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# Difference between revisions of "1978 AHSME Problems/Problem 5"

## Problem 5

Four boys bought a boat for $\textdollar 60$. The first boy paid one half of the sum of the amounts paid by the other boys; the second boy paid one third of the sum of the amounts paid by the other boys; and the third boy paid one fourth of the sum of the amounts paid by the other boys. How much did the fourth boy pay?

$\textbf{(A) }\textdollar 10\qquad \textbf{(B) }\textdollar 12\qquad \textbf{(C) }\textdollar 13\qquad \textbf{(D) }\textdollar 14\qquad \textbf{(E) }\textdollar 15$

## Solution 1

If the first boy paid one half of the sum of the amounts paid by the other boys, that means that he paid $\frac{1}{3}$ of the total. If the second boy paid one third of the sum of the amounts paid by the other boys, that means that he paid $\frac{1}{4}$ of the total. If the third boy paid one fourth of the sum of the amounts paid by the other boys, that means that he paid $\frac{1}{5}$ of the total.

Summing it up, we get $\textdollar 20 + \textdollar 15 + \textdollar 12 = \textdollar 47$. Therefore, our answer is $\textdollar 60 - \textdollar 47 = \boxed{\textbf{(C) }\textdollar 13}$ ~awin

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