FIITJEE SUBJECT:MATHEMATICS (CBSE) CLASS 10 SOLUTION
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1 FIITJEE SUBJECT:MTHEMTICS (CBSE) CLSS 10 SOLUTION 1. SECTION HCF of 55 and 99 is (99 551) So m =. cubic polynomial with zeros -, - and -1 is f(x) = k(x+)(x+)(x+1) f(x) = K ( x 6x 11x 6). If x, x+10 and x+ are in.p. Then x+10-x = x+-x-10 -x+10 = x-8 x = 18 x = 6 4. If coins are tossed simultaneously, then the probability of getting at least two heads = The mode of frequency distribution can be determined graphically from Histogram FIITJEE Ltd. Kanpur Centre 16/116 Karmin rcade (opp to ITO) Civil line, Kanpur. Mob ; Ph:
2 6. To have infinitely many solutions 7 a b a b 1 1 a b a b ab6 ab9 So a 15 a 5 b 1 7. The smallest number exactly divisible by 50 and 468 = L.C.M. of 50, 468 = 4680 So the required number is = D E B C rea of DE = 1 r of DE r of BC rea of BC 1 s DE BC so DE ~ BC ( similarity). The ratio of areas of two similar triangles is same as the ratio of squares of their corresponding sides r of DE D r of BC B 1 D B D 1 B B D 1 FIITJEE Ltd. Kanpur Centre 16/116 Karmin rcade (opp to ITO) Civil line, Kanpur. Mob ; Ph:
3 B 1 1 D 1 BD D D BD D 1 or 1 BD 1 9. Total number of elementary events = 66=6 Let be the event of getting a total more than 7 (,6), (6,), (,5), (5,),(4,4), (6,), (,6),(4,5),(5,4),(6,4),(4,6),(5,5),(6,5),(5,6),(6,6) Elementary events favorable to event = 15 Hence required probability = Let the three given points be (,1) B(1,) C(-,-4) Using distance formula 11. B= BC= C= (1) (1 ) 5 (1 ) ( 4) 45 ( ) ( 4 1) 50 Here B BC C So BC is a right angled triangle having C as hypotenuse and So it can form a rectanlge. BC 90 o P O GIVEN : circle with centre O and tangents from P as P and PB To prove : P=PB Proof : In PO and PBO O=BO (radii) OP= OBP= 90 o (radius tangent) PO=PO (common) PO PBO (RHS congrnency) P=PB B FIITJEE Ltd. Kanpur Centre 16/116 Karmin rcade (opp to ITO) Civil line, Kanpur. Mob ; Ph:
4 1. Let C be the ladder and B be the wall BC=cm CB 60 o Cos 60 o = BC C 1 C 60 O C = 4 metres C m B 1. If 5 5 is a rational number. Then there exist co-prime integers a and b such that a b a 5 b a a 5 (squaring both sides) b b a a b b a b a b b a b ab So is a rational number [ a,b are integers This contradicts the fact Hence, 5 is irrational. a b is rational] ab is irrational. So our assumption is wrong O 0 O 0 O 45 O D 1km C B Let C and D be the kilometres stone and B be the foot of the hill. tan 45 o = B BC 1= B BC FIITJEE Ltd. Kanpur Centre 16/116 Karmin rcade (opp to ITO) Civil line, Kanpur. Mob ; Ph:
5 B=BC tan0 o = B BD 1 = B DC CB 1 B 1 B B+1= B B( -1)=1 1( 1) B= ( 1)( 1) B= 1 1 B= B=1.66 So, the height of the hill is 1.66 km O C B Radius r = OB= cm i) Let h be the height of the cone h= 5 h= 4 cm ii) Volume of solid = 1 r h r = 1 (9 4 7) = 1 90 =0 cm = 0.14 cm =94.0 cm. FIITJEE Ltd. Kanpur Centre 16/116 Karmin rcade (opp to ITO) Civil line, Kanpur. Mob ; Ph:
6 tan sin tan sin sin sin = cos sin 1 cos sin (sec 1) = sin (sec 1) = sec 1 sec 1 1 cos 1 cos (1 cos ) 1 cos = 1 cos 1 cos 1 cos s in Or tan cot 1cot 1tan sin cos = cos sin cos sin 1 1 sin cos sin cos = cos (sin cos ) sin (cos sin ) sin cos = sin cos (sin cos ) (sin cos )(sin cos sin cos ) = sin cos (sin cos ) 1 sincos = sincos =seccosec+1 =R.H.S. f ( x) x px p c p, ( p c) ( 1)( 1) 1 p c p 1 1c Or FIITJEE Ltd. Kanpur Centre 16/116 Karmin rcade (opp to ITO) Civil line, Kanpur. Mob ; Ph:
7 4x ( a b ) x a b 0 4x a x b x a b 0 x( x a ) b ( x a ) 0 ( x a )( x b ) 0 x a 0 or x b 0 a b x or x 18.,15,7,9, _ First term a= 1 Common difference d=1 Let nth term is 1 more than 54 th term Tn T54 1 a+(n-1)d=a+5d+1 +(n-1)1= n-1=66+1 1n=780 n=65 or n 5n Sn S5 S S4 5 = 94 T S S = = BC DE ( criterionof similarity ) B BC D DE 1 9 ( BC C B ) D D 4cm FIITJEE Ltd. Kanpur Centre 16/116 Karmin rcade (opp to ITO) Civil line, Kanpur. Mob ; Ph:
8 0. 1 reaof DE 4 6 cm D R S Q P B BCD is a parallelogram which touches the circle at P,Q,R,S. Tangents from a point outside a circle are equal So P=S BP=BQ CR=CQ DR=DS P+BP+CR+DR=S+BQ+CQ+DS (P+BP)+(CR+DR)=(S+DS)+(BQ+CQ) B+CD=D+BC B=BC (B=CD, D=BC) B=BC So, BCD is a rhombus. 1. Length of the minute hand R = 6 cm Length of the hour hand r = 4 cm Distance covered by minute hand in1 hour = R = 6 7 = 64 7 cm Distance covered by minute hand in days = cm Distance covered by the hour hand in 1 hours = r = 4 7 = Distance covered by hour hand in days = = cm Sum of distances travelled by their tip in days = FIITJEE Ltd. Kanpur Centre 16/116 Karmin rcade (opp to ITO) Civil line, Kanpur. Mob ; Ph:
9 4. On dividing x x 8x ax b by x x 7 4 x 1 x x 8x ax b x 4 x x 7x ax x x 7x (a 1)x b 7x 0 7 (a 1)x (b 7) (a-1)x + (b+7) must be 0 So a-1 = 0 b+7 = 0 a = 1 b =-7. Height of bucket h = cm Radius of bucket r = 18 cm Height of cone H = 4 cm Let the radius of cone = R Volume of bucket = Volume of cone 1 r h R H R 4 R R 18 R 6cm Let the slant height be L L R H L L 4.66cm L 4.7 cm = = 1910 cm 7 x 1 remainder must be 0 FIITJEE Ltd. Kanpur Centre 16/116 Karmin rcade (opp to ITO) Civil line, Kanpur. Mob ; Ph:
10 4. Class Frequency(f) Mid value Cumulative f(x) Interval frequency(f) Mean = Modal class is f 64 fx 1400 fx 1400 f 64 Median class is50 60 N F Median L h f f f1 Mode = L h f f1 f a P Q x B h C h+a D FIITJEE Ltd. Kanpur Centre 16/116 Karmin rcade (opp to ITO) Civil line, Kanpur. Mob ; Ph:
11 Let QC be the surface of the lake and let P be the point of observation such that PQ=h Let D be the reflection of the cloud in the lake. Then CD=C Let PB D PB=, BPD= PB= x, B=a In PB In PBD h a tan x a h xtan tan x x a xtan x tan h x tan h x tan tan 6. P xsec hsec tan tan Distance of the cloud from the point of observation is hsec tan tan FIITJEE Ltd. Kanpur Centre 16/116 Karmin rcade (opp to ITO) Civil line, Kanpur. Mob ; Ph:
12 FIITJEE Ltd. Kanpur Centre 16/116 Karmin rcade (opp to ITO) Civil line, Kanpur. Mob ; Ph:
13 o 7. Given right-angled triangle BC in which B 90 To Prove (Hypotenuse) = (Base) + (Perpendicular) i.e., C = B + BC Construction From B draw BD C. Proof In triangles DB and BC, we have o DB BC Each equal to 90 nd, [Common] So, by - similarity criterion, we have DB BC D B In similar triangle corresponding B C sides are proportional B D C...(i) In triangles BDC and BC, we have CDB BC and, C C So, by - similarity criterion, we have BDC BC DC BC In similar triangles corresponding BC C sides are proportional BC C DC...(ii) dding equations (i) and (ii), we get B BC D C C DC B BC C(D DC) B BC C C B BC C Hence, C B BC B D BD (1) CD C D () On adding (1) and () B CD C BD B D C (1) x y x y FIITJEE Ltd. Kanpur Centre 16/116 Karmin rcade (opp to ITO) Civil line, Kanpur. Mob ; Ph:
14 () x y x y On multiplying (1) by 4 and () by x y x y x y x y 11 On subtraction 1 x y so x y 11 () Putting x y 11 in (1) x y 5 (4) n n Eq () Eq (4) x 16 x 8 So y Or 1 ( x)( x) ( x 5x 6) x 10x 9 0 Compairing it to the standard form of quadratic equation ax bx c 0 a b 10 c 9 Using quadratic formula x b b 4ac a x x 9. 1,,5, _ 49 Total number of elementary events = 5 FIITJEE Ltd. Kanpur Centre 16/116 Karmin rcade (opp to ITO) Civil line, Kanpur. Mob ; Ph:
15 i) The probability that the number drawn is divisible by = 16 5 ii) The probability that the number drawn is composite = 11 5 iii) The probability that the number drawn is not a perfect square = 18 5 iv) The probability that the number drawn is multiple of and 5 = 5 0. Diagonals of a parallelogram bisect each other So midpoint of C is same as midpoint of BD 1k 4 1 k k (-4,-)D (1,-) O B(,) C(k,) Length B (1 ) ( ) rea of 1 BC 1( ) ( ) ( ) 1 1 B h h 4 4 h 6 FIITJEE Ltd. Kanpur Centre 16/116 Karmin rcade (opp to ITO) Civil line, Kanpur. Mob ; Ph:
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