Class-10 - Mathematics - Solution

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1 SECTION-A 1. Using, dividend = divisor quotient + reainder. 3. = = a a 5 BC BC tana cotc AB AB Using PT : BC Given A.P.:, 8, 18 3,... So, d Class-10 - Matheatics - Solution a1 b1 c1 5. For no solution: a b c 6. b 4ac ( ) SECTION-B cos (1 sin ) cos (1 sin ) 7. L.H.S : (1 sin ) (1 sin ) 1 sin 8. HCF LCM = Product of both nubers LCM = LCM , 18 5 cos (1 sin ) 1 sin 1 sin sec tan R.H.S cos cos cos cos [] [] [] So, p(x) x ( 18)x 7 ( 5) x 18x Since, A = B AC = BC Now, AC A = BC BE C = CE A BE Since, E AB (Converse of B.P.T.) C CE 11. For equal roots : = 0 [] [] k k k = ±4 3 = ±1 1. Required nubers are : 1, 18, [] So, 13. Since, n SECTION-C tana tanb tan(a B) tan45 1 tana tanb A + B = 45 1 Half Yearly Exa Class-10 (Sol.)

2 14. Since, AOB ~ CO (AA siilarity) so, ar(aob) AB (C) 4C 4 4 :1 ar(co) C (C) C So, Sn 5n 3n S a 1 1 S a S a So, given AP is : 8, 18, 8,... Thus, an 8 (n1) n 10 10n 16. Let the speed of the fast train be x k/h Speed of the slow train would be (x 10)k/h Now, tie taken by fast train = 600 h x A.T.Q : and, tie taken by slow train = (x 10) x 600(x x 10) 3 x(x 10) 600 h (x 10) (x 10x) x 10x x 10x 000 x 10x x 50x 40x x(x 50) 40(x 50) 0 (x 40)(x 50) 0 Since, x for speed. x 50 0 x 50 k / h Thus, speed of fast train is 50 k/h and, speed of slow train is 40 k/h. 17. Using division algorith : 4 3 x x 3x x 0 g(x) (x 1) g(x)(x 1) x x 3x x Now, Thus, g(x) x x x 3x x (x 1) x + x + x 4 + x 3 + 3x + x + x 4 + x x 3 + x + x + x 3 g(x) x x + x x + x + 0 Half Yearly Exa Class-10 (Sol.)

3 18. Given : In right triangles ABC and PQR, sin B = sin Q A P C B R Q R.T.P: B = Q Proof : Since, sin B = sin Q AC PR AB AC K (Let)...(1) AB PQ PQ PR AB K PQ, AC K PR Now, BC AB AC (K PQ) (KPR) K (PQ PR ) K QR PQ PR PQ PR (PQ PR ) Fro equation (1) and (), AB AC BC PQ PR QR ABC ~ PQR (SSS siilarity) B = Q 19. Given equations are : (CPST) (a + b)x + (a b)y = 0 (a b)x + (a + b)y 3 = 0 Using cross ultiplication ethod : K...() (a b) (a + b) (a b) (a + b) 3 (a b) (a + b) 3(a b) ( )(a b) ( )(a b) ( 3)(a b) (a b)(a b) (a b)(a b) ( 6a 3b) ( 4a b) ( a 4b) ( 3a 6b) (a ab 4ab b ) (a ab 4ab b ) 6a 3b 4a b a 4b 3a 6b a 5ab b (a 5ab b ) a 5b a 10b a 5ab b a 5ab b a 5b a 10b 10ab 5b a a 10b x, y 10ab 10ab 3 Half Yearly Exa Class-10 (Sol.)

4 0. Let 'a' be any positive integer and b = 5 So, r = 0, 1,, 3, 4 because, 0 r b (I) For r = 0 a = 5q a 5q 5 (II) For r = 1 a = 5q+1 a 5q 10q 1 (5 + 1) (III) For r = a = 5q + (IV) For r = 3 a = 5q + 3 a 5q 0q 4 (5 + 4) a 5q 30q 9 (5 + 4) (V) For r = 4 a = 5q + 4 a 5q 40q 16 (5 + 1) So, no value of 'a ' is in the for of 5 + or Let the nuber of rows be x and the nuber of students in each row be y, then the nuber of students = xy When one student is added to each row, the nuber of rows becoe x. Hence, we ust have (x ) (y + 1) = nuber of students = xy nuber of rows nuber of students in each row xy + x y = xy x y = 0...(1) Again, when nuber of students in each row is decreased by 1, then the nuber of rows becoes x + 3. So, we ust have (x + 3) (y 1) = nuber of students = xy xy x + 3y 3 = xy x + 3y 3 = 0...() Adding (1) and (), we get y + 3y 3 = 0 y = 5 and then fro (1), x 5 = 0 x 10 = 0 x = 1 Hence, the nuber of students = xy = 1 5 = 60.. Given : In ABC, A is the bisector of A E A x x B C R.T.P : AB B AC C Construction: Extend BA to a point E such that A EC 4 Half Yearly Exa Class-10 (Sol.)

5 Proof : Since, A EC and AC is as transversal. So, AC = ACE = x (let) (alternate int. angles) and BA = AEC = x (corresponding angles) ACE = AEC AE = AC Now, in BEC, using B.P.T. : BA B AE C but, AE = AC AB B Hence, proved. AC C SECTION - 3. Given : ABC and PQR, A P B C Q S R E T AB AC A PQ PR PS R.T.P: ABC ~ PQR Construction : Extend A to a point E such that A= E and PS to a point T such that PS = TS Proof : In s AB and EC, B = C A = E AB = EC (Given) (By construction) (V.O. A.) AB EC (SAS congruence rule) So, AB = CE (CPCT) Siilarly, PQS TRS PQ = RT (CPCT) Now, AB AC A (given) PQ PR PS CE AC A AE RT PR PS PT ACE ~ PRT (SSS siilarity rule) AC = SPR (CPST)...(1) Siilarly, BA = QPS...() Adding equation (1) and (), A = P 5 Half Yearly Exa Class-10 (Sol.)

6 Now, in s ABC and PQR, AB AC and A = P PQ PR ABC ~ PQR (SAS siilarity rule) tana cot A 4. L.H.S. : cosa 1 cota 1 tana cosa 1 1 ( ) ( ) sin A ( ) ( ) 3 3 sin A cos a ( cosa)(sin A ) cosa( cosa) cosa(cosa) 1 cosa 1 cosa cosa cosa = cosec A sec A + 1 or 1 + sec A cosec A 5. 1 for x kx (1) 1 k k k (8 ) k 16 for x 8x k 0...() ( 8) 4 1 k 64 4k k k 16 fro (1) and (), k = Given : In ABC, C = 90, C AB R.T.P. : CB CA B A Proof : AC ~ ABC AC C A AB BC AC CA AB A...(1) Now, BC ~ BAC BC C B BA AC BC CB BA B...() ividing eqn. () by eqn. (1) : (AA siilarity) (AA siilarity) CB BAB B CA BA A A Hence, proved. 7. Let us assue to contrary that 7 is a rational nuber. 7 r (where r is a rational) 7 r 7 r 6 Half Yearly Exa Class-10 (Sol.)

7 Here, 7 r represents a rational, thus will also be a rational. But, this contradicts the fact that So, our assuption is contradicted. Hence, 7 is an irrational. is irrational. 8. Let the work done by a woen be in x days and by a an be in y days. A.T.P : and (1) x y () x y 3 By ultiplying eqn. (1) by 3 and eqn. () by and subtracting the, x y x y y y 36 days and 9. Given : x x = 18 days. s s n [a ( 1)d] n [a (n1)d] n 1 a ( 1)d a (1)d a (n1)d n 1 n a (n1)d (By putting = 1, n = n 1) 1 a ( )d 1 1 a (n )d n1 a ( 1)d 1 a (n1)d n A + B + C = 180 B + C = 180 A L.H.S : 180 A A 180 A A A A A A sin cos cos sin sin 90 cos cos 90 sin A A A A cos cos sin sin A A cos sin = 1 7 Half Yearly Exa Class-10 (Sol.)

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