4 = 0 Quadratic formula x= b b 4ac a PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, 14
L.K. Gupta (Mathematic Classes) www.poineermathematics.com. MOBILE: 98155771, 461771 x x x 19 (19) 4 4 4 19 61 6 4 19 5 4 4 14 x, 4 4 7 x 6, 6. Total no. of trees planted = [ + 4 + 6 +.4] = 1 [ 1 1 ] = 1[4 + ] [Sum of n terms of on A.P. = n [A + (n 1) D] = 1 6 = 1 7. Let BD be the flagstaff BC be the tower In ABC BC AC BC 10 tan 45 1 0 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, 15
L.K. Gupta (Mathematic Classes) www.poineermathematics.com. MOBILE: 98155771, 461771 BC = 10 m In ADC DC tan 60 0 AC DB = DC BC = 10 10 = 10( 1) = 10(1.7 1) = 10(0.7) DC 10 DC 10 m = 87.60 m 8. Total cards = 5 Cards removed = 4 5 4 = 48 Cards removed red queens blocks Jacks (1) Probability of a king = No. of favourable cases Total no.of cases Outcomes = 4 = 4 1 48 1 () Probability of a red colour card Outcomes = 4 = 4 1 48 () Probability of a face card Outcomes = 8 = 8 1 48 6 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, 16
L.K. Gupta (Mathematic Classes) www.poineermathematics.com. MOBILE: 98155771, 461771 (4) Probability of a queen Outcomes = = 1 48 4 9. A(, 5), B(, 7), C(1, 8) and D(6, ) Area = ACD 1 x 1 (y y ) x (y y 1 ) x (y 1 y ) 1 ( )(( 8 )) 1( 5) 6( 5( 8)) 1 ( )( 11) 1( ) 6(1) = 1 109 Area of ABC 109 cm 1 ( )(( 7) ( 8)) ( )(( 8) 5) 1(5 ( 7)) 1 ( )(1) ( )( 1) 1(1) = 1 1 6 1 (5) = 5 cm ar of quad. ABCD PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, 17
L.K. Gupta (Mathematic Classes) www.poineermathematics.com. MOBILE: 98155771, 461771 = area of ACD + area of ABC 109 5 144 7 cm 0. Let the speed of the stream = x km/hr Upstream speed = (18 x) km/hr Downstream speed = (18 + x) km/hr Speed time Dis tance time Dis tance Speed Upstream time = 1 + Downstream time 4 4 1 18 x 18 x 4 4 18 x 18 x 1 18 x 18 x 4 1 18 x 18 x 4 x = 4 x x + 48x 4 = 0 x 48 04 4 4 x x 48 600 48 60 48 60 48 60 x, x = 6 km/hr Speed of the stream = 6 km/hr PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, 18
L.K. Gupta (Mathematic Classes) www.poineermathematics.com. MOBILE: 98155771, 461771 1. To is the angle bisector of OT PQ PTQ OT bisects PQ PR = RQ = 8 cm Also TPO = 90 0 In PRO OP = OR + PR (10) = OR + 8 OR = 6 cm Let TP = x, TR = y In PRT PT = PR + TR x = 8 + y..(1) In PTO OT = TP + OP (6 + y) = x + 10 6 + y + 1y = x + 100 x = y + 1y 64..() Solving (1) & () y + 1y 64 = 64 + y 1y = 18 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, 19
L.K. Gupta (Mathematic Classes) www.poineermathematics.com. MOBILE: 98155771, 461771 y = 18 1 y x = 64 + y x = 64 + x = 64 + 104 9 64 9 104 x 9 576 104 x 9 1600 x 9 x TP 40 cm 40 cm. To Prove: OP AB Construction : Take any point Q, other than P on the tangent AB.Join OQ. Suppose OQ meets the circle at R. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, 0
L.K. Gupta (Mathematic Classes) www.poineermathematics.com. MOBILE: 98155771, 461771 Proof : We know that among all line segments joining the point O to a point on AB, the shortest one is perpendicular to AB. So to prove OP OP is shorter than any other segment joining O to any point of AB OP = OR (Radii of the same circle) OQ = OR + RQ OQ > OR OQ > OP OR < OQ AB, it is sufficient to prove that Thus OP is shorter than any other segment joining O to any point of AB OP AB. Radius of spherical marbles = 0.7 cm Radius of cylinder =.5 cm Volume of one spherical marble = 4 7 π 10 4 πr Volume of 150 spherical marbles = Volume of water risen in the cylinder 4 7 7 150 π π h 10 4 7 1 150 h 10 4 4 7 150 4 h 10 5 16 7 h 100 560 h 100 h 5.6 cm 4. r1 = 8 cm r = 0 cm PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, 1
L.K. Gupta (Mathematic Classes) www.poineermathematics.com. MOBILE: 98155771, 461771 h = 4 cm Volume of frustum = = 1 πh r 1 r r 1 r 1 4 8 0 8 0 7 8 64 400 160 7 = 176 64 7 = 15.689 litres = 15689.14 cm 1 litre of milk costs Rs 1 15.689 litres. 1 15.689 1 = Rs 9.47 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0,