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H.C.F & L.C.M. – Practice Problems

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H.C.F & L.C.M. – Practice Problems

Q 1 – Find the lowest common multiple of 24, 36 and 40?

A – 120

B – 360

C – 240

D – 480

Answer : B

Explanation

L.C.M = 2 x 2 x 2 x 3 x 3 x 5 = 360

Q 2 – The smallest fraction, which each of 6⁄7, 5⁄14, 10⁄21 will divide exactly, is?

A – 38⁄98

B – 60⁄147

C – 50⁄294

D – 30⁄7

Answer : D

Explanation

Required fraction = L.C.M of 6⁄7, 5⁄14, 10⁄21 = L.C.M of 6, 5, 10⁄H.C.F of 7, 14, 21 = 30⁄7

Q 3 – The least number, which when divided by 12,15,20 and 54 leaves in each case a remainder of 8, is?

A – 500

B – 548

C – 544

D – 536

Answer : B

Explanation

Required number = (L.C.M of 12, 15, 20, 54) + 8 = 540 + 8 = 548.

Q 4 – 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 – 11⁄120

B – 55⁄601

C – 601⁄55

D – 120⁄11

Answer : A

Explanation

Let the numbers be a and b. Then a + b = 55 and a x b = 5 x 120 = 600
therefore Required sum = 1⁄a + 1⁄b = a + b⁄a x b
55⁄600 = 11⁄120

Q 5 – L.C.M of two prime numbers a and b(a>b) is 161. The value of 3b – a is?

A – -1

B – -2

C – 1

D – 2

Answer : B

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 a x b = 161
now co-primes with product 161 are (1, 161) (7,23).
Since a and b are prime numbers and a > b, we have a = 23 and b = 7.
Therefore  3b - a = (3 x 7) - 23 = -2.

Q 6 – The number of prime factors in the expression 25x26x36x53

A – 20

B – 14

C – 13

D – 8

Answer : A

Explanation

2, 3, 5 are the prime numbers and the given expression is 25x26x36x53 (25x66x53).
So the total would be the sum of powers in the expression that is 5+6+6+3=20

Q 7 – The G.C.D. of 0.93, 0.60 and 0.75 is:

A – 0.03

B – 0.06

C – 0.09

D – 0.12

Answer : A

Explanation

Given numbers are 0.93, 0.60 and 0.75.   H.C.F. of 93, 60 & 75 is 3.
 H.C.F. of given numbers = 0.03

Q 8 – If two numbers are greater than 13 and the H.C.F of two numbers be 13, L.C.M 273, then the sum of the numbers is:

A – 286

B – 130

C – 288

D – 290

Answer : B

Explanation

Let the number be 13 a and 13 b, where a and b are co-primes.
Then, 13a * 13b=   (13* 273)  ⇒ab= 21
Two co-primes with product 21are 3 and 7.
∴ numbers are (13*3, 13*7) i.e , 39 and 91.
Their sum = (39+91) = 130

Q 9 – Having two numbers in the ratio 5:7 and their L.C.M as 315, then their product is:

A – 2358

B – 2385

C – 2538

D – 2835

Answer : A

Explanation

Let the two numbers be 5x and 7x . Then, their L.C.M is  35 x.
35x =  315    ⇒   x= 9
∴ The numbers are 45 and 63 and their product is 2835.

Q 10 – If the product of two co-primes is 117 then their L.C.F should be

A – 1

B – 117

C – equal to their H.C.F

D – cannot be calculated

Answer : B

Explanation

H.C.F of co-primes=1
H.C.F * L.C.M = Their product = 117
∴ 1* L.C.M = 117 ⇒L.C.M = 117
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