L.C.M Search

Q:

Find the G.C.F. & L.C.M. of 20 and 28.

A) 20,280	B) 5,280
C) 10,140	D) 4,140

Answer & Explanation Answer: B) 5,280

Explanation:

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Filed Under: Inventions
Exam Prep: Bank Exams

Q:

Find the L.C.M of 4/5, 2/3 and 5/7.

A) 25	B) 20
C) 40	D) 30

Answer & Explanation Answer: B) 20

Explanation:

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Filed Under: HCF and LCM
Exam Prep: Bank Exams

Q:

H.C.F of 3240, 3600 and a third number is 36 and their L.C.M is $2^{4} x 3^{5} x 5^{2} x 7^{2 . What is the third number?}$

A) 47628	B) 49874
C) 24157	D) 42146

Answer & Explanation Answer: A) 47628

Explanation:

hcf_of_three_numbers_3240,_3600_and_third_number_is_361542439249.jpg image

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Filed Under: HCF and LCM
Exam Prep: AIEEE , Bank Exams , CAT , GATE
Job Role: Analyst , Bank Clerk , Bank PO

Q:

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) 13/125	B) 14/57
C) 11/120	D) 16/41

Answer & Explanation Answer: C) 11/120

Explanation:

Let the numbers be a and b.
We know that product of two numbers = Product of their HCF and LCM
Then, a + b = 55 and ab = 5 x 120 = 600.
=> The required sum = (1/a) + (1/b) = (a+b)/ab
=55/600 = 11/120

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Filed Under: HCF and LCM
Exam Prep: AIEEE , Bank Exams , CAT
Job Role: Bank Clerk , Bank PO

Q:

Three numbers are in the ratio of 3:4:5 and their L.C.M is 3600.Their HCF is:

A) 40	B) 60
C) 100	D) 120

Answer & Explanation Answer: B) 60

Explanation:

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

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

So, 60x=3600 or x=60.

Therefore, The numbers are (3 x 60), (4 x 60), (5 x 60).

Hence,required H.C.F=60

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Filed Under: HCF and LCM
Exam Prep: CAT
Job Role: Bank PO

Q:

L.C.M of two prime numbers x and y (x>y) is 161. The value of 3y-x is :

A) -2	B) -1
C) 1	D) 2

Answer & Explanation Answer: A) -2

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.

Therefore, 3y-x = (3 x 7)-23 = -2

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Filed Under: HCF and LCM

Q:

The H.C.F and L.C.M of two numbers are 11 and 385 respectively. If one number lies between 75 and 125 , then that number is

A) 77	B) 88
C) 99	D) 110

Answer & Explanation Answer: A) 77

Explanation:

Product of numbers = 11 x 385 = 4235

Let the numbers be 11a and 11b . Then , 11a x 11b = 4235 => ab = 35

Now, co-primes with product 35 are (1,35) and (5,7)

So, the numbers are ( 11 x 1, 11 x 35) and (11 x 5, 11 x 7)

Since one number lies 75 and 125, the suitable pair is (55,77)

Hence , required number = 77

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Filed Under: HCF and LCM

Q:

Product of two co-prime numbers is 117. Their L.C.M should be

A) 1	B) 117
C) Equal to their H.C.F	D) cannot be calculated

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