The least number which when divided by 12, 16, 18, 30 leaves remainder 4 in each case but it is completely dividable by 7 is

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2488

2884

8824

4884

answer is C.

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Detailed Solution

$∴$ The L.C. M of 12, 16, 18 and 30 $= 2 \times 2 \times 3 \times 4 \times 3 \times 5 = 720$

Hence the number can be written in the term $(720 \times K + 4)$ which is divisible by 7.

If $K = 1, n u m b e r = (720 \times 1) + 4 = 724$ which is not divisible by ‘7’

If $K = 2, n u m b e r = (720 \times 2) + 4 = 1444$ which is not divisible by ‘7’

If $K = 3, n u m b e r = (720 \times 3) + 4 = 2164$ which is not divisible by ‘7’

If $K = 4, n u m b e r = (720 \times 4) + 4 = 2884$ which is divisible by ‘7’

$∴$ The required number is $= (720 \times 4) + 4 = 2884$

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