Discussion
Home ‣ Aptitude ‣ Problems on H.C.F and L.C.M Comments
- Question
The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:
Options- A. 123
- B. 127
- C. 235
- D. 305
- Correct Answer
- 127
Explanation
Required number = H.C.F. of (1657 - 6) and (2037 - 5)= H.C.F. of 1651 and 2032 = 127.
- 1. 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:
Options- A. 276
- B. 299
- C. 322
- D. 345 Discuss
- 2. Which of the following has the most number of divisors?
Options- A. 99
- B. 101
- C. 176
- D. 182 Discuss
- 3. The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:
Options- A. 504
- B. 536
- C. 544
- D. 548 Discuss
- 4. Which of the following fraction is the largest?
Options- A. 7⁄8
- B. 13⁄16
- C. 31⁄40
- D. 63⁄80 Discuss
- 5. Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is:
Options- A. 75
- B. 81
- C. 85
- D. 89 Discuss
- 6. 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:
Options- A. 279
- B. 283
- C. 308
- D. 318 Discuss
- 7.
(469 + 174)2 - (469 - 174)2 =? (469 x 174)
Options- A. 2
- B. 4
- C. 295
- D. 643 Discuss
- 8. A man has Rs.480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has?
Options- A. 45
- B. 60
- C. 75
- D. 90 Discuss
- 9. There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is:
Options- A. 20
- B. 80
- C. 100
- D. 200 Discuss
- 10. David gets on the elevator at the 11th floor of a building and rides up at the rate of 57 floors per minute. At the same time, Albert gets on an elevator at the 51st floor of the same building and rides down at the rate of 63 floors per minute. If they continue travelling at these rates, then at which floor will their paths cross?
Options- A. 19
- B. 28
- C. 30
- D. 37 Discuss
Problems on H.C.F and L.C.M
Search Results
Correct Answer: 322
Explanation:
Clearly, the numbers are (23 x 13) and (23 x 14).
∴ Larger number = (23 x 14) = 322.
Correct Answer: 176
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.
Correct Answer: 548
Explanation:
Required number = (L.C.M. of 12, 15, 20, 54) + 8
= 540 + 8
= 548.
Correct Answer: 7⁄8
Explanation:
L.C.M. of 8, 16, 40 and 80 = 80.
| 7 | = | 70 | ; | 13 | = | 65 | ; | 31 | = | 62 |
| 8 | 80 | 16 | 80 | 40 | 80 |
| Since, | 70 | > | 65 | > | 63 | > | 62 | , so | 7 | > | 13 | > | 63 | > | 31 |
| 80 | 80 | 80 | 80 | 8 | 16 | 80 | 40 |
| So, | 7 | is the largest. |
| 8 |
Correct Answer: 85
Explanation:
Since the numbers are co-prime, they contain only 1 as the common factor.
Also, the given two products have the middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29;
| First number = | ❨ | 551 | ❩ | = 19; Third number = | ❨ | 1073 | ❩ | = 37. |
| 29 | 29 |
∴ Required sum = (19 + 29 + 37) = 85.
Correct Answer: 308
Explanation:
| Other number = | ❨ | 11 x 7700 | ❩ | = 308. |
| 275 |
Correct Answer: 4
Explanation:
| Given exp. = | (a + b)2 - (a - b)2 |
| ab |
| = | 4ab |
| ab |
= 4 (where a = 469, b = 174.)
Correct Answer: 90
Explanation:
Let number of notes of each denomination be x
Then x + 5x + 10x = 480
⟹ 16x = 480
∴ x = 30.
Hence, total number of notes = 3x = 90.
Correct Answer: 100
Explanation:
Let the number of students in rooms A and B be x and y respectively.
Then, x - 10 = y + 10 ⟹ x - y = 20 .... (i)
and x + 20 = 2(y - 20) ⟹ x - 2y = -60 .... (ii)
Solving (i) and (ii) we get: x = 100 , y = 80.
∴ The required answer A = 100.
Correct Answer: 30
Explanation:
Suppose their paths cross after x minutes.
Then, 11 + 57x = 51 - 63x ⟺ 120x = 40
| x = | 1 |
| 3 |
| Number of floors covered by David in (1/3) min. = | ❨ | 1 | x 57 | ❩ | = 19. |
| 3 |
So, their paths cross at (11 +19) i.e., 30th floor.
Comments
There are no comments.More in Aptitude:
Programming
Copyright ©CuriousTab. All rights reserved.