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- Problems on H.C.F and L.C.M
6.
The least multiple of 7, which leaves a remainder of 4. when divided by 6, 9, 15 and 18 is:
- A.74
- C.184
- B.94
- D.364
- Answer & Explanation
- Report
Answer : [D]
Explanation :
Explanation :
| L.C.M. of 6, 9, 15 and 18 is 90. | |||
| Let required number be 90k + 4, which is a multiple of 7. | |||
| Least value of k for which (90k + 4) is divisible by 7 is k = 4 | |||
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7.
L.C.M. of two prime numbers x and y (x > y) is 161 The value of 3y – x is:
- A.-2
- C.1
- B.-1
- D.2
- Answer & Explanation
- Report
Answer : [A]
Explanation :
Explanation :
| H.C.F. of two prime numbers is 1. Product of numbers = (1 x 161) = 161 | ||||
| Let the numbers be a and b. Then, ab = 161. | ||||
| Now, co-primes with product 161 are (1, 161) and (7, 23). | ||||
| Since x and y are prime numbers and x > y, we have x = 23 and y = 7. | ||||
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8.
The L.C.M. of two numbers is 48. The numbers are in the ratio 2:3. the sum of the numbers is:
- A.28
- C.40
- B.32
- D.64
- Answer & Explanation
- Report
Answer : [C]
Explanation :
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 |
9.
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
- C.2523
- B.1683
- D.3363
- Answer & Explanation
- Report
Answer : [B]
Explanation :
Explanation :
| L.C.M. of 5, 6, 7, 8 = 840. | |||
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| Least value of k for which (840k + 3) is divisible by 9 is k = 2. | |||
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10.
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
- C.322
- B.299
- D.345
- Answer & Explanation
- Report
Answer : [C]
Explanation :
Explanation :
| Clearly, the numbers are (23 x 13) and (23 x 14). | ||||
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