1955 AHSME Problems/Problem 35

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Three boys agree to divide a bag of marbles in the following manner. The first boy takes one more than half the marbles. The second takes a third of the number remaining. The third boy finds that he is left with twice as many marbles as the second boy. The original number of marbles:

$\textbf{(A)}\ \text{is none of the following}\qquad\textbf{(B)}\ \text{cannot be determined from the given data}\\ \textbf{(C)}\ \text{is 20 or 26}\qquad\textbf{(D)}\ \text{is 14 or 32}\qquad\textbf{(E)}\ \text{is 8 or 38}$

Solution

[asy] draw((0,0)--(13,0)); dot((0,0)); dot((7,0)); dot((9,0)); dot((13,0)); [/asy] The line that represents the marbles is split into three sections: $3x+1,  x, \text{and } 2x$. Since the amount of marbles each boy relies on the variable $x$, which cannot be solved for, the original amount of marbles $\boxed{\textbf{(B) } \text{cannot be determined from the given data}}$.

See Also

Go to the rest of the 1955 AHSME Problems

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