Problems on H.C.F and L.C.M

Question: Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:

[A].

4

[B].

5

[C].

6

[D].

8

Answer: Option A

Explanation:

N = H.C.F. of (4665 – 1305), (6905 – 4665) and (6905 – 1305)

  = H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

Question:
Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

[A].

4

[B].

7

[C].

9

[D].

13

Answer: Option A

Explanation:

Required number = H.C.F. of (91 – 43), (183 – 91) and (183 – 43)

     = H.C.F. of 48, 92 and 140 = 4.

Question: The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is:

[A].

15 cm

[B].

25 cm

[C].

35 cm

[D].

42 cm

Answer: Option C

Explanation:

Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm.

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:

[A].

276

[B].

299

[C].

322

[D].

345

Answer: Option C

Explanation:

Clearly, the numbers are (23 x 13) and (23 x 14).

Larger number = (23 x 14) = 322.

Question:
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: Option C

Explanation:

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

Question:
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: Option C

Explanation:

Other number =		11 x 7700		= 308.
		275

Question: Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:

[A].

40

[B].

80

[C].

120

[D].

200

Answer: Option A

Explanation:

Let the numbers be 3x, 4x and 5x.

Then, their L.C.M. = 60x.

So, 60x = 2400 or x = 40.

The numbers are (3 x 40), (4 x 40) and (5 x 40).

Hence, required H.C.F. = 40.

Question:
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: Option C

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.

Question:
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:

[A].

75

[B].

81

[C].

85

[D].

89

Answer: Option C

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.

Question: The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:

[A].

1

[B].

2

[C].

3

[D].

4

Answer: Option B

Explanation:

Let the numbers 13a and 13b.

Then, 13a x 13b = 2028

ab = 12.

Now, the co-primes with product 12 are (1, 12) and (3, 4).

[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]

So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).

Clearly, there are 2 such pairs.