Sample Question Paper 7

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4 0th Mathematics Solution Sample paper -0 Sample Question Paper 7 SECTION The equation of one line 4x + y 4. We know that if two lines a x + b y + c 0 and a x + b y + c 0 are parallel, then a b c ; so there can be infinite such lines. a b c One of the examples of such a parallel line is given below. Second parallel line is x 9y. where C 0.. or similar triangles, we know that area of BC area of DE 80 area of DE area of DE BC E ( ) 4 6 ( 5) cm. 4. The number of families having income range from ` 6000 to ` (The class has frequency 9) SECTION B 5. g(x) x + x +, f(x) x 4 + 5x 7x + x + 4 ½ x 4x+ 4 x + x + x + 5x 7x + x 4+ x + 9x + x 4 ( ) ( ) ( ) 4 x 0x + x x x x 4 4 (+) (+) (+) x + 6x + 4 x + 6x + ( ) ( ) ( ) Remainder, r (x) ½ r(x) 0, g(x) is not a factor of p (x). ½ 6. Condition for unique solution, a b a b B C 4 cm E 5 cm D

5 C.B.S.E. (CCE) Term-I, Mathematics, Class - X k k k 6 k ± Since, O OB OC OD In OD and COB O OC OD OB OD COB (Vertically opposite angles) OD ~ COB (SS similarity) C and B D. 8. cos ( B) sin ( + B) cos 0º B 0º...(i) ½ sin 60º + B 60º...(ii) ½ dding equations (i) and (ii), 90º 45º ½ rom (ii), B 60º 60º 45º 5º ½ 9. LHS cos + cos cos cos + cos cos ( cos ) ( cos ) ( cos ) sin 0. The multiples of 5, according to the problem are : Mean cos cos sin sin sin cosec cot RHS. Proved. 5, 5, 5, 5, 45 ½ ½ SECTION C. The greatest number of cartons in each stack is the HC of 44 and HC 8 The greatest number of cartons 8.. Let be a rational number a. (a and b are integers and co-primes and b 0) b

6 Sample Question Paper-7 Solutions On squaring both the sides, a b b a a is divisible by a is divisible by We can write a c for some integer c a 9c b 9c b c b is divisible by b is divisible by rom (i) and (ii), we get as a factor of a and b which is contradicting the fact that a and b are coprimes. Hence our assumption that is an rational number is false. So is irrational number.. Let f(x) 4x + 4x ; since and are zeroes of f(x) We must have lso, f 0; f f I K J 4 I 4 K J + I K J H G I K J I K J + 4 I K J...(i)...(ii) f f 0, are zeroes of polynomial 4x + 4x Now Sum of zeroes 4 coeff. of x ½ 4 coeff. of x constant term Product of zeroes 4 coeff. of x Relation between zeroes and coeff. of polynomial is verfied. 4. Let the fixed charge of taxi be Rs. x per km and the running charge be ` y per km. ccording to the question, x + 0y 75 x + 5y 0 Subtracting equation (ii) from equation (i), we get 5y 5 y 7 Putting y 7 in equation (i), we get x 5 Total charges for travelling a distance of 5 km x + 5y ` ( ) ` (5 + 75) ½...(i)...(ii) ½ ` 80 ½

7 4 C.B.S.E. (CCE) Term-I, Mathematics, Class - X 5. We have, PQR ~ PB ( P is common and P PQ PB PR ) area PQR area PB area PB I K J P k ki 4k k K J PQ P 4 area PB 8 cm k area of QRB area of PQR area of PB 8 4 cm Q R 6. D, E and are mid-points of BC, C and B respectively. (Given) BDE and DCE are parallelograms. ( line joining mid point of two sides of a is parallel to the third side and is one half of it) B E B D C In triangles BC and DE, B E and C (Opp. angles of a parallelogram) BC ~ DE ( Similarty) ar DE DE ar BC B DE (DE) (DE B, B B) ar DE DE ar BC 4DE 4 7. LHS 8. sin θ cos θ sinθ + cosθ + sin θ + cos θ sinθ cosθ (sin θ cos θ) + (sin θ + cos θ) sin θ cos θ (sin θ + cos θ) sin θcos θ + (sin θ + cos θ) + sin θcosθ sin θ ( sin θ) + sin θ + sin θ RHS. Proved. sin θ sec 4º.sin 49º + cos 9º.cos ec 6º (tan 0º.tan 60º.tan 70º) (sin º + sin 59º) cosec (90º 4º)sin49º + cos9º.sec (90º 6º) [tan0º. cot(90º 70º)] [sin º + cos (90º 59º)] [ cosec (90 θ) sec θ, sin (90 θ) cos θ] cosec 49º.sin49º + cos9º.sec9º [tan0º.cot0º] (sin º + cos º) + 0

8 Sample Question Paper-7 Solutions 5 9. C.I f c. f Here, N 00 N 50. So, median class is Median l N I c. f. f h ½ KJ 50 46I 8 K J Median weight 9 gm. ½ 0. Modal class : 0 0 Here l 0, f 40, f 0 4, f 6, h 0 ( f f0) Mode l + h f f0 f ( 40 4) SECTION D. ny positive integer is of the form q or q +, for some integer q. When which is divisible by. n q n n q(q ) m, when m q(q ) When n q + which is divisible by. n n (q + ) (q + ) q(q + ) m, when m q(q + ) Hence, n n is divisible by for every positive integer n.. Since α and β are the zeroes of polynomial x + x +. Hence, α + β and αβ Now for the new polynomial, Sum of the zeroes α + β + α + β ( α + β αβ ) + ( + α β αβ ) ( + α)( + β)

9 6 C.B.S.E. (CCE) Term-I, Mathematics, Class - X αβ + α + β + αβ + 4 Sum of zeroes Product of zeroes α βi ( α )( β ) + α β ( + α)( + β) I K J + α β + αβ ( α + β) + αβ + α + β + αβ + ( α + β) + αβ KJ Product of zeroes Hence, Required polynomial x (Sum of zeroes)x + Product of zeroes. x + y y x x x +. x 0 6 y 4 0 ½ y x y x + x 5 y 0 ½ y x' 5 (0, 4) D 4 (, 0) B x y y x (5, ) (, ) C (6, 0) x y' Plotting the above points we get the graph of the equations x + y and y x. Clearly, the two lines intersect at the point (, ). gain the required coordinates of vertices of the triangle BC are (, ), B (, 0) and C(6, 0).

10 Sample Question Paper-7 Solutions 7 4. Let the number of red balls be x and white balls be y. ccording to the question, y x or x y 0...(i) and (x + y) 7y 6 or, x 4y 6...(ii) Multiplying eqn. (i) by and eqn. (ii) by and then subtracting, we get 6x 9y 0 6x 8y y Subtracting from (i), y x 6 0 x 8 Hence, number of red balls 8 and number of white balls. x 8, y 5. Given : In BC and DE, P and DQ are medians, such that B DE BC E P DQ D...(i) B P C E Q To prove : BC ~ DE Proof : rom (), B DE BC E P DQ B DE BP EQ P DQ BP ~ DEQ [SSS similarity] B E In BC and DE, B DE BC E and B E, (By SS criterion) BC ~ DE. Proved. 6. (i) Let B be the vertical tree and C be its shadow. lso, let DE be the vertical tower and D be its shadow. Join BC and E. Let DE x. B E x C D 8 40

11 8 C.B.S.E. (CCE) Term-I, Mathematics, Class - X We have B m ½ C 8 m and D 40 m ½ In BC and DE, we have D 90º and C ½ Therefore by criterion of similarity, we have BC ~ DE B DE C D 8 x x 8 x 60 m. (ii) Similarity of s and heights and distances. (iii) Growing more and more trees will help to save and protect our enviornment. Trees give us so many things including shade. 7. LHS cos θ sin θ + tanθ sinθ cos θ ½ cos θ sin θ + sin θ sin θ cosθ cosθ cos θ cos sin sin θ θ θ cosθ sinθ cos θ sin θ cosθ sinθ 8. LHS (cosθ sin θ)(cos θ + sin θ + sin θcos θ) (cosθ sin θ) ( sin θ+ cos θ ) + sin θ cos θ RHS. Proved. secq+ tanq tanq secq+ (secq+ tan q) (sec q tan q) tanq secq+ (secq+ tan q) (secq+ tan q)(secq tan q) (tanq secq+ ) (secq+ tan q)( secq+ tan q) (tanq secq+ ) sec θ + tan θ + sin cosq cos q q + sin q cos q + sinq sin q cosq sin q

12 Sample Question Paper-7 Solutions 9 sin q cos q( sin q ) cos q cos q( sin q) cos q sin q RHS. Proved tan sec We have cos tan + cot cosec...(i) ½ sin gain rom (i) and (ii), we have tan cot I K J G H tan G tan I KJ I K J tan tan tan ( tan ) tan...(ii) + tan + cot tan cot tan 0. C.I. f c.f x 5 + x x 0 40 y + x + y x + y x + y Proved. ½ Here from table, N + x + y 40 x + y 8...(i) Since, median, Median class is Median l ( 9 x) 0 y N c.f. h f 0 ( + x) y I K J 0 y 90 0x rom (i), 0x + y 90...(ii) x + y 8 On subtraction, 9x 7

13 0 C.B.S.E. (CCE) Term-I, Mathematics, Class - X x rom (i), y Let assumed mean, a and h 00 Life time x i u i x i - a f h i f i u i (in hrs) Total Sf i 00 Sf i u i 57 Mean, x a + verage life time is 678 hours. Σfiui h Σfi I KJ