maths super 10

Sample Paper 9 ANSWERKEY 1 (b) 2 (b) 3 (b) 4 (b) 5 (d) 6 (a) 7 (b) 8 (c) 9 (c) 10 (b) 11 (a) 12 (a) 13 (b) 14 (b) 15 (b) 16 (b) 17 (c) 18 (c) 19 (c) 20 (a) 21 (c) 22 (c) 23 (d) 24 (a) 25 (b) 26 (b) 27 (d) 28 (b) 29 (b) 30 (d) 31 (d) 32 (c) 33 (b) 34 (b) 35 (b) 36 (a) 37 (b) 38 (b) 39 (a) 40 (a) 41 (b) 42 (a) 43 (c) 44 (a) 45 (b) 46 (a) 47 (c) 48 (b) 49 (b) 50 (a) 1. (b) Principal of similarity of figures. ( ) ( ) BC = − 3a + a 2 + 3a +1 2 2. (b) Let the quadratic polynomial be ax2 + bx + c, and its zeroes be a and b. −b We have α+β = −3 = a and αβ= 2= c ( ) ( ) ⇒ BC = a2 1− 3 2 + a2 3 +1 2 a 3. (b) Area of rectangle = 28 cm × 23 cm = 644 cm2 ( ) ( ) ⇒ BC= a 1− 3 2 + 1+ 3 2 Radius of semicircle = 28 cm ÷ 2 = 14 cm Radius of quadrant = 23 cm – 16 cm = 7 cm ⇒ BC= a 1+ 3 − 2 3 +1+ 3 + 2 3 Area of unshaded region = 12 × 22 ×14cm ×14cm  ⇒ =BC a=8 2 2a 7  2 3a − a 2 ( ) ( )  1 22  and AC = − 3a − a +  4 7  + 2 × × × 7cm × 7cm = 385 cm2 ( ) ( ) ⇒ =AC a2 3 +1 2 + a2 3 −1 2 Shaded area = 644 cm2 – 385 cm2 ( ) ( ) ⇒ A=C a 3 +1 2 + 3 −1 2 = 259 cm2 If a = 1, then b = 3 and c = 2. So, one quadratic polynomial which fits the given ⇒ AC= a 3 +1+ 2 3 + 3 +1− 2 3 conditions is x2 + 3x + 2. =a 8 = 2 2a 4. (b) Let the lady has x coins of 25 p and y coins of 50 p. Then, according to problem Clearly, we have AB = BC = AC Hence, the triangle ABC formed by the given points is an x + y = 40 ......... (i) equilateral triangle. 25 x + 50 y = 1250 ......... (ii) 6. (a) one Solving for x & y we get 1 7. (b) secθ = cos q and maximum value of cos q is 1 x = 30 (25 p coins) & y = 10 ( 50 p coins) ( )5. (d) Let A(a, a), B(–a, –a) and − 3a, 3a be the given 1 sec θ points. Then, we have ⇒ Maximum value of is 1 AB = (−a − a)2 + (−a − a)2 8. (c) Let ABC be an isosceles triangle, where base AB = a and equal sides AC = BC = b. Let CD be the perpendicular = 4a2 + 4a2 = 2 2a on AB.

S-34 Mathematics C 14. (b) AB = (9 − 9)2 + (6 − 0)2 =6 bb BC = (−9 − 9)2 + (6 − 6)2 =18 AD B CD = (−9 + 9)2 + (0 − 6)2 =6 So, AD = DB = 1 AB = a DA = (9 + 9)2 + (0 − 0)2 =18 2 2 AC = (−9 − 9)2 + (6 − 0=)2 324 + 36 Altitude, CD = height of the = =360 6 10 DABC is given by BC = (−9 − 9)2 + (0 − 6=)2 324 + 36 h = AC2 − AD2 = =360 6 10 NB =⇒ h 1 4b2 − a2 N 2 24 m Area of the ∆ABC =12 base × altitude 15. (b) O 10 m A E W =12 × a × 1 4b2 − a2 = a 4b2 − a2 . 2 4 9. (c) We have, p(x) = x2 –10x –75 S = x2 – 15x + 5x – 75 = x(x – 15) + 5(x –15) = (x –15) (x + 5) Let the initial position of the man be at O and his final \\ p(x) = (x –15) (x + 5) So, p(x) = 0 when x = 15 or x = –5. Therefore required position be B, since the man goes to 10 m due east and zeroes are 15 and –5. then 24 m due north. 10. (b) A(–4, 0), B(4, 0), C(0, 3) Therefore DAOB is a right angled triangle at angle A. AB = (4 + 4)2 + (0 − 0=)2 (8=)2 8 \\ DAOB OB2 = OA2 + AB2 = (10)2 + (24)2 BC = (0 − 4)2 + (3 − 0)2 = 100 + 576 OB = 676 = 16 + 9= 25= 5 16 + 9= 25 = 5 OB = 26m CA = (−4 − 0)2 + (0 − 3)2 = BC = CA ⇒ DABC is isosceles. Hence, the man is at a distance of 26 m from the s tarting 11. (a) The circle is divided into 18 equal sectors point. 360° 16. (b) Let length and breadth be x cm and y cm respectively. Central angle in each sector = 18 = 20° According to problem, 2 (x + y) = 40 .... (i) y 2 .... (ii) Area of each sector = θ × πr 2 and x = 3 360° on solving, x = 12, y = 8 20° =360° × 3.14 × 4 × 4 = 2.79 \\ Length = 12 cm and breadth = 8cm. Area of shaded portion = 9 × 2.79 = 25.12 17. (c) tan A = 3 = P 4 b 12. (a) Product 13. (b) n(S) = 6 × 6 = 36, E = {(1, 1), (2, 2), (3, 3), (4, 4), h = P2 + b2 = 9 +16 (5, 5), (6, 6)} P 3 h 5 n(E) = 6 36=6 sin A = = nn((ES=)) 1 P(E=) 6 18. (c) 3 = 0.6 where as other numbers have non-terminating 5 decimals.

Solutions S-35 19. (c) Let the medians through C meets AB at D. \\ ∠BCA = ∠DEA = q A(2, 2) Clearly, DABC and DADE are similar D \\ AC = AB AE AD ⇒ 24 = 15 ⇒ h = 15 ×16 = 240 = 10 16 h 24 24 Hence, height of the telephone pole = 10 cm. 24. (a) Polynomial p(x) has four real zeros. B(–4, –4) C(5, –8) 25. (b) sin q + 2 cos q = 1 ⇒ (sin q + 2 cos q)2 = 1 ⇒ sin2 q + 4 cos2 q + 4 sin q cos q = 1 Coordinates of D are ⇒ 1 – cos2 q + 4 (1 – sin2 q + 4 sin q cos q = 1  −4 + 2 , −4 + 2  = ( −1, −1) ⇒ 4 sin2 q + cos2 q – 4 sin q cos q = 4 2 2  ⇒ (2 sin q – cos q)2 = 4 Length of CD = 36 + 49 =85. ⇒ 2 sin q – cos q = 2 [Q 2 sin q – cos q ≠ –2] = (5 +1)2 + (−8 +1)2 26. (b) The system of simultaneous equations a1x + b1y + c1 = 0 and iaf2xaa12+ ≠b2bby12 + c2 = 0, have exactly one (unique) solution . 20. (a) Let the number of blue balls = x \\ Total number of balls = 5 + x 27. (d) The L.C.M. of 16, 20 and 24 is 240. The least multiple of 240 but it is not a perfect square. x P (blue ball) = 5+ x Similarly 2400 is also ruled out because it is also not a perfect square. 1600 is divided by 16 and 20 but not by 24. 5 Therefore 3600 is least number which is a perfect square P (red ball) = 5+ x and divisible by 16, 20, 24. 28. (b) The centre of the circle is the midpoint of the diameter. Given that P (blue) = 2 • P (red) x5 So coordinates of centre = midpoint of AB 5+ x = 2 × 5+ x =  −2 + 4 , 3 − 5  =  2 , −2  = (1, – 1) x 10  2 2   2 2  5+ x = 5+ x 21. (c) Total number of outcomes are {HH, HT, TH, TT}. 29. (b) Since, the graph of y = f(x) is a parabola, therefore f(x) is quadratic. The outcomes favourable to the event ‘atmost one head’ 30. (d) 1+ sin2 q = 3 sin q cos q are HT, TH and TT. [Q 1 = cos2 q + sin2 q] n(E) =43 ⇒ cos2 q + 2 sin2 q = 3 sin q cos q ∴ P(E) = n(s) ⇒ cos2 q – 3 sin q cos q + 2 sin2 q = 0 ⇒ cos q – sin q) (cos q – 2 sin q) = 0 22. (c) a1 ≠ b1 ⇒ 3 ≠ −2 ⇒ cos q – sin q = 0 or cos q – 2 sin q = 0 a2 b2 2m − 5 7 ⇒ sin q = cos q or 2 sin q = cos q or –4m + 10 ≠ 21 ⇒= csoinsθθ 1=or 2 csoinsθθ 1 or –4m ≠ 11 or m ≠ −141 ⇒ tan q = 1 or 2 tan q = 1 23. (d) Let h metres be the height of the telephone pole. Since Thus, tan q = 1 or tan q = 1 . time is the same in both the cases. 2 B 31. (d) By Pythagoras theorem in DBAC, we have C D 15 m q q h O C E 16 m A r 24 m A B

S-36 Mathematics BC2 = AB2 + AC2 = 62 + 82 = 100  ⇒ BC = 10 CM 37. (b) –2, 1, 3 A Now, Area of DABC = Area of DOAB + Area of DOBC + Area of DOCA 1 ⇒ 1 AB × AC = 1 2 BC × r + 1 38. (b) 2 2 AB × r + 2 CA × r ⇒ 11 × 6 × 8 = 2 (6 × r) + (10 × r) + 2 (8 × r) ⇒ 48 = 24 r ⇒ r = 2 cm B ED C 32. (c) (– 1)n + (– 1)4n = 0 will be possible, when n is any odd In DAEB, ∠AEB = 90° ....(i) natural number i.e., \\ AB2 = AE2 + BE2 33. (b) O In DAED, ∠AED = 90° 28 \\ AD2 = AE2 + DE2 45 28 ⇒ AE2 = (AD2 – DE2) \\ AB2 = (AD2 – DE2) + BE2 A B = (AD2 – DE2) + (BD – DE2) C = (AD2 – DE2) +  1 BC − DE 2  2  Area of sector OACB 45 × 22 × 28 × 28 = 308 cm27 1 =360 7 = AD2 + 4 BC2 – BC × DE 39. (a) The numbers common to given numbers are 22, 5 and 1 Area (DAOB) = 2 (28) (28) sin 45° 72. \\ H.C.F. = 22 × 5 × 72 = 980. = 14 × 28 × = 277.19 40. (a) Perimeter of sector = 25 cm Area (minor segment) = 308 – 277.19 = 30.81 θ 34. (b) Let the required ratio be K : 1 2r + 360 × 2r = 25 \\ The co–ordinates of the required point on the y–axis is 90 25 x = K(−4) + 3(1) ; y = K(2) + 5(1) ⇒ 2r + 360 × 2 × 7 × r = 25 K +1 K +1 11 22 Since, it lies on y – axis ⇒ 7 2r + r = 25 ⇒ 7 r = 25 ⇒ r = 7 \\ Its x–cordinates = 0 Area of minor segment =  πθ − Sinθ  r 2  360° 2  −4K + 3 ∴ K +1 =0 ⇒ –4K + 3 = 0  22 90 sin 90  (7)2 = 7 360° 2  3 × − 4 ⇒ K= = 1141 1  4 2  14 3 − × 49 = × 49 = 14 cm2. ⇒ Required ratio = 4 : 1 41. (b) 42. (a) 43. (c) 44. (a) 45. (b) \\ ratio = 3 : 4 35. (b) Consistent system 46. (a) As three faces are marked with number ‘2’, so number tan θ sec θ −1 36. (a) tan θ + sec θ +1 of favourable cases = 3. − 63= 1 (tan θ + sec θ) − (sec2 θ − tan2 θ) \\ Required probability, P(2)= 2 tan θ − secθ +1 = (tan secθ)[1− (secθ − tan θ)] 47. (c) No. of favourable cases = No. of events of getting the tan θ − secθ +1 = θ + number 1 + no. of events of getting the number 3 = 2 + 1 = 3 (tan θ + secθ)(1− secθ + tan θ) \\ Required probability, P(1 or 3)= 63= 1 (1+ tan θ − secθ) 2 = 48. (b) Only 1 face is marked with 3, so there are 5 faces = tan q + sec q which are not marked with 3. 5 6 = 1 + sin θ \\ Required probability, P (not 3) = cos θ 49. (b) 50. (a)

Sample Paper 10 ANSWERKEY 1 (c) 2 (b) 3 (a) 4 (b) 5 (c) 6 (d) 7 (d) 8 (b) 9 (a) 10 (d) 11 (b) 12 (d) 13 (a) 14 (a) 15 (b) 16 (d) 17 (b) 18 (c) 19 (b) 20 (d) 21 (b) 22 (c) 23 (d) 24 (d) 25 (b) 26 (b) 27 (d) 28 (c) 29 (a) 30 (b) 31 (a) 32 (a) 33 (d) 34 (c) 35 (a) 36 (b) 37 (b) 38 (c) 39 (a) 40 (d) 41 (c) 42 (c) 43 (c) 44 (d) 45 (c) 46 (c) 47 (a) 48 (d) 49 (a) 50 (b) 1. (c) Let a and b be the zeroes of the quadratic polynomial. 3 a2 4 we have a = 8 and b = 10 ⇒ =121 3 Sum of zeroes = a + b = 8 + 10 = 18 ⇒ a2 = 484 Product of zeroes = ab = 8 × 10 = 80. ⇒ a = 22 cm \\ The required quadratic polynomial = x2 – (Sum of the zeroes)x + Product of the zeroes Perimeter of equilateral D = 3a = x2 – 18x + 80 = 3 (22) Any other quadratic polynomial that fits these condition = 66 cm will be of the form Since the wire is bent into the form of Q circle, So k (x2 – 18x + 80), where k is a real. perimeter of circle = 66 cm ( )2. (b) A(3, –3), B(–3, 3), −3 3, −3 3 ⇒ 2pr = 66 ⇒ 2 × 22 × r =66 7 AB = (−6)2 + (6)2 = 36 + 3=6 =72 6 2 ⇒r= 66 × 1 × 7 2 22 ( ) ( )BC = −3 3 + 3 2 + −3 3 − 3=2 =72 6 2 ⇒ r = 10.5 cm So Area enclosed by circle = pr2 3−3 2 + 2 ( ) ( )AC =−3 −3 3 + 3 =72 6 2  22 10.510.5 = 7 \\ DABC is equilateral triangle. = 22 × 1.5 × 10.5 = 346.5 cm2 3. (a) Let the two numbers be x and y (x > y). Then, 5. (c) 1st wheel makes 1 revolutions per sec x – y = 26 ...(i) 6 x = 3y ...(ii) 2nd wheel makes 10 revolutions per sec Substituting value of x from (ii) in (i) 3y – y = 26 4 3rd wheel makes 10 revolutions per sec 2y = 26 y = 13 5 3 Substituting value of y in (ii) x = 3 × 13 = 39 In other words 1st, 2nd and 3rd wheel take 1, and Thus, two numbers are 13 and 39. 4. (b) Area of equilateral triangle  3 a 2 seconds respectively to complete one revolution. 4

S-38 Mathematics 5 5 = L.C.M of 1, 5, 5 = 5 = (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3) = 6 3 2 H.C.F of 1, 3, 2 L.C.M of 1, and No. of ways of getting 11 = (5, 6), (6, 5) = 2 Hence, after every 5 seconds the red spots on all the three No. of favourable ways = 1 + 2 + 4 + 6 + 2 = 15 wheels touch the ground. No. of exhaustive ways = 6 × 6 = 36 a2 − b2 Probability of the sum as a prime a2 + b2 6. (d) sin θ = = 13=56 5 12 A 9. (a) Given, AB = 2DE and DABC ~ DDEF Hence, area(∆ABC) = AB2 area(∆DEF ) DE 2 a² + b² a² – b² or area(5∆6D=EF ) 4=DDEE22 4 [Q AB = 2DE] B C area (DDEF=) 54=6 14 sq.cm. Since, sin θ = perpendicular 10. (d) Given : length of the sheet = 11 cm base \\ AC a2 − b2 Breadth of the sheet = 2cm AB a2 + b2 = Diameter of the circular piece = 0.5 cm Now in D ABC , Radius of the circular piece ∠ B = q and ∠ C = 90° = 0.5 = 0. 25 cm 2 (a2 + b2)2 = BC2 + (a2 – b2)2 Now, area of the sheet = length × breadth \\ BC = 2ab = 11 × 2 = 22 cm2. a2 + b2 cosec q= a2 − b2 , Area of a circular disc = pr2 BA=CC 2ab = 22 × (0.25)2 cm2 a2 − b2 7 co=t θ Number of circular discs formed Area of the sheet cosec q +=cot q a2 + b2 + 2ab = a + b a2 − b2 a2 − b2 a − b = Area of one disc 1 2 = 22 = 22× 7 = 112 7. (d) Q (3,4) 22× 0.0625 R 22 × ( 0.25 )2 P (7, –6) 7 Coordinate of R = 7 (2) + 3 (1) , −6(2) + 4(1)  Hence, 112 discs can be formed.  1+ 2 1+ 2  11. (b) a = x3y2 =  17 , −8  = x × x × x × y × y  3 3  b = xy3 = x × y × y × y Thus, the point R lies in IV quadrant. ⇒ HCF (a, b) = xy2 8. (b) The sum of the two numbers lies between 2 and 12. 12. (d) 1 tan q + 1 cot q q So the primes are 2, 3, 5, 7, 11. − cot q − tan No. of ways for getting 2 = (1, 1) = 1 sin q cos q No. of ways of getting 3 = (1, 2), (2, 1) = 2 = cos q + sin q No. of ways of getting 5 = (1, 4), (4, 1), cos q sin q 1 − sin q 1 − cos q (2, 3), (3, 2) = 4 No. of ways of getting 7

Solutions S-39 = q. sin q cos q  + q. cos q sin q  A sin q − q  cos q − q  Py Q cos sin 10 cm 10 cm sin cos = cos sin2 q cos q) + sin cos2 q sin q) 2x 2x q(sin q − q(cos q − = cos sin2 q cos q) − sin cos2 q q) B y C q(sin q − q(sin q − cos SR 12 cm sin2 q× sin q − cos2 q× cos q 15. (b) Let the common factor be x – k sin q cos q(sin q − cos q) = we have, f(k) = g(k) = 0 = sin3 q − cos3 q ⇒ k2 + 5k + p = k2 + 3k + q q cos q(sin q − cos sin q) k = q − p 2 (sin q − cos q)(sin2 q + cos2 q + sin q cos q) = sin q cos q(sin q − cos q) substituting “k” in x2 + 5x + p = 0 x2 + 5x + p = 0 = 1+ sin q cos q q − p 2 q − p sin q cos q 2  2 =1+ sec θ cosec θ  +   + p =0    –k So, 2 =1 \\ (p – q)2 = 2 (3p – 5q) \\ k=2 16. (d) X = (April, June, September, November) 13. (a) Let present age of Nuri = x years Hence, n(X) = 4 x2 =95 ⇒ =±  1  Let present age of Sonu = y years ( )17. (b) x 3  5 = irrational Five years ago, x – 5 = 3(y – 5) 18. (c) PQ = 13 ⇒ PQ2 = 169 x – 5 = 3y – 15 ⇒ (x – 2)2 + (–7 – 5)2 = 169 x – 3y = –10 ...(i) ⇒ x2 – 4x + 4 + 144 = 169 Ten years later, ⇒ x2 – 4x – 21 = 0 (x + 10) = 2(y + 10) ⇒ x2 – 7x + 3x – 21 = 0 x + 10 = 2y + 20 ⇒ (x – 7) (x + 3) = 0 x – 2y = 10 ...(ii) ⇒ x = 7, –3 Subtracting (ii) from (i), we get 19. (b) B – y = – 20 ⇒ y = 20 Substituting y = 20 in (ii), we get x – 2 × 20 = 10 D ⇒ x = 50 A C So, present age of Nuri is 50 years and present age of Sonu is 20 years 14. (a) Using Pythagoras theorem in DABL we have P AL = 8cm, We have two chord AB and CD when produced Also, DBPQ ~ DBAL meet outside the circle at P. \\ BQ = BL ⇒ 6− x = 6 or x= 6 − 3 y Since in a cyclic quadrilateral the exterior angle is PQ AL y 8 4 equal to the interior opposite angle, \\ ∠PAC = ∠PDB ....(i) From (1) and (2) and using AA similarity we have DPAC ~ DPDB

S-40 Mathematics \\ Their corresponding sides are proportional. Area (DABE) = 1 Area (DACD) 2 ⇒ PA = PC Thus PD PB Thus reqd. ratio is 1 : 2. ⇒ PA.PB = PC.PD. 23. (d) Area of circle A = 3.14 × 10 × 10 = 314 Area of circle B = 3.14 × 8 × 8 = 200.96 20. (d) we have, tan q =1 a sin φ φ − a cos 1 Area of Q = 8 × Area of B ⇒ =cot q a 1 φ − cot φ sin 1 1 = 8 × 200.96 = 25. 12 ⇒ cot q + cot φ = a sin φ ...(i) Area of P Now, = Area of Q tan φ =1 b sin q q − b cos 5 ⇒ =cot φ 1 − cot q ⇒ Area of P = 4 × Area of Q sin b q 5 4 ⇒ cot φ + cot q = 1 ...(ii) = × 25.12 = 31.4 sin b q Area of square = 7 × 7 = 48 From (i) and (ii), we have Required Area 1 = 1 = (314 + 200.96 + 49 – 25.12 – 31.4) a sin b sin φ q = 507.44 cm2 a sin q 24. (d) All the properties are satisfied by real numbers. b sin φ ⇒ = 25. (b) A(0, 4), B(0, 0), C(3, 0) 21. (b) Let f(x) = xn + yn. AB = (0 − 0)2 + (0 − 4)2 =4 Divisible by (x + y) means f(–y) = 0. BC = (3 − 0)2 + (0 − 0)2 =3 So, (–y)n + yn = 0. CA = (0 − 3)2 + (4 − 0)2 =5 This is possible only when “n” is an odd number. 22. (c) D AB + BC + CA = 12 26. (b) Let the two parts be x and y. We have, EA x + y = 62 ...(i) x ...(ii) ...(i) 4 = 2 2y 3 x 5 2x 15x – 16y = 0 By solving (i) and (ii) we get x = 32, y = 30 Bx C 27. (d) Here, (p + 2 )  q − 1  = pq − 5  2  Let AB = BC = x. Since DABC is right-angled with ⇒ pq − 1 p + 2q −1= pq − 5 ...(ii) 2 ∠B = 90° \\ AC2 = AB2 + BC2 = x2 + x2 = 2x2 ⇒ − p + 2q =−4 ...(iii) 2 ⇒ AC = 2x Since DABE ~ DACD ⇒ p − 2q =4 2 \\ Area ((DDAACB=DE )) AA=CB22 2=xx22 1 . Area 2 1 also, (p − 2)  q − 2  = pq − 5  

Solutions S-41 ⇒ pq − 1 p − 2q + 1= pq − 5 p = 4 and 2q = 2p – 4 2 ⇒ 2q = 8 – 4 = 4 Now, q = 2 ⇒ − 1 p − 2q =−6 ...(iv) ⇒ p + q = 4 + 2 = 6 2 ⇒ p – q = 4 – 2 = 2 By adding (iii) and (iv), we get 32. (a) An irrational number. p = 10 = p2 − 2q =4 33. (d) cos2 q = 3 ⇒ cos2 q =3 cot2 q − cos2 q cos2 q − cos2 q sin2 q or 10 − 2q =4 ⇒ cos2 q× sin2 q =3 2 cos2 q − sin2 q cos2 q 1 ⇒ 5 – 4 = 2q ⇒ q = 2 sin2 q cos2 q 1 ⇒ cos2 q(1− sin2 q) =3 2 Hence, solution set (p,q) = 10,   sin2 q 28. (c) Let cos q + 3 si=n q 2sin q ⇒ cos2 q =3 ⇒ tan2q = 3 ⇒ tan q = 3 Multiplying both sides by 2 + 3 , we get tan q = tan 60º ⇒ q = 60º (acute angle) ⇒ cos q= 2sin q − 3 sin q= (2 − 3)sin q 34. (c) (2 + 3) cos q= (2 + 3)(2 − 3)sin q d 5m ⇒ (2 + 3) c=os q {(2)2 − ( 3)2}sin q 12 m 11 m 6m ⇒ 2cos q + 3 cos q= (4 − 3)sin q ⇒ 2cos q + =3 cos q (4 – 3)sin q ⇒ sin q − 3 co=s q 2cos q Using pythagoras theorem, d= 122 + 52= 144 + 25= 169 29. (a) Substituting the given zeros in (x – a) (x – b), we get \\ d = 13 m So, distance between the tops of poles is 13 m.  x − 1  x + 2  35. (a) Radius of circle = 14 cm ÷ 2 = 7 cm 3  5  One side of the figure opposite to 35cm = 35cm –7cm = 28cm = 1 15x2 + x − 2 15 35 m 30. (b) S = {(1, 1), ....., (1, 6), (2, 1),.....,(2, 6), (3, 1), ...., (3, 6), (4, 1), ....., (4, 6), (5, 1),...., (5, 6), (6, 1),....., (6, 6)} n(S) = 36 7 cm Let E be the event that both dice show different numbers. 7 cm O E {(1, 2), (1, 3),...., (1, 6), (2, 1), (2, 3), (2, 4),...., (2, 6), (3, 1), (3, 2), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 28 cm 3), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (6, 1), 14 cm 7 cm (6, 2), (6, 3), (6, 4), (4, 5)} n(E) = 30 14 cm P (E=) n ((ES=)) 33=60 5 Perimeter of the two sectors of circle n 6 = 1 × 22 × 14cm = 22cm 2 7 31. (a) A (3p, 4) P (5, p) B (–2, 2q) \\ Total perimeter = 134 cm The perimeter of the given figure is 134 cm. Since, P (5, p) is the mid point of AB \\ 5 = 3p - 2 and p = 4 + 2q 22

S-42 Mathematics 36. (b) Product of zeroes = 161 = 7 Also 27a × 27b = 27 × 162. 23 ⇒ ab = 6 ...(ii) ⇒ 2 × product of zeroes = 14p (a – b)2 = (a + b)2 – 4ab ⇒ 2 × 7 = 14p ⇒ a – b = 1 14 Solving (i) and (iii), we get 14 \\ p= ⇒ p = 1 a = 3, b = 2 So numbers are 27 × 3, 27 × 2 i.e., 81, 54 37. (b) Suppose the required ratio is m1 : m2 Then, using the 43. (c) H.C.F. of two co-prime natural number is 1. section formula, we get 44. (d) LCM =HCF –2 = m1 (4)+ m2 (–3) ⇒ two numbers are equal. m1 + m2 45. (c) Clearly, LCM = (LCM of p and p3) ⇒ – 2m1 – 2m2 = 4m1 – 3m2 (LCM of q2 and q) = p3q2 ⇒ m2 = 6 m1 ⇒ m1 :m2 =1:6 46. (c) Area of minor sector OAPB 38. (c) If the sum of 3 prime is even, then one of the numbers = q × p=r 2 90 × 3.14 × (10)2 must be 2. 360 360 Let the second number be x. Then as per the given condition, = 78.5 cm2 x + (x + 36) + 2 = 100 ⇒ x = 31 47. (a) Area of minor segment APB So, the number are 2, 31, 67. Hence largest number is 67. =  pq − sin q cos q  r 2  360 2 2  =  3.14 × 90 − sin 45° cos 45°  (10)2  360  ar (DABC) BC2 39. (a) ar (DDEF) = EF2 = 28.5 cm2 48. (d) Area of the major sector OAQB ⇒ ar (DAB=C)  2.1 2 ar (DDE=F) 9cm2 = Area of circle – Area of minor sector OAPB.  2.8  × = (314 – 78.5)cm2 = 235.5 cm2 49. (a) Area of major segment AQB 40. (d) k ≠ −1 ⇒ k ≠ 3 . = Area of the circle – Area of the minor segment APB 6 −2 = (3.14 × 10 × 10 – 28.5) cm2 41. (c) H.C.F. = 16 and Product = 3072 = 285. 5 cm2 L.C.M. = PHr o.Cd=.uFc.t 3=10672 192 50. (b) Length of arc APB 90 22 = 360 × 2 × 7 ×10 42. (c) H.C.F. of two numbers is 27 = 15.71 cm So let the numbers are 27a and 27b Now 27a + 27b = 135 ...(i) ⇒ a + b = 5