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HCF And LCM - MCQ 1


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10 Questions MCQ Test Quantitative Techniques for CLAT | HCF And LCM - MCQ 1

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HCF And LCM - MCQ 1 - Question 1

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together?

Detailed Solution for HCF And LCM - MCQ 1 - Question 1

L.C.M. of 2, 4, 6, 8, 10, 12 is 120.So, the bells will toll together after every 120 seconds, i.e, 2 minutes.In 30 minutes, they will toll together 30/2 + 1 = 16

HCF And LCM - MCQ 1 - Question 2

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

Detailed Solution for HCF And LCM - MCQ 1 - Question 2

L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
Required number = (90 x 4) + 4 = 364.

HCF And LCM - MCQ 1 - Question 3

The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 50, 62, 98 and 130 as remainders respectively, is:

Detailed Solution for HCF And LCM - MCQ 1 - Question 3

Here (48 – 38) = 10, (60 – 50) = 10, (72 – 62) = 10, (108 – 98) = 10 & (140 – 130) = 10.
Required number = (L.C.M. of 48, 60, 72, 108, 140) – 10
= 15120 – 10 = 15110

HCF And LCM - MCQ 1 - Question 4

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:

Detailed Solution for HCF And LCM - MCQ 1 - Question 4

Other number =[11 x 7700]/275 = 308

HCF And LCM - MCQ 1 - Question 5

A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds, all starting at the same point. After what time will they meet again at the starting point?

Detailed Solution for HCF And LCM - MCQ 1 - Question 5

L.C.M. of 252, 308 and 198 = 2772.So, A, B and C will again meet at the starting point in 2772 see i.e., 46 min. 12 sec

HCF And LCM - MCQ 1 - Question 6

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

Detailed Solution for HCF And LCM - MCQ 1 - Question 6

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.

HCF And LCM - MCQ 1 - Question 7

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

Detailed Solution for HCF And LCM - MCQ 1 - Question 7

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

HCF And LCM - MCQ 1 - Question 8

The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:

Detailed Solution for HCF And LCM - MCQ 1 - Question 8

Let the numbers be 37a and 37b. Then, 37a x 37b = 4107
ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
Greater number = 111.

HCF And LCM - MCQ 1 - Question 9

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

Detailed Solution for HCF And LCM - MCQ 1 - Question 9

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.

HCF And LCM - MCQ 1 - Question 10

The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:

Detailed Solution for HCF And LCM - MCQ 1 - Question 10

L.C.M. of 5, 6, 4 and 3 = 60. On dividing 2497 by 60, the remainder is 37. Number to be added = (60 – 37) = 23

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