Hello Aspirants. Welcome to Online Quantitative Aptitude Section with explanation in AffairsCloud.com. Here we are creating question sample in LCM & HCF, which is common for all the IBPS,SBI exam and other competitive exams. We have included Some questions that are repeatedly asked in bank exams !!!

**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?**

A.8

B.11

C.13

D.16

E.None of these

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**Answer – D (16)**

Explanation – 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

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

A.68

B.98

C.180

D.364

E.None of these

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**Answer – D (364)**

Explanation – 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.

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

A.11115

B.15110

C.15130

D.15310

E.None of these

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**Answer – B (15110)**

Explanation – 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

[/su_accordion]**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.269

B.275

C.308

D.310

E.None of these

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**Answer – C (308)**

Explanation –

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

[/su_accordion]**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?**

A.15 minutes 15 seconds

B.42 minutes 30 seconds

C.42 minutes

D.46 minutes 12 seconds

E.None of these

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**Answer – D (46 minutes 12 seconds)**

Explanation – 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

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

A.30

B.22

C.40

D.60

E.None of these

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**Answer – C (40)**

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.

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

A.534

B.486

C.544

D.548

E.None of these

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**Answer – D (548)**

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

= 540 + 8

= 548.

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

A.124

B.100

C.111

D.175

E.None of these

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**Answer – C (111)**

Explanation – 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.

[/su_accordion]**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.5

E.None of these

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**Answer – B (2)**

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).

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

A.10

B.14

C.23

D.30

E.None of these

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**Answer – C (23)**

Explanation – 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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