Number 136 is added to 5B7 and the sum obtained is 7A3, where A and B are integers. It is given that 7A3 is exactly divisible by 3. The only possible value of B is:

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Asked: March 18, 20232023-03-18T04:16:08+05:30
2023-03-18T04:16:08+05:30In: CSAT

14.29%(a) 2 ( 1 voter )

0%(b) 5 ( 0 voters )

28.57%(c) 7 ( 2 voters )

57.14%(d) 8 ( 4 voters )

Based On 7 Votes

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This answer was edited.Ans (d) 8

‘7A3’ is divisible by 3 only when the sum of its digits is divisible by 3.

Rule of Divisible of 3

7 + A + 3 = 10 + A

Now, (10 + A) would be divisible by 3, for A = 2, 5, 8

As per the question,

136 + 5B7 = 7A3

5B7 = 7A3 – 136

For A = 2,

5B7 = 723 – 136 = 587

B = 8

Hence, B = 8.