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:

• A. 2098
• B. 1695
• C. 5675
• D. 1687.5

Answer 1

The correct option is B

1695

$\begin{array}{ccc}1& & \\ 2& 3& \\ 4& 5& 6\end{array}$
This equation is based on the $\frac{n(n+1)}{2}$ pattern, where the last coin in the nth box will have the value $\frac{n(n+1)}{2}$.
Thus the last coin in the ${15}^{th}$ box will have a value $=\frac{15\times 16}{2}=120$
The first coin will have a value = 106
Thus the sum $=\frac{15}{2}(106+120)=1695.$

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.