1951 AHSME Problems/Problem 13

Revision as of 20:16, 10 April 2013 by Bobthesmartypants (talk | contribs) (Solution)

Problem

$A$ can do a piece of work in $9$ days. $B$ is $50\%$ more efficient than $A$. The number of days it takes $B$ to do the same piece of work is:

$\textbf{(A)}\ 13\frac {1}{2} \qquad\textbf{(B)}\ 4\frac {1}{2} \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ \text{none of these answers}$

Solution

Because $B$ is $50\%$ more efficient, he can do $1.5$ pieces of work in $9$ days. This is equal to 1 piece of work in $\textbf{(C)}\ 6$ days

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AHSME Problems and Solutions