Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case. A. 4 B. 7 C. 9 D. 13 Answer: Option A Explanation: Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43) = H.C.F. of 48, 92 and 140 = 4. The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is: A. 276 B. 299 C. 322 D. 345 Answer: Option C Explanation: Clearly, the numbers are (23 x 13) and (23 x 14). Larger number = (23 x 14) = 322. Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ? A. 4 B. 10 C. 15 D. 16 Answer: Option D Explanation: L.C.M. of 2, 4, 6, 8, 10, 12 is 120. So, the bells will toll together after every 120 seconds(2 minutes). In 30 minutes, they will toll together= 30/2+ 1 = 16 times. Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is: A. 4 B. 5 C. 6 D. 8 Answer: Option A Explanation: N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305) = H.C.F. of 3360, 2240 and 5600 = 1120. Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4 The greatest number of four digits which is divisible by 15, 25, 40 and 75 is: A. 9000 B. 9400 C. 9600 D. 9800 Answer: Option C Explanation: Greatest number of 4-digits is 9999. L.C.M. of 15, 25, 40 and 75 is 600. On dividing 9999 by 600, the remainder is 399. Required number (9999 - 399) = 9600. The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is: A. 101 B. 107 C. 111 D. 185 Answer: Option C Explanation: Let the numbers be 37a and 37b. Then, 37a x 37b = 4107 ab = 3. Now, co-primes with product 3 are (1, 3). So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111). Greater number = 111. Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is: A. 40 B. 80 C. 120 D. 200 Answer: Option A Explanation: Let the numbers be 3x, 4x and 5x. Then, their L.C.M. = 60x. So, 60x = 2400 or x = 40. The numbers are (3 x 40), (4 x 40) and (5 x 40). Hence, required H.C.F. = 40. The G.C.D. of 1.08, 0.36 and 0.9 is: A. 0.03 B. 0.9 C. 0.18 D. 0.108 Answer: Option C Explanation: Given numbers are 1.08, 0.36 and 0.90. H.C.F. of 108, 36 and 90 is 18, H.C.F. of given numbers = 0.18. The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is: A. 1 B. 2 C. 3 D. 4 Answer: Option B Explanation: Let the numbers 13a and 13b. Then, 13a x 13b = 2028 ab = 12. Now, the co-primes with product 12 are (1, 12) and (3, 4). [Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ] So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4). Clearly, there are 2 such pairs. The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is: A. 74 B. 94 C. 184 D. 364 Answer: Option D Explanation: L.C.M. of 6, 9, 15 and 18 is 90. Let required number be 90k + 4, which is multiple of 7. Least value of k for which (90k + 4) is divisible by 7 is k = 4. Required number = (90 x 4) + 4 = 364. Find the lowest common multiple of 24, 36 and 40. A. 120 B. 240 C. 360 D. 480 Answer: Option C Explanation: 2 | 24 - 36 - 40 -------------------- 2 | 12 - 18 - 20 -------------------- 2 | 6 - 9 - 10 ------------------- 3 | 3 - 9 - 5 ------------------- | 1 - 3 - 5 L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 = 360. The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is: A. 3 B. 13 C. 23 D. 33 Answer: Option C Explanation: L.C.M. of 5, 6, 4 and 3 = 60. On dividing 2497 by 60, the remainder is 37. Number to be added = (60 - 37) = 23. The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is: A. 1677 B. 1683 C. 2523 D. 3363 Answer: Option B Explanation: L.C.M. of 5, 6, 7, 8 = 840. Required number is of the form 840k + 3 Least value of k for which (840k + 3) is divisible by 9 is k = 2. Required number = (840 x 2 + 3) = 1683. A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ? A. 26 minutes and 18 seconds B. 42 minutes and 36 seconds C. 45 minutes D. 46 minutes and 12 seconds Answer: Option D Explanation: L.C.M. of 252, 308 and 198 = 2772. So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec. The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is: A. 279 B. 283 C. 308 D. 318 Answer: Option C Explanation: Other number =11 x 7700/275= 308. What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ? A. 196 B. 630 C. 1260 D. 2520 Answer: Option B Explanation: L.C.M. of 12, 18, 21 30 2 | 12 - 18 - 21 - 30 ---------------------------- = 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3 | 6 - 9 - 21 - 15 ---------------------------- Required number = (1260 ÷ 2) | 2 - 3 - 7 - 5 = 630. The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is: A. 12 B. 16 C. 24 D. 48 Answer: Option D Explanation: Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4. So, the numbers 12 and 16. L.C.M. of 12 and 16 = 48. The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is: A. 1008 B. 1015 C. 1022 D. 1032 Answer: Option B Explanation: Required number = (L.C.M. of 12,16, 18, 21, 28) + 7 = 1008 + 7 = 1015 252 can be expressed as a product of primes as: A. 2 x 2 x 3 x 3 x 7 B. 2 x 2 x 2 x 3 x 7 C. 3 x 3 x 3 x 3 x 7 D. 2 x 3 x 3 x 3 x 7 Answer: Option A Explanation: Clearly, 252 = 2 x 2 x 3 x 3 x 7. The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is: A. 15 cm B. 25 cm C. 35 cm D. 42 cm Answer: Option C Explanation: Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm. Find the highest common factor of 36 and 84. A. 4 B. 6 C. 12 D. 18 Answer: Option C Explanation: 36 = 22 x 32 84 = 22 x 3 x 7 H.C.F. = 22 x 3 = 12. The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is: A. 504 B. 536 C. 544 D. 548 Answer: Option D Explanation: Required number = (L.C.M. of 12, 15, 20, 54) + 8 = 540 + 8 = 548. The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is: A. 123 B. 127 C. 235 D. 305 Answer: Option B Explanation: Required number = H.C.F. of (1657 - 6) and (2037 - 5) = H.C.F. of 1651 and 2032 = 127. Which of the following has the most number of divisors? A. 99 B. 101 C. 176 D. 182 Answer: Option C Explanation: 99 = 1 x 3 x 3 x 11 101 = 1 x 101 176 = 1 x 2 x 2 x 2 x 2 x 11 182 = 1 x 2 x 7 x 13 So, divisors of 99 are 1, 3, 9, 11, 33, .99 Divisors of 101 are 1 and 101 Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176 Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182. Hence, 176 has the most number of divisors. The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is: A. 28 B. 32 C. 40 D. 64 Answer: Option C Explanation: Let the numbers be 2x and 3x. Then, their L.C.M. = 6x. So, 6x = 48 or x = 8. The numbers are 16 and 24. Hence, required sum = (16 + 24) = 40. 2, 1, (1/2), (1/4), ... What number should come next? A. (1/3) B. (1/8) C. (2/8) D. (1/16) Answer: Option B Explanation: This is a simple division series; each number is one-half of the previous number. In other terms to say, the number is divided by 2 successively to get the next result. 4/2 = 2 2/2 = 1 1/2 = 1/2 (1/2)/2 = 1/4 (1/4)/2 = 1/8 and so on. 7, 10, 8, 11, 9, 12, ... What number should come next? A. 7 B. 10 C. 12 D. 13 Answer: Option B Explanation: This is a simple alternating addition and subtraction series. In the first pattern, 3 is added; in the second, 2 is subtracted. 36, 34, 30, 28, 24, ... What number should come next? A. 20 B. 22 C. 23 D. 26 Answer: Option B Explanation: This is an alternating number subtraction series. First, 2 is subtracted, then 4, then 2, and so on. Look at this series: 22, 21, 23, 22, 24, 23, ... What number should come next? A. 22 B. 24 C. 25 D. 26 Answer: Option C Explanation: In this simple alternating subtraction and addition series; 1 is subtracted, then 2 is added, and so on. 53, 53, 40, 40, 27, 27, ... What number should come next? A. 12 B. 14 C. 27 D. 53 Answer: Option B Explanation: In this series, each number is repeated, then 13 is subtracted to arrive at the next number. 21, 9, 21, 11, 21, 13, 21, ... What number should come next? A. 14 B. 15 C. 21 D. 23 Answer: Option B Explanation: In this alternating repetition series, the random number 21 is interpolated every other number into an otherwise simple addition series that increases by 2, beginning with the number 9. 58, 52, 46, 40, 34, ... What number should come next? A. 26 B. 28 C. 30 D. 32 Answer: Option B Explanation: This is a simple subtraction series. Each number is 6 less than the previous number. 3, 4, 7, 8, 11, 12, ... What number should come next? A. 7 B. 10 C. 14 D. 15 Answer: Option D Explanation: This alternating addition series begins with 3; then 1 is added to give 4; then 3 is added to give 7; then 1 is added, and so on. 8, 22, 8, 28, 8, ... What number should come next? A. 9 B. 29 C. 32 D. 34 Answer: Option D Explanation: This is a simple addition series with a random number, 8, interpolated as every other number. In the series, 6 is added to each number except 8, to arrive at the next number. 31, 29, 24, 22, 17, ... What number should come next? A. 15 B. 14 C. 13 D. 12 Answer: Option A Explanation: This is a simple alternating subtraction series, which subtracts 2, then 5. 664, 332, 340, 170, ____, 89, ... What number should fill the blank? A. 85 B. 97 C. 109 D. 178 Answer: Option D Explanation: This is an alternating division and addition series: First, divide by 2, and then add 8. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is: A. 15360 B. 153600 C. 30720 D. 307200 Answer: Option B Explanation: Perimeter = Distance covered in 8 min. =12000x 8/ 60 m = 1600 m. Let length = 3x metres and breadth = 2x metres. Then, 2(3x + 2x) = 1600 or x = 160. Length = 480 m and Breadth = 320 m. Area = (480 x 320) m2 = 153600 m2. A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road? A. 2.91 m B. 3 m C. 5.82 m D. None of these Answer: Option B Explanation: Area of the park = (60 x 40) m2 = 2400 m2. Area of the lawn = 2109 m2. Area of the crossroads = (2400 - 2109) m2 = 291 m2. Let the width of the road be x metres. Then, 60x + 40x - x2 = 291 x2 - 100x + 291 = 0 (x - 97)(x - 3) = 0 x = 3. The diagonal of a rectangle is 41 cm and its area is 20 sq. cm. The perimeter of the rectangle must be: A. 9 cm B. 18 cm C. 20 cm D. 41 cm Answer: Option B Explanation: whole square root l2 + b2 = whole square root 41. Also, lb = 20. (l + b)2 = (l2 + b2) + 2lb = 41 + 40 = 81 (l + b) = 9. Perimeter = 2(l + b) = 18 cm. The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is: A. 1520 m2 B. 2420 m2 C. 2480 m2 D. 2520 m2 Answer: Option D Explanation: We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103. Solving the two equations, we get: l = 63 and b = 40. Area = (l x b) = (63 x 40) m2 = 2520 m2. The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ 26.50 per metre is Rs. 5300, what is the length of the plot in metres? A. 40 B. 50 C. 120 D. Data inadequate Answer: Option E Explanation: Let breadth = x metres. Then, length = (x + 20) metres. Perimeter =5300/26.,50 m = 200 m. 2[(x + 20) + x] = 200 2x + 20 = 100 2x = 80 x = 40. Hence, length = x + 20 = 60 m. A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required? A. 34 B. 40 C. 68 D. 88 Answer: Option D Explanation: We have: l = 20 ft and lb = 680 sq. ft. So, b = 34 ft. Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft. A tank is 25 m long, 12 m wide and 6 m deep. The cost of plastering its walls and bottom at 75 paise per sq. m, is: A. Rs. 456 B. Rs. 458 C. Rs. 558 D. Rs. 568 Answer: Option C Explanation: Area to be plastered = [2(l + b) x h] + (l x b) = {[2(25 + 12) x 6] + (25 x 12)} m2 = (444 + 300) m2 = 744 m2. Cost of plastering = Rs. 744 x75/100= Rs. 558. A, B and C jointly thought of engaging themselves in a business venture. It was agreed that A would invest Rs. 6500 for 6 months, B, Rs. 8400 for 5 months and C, Rs. 10,000 for 3 months. A wants to be the working member for which, he was to receive 5% of the profits. The profit earned was Rs. 7400. Calculate the share of B in the profit. A. Rs. 1900 B. Rs. 2660 C. Rs. 2800 D. Rs. 2840 Answer: Option B Explanation: For managing, A received = 5% of Rs. 7400 = Rs. 370. Balance = Rs. (7400 - 370) = Rs. 7030. Ratio of their investments = (6500 x 6) : (8400 x 5) : (10000 x 3) = 39000 : 42000 : 30000 = 13 : 14 : 10 Bs share = Rs.7030 x14/ 37 = Rs. 2660. A, B, C subscribe Rs. 50,000 for a business. A subscribes Rs. 4000 more than B and B Rs. 5000 more than C. Out of a total profit of Rs. 35,000, A receives: A. Rs. 8400 B. Rs. 11,900 C. Rs. 13,600 D. Rs. 14,700 Answer: Option D Explanation: Let C = x. Then, B = x + 5000 and A = x + 5000 + 4000 = x + 9000. So, x + x + 5000 + x + 9000 = 50000 3x = 36000 x = 12000 A : B : C = 21000 : 17000 : 12000 = 21 : 17 : 12. As share = Rs.35000 x 21/ 50= Rs. 14,700. A starts business with Rs. 3500 and after 5 months, B joins with A as his partner. After a year, the profit is divided in the ratio 2 : 3. What is Bs contribution in the capital? A. Rs. 7500 B. Rs. 8000 C. Rs. 8500 D. Rs. 9000 Answer: Option D Explanation: Let Bs capital be Rs. x. Then,3500 x 12/7x =126000 14x = 126000 x = 9000. An accurate clock shows 8 oclock in the morning. Through how may degrees will the hour hand rotate when the clock shows 2 oclock in the afternoon? A. 144º B. 150º C. 168º D. 180º Answer: Option D Explanation: Angle traced by the hour hand in 6 hours =(360x 6/12)º = 180º. How many times are the hands of a clock at right angle in a day? A. 22 B. 24 C. 44 D. 48 Answer: Option C Explanation: In 12 hours, they are at right angles 22 times. In 24 hours, they are at right angles 44 times. How many times in a day, are the hands of a clock in straight line but opposite in direction? A. 20 B. 22 C. 24 D. 48 Answer: Option B Explanation: The hands of a clock point in opposite directions (in the same straight line) 11 times in every 12 hours. (Because between 5 and 7 they point in opposite directions at 6 oclock only). So, in a day, the hands point in the opposite directions 22 times
Posted on: Mon, 08 Dec 2014 08:32:50 +0000
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