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Home ‣ Aptitude ‣ Problems on H.C.F and L.C.M Comments
- Question
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
- Correct Answer
- 322
Explanation
Clearly, the numbers are (23 x 13) and (23 x 14).∴ Larger number = (23 x 14) = 322.
- 1. Which of the following has the most number of divisors?
Options- A. 99
- B. 101
- C. 176
- D. 182 Discuss
- 2. 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
- 3. Which of the following fraction is the largest?
Options- A. 7⁄8
- B. 13⁄16
- C. 31⁄40
- D. 63⁄80 Discuss
- 4. 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
- 5. The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:
Options- A. 101
- B. 107
- C. 111
- D. 185 Discuss
- 6. 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 Discuss
- 7. 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
- 8.
(469 + 174)2 - (469 - 174)2 =? (469 x 174)
Options- A. 2
- B. 4
- C. 295
- D. 643 Discuss
- 9. 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
- 10. 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
Problems on H.C.F and L.C.M
Search Results
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: 111
Explanation:
Let the numbers be 37 a and 37 b
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.
Correct Answer: 127
Explanation:
Required number = H.C.F. of (1657 - 6) and (2037 - 5)
= H.C.F. of 1651 and 2032 = 127.
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.
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