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

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 again at the starting point ?

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Question:
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 again at the starting point ?

[A].

26 minutes and 18 seconds

[B].

42 minutes and 36 seconds

[C].

45 minutes

[D].

46 minutes and 12 seconds

Answer: Option D

Explanation:

L.C.M. of 252, 308 and 198 = 2772.

So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.

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 ?

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Question:
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].

4

[B].

10

[C].

15

[D].

16

Answer: Option D

Explanation:

L.C.M. of 2, 4, 6, 8, 10, 12 is 120.

So, the bells will toll together after every 120 seconds(2 minutes).

In 30 minutes, they will toll together 30 + 1 = 16 times.
2

The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:

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Question:
The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:

[A].

1677

[B].

1683

[C].

2523

[D].

3363

Answer: Option B

Explanation:

L.C.M. of 5, 6, 7, 8 = 840.

Required number is of the form 840k + 3

Least value of k for which (840k + 3) is divisible by 9 is k = 2.

Required number = (840 x 2 + 3) = 1683.

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

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

[A].

74

[B].

94

[C].

184

[D].

364

Answer: Option D

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.

The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

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Question:
The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

[A].

9000

[B].

9400

[C].

9600

[D].

9800

Answer: Option C

Explanation:

Greatest number of 4-digits is 9999.

L.C.M. of 15, 25, 40 and 75 is 600.

On dividing 9999 by 600, the remainder is 399.

Required number (9999 – 399) = 9600.

What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ?

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Question: What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ?

[A].

196

[B].

630

[C].

1260

[D].

2520

Answer: Option B

Explanation:

L.C.M. of 12, 18, 21 30 2 | 12 – 18 – 21 – 30
—————————-
= 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3 | 6 – 9 – 21 – 15
—————————-
Required number = (1260 2) | 2 – 3 – 7 – 5

= 630.