Problems On Hcf And Lcm

If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:

A:

55

601

B:

601

55

C:

11

120

D:

120

11

Answer: C

Let the numbers be a and b.

Then, a + b = 55 and ab = 5 x 120 = 600.

The required sum =	1	+	1	=	a + b	=	55	=	11
	a		b		ab		600		120

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The H.C.F. of	9	,	12	,	18	and	21	is:
	10		25		35		40

A:

3

5

B:

252

5

C:

3

1400

D:

63

700

Answer: C

Required H.C.F. =	H.C.F. of 9, 12, 18, 21	=	3
	L.C.M. of 10, 25, 35, 40		1400

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The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:

A:

28

B:

32

C:

40

D:

64

Answer: C

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.

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Which of the following has the most number of divisors?

A:

99

B:

101

C:

176

D:

182

Answer: C

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.

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The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:

A:

123

B:

127

C:

235

D:

305

Answer: B

Required number = H.C.F. of (1657 - 6) and (2037 - 5)

= H.C.F. of 1651 and 2032 = 127.

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Questions for: Problems On Hcf And Lcm