Problems On Hcf And Lcm

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: A

Clearly, 252 = 2 x 2 x 3 x 3 x 7.

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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: B

Required number = (L.C.M. of 12,16, 18, 21, 28) + 7

= 1008 + 7

= 1015

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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: D

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.

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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

L.C.M. of 12, 18, 21 30 2 | 12 - 18 - 21 - 30

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= 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3 | 6 - 9 - 21 - 15

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Required number = (1260 ÷ 2) | 2 - 3 - 7 - 5

= 630.

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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: C

Other number =		11 x 7700		= 308.
		275

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Questions for: Problems On Hcf And Lcm