Problems on HCF and LCM – Aptitude Questions and Answers | Latest Online Problems on HCF and LCM MCQ Aptitude Test
Problems on HCF and LCM – Aptitude Questions and Answers | Online Problems on HCF and LCM MCQ Aptitude Test Quiz Name Problems on HCF... explore below more
Problems on HCF and LCM – Aptitude Questions and Answers | Online Problems on HCF and LCM MCQ Aptitude Test
Quiz Name
Problems on HCF and LCM
Category
Online Aptitude Test
Number of Questions
30
Time
30 Minutes
Exam Type
MCQ (Multiple Choice Questions)
1. Determine the largest integer that will divide 43, 91, and 183 in such a way that the remainders are the same in each case. a. 4 b. 7 c. 9 d. 13
Answer: A
Explanation: H.C.F. of (91 – 43), (183 – 91), and (183 – 91) are required (183 – 43)= 48, 92, and 140 H.C.F. = 4.
2. The H.C.F. of two integers is 23, and the L.C.M.’s other two components are 13 and 14. The greater of the two figures is: a. 276 b. 299 c. 322 d. 345
Answer: C
Explanation: The figures are clearly (23 x 13) and (23 x 14).(23 x 14) = 322, which is the largest.
3. Six bells start tolling at the same time and at 2, 4, 6, 8, 10, and 12 second intervals, respectively. How many times do they toll jointly in 30 minutes? a. 4 b. 10 c. 15 d. 16
Answer: D
Explanation: The L.C.M. of two, four, six, eight, ten, and twelve equals 120.As a result, every 120 seconds, the bells will chime together (2 minutes).They will toll together 30/2+ 1 = 16 times in 30 minutes.
4. Let N be the largest integer that divides 1305, 4665, and 6905 without leaving the same residual in each case. Then in N, the total of the digits is: a. 4 b. 5 c. 6 d. 8
Answer: A
Explanation: N = HCF of (4665 – 1305), (6905 – 4665), and (6905 – 4665). (6905 – 1305),3360, 2240, and 5600 H.C.F. = 1120,In N, the sum of the digits is ( 1 + 1 + 2 + 0 ) = 4.
5. The greatest four-digit number that is divisible by 15, 25, 40, and 75 is: a. 9000 b. 9400 c. 9600 d. 9800
Answer: C
Explanation: 9999 is the largest number of four digits.The L.C.M. for the ages of 15, 25, 40, and 75 is 600.The remainder is 399 when you divide 9999 by 600.9600 is the required number (9999 – 399).
6. The sum of two integers equals 4107. If these numbers have an H.C.F. of 37, the bigger number is: a. 101 b. 107 c. 111 d. 185
Answer: C
Explanation: Let’s call them 37a and 37b.As a result, 37a + 37b = 4107,ab = three.Co-primes with product 3 are now available (1, 3).As a result, the needed numbers are (37 x 1, 37 x 3), or (37 x 1, 37 x 3). (37, 111).111 is a higher number.
7. The L.C.M. of three numbers is 2400, and their ratio is 3: 4: 5. Their H.C.F. is as follows: a. 40 b. 80 c. 120 d. 200
Answer: A
Explanation: 3x, 4x, and 5x are the numbers to use.Their L.C.M. then becomes 60x.As a result, 60x = 2400, or x = 40..(3 x 40), (4 x 40), and (5 x 40) are the numbers (5 x 40).As a result, the necessary H.C.F. is 40.
8. The G.C.D. for the numbers 1.08, 0.36, and 0.9 is: a. 0.03 b. 0.9 c. 0.18 d. 0.108
Answer: C
Explanation: 1.08, 0.36, and 0.90 are the numbers given. 18 is the H.C.F. of 108, 36, and 90.The H.C.F. of the numbers presented is 0.18.
9. The H.C.F. of two integers is 13 and their product is 2028. The total number of such pairings is as follows: a. 1 b. 2 c. 3 d. 4
Answer: B
Explanation: Let’s use the numbers 13a and 13b as an example.As a result, 13a x 13b = 2028.ab = twelve.(1, 12) and (1, 12) are now the co-primes for product 12. (3, 4). [Note: If two numbers a and b have no common positive factor other than 1 or, equivalently, if their largest common divisor is 1, they are said to be coprime or comparatively prime.].As a result, the needed numbers are (13 x 1, 13 x 12) and (13 x 1, 13 x 12). (13 x 3, 13 x 4).There are clearly two such pairings.
10.When divided by 6, 9, 15, and 18, the least multiple of 7 that leaves a residual of 4 is: a. 74 b. 94 c. 184 d. 364
Answer: D
Explanation: The L.C.M. for the numbers 6, 9, 15, and 18 equals 90.Allow 90k + 4 to be the needed number, which is a multiple of 7.k = 4 is the smallest number of k for which (90k + 4) is divisible by 7.(90 x 4) + 4 Equals 364, which is the required number.
11. Find the smallest common multiple of the numbers 24, 36, and 40. a. 120 b. 240 c. 360 d. 480
14. The smallest number that leaves a remnant of 3 when divided by 5, 6, 7, and 8, but no remainder when divided by 9, is: a. 1677 b. 1683 c. 2523 d. 3363
Answer: B
Explanation: 5, 6, 7, 8 = 840 L.C.M.The number must be in the format 840k + 3.k = 2 is the smallest number of k for which (840k + 3) is divisible by 9.(840 x 2 + 3) = 1683 is the required number.
15. A, B, and C all begin running in the same direction around a circular stadium at the same time. Starting at the same position, A completes a round in 252 seconds, B in 308 seconds, and C in 198 seconds. When will they return to their original starting point? a. 26 minutes and 18 seconds b. 42 minutes and 36 seconds c. 45 minutes d. 46 minutes and 12 seconds
Answer: D
Explanation: 252, 308, and 198 L.C.M. = 2772.As a result, A, B, and C will reunite at the beginning location in 2772 seconds, or 46 minutes and 12 seconds.
16. Two numbers have an H.C.F. of 11 and an L.C.M. of 7700. In the case when one of the numbers is 275, the other is: a. 279 b. 283 c. 308 d. 318
Answer: C
Explanation: Other number = 11 x 7700 divided by 275 is 308.
17. 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: B
Explanation: 12 x 3 x 2 x 3 x 7 x 5 = 1260 L.C.M. = 2 x 3 x 2 x 3 x 7 x 5 = 1260,(1260 2) is the required number.,equals 630.
18. The H.C.F. of two numbers is 4 and their ratio is 3:4. Their L.C.M. is as follows: a. 12 b. 16 c. 24 d. 48
Answer: D
Explanation: Let’s say 3x and 4x are the numbers. Then their H.C.F. is equal to x. As a result, x = 4.As a result, the numerals 12 and 16 are used.12 and 16 L.C.M. = 48.
19. The smallest integer that is divisible by 12, 16, 18, 21, and 28 when reduced by 7 is: a. 1008 b. 1015 c. 1022 d. 1032
Answer: B
Explanation: (L.C.M. of 12,16,18,21,28) + 7 is the required number. 7 +1008=1015.
20. As a product of primes, 20. 252 may be written 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: A
Explanation: 252 clearly equals 2 x 2 x 3 x 3 x 7
21. The maximum length that may be utilised to precisely measure the lengths 7 m, 3 m 85 cm, and 12 m 95 cm is: a. 15 cm b. 25 cm c. 35 cm d. 42 cm
Answer: C
Explanation: H.C.F. of 700 cm, 385 cm, and 1295 cm = 35 cm is the required length.
22. Three co prime integers exist such that the product of the first two is 551 and the product of the final two is 1073. The three digits add up to: a. 75 b. 81 c. 85 d. 89
Answer: C
Explanation: Because the numbers are co-prime, the common factor is simply one.Furthermore, the middle number is shared by the two items.As a result, the middle number equals the H.C.F. of 551 and 1073 = 29;The first number is 551/29, which equals 19, and the third number is 1073/29, which equals 37,19 + 29 + 37 = 85 is the required total.
23. Find the 36th and 84th greatest common factor. a. 4 b. 6 c. 12 d. 18
Answer: C
Explanation: 36 = 22 x 3 x 2 x 2 x 2 x 2 x 2 x 2 x 2,2 to the power of 84 2 x 3 x 7 x 2 x 3 x 7 x 2 x,2 to the power 2 × 3 Equals 12 H.C.F.
24. Which of the fractions below is the largest? a. 7/8 b. 13/16 c. 31/40 d. 63/80
Answer: A
Explanation: 8, 16, 40, and 80 = 80 L.C.M.7/8 = 70/80; 13/16 = 65/80; 31/40 = 62/80Since 70/80>65/80>63/80>62/8, 7/8>13/16 >63/80 >31/40 is the result.As a result, 7/8 is the largest.
25. The smallest number that leaves an 8-digit residual when divided by 12, 15, 20, and 54 is: a. 504 b. 536 c. 544 d. 548
Answer: D
Explanation: (L.C.M. of 12, 15, 20, 54) + 8 is the required number.540 + 8 =The number 548.
26. The largest number, when divided by 1657 and 2037, yields 6 and 5 as remainders, is: a. 123 b. 127 c. 235 d. 305
Answer: B
Explanation: H.C.F. of (1657 – 6) is required, and (2037 – 5)= 1651 and 2032 H.C.F. = 127
27. Which of the following has the most divisors? Which of the following has the most divisors? a. 99 b. 101 c. 176 d. 182
Answer: C
Explanation: 1 x 3 x 3 x 11 = 99,1 x 101 = 101,176 = 1 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x1 × 2 x 7 x 13 = 182,As a result, the divisors of 99 are 1, 3, 9, 11, 33, and so on.101’s divisors are 1 and 101.1, 2, 4, 8, 11, 16, 22, 44, 88, and 176 are the divisors of 176.182 has divisors of 1, 2, 7, 13, 14, 26, 91, and 182.As a result, 176 has the most divisors.
28. When two numerals are added together, the L.C.M. is 48. The numbers are in a 2:3 ratio. The total of the numbers is as follows: a. 28 b.32 c. 40 d. 64
Answer: C
Explanation: Let’s say 2x and 3x are the numbers.Then their L.C.M. is equal to 6x.As a result, 6x = 48 or x = 8.16 and 24 are the numbers.As a result, the needed total is (16 + 24) = 40.
29. The HCF for 9/10, 12/25, 18/35, and 21/40 is: a. 3/5 b. 252/5 c. 3/1400 d. 63/700
30. If the total of two integers is 55, and their H.C.F. and L.C.M. are 5 and 120, respectively, the sum of the reciprocals of the numbers equals: a. 55/601 b. 601/55 c. 11/120 d. 120/11
Answer: C
Explanation: Let’s call the numbers a and b.Then ab = 5 x 120 = 600 and a + b = 55.1/a + 1/b = a + b/ab = 55/600 = 11/120 is the needed amount.
