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Question

There are several boxes with coins bearing a value (1 to n) each. The coins in different boxes are grouped as follows:- 1(first box), 2+3(second box), 4+5+6(third box), 7+8+9+10(fourth box), ..... and so on. The value of the coins in the 15th box is:

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Solution

The correct option is **B**

1695

123456

This equation is based on the n(n+1)2 pattern, where the last coin in the nth box will have the value n(n+1)2.

Thus the last coin in the 15th box will have a value =15×162=120

The first coin will have a value = 106

Thus the sum =152(106+120)=1695.

As there are 15 coins in the 15th box and it is 15 numbers ... it has to be a multiple of 15. Only option (b) satisfies this condition.

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