Page 1 QA 1 to 30 30 EU + RC 31 to 60 30 DI + DS + AR 61 to 90 30 T otal 90 T otal questions T otal attempted T otal correct T otal wrong Net Score Time T aken Question number 1 1 2 3 3 2 4 4 5 1 6 2 7 2 8 4 9 4 10 3 11 4 12 1 13 2 14 3 15 2 16 1 17 2 18 4 19 4 20 3 21 3 22 2 23 1 24 1 25 1 26 3 27 4 28 4 29 3 30 2 31 3 32 2 33 2 34 3 35 2 36 4 37 2 38 2 39 1 40 3 41 2 42 3 43 1 44 4 45 4 46 3 47 1 48 1 49 1 50 2 51 3 52 1 53 2 54 3 55 2 56 1 57 3 58 3 59 4 60 4 61 3 62 4 63 3 64 3 65 1 66 2 67 4 68 1 69 2 70 2 71 4 72 3 73 1 74 3 75 1 76 2 77 3 78 2 79 4 80 4 81 4 82 1 83 2 84 3 85 3 86 3 87 3 88 1 89 2 90 4 Page 2 QA 1 to 30 30 EU + RC 31 to 60 30 DI + DS + AR 61 to 90 30 T otal 90 T otal questions T otal attempted T otal correct T otal wrong Net Score Time T aken Question number 1 1 2 3 3 2 4 4 5 1 6 2 7 2 8 4 9 4 10 3 11 4 12 1 13 2 14 3 15 2 16 1 17 2 18 4 19 4 20 3 21 3 22 2 23 1 24 1 25 1 26 3 27 4 28 4 29 3 30 2 31 3 32 2 33 2 34 3 35 2 36 4 37 2 38 2 39 1 40 3 41 2 42 3 43 1 44 4 45 4 46 3 47 1 48 1 49 1 50 2 51 3 52 1 53 2 54 3 55 2 56 1 57 3 58 3 59 4 60 4 61 3 62 4 63 3 64 3 65 1 66 2 67 4 68 1 69 2 70 2 71 4 72 3 73 1 74 3 75 1 76 2 77 3 78 2 79 4 80 4 81 4 82 1 83 2 84 3 85 3 86 3 87 3 88 1 89 2 90 4 1. 1 3333 x 16171819 =+ + + is even number Therefore, 2 divides x. 33 2 2 a b (a b)(a â€“ ab b ) += + + ? a + b always divides a 3 + b 3 . Therefore, 16 3 + 19 3 is divisible by 35. 18 3 + 17 3 is divisible by 35. Thus, x is divisible by 70. Hence, option (1) is the correct choice. 2. 3 AB C D â€“20 20 90 â€“90 â€“10 10 â€“50 50 â€“100 100 110 â€“110 Total +60 30 â€“40 â€“50 D gets emptied first, it gets emptied in 20 minutes. Hence, option (3) is the correct answer. 3. 2 A C D B Shaded area = 2 × (area of sector ADC â€“ area of ? ADC) 2 1 21 â€“ 11 42 p?? =× × × × ?? ?? â€“ 1 2 p = Hence option (2) 4. 4 Let r be the radius of the two circular tracks. ? The rectangle has dimensions 4r × 2r. B rr r r r r A A covers a distance of 2r + 2r + 4r + 4r = 12 r B covers a distance of 2r 2r 4 r p+ p = p Time taken by both of them is same. ? BA BA 4r 12r SS SS 3 pp =? = ?Required percentage BA A S â€“ S 100 S =× â€“ 3 100 4.72%. 3 p =× = 5. 1 Let there be m boys and n girls n 2 n(n â€“ 1) C45 n(n â€“ 1) 90 n 10 2 == ? = ? = m 2 m(m â€“ 1) C 190 190 m(m â€“ 1) 380 m 20 2 =? =? = ? = Number of games between one boy and one girl = 10 20 11 C C 10 20 200 ×=× = Hence option (1) Questions 6 and 7: A B C 2.5 km 5 km Ram: A (9:00 AM) C (9:30) A (11:00 A M ) C (10:30) B (10:00 AM) B (10:00 AM) @ 5km/h Shyam: A (9:45 AM) C (10:00) AM A (10:45 AM ) C (10:30) B (10:15 AM) B (10:15 AM) @ 10 km/h 6. 2 It is clear that Ram and Shyam shall meet each other between C & B, sometime after 10:00 AM. At 10:00 AM they are moving as shown below: Shyam @ 10 km /h C B 2.5 km Ram @ 5 km/h Page 3 QA 1 to 30 30 EU + RC 31 to 60 30 DI + DS + AR 61 to 90 30 T otal 90 T otal questions T otal attempted T otal correct T otal wrong Net Score Time T aken Question number 1 1 2 3 3 2 4 4 5 1 6 2 7 2 8 4 9 4 10 3 11 4 12 1 13 2 14 3 15 2 16 1 17 2 18 4 19 4 20 3 21 3 22 2 23 1 24 1 25 1 26 3 27 4 28 4 29 3 30 2 31 3 32 2 33 2 34 3 35 2 36 4 37 2 38 2 39 1 40 3 41 2 42 3 43 1 44 4 45 4 46 3 47 1 48 1 49 1 50 2 51 3 52 1 53 2 54 3 55 2 56 1 57 3 58 3 59 4 60 4 61 3 62 4 63 3 64 3 65 1 66 2 67 4 68 1 69 2 70 2 71 4 72 3 73 1 74 3 75 1 76 2 77 3 78 2 79 4 80 4 81 4 82 1 83 2 84 3 85 3 86 3 87 3 88 1 89 2 90 4 1. 1 3333 x 16171819 =+ + + is even number Therefore, 2 divides x. 33 2 2 a b (a b)(a â€“ ab b ) += + + ? a + b always divides a 3 + b 3 . Therefore, 16 3 + 19 3 is divisible by 35. 18 3 + 17 3 is divisible by 35. Thus, x is divisible by 70. Hence, option (1) is the correct choice. 2. 3 AB C D â€“20 20 90 â€“90 â€“10 10 â€“50 50 â€“100 100 110 â€“110 Total +60 30 â€“40 â€“50 D gets emptied first, it gets emptied in 20 minutes. Hence, option (3) is the correct answer. 3. 2 A C D B Shaded area = 2 × (area of sector ADC â€“ area of ? ADC) 2 1 21 â€“ 11 42 p?? =× × × × ?? ?? â€“ 1 2 p = Hence option (2) 4. 4 Let r be the radius of the two circular tracks. ? The rectangle has dimensions 4r × 2r. B rr r r r r A A covers a distance of 2r + 2r + 4r + 4r = 12 r B covers a distance of 2r 2r 4 r p+ p = p Time taken by both of them is same. ? BA BA 4r 12r SS SS 3 pp =? = ?Required percentage BA A S â€“ S 100 S =× â€“ 3 100 4.72%. 3 p =× = 5. 1 Let there be m boys and n girls n 2 n(n â€“ 1) C45 n(n â€“ 1) 90 n 10 2 == ? = ? = m 2 m(m â€“ 1) C 190 190 m(m â€“ 1) 380 m 20 2 =? =? = ? = Number of games between one boy and one girl = 10 20 11 C C 10 20 200 ×=× = Hence option (1) Questions 6 and 7: A B C 2.5 km 5 km Ram: A (9:00 AM) C (9:30) A (11:00 A M ) C (10:30) B (10:00 AM) B (10:00 AM) @ 5km/h Shyam: A (9:45 AM) C (10:00) AM A (10:45 AM ) C (10:30) B (10:15 AM) B (10:15 AM) @ 10 km/h 6. 2 It is clear that Ram and Shyam shall meet each other between C & B, sometime after 10:00 AM. At 10:00 AM they are moving as shown below: Shyam @ 10 km /h C B 2.5 km Ram @ 5 km/h Fig. at 10:00 AM From now, time taken to meet = () 2.5 10 5 + × 60 min = 10 minutes So, they meet each other at 10:10 AM. 7. 2 It is clear from the diagram that at 10:30; Shyam overtakes Ram. Alternate: At 10:15 the situation is as show: A B C D 1.25 km Shyam at B m oving @ 10 km /h Ram at D moving @ 5 km/h Time taken for Shyam to overtake Ram = () 1.25 10 5 - × 60 = 15 min. ? Shyam overtakes Ram at 10:30 AM. 8. 4 R = () () 65 65 65 65 65 64 64 64 64 64 1 30 30 1 30 30 1 30 30 30 1 1 30 30 1 30 ?? -- ?? -- ?? = +- ?? +- ?? ?? ? R = 65 65 64 64 1 11 36 30 30 1 11 30 ?? ?? ?? -- ?? ?? ?? ?? ?? ?? +- ?? ?? ?? ?? ? R = 30 () () 65 64 10.96 10.96 ?? - ?? ?? +?? ?? In () () 65 64 10.96 , 10.96 - + numerator is only slightly less then 1. and denominator is only slightly more than 1. Hence, R is certainly greater than 1. 9. 4 Case I:Chords on same side of the centre. A C 12 D 4 B O 12 20 20 16 OB 2 = OA 2 â€“ AB 2 = 20 2 â€“ 16 2 = 144 OB = 12 OD 2 = 20 2 â€“ 12 2 = 400 â€“ 144 = 256 OD = 16 BD = 4 cm Case II: Chords on opposite side of the centre. A B C D P Q O AB = 32 cm CD = 24 cm OP = 22 AO AP - = () ( ) 22 20 16 - OP = 12 cm & OQ = () () 22 OC CQ - = () ( ) 22 20 12 - OQ = 16 cm Distance = PQ = 12 + 16 = 28 cm. 10. 3 y 2 = x 2 2x 2 â€“ 2kx + k 2 â€“ 1 = 0 D = 0 ? 4k 2 = 8k 2 â€“ 8 ? 4k 2 = 8 ?k 2 = 2 ? k = ± 2 . k = + 2 gives the equation = 2x 2 â€“ 22x +1 = 0; Its root is b1 , 2a 2 - =+ k = â€“ 2 gives the equation 2x 2 + 22x + 1 = 0. Its root is â€“ 1 2 this root is â€“ve, will reject k = â€“ 2 . Only answer is k = + 2 . Alternate: Graph based. x 2 â€“ y 2 = 0 & (x â€“ k) 2 + y 2 = 1 are plotted below. We are solving for a unique positive x. x 2 â€“ y 2 = 0 is a pair of straight lines y = x & y = â€“x (x â€“ k) 2 + y 2 = 1 is a circle with center (k, 0) & radius 1. Page 4 QA 1 to 30 30 EU + RC 31 to 60 30 DI + DS + AR 61 to 90 30 T otal 90 T otal questions T otal attempted T otal correct T otal wrong Net Score Time T aken Question number 1 1 2 3 3 2 4 4 5 1 6 2 7 2 8 4 9 4 10 3 11 4 12 1 13 2 14 3 15 2 16 1 17 2 18 4 19 4 20 3 21 3 22 2 23 1 24 1 25 1 26 3 27 4 28 4 29 3 30 2 31 3 32 2 33 2 34 3 35 2 36 4 37 2 38 2 39 1 40 3 41 2 42 3 43 1 44 4 45 4 46 3 47 1 48 1 49 1 50 2 51 3 52 1 53 2 54 3 55 2 56 1 57 3 58 3 59 4 60 4 61 3 62 4 63 3 64 3 65 1 66 2 67 4 68 1 69 2 70 2 71 4 72 3 73 1 74 3 75 1 76 2 77 3 78 2 79 4 80 4 81 4 82 1 83 2 84 3 85 3 86 3 87 3 88 1 89 2 90 4 1. 1 3333 x 16171819 =+ + + is even number Therefore, 2 divides x. 33 2 2 a b (a b)(a â€“ ab b ) += + + ? a + b always divides a 3 + b 3 . Therefore, 16 3 + 19 3 is divisible by 35. 18 3 + 17 3 is divisible by 35. Thus, x is divisible by 70. Hence, option (1) is the correct choice. 2. 3 AB C D â€“20 20 90 â€“90 â€“10 10 â€“50 50 â€“100 100 110 â€“110 Total +60 30 â€“40 â€“50 D gets emptied first, it gets emptied in 20 minutes. Hence, option (3) is the correct answer. 3. 2 A C D B Shaded area = 2 × (area of sector ADC â€“ area of ? ADC) 2 1 21 â€“ 11 42 p?? =× × × × ?? ?? â€“ 1 2 p = Hence option (2) 4. 4 Let r be the radius of the two circular tracks. ? The rectangle has dimensions 4r × 2r. B rr r r r r A A covers a distance of 2r + 2r + 4r + 4r = 12 r B covers a distance of 2r 2r 4 r p+ p = p Time taken by both of them is same. ? BA BA 4r 12r SS SS 3 pp =? = ?Required percentage BA A S â€“ S 100 S =× â€“ 3 100 4.72%. 3 p =× = 5. 1 Let there be m boys and n girls n 2 n(n â€“ 1) C45 n(n â€“ 1) 90 n 10 2 == ? = ? = m 2 m(m â€“ 1) C 190 190 m(m â€“ 1) 380 m 20 2 =? =? = ? = Number of games between one boy and one girl = 10 20 11 C C 10 20 200 ×=× = Hence option (1) Questions 6 and 7: A B C 2.5 km 5 km Ram: A (9:00 AM) C (9:30) A (11:00 A M ) C (10:30) B (10:00 AM) B (10:00 AM) @ 5km/h Shyam: A (9:45 AM) C (10:00) AM A (10:45 AM ) C (10:30) B (10:15 AM) B (10:15 AM) @ 10 km/h 6. 2 It is clear that Ram and Shyam shall meet each other between C & B, sometime after 10:00 AM. At 10:00 AM they are moving as shown below: Shyam @ 10 km /h C B 2.5 km Ram @ 5 km/h Fig. at 10:00 AM From now, time taken to meet = () 2.5 10 5 + × 60 min = 10 minutes So, they meet each other at 10:10 AM. 7. 2 It is clear from the diagram that at 10:30; Shyam overtakes Ram. Alternate: At 10:15 the situation is as show: A B C D 1.25 km Shyam at B m oving @ 10 km /h Ram at D moving @ 5 km/h Time taken for Shyam to overtake Ram = () 1.25 10 5 - × 60 = 15 min. ? Shyam overtakes Ram at 10:30 AM. 8. 4 R = () () 65 65 65 65 65 64 64 64 64 64 1 30 30 1 30 30 1 30 30 30 1 1 30 30 1 30 ?? -- ?? -- ?? = +- ?? +- ?? ?? ? R = 65 65 64 64 1 11 36 30 30 1 11 30 ?? ?? ?? -- ?? ?? ?? ?? ?? ?? +- ?? ?? ?? ?? ? R = 30 () () 65 64 10.96 10.96 ?? - ?? ?? +?? ?? In () () 65 64 10.96 , 10.96 - + numerator is only slightly less then 1. and denominator is only slightly more than 1. Hence, R is certainly greater than 1. 9. 4 Case I:Chords on same side of the centre. A C 12 D 4 B O 12 20 20 16 OB 2 = OA 2 â€“ AB 2 = 20 2 â€“ 16 2 = 144 OB = 12 OD 2 = 20 2 â€“ 12 2 = 400 â€“ 144 = 256 OD = 16 BD = 4 cm Case II: Chords on opposite side of the centre. A B C D P Q O AB = 32 cm CD = 24 cm OP = 22 AO AP - = () ( ) 22 20 16 - OP = 12 cm & OQ = () () 22 OC CQ - = () ( ) 22 20 12 - OQ = 16 cm Distance = PQ = 12 + 16 = 28 cm. 10. 3 y 2 = x 2 2x 2 â€“ 2kx + k 2 â€“ 1 = 0 D = 0 ? 4k 2 = 8k 2 â€“ 8 ? 4k 2 = 8 ?k 2 = 2 ? k = ± 2 . k = + 2 gives the equation = 2x 2 â€“ 22x +1 = 0; Its root is b1 , 2a 2 - =+ k = â€“ 2 gives the equation 2x 2 + 22x + 1 = 0. Its root is â€“ 1 2 this root is â€“ve, will reject k = â€“ 2 . Only answer is k = + 2 . Alternate: Graph based. x 2 â€“ y 2 = 0 & (x â€“ k) 2 + y 2 = 1 are plotted below. We are solving for a unique positive x. x 2 â€“ y 2 = 0 is a pair of straight lines y = x & y = â€“x (x â€“ k) 2 + y 2 = 1 is a circle with center (k, 0) & radius 1. y x y = â€“x y = +x (2, 0) (1) k = 2; clearly, no solution (2) k = 0 y x y = â€“x y = +x x = +a x = â€“a x = a, â€“a are its two solutions. â€“ rejected. (3) k = 2 + unique value of x & a positive one as shown. y x y = â€“x y = +x 1 ( 2, 0) x =ß 1 (4) k = â€“ 2 , also gives the unique value of x but it is negative one. y x 1 (â€“ 2, 0) x = â€“ß 2 11. 4 If p = 1! = 1, then p + 2 = 3 when divided by 2! will give a remainder of 1. If p = 1! + 2 × 2! = 5, then p + 2 = 7 when divided by 3! will give a remainder of 1. Hence, p = 1! + (2 × 2!) + (3 × 3!) + â€¦ + (10 × 10!) when divided by 11! leaves a remainder 1. Alternative method: P = 1 + 2.2! + 3.3!+ â€¦.10.10! = (2 â€“1)1! + ( 3 â€“ 1)2! + (4 â€“ 1)3! + â€¦.(11 â€“ 1)10! =2! â€“ 1! + 3! â€“ 2! + â€¦.. 11! â€“10! = 1 + 11! Hence, the remainder is 1. 12. 1 (0, 41) (41, 0) A B x y (0, 0) equation of line = x + y = 41. If the (x, y) co-ordinates of the points are integer, their sum shall also be integers so that x + y = k (k, a variable) as we have to exclude points lying on the boundary of triangle; k can take all values from 1 to 40 only. k = 0 is also rejected as at k = 0 will give the point A; which canâ€™t be taken. Now, x + y = k, (k = 1, 2, 3, ... 40) with k = 40; x + y = 40; taking integral solutions. We get points (1, 39), (2, 38); (3, 37) ...(39, 1) i.e. 39 points x + y = 40 will be satisfied by 39 points. Similarly, x + y = 39 is satisfied by 38 points. x + y = 38 by 37 points. x + y = 3 by 2 points. x + y = 2 is satisfied by 1 point. x + y = 1 is not satisfied by any point. So, the total no. of all such points is: 39 + 38 + 37 + 36 + ... 3 + 2 + 1 = 39 40 2 × = 780 points. 13. 2 Let A = abc. Then, B = cba. Given, B > A ? c > a As B â€“ A = (100c + 10b + a) â€“ (100a + 10b + 1) ? B â€“ A = 100 (c â€“ a) + (a â€“ c) ? B â€“ A = 99 (c â€“ a). Also, (B â€“ A) is divisible by 7. But, 99 is not divisible by 7 (no factor like 7 or 7 2 ). Therefore, (c â€“ a) must be divisible by 7 {i.e., (c â€“ a) must be 7, 7 2 , etc.}. Since c and a are single digits, value of (c â€“ a) must be 7. The possible values of (c, a) {with c > a} are (9, 2) and (8, 1). Thus, we can write A as: A : abc = 1b8 or 2b9 As b can take values from 0 to 9, the smallest & largest possible value of A are: A min = 108 & A max = 299 Only option (b) satisfies this. Hence, (2) is the correct option. 14. 3 a 1 = 1, a n+1 â€“ 3a n + 2 = 4n a n+1 = 3a n + 4n â€“ 2 when n = 2 then a 2 = 3 + 4 â€“ 2 = 5 when n = 3 then a 3 = 3 × 5 + 4 × 2 â€“ 2 = 21 from the options, we get an idea that a n can be expressed in a combination of some power of 3 & some multiple of 100. (1) 3 99 â€“ 200; tells us that a n could be: 3 nâ€“1 â€“ 2 × n; but it does not fit a 1 or a 2 or a 3 Page 5 QA 1 to 30 30 EU + RC 31 to 60 30 DI + DS + AR 61 to 90 30 T otal 90 T otal questions T otal attempted T otal correct T otal wrong Net Score Time T aken Question number 1 1 2 3 3 2 4 4 5 1 6 2 7 2 8 4 9 4 10 3 11 4 12 1 13 2 14 3 15 2 16 1 17 2 18 4 19 4 20 3 21 3 22 2 23 1 24 1 25 1 26 3 27 4 28 4 29 3 30 2 31 3 32 2 33 2 34 3 35 2 36 4 37 2 38 2 39 1 40 3 41 2 42 3 43 1 44 4 45 4 46 3 47 1 48 1 49 1 50 2 51 3 52 1 53 2 54 3 55 2 56 1 57 3 58 3 59 4 60 4 61 3 62 4 63 3 64 3 65 1 66 2 67 4 68 1 69 2 70 2 71 4 72 3 73 1 74 3 75 1 76 2 77 3 78 2 79 4 80 4 81 4 82 1 83 2 84 3 85 3 86 3 87 3 88 1 89 2 90 4 1. 1 3333 x 16171819 =+ + + is even number Therefore, 2 divides x. 33 2 2 a b (a b)(a â€“ ab b ) += + + ? a + b always divides a 3 + b 3 . Therefore, 16 3 + 19 3 is divisible by 35. 18 3 + 17 3 is divisible by 35. Thus, x is divisible by 70. Hence, option (1) is the correct choice. 2. 3 AB C D â€“20 20 90 â€“90 â€“10 10 â€“50 50 â€“100 100 110 â€“110 Total +60 30 â€“40 â€“50 D gets emptied first, it gets emptied in 20 minutes. Hence, option (3) is the correct answer. 3. 2 A C D B Shaded area = 2 × (area of sector ADC â€“ area of ? ADC) 2 1 21 â€“ 11 42 p?? =× × × × ?? ?? â€“ 1 2 p = Hence option (2) 4. 4 Let r be the radius of the two circular tracks. ? The rectangle has dimensions 4r × 2r. B rr r r r r A A covers a distance of 2r + 2r + 4r + 4r = 12 r B covers a distance of 2r 2r 4 r p+ p = p Time taken by both of them is same. ? BA BA 4r 12r SS SS 3 pp =? = ?Required percentage BA A S â€“ S 100 S =× â€“ 3 100 4.72%. 3 p =× = 5. 1 Let there be m boys and n girls n 2 n(n â€“ 1) C45 n(n â€“ 1) 90 n 10 2 == ? = ? = m 2 m(m â€“ 1) C 190 190 m(m â€“ 1) 380 m 20 2 =? =? = ? = Number of games between one boy and one girl = 10 20 11 C C 10 20 200 ×=× = Hence option (1) Questions 6 and 7: A B C 2.5 km 5 km Ram: A (9:00 AM) C (9:30) A (11:00 A M ) C (10:30) B (10:00 AM) B (10:00 AM) @ 5km/h Shyam: A (9:45 AM) C (10:00) AM A (10:45 AM ) C (10:30) B (10:15 AM) B (10:15 AM) @ 10 km/h 6. 2 It is clear that Ram and Shyam shall meet each other between C & B, sometime after 10:00 AM. At 10:00 AM they are moving as shown below: Shyam @ 10 km /h C B 2.5 km Ram @ 5 km/h Fig. at 10:00 AM From now, time taken to meet = () 2.5 10 5 + × 60 min = 10 minutes So, they meet each other at 10:10 AM. 7. 2 It is clear from the diagram that at 10:30; Shyam overtakes Ram. Alternate: At 10:15 the situation is as show: A B C D 1.25 km Shyam at B m oving @ 10 km /h Ram at D moving @ 5 km/h Time taken for Shyam to overtake Ram = () 1.25 10 5 - × 60 = 15 min. ? Shyam overtakes Ram at 10:30 AM. 8. 4 R = () () 65 65 65 65 65 64 64 64 64 64 1 30 30 1 30 30 1 30 30 30 1 1 30 30 1 30 ?? -- ?? -- ?? = +- ?? +- ?? ?? ? R = 65 65 64 64 1 11 36 30 30 1 11 30 ?? ?? ?? -- ?? ?? ?? ?? ?? ?? +- ?? ?? ?? ?? ? R = 30 () () 65 64 10.96 10.96 ?? - ?? ?? +?? ?? In () () 65 64 10.96 , 10.96 - + numerator is only slightly less then 1. and denominator is only slightly more than 1. Hence, R is certainly greater than 1. 9. 4 Case I:Chords on same side of the centre. A C 12 D 4 B O 12 20 20 16 OB 2 = OA 2 â€“ AB 2 = 20 2 â€“ 16 2 = 144 OB = 12 OD 2 = 20 2 â€“ 12 2 = 400 â€“ 144 = 256 OD = 16 BD = 4 cm Case II: Chords on opposite side of the centre. A B C D P Q O AB = 32 cm CD = 24 cm OP = 22 AO AP - = () ( ) 22 20 16 - OP = 12 cm & OQ = () () 22 OC CQ - = () ( ) 22 20 12 - OQ = 16 cm Distance = PQ = 12 + 16 = 28 cm. 10. 3 y 2 = x 2 2x 2 â€“ 2kx + k 2 â€“ 1 = 0 D = 0 ? 4k 2 = 8k 2 â€“ 8 ? 4k 2 = 8 ?k 2 = 2 ? k = ± 2 . k = + 2 gives the equation = 2x 2 â€“ 22x +1 = 0; Its root is b1 , 2a 2 - =+ k = â€“ 2 gives the equation 2x 2 + 22x + 1 = 0. Its root is â€“ 1 2 this root is â€“ve, will reject k = â€“ 2 . Only answer is k = + 2 . Alternate: Graph based. x 2 â€“ y 2 = 0 & (x â€“ k) 2 + y 2 = 1 are plotted below. We are solving for a unique positive x. x 2 â€“ y 2 = 0 is a pair of straight lines y = x & y = â€“x (x â€“ k) 2 + y 2 = 1 is a circle with center (k, 0) & radius 1. y x y = â€“x y = +x (2, 0) (1) k = 2; clearly, no solution (2) k = 0 y x y = â€“x y = +x x = +a x = â€“a x = a, â€“a are its two solutions. â€“ rejected. (3) k = 2 + unique value of x & a positive one as shown. y x y = â€“x y = +x 1 ( 2, 0) x =ß 1 (4) k = â€“ 2 , also gives the unique value of x but it is negative one. y x 1 (â€“ 2, 0) x = â€“ß 2 11. 4 If p = 1! = 1, then p + 2 = 3 when divided by 2! will give a remainder of 1. If p = 1! + 2 × 2! = 5, then p + 2 = 7 when divided by 3! will give a remainder of 1. Hence, p = 1! + (2 × 2!) + (3 × 3!) + â€¦ + (10 × 10!) when divided by 11! leaves a remainder 1. Alternative method: P = 1 + 2.2! + 3.3!+ â€¦.10.10! = (2 â€“1)1! + ( 3 â€“ 1)2! + (4 â€“ 1)3! + â€¦.(11 â€“ 1)10! =2! â€“ 1! + 3! â€“ 2! + â€¦.. 11! â€“10! = 1 + 11! Hence, the remainder is 1. 12. 1 (0, 41) (41, 0) A B x y (0, 0) equation of line = x + y = 41. If the (x, y) co-ordinates of the points are integer, their sum shall also be integers so that x + y = k (k, a variable) as we have to exclude points lying on the boundary of triangle; k can take all values from 1 to 40 only. k = 0 is also rejected as at k = 0 will give the point A; which canâ€™t be taken. Now, x + y = k, (k = 1, 2, 3, ... 40) with k = 40; x + y = 40; taking integral solutions. We get points (1, 39), (2, 38); (3, 37) ...(39, 1) i.e. 39 points x + y = 40 will be satisfied by 39 points. Similarly, x + y = 39 is satisfied by 38 points. x + y = 38 by 37 points. x + y = 3 by 2 points. x + y = 2 is satisfied by 1 point. x + y = 1 is not satisfied by any point. So, the total no. of all such points is: 39 + 38 + 37 + 36 + ... 3 + 2 + 1 = 39 40 2 × = 780 points. 13. 2 Let A = abc. Then, B = cba. Given, B > A ? c > a As B â€“ A = (100c + 10b + a) â€“ (100a + 10b + 1) ? B â€“ A = 100 (c â€“ a) + (a â€“ c) ? B â€“ A = 99 (c â€“ a). Also, (B â€“ A) is divisible by 7. But, 99 is not divisible by 7 (no factor like 7 or 7 2 ). Therefore, (c â€“ a) must be divisible by 7 {i.e., (c â€“ a) must be 7, 7 2 , etc.}. Since c and a are single digits, value of (c â€“ a) must be 7. The possible values of (c, a) {with c > a} are (9, 2) and (8, 1). Thus, we can write A as: A : abc = 1b8 or 2b9 As b can take values from 0 to 9, the smallest & largest possible value of A are: A min = 108 & A max = 299 Only option (b) satisfies this. Hence, (2) is the correct option. 14. 3 a 1 = 1, a n+1 â€“ 3a n + 2 = 4n a n+1 = 3a n + 4n â€“ 2 when n = 2 then a 2 = 3 + 4 â€“ 2 = 5 when n = 3 then a 3 = 3 × 5 + 4 × 2 â€“ 2 = 21 from the options, we get an idea that a n can be expressed in a combination of some power of 3 & some multiple of 100. (1) 3 99 â€“ 200; tells us that a n could be: 3 nâ€“1 â€“ 2 × n; but it does not fit a 1 or a 2 or a 3 (2) 3 99 + 200; tells us that a n could be: 3 nâ€“1 + 2 × n; again, not valid for a 1 , a 2 etc. (3) 3 100 â€“ 200; tells 3 n â€“ 2n: valid for all a 1 , a 2 , a 3 . (4) 3 100 + 200; tells 3 n + 2n: again not valid. so, (3) is the correct answer. 15. 2 right m ost digit (RM D) left m ost digit (LM D) odd positions can be counted in 2 ways. (i) Counting from the LMD-end: odd positions We have 1, 2, 3, 4 & 5 to be filled in these blocks. Odd nos. (1, 3, 5) to be filled in at odd positions. Other places are to be filled by even nos. (2 or 4) Letâ€™s count, how many such nos. are there with 2 at the unitâ€™s digit odd 2 Odd nos. can be filled in 3 P 2 = 6 way. The remaining two places are to be filled by 2 nos. (one odd no. left out of 1, 3, 5 & one even i.e. 4) in = 2 ways. So, there are 6 × 2 = 12 number with 2 at the rightmost place. Similarly; there are 12 such nos. with 4 at the rightmost digits. The sum of rightmost digits in all such number = 12(2 + 4) = 72 (ii) Now counting from the RMD-end. Letâ€™s place 1 at the units place and check, how many nos. are possible with (1, 3) at the odd positions: 5 & (2 or 4) 3 1 (4 or 2) No. of such cases = 2 × 2 = 4 ways. 5 & (2 or 4) 1 (4 or 2) 3 Here again no. of ways = 2 × 2 = 4 ways So, there are 4 + 4 = 8 nos, in which (1, 3) are at odd positions. Similarly there are 8 nos. in which (1, 5) are at odd positions. So, in all there are 16 nos. where 1 is at unitâ€™s place. Similarly there are 16 nos. with 3 at unitâ€™s place and 16 more with 5 at unitâ€™s place. Summing up all the odd unitâ€™s digits = 16(1 + 3 + 5) = 144 From (i) and (ii) we can now, sum up all (even or odd) nos. at units place = 72 +144 = 216 Hence answer is (2) 16. 1 ((30) 4 ) 680 = (8100) 680 . Hence, the right most non-zero digit is 1. 17. 2 1cm 60° 60° 60° 90° A B C D P Q 90° Drawn figure since it have not to be within distance of 1 m so it will go along APQD, which is the path of minimum distance. AP = 90 21 360 2 p ×p× = Also AP = QD = 2 p So the minimum distance = AP + PQ + QD = 11 22 pp ++ = +p 18. 4 P = xy xy log log yx ?? ?? + ?? ?? ?? ?? = xx yy log x â€“logy logy â€“ log x + = xy 2 â€“ log y â€“ log x Let x tlog y = 2 11 P2â€“â€“ t â€“ t t t ?? == - ?? ?? which can never be positive. Out of given options, it canâ€™t assume a value of +1. 19. 4 It is given that 10 < n < 1000. Let n be a two digit number. Then, n = 10a + b ? p n = ab, s n = a + b Then, ab + a + b = 10a + b ? ab = 9a ? b = 9 ? There are 9 such numbers 19, 29, 33, â€¦, 99. Now, let n be a three digit number. ? n = 100a + 10b + c ? p n = abc, s n = a + b + cRead More

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