Math Basic

SOLUTION 9 SAMPLE PAPER - 9 SECTION - A 1. (a) 1 = 254 = 0.254 (10)3 Explanation: As 6n always end with 6 and 5n always end with 5, So 6n – 5n always end with 6. (a) parallel 6 – 5 i.e., 1. E xplanation: By Thales theorem. 2. (d) k ≠ 3 7. (b) –1, 7 E xplanation: For unique solution, we have E xplanation: As, O is the centre of circle and A, B are points on its circumference. a1 ¹ b1 a2 b2 \\ OA = OB ⇒ k ¹ −1 or OA2 = OB2 6 −2 ⇒ (2 + 1)2 + (–3y – y)2 = (2 – 5)2 + (–3y – 7)2 ⇒ 9 + 16y2 = 9 + 9y2 + 49 + 42y ⇒ k ¹ 1 ⇒ 7y2 – 42y – 49 = 0 62 ⇒ y2 – 6y – 7 = 0 ⇒ k ¹ 6 = 3 ⇒ y2 – 7y + y – 7 = 0 2 ⇒ (y – 7) (y + 1) = 0 ⇒ k ¹ 3 ⇒ y = –1, 7 3. (a) Linear 8. (a) 5 9 4. (c) 10.5 cm PS = 3 and Explanation: Total number of possible cases = SQ 5 x × y = 9. Explanation: Given: ST || QR, PR = 28 cm. \\ Favourable cases = {(1, 1), (1, 4), (2, 1) (2, 4), (3, 1)} B y using basic proportionality theorem, we get : \\ P(xy < 9) = 5 9 PS = PT 9. (a) 4 SQ TR ⇒ PS = PT Explanation: Here, SQ PR – PT 12x + 17y = 53 and 17x + 12y = 63 ⇒ 3 = PT On adding (i) and (ii), we get ...(i) 5 28 – PT 29x + 29y = 116 ...(ii) x + y = 4 ⇒ 3(28 – PT) = 5PT ⇒ 84 = 5PT + 3PT \\ PT = 84 = 10.5 8 10. (a) 4 H ence, the length of PT is 10.5 cm. Explanation: The coordinates of point P is 5. (c) Terminating (–3, 4) i.e. x = –3 and y = 4 Explanation: Here, rational number is 127 .·. Distance of point P from x-axis is 4 units. 22 × 53 11. (a) 1, 3 and 5 = 127 × 2 23 × 53 Explanation: As they are parabolic in shape, representing a quadratic polynomial. Sample Paper 9 109

12. (a) 4 ⇒ (x – 2) = +6 ⇒ x = 2 + 6 ⇒ x = 8, –4 Explanation: Case 1 : P(getting a non defective pens is) 17. (b) 21 26 ⇒ x = y ...(i) 40 Case II. Number of non-defective pens E xplanation: We know that, in English alphabet, there are 26 letters (5 vowels + 21 consonants). = x + 20 \\ Total number of pens = 60 S o, total number of outcomes = 26 \\ P(getting a non-defective pen) and number of favourable outcomes = 21 ⇒ x + 20 = 4y...(ii) H ence, required probability = 21 60 26 From (i) and (ii), 18. (a) (0, –10) 4x = x + 20 E xplanation: Let the coordinates of P and Q be 40 60 (0, y) and (x, 0) respectively. ⇒ 6x = x + 20 ⇒ x = 4 a (2, –5) is the mid-point of PQ. 13. (c) 1 \\ 2 = x + 0 , –5 = 0+y 2 2 Explanation: Here, ⇒ x = 4, y = –10 a = sec q – tan q and b = sec q + tan q \\ Points are P(0, –10) and Q(4, 0). \\ ab = (sec q – tan q)(sec q + tan q) = sec2 q – tan2 q = 1 19. (d) 60° Explanation: We have, 14. (a) 10 cm 4 – sin2 45° = 3.5 cot x. tan 60° Explanation: Since, DABC ~ DLMN 4–e 1 2 2 Perimeter of TABC AB o Perimeter of TLMN LM \\ = ⇒ cot x 3 = 3.5 ⇒ 60 = AB 4 – 1 48 8 3 2 ⇒ = 3.5 [By property of similar triangles] cot x ⇒ AB = 60 × 8 = 10 cm ⇒ 3.5 x = 3.5 48 3 cot 15. (b) 1000 ⇒ 3 cot x = 1 Explanation: Since, ⇒ cot x = 1 = cot 60° 3 HCF × LCM = Product of two numbers .·. a × b = 5 × 200 ⇒ x = 60° = 1000 20. (b) 4663 16. (b) –4, 8 E xplanation: We have, 520 = 2 × 2 × 2 × 5 × 13 Explanation: 468 = 2 × 2 × 3 × 3 × 13 Numbers exactly divisible by 520 and 468 PQ = 10 = LCM (520, 468) ⇒ PQ2 = 100 = 23 × 32 × 5 × 13 = 4680 ⇒ (x – 2)2 + (5 + 3)2 = 100 \\ Required number = LCM (520, 468) – 17 ⇒ (x – 2)2 = 100 – 64 = 4680 – 17 = 4663. ⇒ (x – 2)2 = 36 SECTION - B 21. (b) 132 Let the second number be x. E xplanation: HCF = 33, LCM = 264. Then, x = HCF × LCM 1st number F irst number is divisible by 2 and gives quotient 33. = 33 × 264 = 132 \\ First number = 2 × 33 = 66 66 110 Mathematics (Basic) Class X

Caution S o, total number of possible outcomes = 5 We observe that x2 < 2 when x takes any one  H ere for finding first number, use the formula, Divisor of the following three values – 1, 0 and 1. = Dividend × Quatient + Remainder 3 22. (d) No solution Hence, P(x2 < 2) = 5 Explanation: As x = 0 is the line parallel to x – 29. (a) 1 axis and x = –2 is the line parallel to y-axis at 6 a distance of 2 units form it. Explanation: Total number of possible outcomes These lines do not meet anywhere. So no solution exist = 6 × 6 = 36 F avourable outcomes (doublet) = 6 i.e. 23. (b) 320 cm {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (5, 5), (6, 6)} E xplanation: Given scale = 1 : 8 \\ P(winning) = 6 = 1 36 6 L et, actual length of car = x cm 30. (a) 2 : 3 T hen 40 = 1 ⇒ x = 320 cm Explanation: Consider 2 triangles as D ABC and x 8 D PQR and DABC ~ D PQR 24. (b) 10 units .·. ar (∆ ABC) = AB2 ar (∆PQR ) PQ2 E xplanation: We have, [by property of similar D] R equired distance, AB = (10 cos θ – 0)2 + (0 – 10 sin θ)2 .·. AB2 = 4 [given] PQ2 9 = 100 cos2 θ + 100 sin2 θ = 100 (cos2 θ + sin2 θ) AB = 2 PQ 3 = 100 ×1 = 10 units ⇒ 25. (a) 2 31. (c) 20 E xplanation: tan (q1 + q2) = 3 [given] Explanation: Smallest composite numbers is 4 ⇒ q1 + q2 = 60° ...(i) and smallest two-digit composite number is 10. 2 Now, 10 = 2 × 5 and 4 = 2 × 2 Also, sin (q1 – q2) = 3 \\ LCM (4, 10) = 2 × 2 × 5 = 20 ⇒ q1 – q2 = 30° ...(ii) 32. (a) 3 On adding Eqn (i) and (ii) we get Explanation: We have cx – y = 2 2.q1 = 90° and 6x – 2y = 4 ...(i) F or infinitely many solutions ...(ii) ⇒ q1 = 45° and q2 = 15° \\ sin 2.q1 + tan 3.q2 = sin 90° + tan 45° c = −1 = 2 6 −2 4 = 1 + 1 = 2 26. (c) (0, 2) c = 1 Þ c=3 6 2 Explanation: The graph of 2x + y = 2 and 2y – x 33. (d) 2 units = 4, shows that the two lines intersect at (0, 2) \\ Solution of given pair of linear equations is Explanation: Here, DE || BC (0, 2). 27. (c) 30° .·. AD = AE BD EC Explanation: Here, 3] ⇒ x = x + 3 3 tan 2q – 3 = 0 3x + 4 3x +19 Þ 3 tan 2q = 3 Þ tan 2q = 3 ⇒ 3x2 + 19x = 3x2 + 4x + 9x + 12 Þ tan 2q = tan 60° [·.· tan 60° = ⇒ 19x = 13x + 12 Þ 2q = 60° ⇒ 6x = 12 Þ q = 30° ⇒ x = 2 units 28. (c) 3 34. (a) (6, –5) 5 E xplanation: Let P(x, y) be the point which E xplanation: Clearly, number x can take any divides the line segment joining the points (8, –9) and (2, 3) in the ratio 1 : 2. one of the five given values. Sample Paper 9 111

\\ Using section formula Þ x + y = 5.5 \\ x = 1×2 + 2×8 Adding eq. (i) and (ii), we get 1+2 2 + 16 18 2x = 6.4 = 3 = 3 =6 Þ x = 3.2 \\ y = 5·5 – 3.2 = 2.3 and y = 1×3 + 2 (–9) 1+2 3 –18 –15 37. (d) 26 units = 3 = 3 = –5 Explanation: The given points are A(–6, 7) and So, the required point is P(6, –5). B(–1, –5). Caution \\ AB = (–6 – (–1))2 + (7 – (–5))2  Be careful while putting the values in the formula, = (–5)2 + (12)2 = 169 = 13 otherwise an error could occur. \\ 2AB = 2 × 13 = 26 units 35. (a) p + q 38. (c) A = 60°, B = 30° Explanation: Here, DABC is right angled at B. A pq Explanation: It is given that sin (A + B) = 1 ⇒ A + B = 90°, as sin 90° = 1 Also, sin (A – B) = 1 2 1 ⇒ A – B = 30°, as sin 30° = 2 CB Therefore, A + B = 90° and A – B = 30° .·. AC = p, AB = q and p – q = 1 Solving, we get, 2A = 120° or A = 60° .·. By pythagoras theorem in DABC, AC2 = AB2 + BC2 ⇒  B = 90° – 60° = 30° BC2 = AC2 – AB2 Therefore, A = 60°, B = 30° = (AC – AB)(AC +AB) 39. (a) Their altitudes have ratio a : b = (p – q) (p + q) Explanation: If the ratio of areas of two similar = p + q triangles is a2 : b2, then their altitudes medians, corresponding sides, perimeters and angle .·. BC = p + q bisectors have a ratio a : b. 36. (d) 2.3 40. (c) 1 Explanation: We have E xplanation: Given, a + b + c = 0 x – y = 0.9  ...(i) For x = 1, p(1) = a + b + c = 0 and 11 = 2 ...(ii) \\ 1 is a zero of p(x) = ax2 + bx + c. \\ x+y x + y = 11 2 41. (c) 21 m SECTION - C Explanation: D istance from A to B In ∆OAQ, = 1 + 2 + 1 OA = 4 m tan 30° = OQ P 11 3 = OQ A B ⇒ OQ = 3 = 1.73 Q 30° O R ∴ Side of equilateral triangle = 4 + 1.73 + 1.73 = 7.46 m Perimeter of equilateral = 7.46 × 3 = 22.38 112 Mathematics (Basic) Class X

N ow, 1 m boundary is left for gate 1.73 × 7.5 × 7.5 ∴ Length of boundary = 22.38 – 1 Cost = 4 – 6 × 3.14 = 21.38 m = 21 m = 24.09 – 18.84 42. (a) 5.25 sq m = 5.22 sq m Explanation: 46. (b) quadratic 25 Area of boundary wall, A = 21 × 100 Explanation: A parabola represents a quadratic polynomial. 21 = 4 = 5.25 sq m 43. (c) ` 525 47. (b) atmost 2 Explanation: Explanation: A quadratic polynomial has C ost of making wall = Area × Rate atmost two zeroes. = 5.25 ×100 = ` 525 48. (a) x2 + x – 2 44. (d) ` 2638 Explanation: A quadratic polynomial is written as Explanation: x2 – (sum of zeroes)x + product of zeroes Area of 6 circles of radius 1 m = 6 × π(1)2 = 6 π sq m So, required polynomial Total cost = 140 × 6 × 3.14 = x2 – (– 1)x + (– 2) = ` 2637.6 = ` 2638 = x2 + x – 2 45. (E) (a) 5.22 sq m Explanation: 49. (a) 5 Area of garden covered with grass = Area of triangle – Area of six circles. Explanation: Let p(x) = (k – 2)x2 – 2x – 5 = 43 × (7.5)2 – 6π S ince (–1) is a zero of the given polynomial. So, p(– 1) = 0 (k – 2) (–1)2 – 2(–1) – 5 = 0 Þ k – 2 + 2 – 5 = 0 Þ k = 5 50. (d) 7 12 Explanation: a + b = – b = 7, ab = c = 12. aa Now 1 + 1 = α+β = 7 αβ αβ 12 Sample Paper 9 113

SOLUTION 10 SAMPLE PAPER - 10 SECTION - A 1. (b) 3 6. (c) terminate after 2 decimal places 2 xEx=plqapnahatisona:dWeceimknaolwextphaantsaionrational number Explanation: G iven, 2x2 – 8x – m which terminates if the prime factorization Its one zero is 5 of q is of the form 2n 5m, where n, m are non- 2 negative integers. Let the other zero be b. \\ 5 + b = −(−8) = 4 Here 129 = 3× 43 = 43 ⇒ 22 60 22 × 3 × 5 22 × 5 b = 4 – 5 = 3 = 43× 5 = 43 × 5 22 22 × 52 102 2. (a) 14, 7 = 215 = 2.15 100 Here, AB = CD \\ 3x – y = 2x + y ...(i) Therefore, the decimal representation of the ⇒3 x – y + 2x – 3 + 2x + y + 2x – 3 = 120 number 129 will terminate after 2 decimal places. 60 ⇒ 9x = 120 + 6 ⇒ x = 126 = 14 7. (c) 3 9 5 Put the value of x in (i), we get Explanation: Here, total outcomes i.e. no. of 14 – 2y = 0 letters in word = 5 ⇒ y = 7 Favourable cases i.e. getting ‘R’ = 3 3. (c) 1 3 \\ P (drawing letter R) = 5 6 Here, total outcomes at A = 6 8. (a) 4 (4π – 3 3) cm2 Favourable outcomes = 4 i.e. 1 3 1 Explanation: We have, R = 4 cm 6 ∴P (getting number 4 in column A) = \\ AB = BC = CA = R 3 = 4 3 cm 4. (c) 4.5 cm  =a R = 2 b and b = 3 a; ` R = aG 3 2 3 E xplanation : In DABC, ∠ADE = ∠ABC \\ Required area = 13 (Area of the circolef – Area DABC) \\ By converse of corresponding angle axiom A DE || BC \\ Using basic proportionality theorem in DABC AD = AE DB CE ⇒ 2 = 3 ⇒ CE = 9 = 4.5 cm O 3 CE 2 BC 5. (c) 5 units Explanation: Radius of circle = Distance between origin O and the point P. = (5 – 0)2 + (0 – 0)2 = 5 units Sample Paper 10 1

\\ Required area = 1 (πR2 – 3 × (Side)22 ∴ DAMN ~ DABC 3 4 ar (∆AMN) AM2 ∴ = = 1 (16π – 3 × ^4 3 h2 2 ar (∆ABC) AB2 3 4 AM = 1 = 1 = 1 ^16π – 12 3h ∴ MB + AM 1 + 2 3 3 = 4 ^4π – 3 3 hcm2 But, AM = 1 3 MB 2 9. (a) –2, –5 ∴ ar( ∆AMN) =  12 = 1  3 9 Explanation : Let ar( ∆ABC ) p(x) = 2x2 + 14x + 20 13. (c) P = 2(x2 + 7x + 10) Explanation: = 2(x2 + 5x + 2x + 10) ∵ (25)2 = (24)2 + (7)2  [by splitting the middle term] ⇒ (QR)2 = (RP)2 + (PQ)2 = 2[x(x + 5) + 2 (x + 5)] ⇒ ∠ P is a right angle = 2(x + 2) (x + 5) [by converse of pythagoras theorem] To determine the zeroes, put p(x) = 0 Q Þ 2(x + 2) (x + 5) = 0 \\ x = –2 and x = –5 Hence, the pair of zeroes of given polynomial is –2 and –5. 10. (c) congruent R P Explanation : If the areas of two similar [Angle opposite to hypotenuse is a right angle] triangles are equal, than their corresponding sides are equal. 14. (c) 0.69 As ar (∆ABC) = AB2 BC2 = AC2 E xplanation: Total number of envelopes in the ar (∆PQR ) PQ2 = QR2 PR2 box = 1000 But ar(DABC) = ar(DPQR) N umber of envelopes containing cash prize = 10 + 100 + 200 = 310 AB = PQ, BC = QR, AC = PR Number of envelopes containing no cash prize 11. (c) – x2 + 2x + 35 = 1000 – 310 = 690 E xplanation: Clearly, other zero = − 35 = – 5 ∴ Required probability = 690 = 0.69 1000 Thus, the zeroes are 7 and –5. 7 15. (b) Infinite solution Hence, the required polynomial is given by 16. (d) DPQR ~ DXZY k(x – 7) (x + 5) Explanation: PQ = PR = QR i.e., k(x2 + 5x – 7x – 35) XZ XY YZ i.e., k(x2 – 2x – 35) ⇒ DPQR ~ DXZY Since, the shape of gate is always in the shape 17. (a) 2 of downward parabola, therefore coefficient of x2 should be negative. Explanation: The smallest number will be 2 . So, putting k = –1, we get the required polynomial Because, 8 × 2 = 16 = 4 , which is rational. as – x2 + 2x + 35. 18. (a) 12. (b) 1 : 9 MN || BC Here, 2 Mathematics (Basic) Class X

Let, AB be a longer tree and it's shadow is BC. \\ AB = (–6 – (–1))2 + (7 – (–5))2 = (–5)2 + (12)2 = 169 = 13 PQ be a smaller tree and it's shadow is QR. \\ 2AB = 2 × 13 = 26 units Now, DABC ~ DPQR 20. (c) 2 \\ AB = BC Explanation: Since, 3 is the least prime factor PQ QR of a. \\ a is an odd number. ⇒ 8= 3 Again, 7 is the least prime factor of b. 4 QR \\ b is also an odd number. \\    ( a + b) is an even number, because sum of ⇒ QR = 8 × 3 = 6m 4 two odds is even. Hence, least factor of (a + b) is 2. 19. (d) 26 units Explanation: The given points are A(–6, 7) and B(–1, –5). SECTION - B 21. (c) 4.5 cm For infinitely many solutions, we have Explanation : In DABC, 8 = 5 = −9 ∠ADE = ∠ABC k 10 −18 \\ By converse of corresponding angle axiom 8 = 1 Þ k = 16 k 2 DE || BC ⇒ \\ Using basic proportionality theorem in DABC 1 AD = AE 27. (b) 2 DB CE E xplanation: Given, sin q = cos q ⇒ 2 = 3 ⇒ CE = 9 = 4.5 cm ⇒ sin θ = 1 ⇒ tan q = 1 3 CE 2 cos θ 22. (b) n ⇒ tan q = tan 45° E xplanation: A polynomial of degree n can ⇒ q = 45° have at most n zeroes. \\ 2 tan2 q + cos2 q – 2 = 2 tan2 45° + cos2 45° –2 23. (a) 6 = 2(1)2 + e 1 2 –2 2 E xplanation: The product of three consecutive o numbers n, n + 1 and n + 2 is divisible by 2 and 3. Therefore, it be divisible by 6. = 2 + 1 –2 2 24. (a) x – 3y = 0 = 1 2 E xplanation: Digit at ten's place be x and digit at one's place be y. 28. (c) 9 \\ As per condition, 20 x = 3y Þ x – 3y = 0 Explanation: Total number of cards is a deck = 52 Number of face cards = 12 25. (b) 55 \\ Number of cards left in the deck = 52 – 12 Explanation: For finding the number of pens = 40 and pencils to be packed, find the HCF of 55 and 110. \\ Total number of possible outcomes = 40 Q 55 = 11 × 5 and 110 = 11 × 5 × 2 Number of black cards and ace in the \\ HCF = 11 × 5 = 55 remaining deck Hence, 55 pens or pencils will be packed in = 20 + 2 = 22 each packet. ∴ Number of favourable outcomes = 40 – 22 = 18 \\ P(neither a black card nor an ace) 26. (b) k = 16 = 18 = 9 40 20 Explanation: We have 8x + 5y – 9 = 0 29. (b) 0.4 and, kx + 10y – 18 = 0 P(E) + P(Ē) = 1 Sample Paper 10 3

⇒ 0.6 + P(E) = 1 Therefore, the power of 2 in the prime ⇒ P(E) = 1 – 0.6 = 0.4 factorization of 792 is 3. 30. (d) 4 : 1 36. (c) − 8 Explanation : Since, DABC and DBDE are two 9 equilateral triangles. Explanation : L et p(x) = 3x2 + 4x + 2 \\ DABC ~ DEBD [By AA similarity criterion] So, sum of zeroes, a + b = − 4 3 ⇒ aarr ((∆∆AEBBDC)) = BC2 = x2 = 4 BD2 x2 1 and product of zeroes, ab = 2 4 3 31. (b) 12 2  − 4  −8 3 3  9 E xplanation: The smallest multiple of 4 is 4 Now, a b2 + ba2 = ab (b + a) = × = and the smallest multiple of 6 is 6. T o find LCM (4, 6), we will first find their prime 37. (a) 4 factors. Q 4 = 22; 6 = 21 × 31 Explanation: Given, polynomial is p(x) = 4x2 + 2x – 5a Therefore, LCM = 22 × 3 = 12 32. (d) 4 Since, 2 is a zero of the polynomial Explanation: Graph of y = p(x) intersects the \\ p(2) = 0 x-axis at four points. So, the number of zeroes for the given graph is 4. Þ 4(2)2 + 2(2) – 5a = 0 [putting x = 2] 33. (b) Infinite Þ 16 + 4 – 5a = 0 Explanation: Here, equations are Þ 5a – 20 = 0 x + 2y – 8 = 0, 2x + 4y – 16 = 0 a = 20 = 4 5 a1 = 1 , Þ a2 2 b1 2 = 1 38. (d) 7 cm b2 4 2 = ; and Explanation: Let C be the circumference and r be the radius of the circle. c1 c2 = −8 = 1 Then, C = 2r + 30 −16 2 ⇒ 2pr = 2r + 30 Since, a1 = b1 = c1 ⇒ 2r (π – 1) = 30 a2 b2 c2 ⇒ 2r c 22 – 1m = 30 \\ The given pair of linear equations has infinitely 7 many solutions. ⇒ 2r c 15 m = 30 7 7 34. (d) 12 cm2 ⇒ r = 7 cm E xplanation: Area of the segment PQR 39. (c) (4, 3) Area to sector OPQRO – Area of DOPR Explanation: We have, x = 4 and y = 3 = θ × πr2 – 1 r2 sin θ 360° 2 \\ Point of intersection is (4, 3). = 30° × 22 × (7)2 – 1 × (7)2 × sin 30° 40. (b) 21 m 360° 7 2 = 1 ×22×7 – 49 = 77 – 49 Explanation: Let the inner radius of the circular 12 4 6 4 path be r and its outer radius be R. = 154 – 147 = 7 cm2 2pR – 2pr = 132 (Given) 12 12 ⇒ 2p(R – r) = 132 35. (c) 3 ⇒ R – r = 132 = 21 Explanation: The prime factors of 792 = 2 × 2 2π × 2 × 3 × 3 × 11 = 23 × 32 × 11 \\ Width of the path = 21 m 4 Mathematics (Basic) Class X

SECTION - C 41. (b) 5 : 3 45. (d) None of these Explanation: Clearly, the coordinates of Explanation: Clearly the coordinates of U and W, C and E are (–4, –3), (1, –3) and (4, –3), G are (–7, 2) and (7, 2) respectively, therefore its respectively. mid-point is (0, 2), which lies on y-axis. Since, WC = 5 units and CE = 3 units A lso, the coordinates of P and L are (–4, 8) and \\ C divides the line joining W and E in the ratio (4, 9) respectively, therefore its mid-point is 5 : 3. c0, 17 m , which also lies on y-axis. 2 A nd, the coordinates of Q and K are (–3, 3) and 42. (c) 2 : 1 (3, 4) resp ectively, therefore, its mid-point is 7 E xplanation: Clearly, the coordinates of P and c0, 2 m which also lies on y-axis. D are (–4, 8) and (5, –4) respectively. 46. (d) 208 Let the x-axis divides the join of P and D at 233 point (x, y) in the ratio of k : 1. Explanation: Applying Pythagoras theorem, k:1 w e get PR2 = PQ2 + QR2 = 1052 + 2082 P D Simplifying we get, (5, –4) (–4, 8) (x, y) PR2 = 11025 + 43264 = 54289 ⇒ PR = 233 m. Then, y = –4k + 8 \\ cos R = Base = QR = 208 k+1 Hypotenuse PR 233 But (x, y) lies on x-axis, therefore y = 0 47. (b) 208 233 ⇒ –4k + 8 = 0 ⇒ 4k = 8 ⇒ k = 2 Perpendicular QR Explanation: sin P = Hypotenuse = PR T hus, the required ratio is 2 : 1. 43. (c) 4 : 7 = 208 233 E xplanation: Clearly, the coordinates of L and U are (4, 9) and (–7, 2) respectively. 48. (b) 233 105 Let the y-axis divides the join of L and U at the Hypotenuse point (x, y) in the ratio k : 1. Explanation: cosec R = Perpendicular k:1 PR 233 PQ 105 = = L (x, y) U 49. (c) –1 (4, 9) (–7, 2) –7k + 4 Explanation: We know, sec2 q – tan2 q = 1 k+1 Then, x = ⇒ sec2 P – tan2 P = 1 or, –(tan2 P – sec2 P) = 1 ⇒ tan2 P – sec2 P = –1 But (x, y) lies on y-axis, therefore x = 0 ⇒ –7k + 4 =0 50. (b) 0 k+1 ⇒ 7k = 4 Explanation: tan P = Perpendicular = QR Base PQ 4 ⇒ k = 7 = 208 105 Base Thus, the required ratio is 4 : 7. and cot R = Perpendicular 44. (b) 5 units = QR = 208 PQ 105 Explanation: Coordinates of K are (3, 4), t herefore its distance from origin Therefore, = 32 + 42 = 9 + 16 = 25 = 5 units tan P – cot R = 208 – 208 =0 105 105 Sample Paper 10 5

SOLUTION 11 SAMPLE PAPER - 11 SECTION - A 1. (d) –1 5. (d) 2 3 Explanation: We have, p(x) = ax2 + bx + c Explanation: Let point Q(–3, k) divides AB in the ratio of p : 1. Put x = – 1, we get p(–1) = a – b + c = (a + c) – b \\ –3 = –2p – 5 p+1 = b – b = 0 (as a + c = b) \\ – 1 is a zero of p(x). ⇒ –3p – 3 = –2p – 5 ⇒ p = 2 2. (d) Unique \\ Ratio is 2 : 1. k = 2×3– 4 = 2 Then, 2+1 3 Here, pair of equations are : x – 2y + 4 = 0 6. (b) 30 seconds and 3x + 4y + 2 = 0 Explanation: To find the time after which the Then; a1 = 1 , b1 = −2 = −1 three lights will glow again, we will find their a2 3 b2 42 LCM by prime factorization. 3 = 31 c1 = 4 = 2 5 = 51 c2 2 1 6 = 21 × 31 \\ a1 ≠ b1 Therefore, LCM = 21 × 31 × 51 = 30 a2 b2 So, that lights of the three strings will glow again together after 30 seconds. 3. (c) 16 7. (c) Sure event 27 E xplanation: Total number of students = 54 E xplanation: The event with probability 1 is called a sure event, which has a surety of Number of girls = 32 occuring. \\ P (getting a girl's name) = 32 = 16 54 27 8. (b) 5.48 cm2 4. (c) 5 E xplanation: Area of shaded portion = Area of E xplanation: Here, equations are: square – 2 × Area of quadrant – Area  of circle 2x + 3y = 5 …(i) 1 22 22 3x + 2y = 10 …(ii) = 8 × 8 – 2 × 4 × 7 × 1.4 × 1.4 – 7 × 4.2 × 4.2 By applying (i) × 2 – (ii) × 3 = 64 – 11 × 0.2 × 1.4 – 22 × 0.6 × 4.2 4x + 6y = 10 = 64 – 3.08 – 55.44 9x + 6y = 30 = 5.48 cm2 –– – 9. (d) – 5x = – 20 ⇒ x = 4 \\ y = – 1 \\ x – y = 4 + 1 = 5 Sample Paper 11 107

\\ x = 2 (0) + 3 (5) = 3 and y = 2 (4) + 3 (0) = 8 Explanation: The graph does not intersect 2+3 2+3 5 x-axis at any point. So, it has no zero. Hence, the required point is c3, 8 m 10. (b) 2a = b 5 Explanation: The given points are O (0, 0), 14. (c) Equally likely A (1, 2) and C (a, b) Explanation: All the events have equal chances Now, 3 points are collinear of occuring. \\ OC = (0 − a)2 + (0 − b)2 = a2 + b2 15. (a) 0 OA = (0 −1)2 + (0 − 2)2 = 1 + 4 = 5 Explanation: If the graph would be parallel to x-axis it would have no zero, as it will not AC = (1 − a)2 + (2 − b)2 intersect x-axis at any point. Since, points are collinear \\ AO + OC = AC 16. (a) 8 units Explanation: Distance PQ = (0 – 0)2 + (–2 – 6)2 5 + a2 + b2 = (1 − a)2 + (2 − b)2 = 8 units On squaring both sides, we get 17. (b) 336 5 + a2 + b2 + 2 5 × a2 + b2 = 1 + a2 – 2a + 4 + b2 – 4b Explanation: Let, the other number be x. Then, by formula, LCM × HCF =54 × x ⇒ 2 5(a2 + b2 ) = – 2(a + 2b) x = 2016 × 9 54 On again squaring, we get = 336 ⇒ 4a2 + b2 – 4ab = 0 18. (b) 7 ⇒ (2a – b) = 0 ⇒ 2a = b Explanation : Here, AB || DC 11. (d) 24 C \\ ABCD forms a trapezium Explanation : Given a and b are the zeroes of D the given polynomial. x–2 x+3 \\ a + b = −  − 5  =5 O 1  +5 x–1 x and ab = k = k A B Since, 1 \\ Also DAOB ~ DCOD (AA similarity) a – b = 1 \\ Þ (a – b)2 = 1 ⇒ OA = OB OC OD Þ (a + b)2 – 4ab = 1 x +5 x –1 Þ (5)2 – 4 × k = 1 x +3 = x –2 Þ 25 – 4k = 1 ⇒ (x + 5)(x – 2) = (x – 1)(x + 3) ⇒ x2 + 3x – 10 = x2 + 2x – 3 Þ 4k = 24 12. (b) 2 ⇒ x = 7 Explanation: Since, A and B lie on the circle 19. (c) 1 : 2 having centre O. ⇒   OA = OB Explanation: Let the required ratio be k : 1. ⇒ (4 – 2)2 + (3 – 3)2 = (x – 2)2 + (5 – 3)2 We know, y-coordinate of any point on x-axis is zero. ⇒ 2 = (x – 2)2 + 4 \\ Using section formula, ⇒ 4 = (x – 2)2 + 4 ⇒ (x – 2)2 = 0 ⇒ x = 2 6kk+–13 = 0 ⇒ 6k – 3 = 0 ⇒ k = 1 2 \\ Required ratio = 1 : 2 13. (c) c3, 8 m 20. (b) HCF 5 Explanation: A largest positive integer that E xplanation: Let P(x, y) be the point which divides given two positive integers completely is called HCF. divides the join of A(5, 0) and B(0, 4) in the ratio 2 : 3 internally. 108 Mathematics (Basic) Class X

SECTION - B 21. (d) 3 2 units Now, sin2 q – cos2 q = 1 – cos2 q – cos2 q Explanation: Coordinates of A (3, 4) and = 1 – 2 cos2 q B(6, 7) 1 2 \\ AB = (6 − 3)2 + (7 − 4)2 = 1 – 2 × e 3 o [Using (i)] = 32 + 32 = 9 + 9 = 1 – 2 = 1 3 3 = 18 = 3 2 27. (a) −4 3 22. (d) None of these Explanation: Let a and b be the zeroes of polynomial kx2 + 4x + 3k. Explanation: a and b are the zeroes of the polynomial x2 + 5x + 8. ab = 8 = 8 According to the question. 1 \\ a + b = ab Hence, none of the given options is correct. −4 = 3k kk 23. (a) terminating Þ Explanation: As 17 = 17 = 17 Þ k = −4 8 23 23 × 5 3 Since, denominator is of the form 2m × 5n, 28. (d) 25 where m and n are integers. So given rational number is a terminating decimal. Explanation: The system of linear equations, representing the given situation, is 24. (c) 15 26 x + y = 50 ...(i) Explanation: Consider the fraction as x . y and x + 2y = 75 ...(ii) A.T.Q., x −1 = 1 On subtracting (i) from (ii), we get y+2 2 y = 25 ⇒ 2x – 2 = y + 2 ⇒ 2x – y = 4 …(i) On substituting y = 25 in (i), we get x − 7 = 1 x = 25 y−2 3 Thus, total number of ` 1 coins is 25. ⇒ 3x – 21 = y – 2 ⇒ 3x – y = 19 …(ii) 29. (b) 12p cm2 On applying (i) - (ii), we get Explanation: Here, AP = PQ = QB = AB = 4 cm 3 x = 15 and y = 26 \\ Required fraction = 15 \\ Area of shaded region = 2× ;2π (4)2 – π × (2)2E 26 2 25. (b) 3 8 = 2 × [8p – 2p] = 12p cm2 26. (b) 1 30. (a) 0.136 3 Explanation: Explanation: We have, We have, 3 tan q = 3 sin q 17 = 17 17 × 23 ⇒ 125 53 23 × 53 3 sin θ = 3 sin q = cos θ ⇒ 3 = 3 136 cos θ = (10)3 = 0.136 ⇒ 3 = cos q 31. (b) 6 \\ 3 Explanation : Quadrilaterals ABCD and PQRS cos q = 1 ...(i) are similar. 3 Sample Paper 11 109

∴ PS = PQ Then, a + b = − (−8k ) = – 2k AD AB 4 (Ratio of corresponding sides are similar) ab = 9 4 ⇒ x = 8 15 20 ⇒ x = 8 ×15 = 6 Since, (a – b)2 = (a + b)2 – 4ab 20 42 = ( – 2k)2 – 4 × 9 4 32. (d) 3 ⇒ 16 = 4k2 – 9 ⇒ Explanation: Here, graph touches or cuts the 4k2 = 25 x-axis at 3 points. So, number of zeroes are 3. 33. (a) (0, 1) ⇒ k2 = 25 4 Explanation: The point at which line cuts the y-axis, x-coordinate is zero. ⇒ k = 25 = ± 5 But 42 ∴ y – 2 × 0 = 1 \\ y = 1 k > 0 37. (d) 1 ∴ Point is (0, 1) k = 5 2 2 34. (a) 9 Explanation: Here, numbers are 84, 63, and 42 So we need to find their HCF. Explanation: Q Total number of numbers o n the wheel = 8 \\ 84  = 22 × 3 × 7 63 = 32 × 7 42 = 2 × 3 × 7 Let E = Event of getting an even number. \\ HCF = 3 × 7 \\ Number of outcomes favourable to E = 4 = 21 Probability that arrow comes at number 8 \\ Number of rows for class fifth = 84 = 4 = P(E) = Outcomes favourable to E1 = 4 = 4 21 Total number of outcomes 8 63 = 3 38. (a) linear polynomial 21 \\ Number of rows for class sixth = Explanation: As it is a straight line \\ Number of rows for class seventh = 42 = 2 39. (b) 1 21 50 \\ Total number of rows = 4 + 3 + 2 = 9 Explanation: Since, total number of tickets = 35. (a) 55 cm2 250 Explanation: Perimeter of a sector of a circle = \\ Total number of outcomes = 250 2 × radius + length (l) of an arc of sector Number of tickets having prize i.e. favourable outcomes = 5 \\ 30 = 2 × 6.4 + l \\ P(getting a prize) = 5 250 ⇒ l = 30 – 12.8 = 17.2 cm Now, area of sector =1 50 = 1 lr = 1 40. (a) 10.5 cm2 2 2 = 1 × 17.2 × 6.4 Explanation: Radius of each quadrant 2 Side of square 7 = 55.04 cm2 = 2 = 2 cm ≅ 55 cm2 (approx) \\ Area of shaded region = Area of square 36. (a) 5  ABCD – 4 × Area of quadrant. 2 = 7 × 7 – 4 × 1 × 22 × 7 × 7 Explanation: Here, equation is 4x2 – 8kx + 9 = 0 4 7 2 2 77 Let, a and b be the zeroes of the given = 49 – 2 = 49 – 38.5 polynomial. = 10.5 cm2 110 Mathematics (Basic) Class X

SECTION - C 41. (d) 2 Scale factor = AC = 1 × 10.8 × 7 AE 2 3 = 37.8 sq. cm = AC = 8 46. (d) 5 AC + CE 8 + 4 13 = 8 = 2 Explanation: To find sin a, we will first find BD. 12 3 It is given that BD – BC = 1 m and CD = 5 m. Therefore, applying Pythagoras theorem in triangle BCD, we get : 42. (d) 12.6 sq. cm BD2 = BC2 + CD2 ⇒ BD2 = (BD – 1)2 + 52 In DABC, using Pythagoras theorem, we have ⇒ BD2 = BD2 – 2BD + 1 + 25 AC2 = BC2 + AB2 AB2 = 64 – 12.96 Solving further, 2BD = 26, or BD = 13 m Therefore, BC = 12 m. In DBCD, sin a = Perpendicular Hypotenuse = 51.04 ⇒ AB = 7.15  7. = CD = 5 BD 13 \\ Area of DABC = 1 × BC × AB 47. (a) 12 2 9 = 1 × 3.6 × 7 Explanation: To find tan b, we will find AE 2 a nd DE (drawn parallel to BC) = 12.6 cm2 A b 43. (d) 10.8 cm Since, DEBC ~ DEAF EC = BC 14 m EA AF DE 4 = 3.6 5m 12 AF ⇒ C aB ⇒ AF = 3.6 × 3 = 10.8 cm We construct DE || BC and we get a rectangle. 44. (d) 37 : 8 sq. cm \\ AB = BE + AE or 14 = 5 + AE Since, with centre E enlargement as done to Therefore, AE = 9 m. DEBC and DEFA. and, DE = BC = 12 m \\ DEBC ~ DEFA Therefore, tan b = Perpendicular Hypotenuse ar(∆EBC) = EC2 ar(∆EFA) AE2 DE = 12 AE 9 ⇒ 4.2 42 48. (b) 197 = 72 ar( ∆EFA ) 122 Explanation: ar (DEFA) = 12 × 12 × 4.2 ⇒ 4×4 sec a = Hypotenuse Base = 3 × 3 × 4.2 = BD = 13 BC 12 = 37.8 sq. cm 45. (d) 37.8 sq. cm and cosec b = Hypotenuse Perpendicular ar of (DBAF) = 1 × AF × AB 2 = AD = 15 DE 12 Sample Paper 11 111

In DAED, by pythagoras theorem \\ sin2a + cos2a = 1 AD2 = AE2 + DE2 = 92 + 122 = 81 + 144 = 225 ⇒ AD 50. (a) 81 144 = 15 m Therefore, sec2a + cosec2 b = d 13 2 + d 15 2 Explanation: cot b = Base 12 12 Perpendicular n n = 169 + 225 = 394 = 197 = AE = 9 144 144 72 DE 12 49. (c) 1 Therefore, c ot2 b = d 9 2 = 81 12 144 Explanation: We know, n sin2 q + cos2 q = 1 112 Mathematics (Basic) Class X