1953 AHSME Problems/Problem 19

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Problem 19

In the expression $xy^2$, the values of $x$ and $y$ are each decreased $25$ %; the value of the expression is:

$\textbf{(A)}\ \text{decreased } 50\% \qquad \textbf{(B)}\ \text{decreased }75\%\\  \textbf{(C)}\ \text{decreased }\frac{37}{64}\text{ of its value}\qquad \textbf{(D)}\ \text{decreased }\frac{27}{64}\text{ of its value}\\ \textbf{(E)}\ \text{none of these}$

Solution

$xy^2$

$(\frac{3x}{4})(\frac{3y}{4})^2$

$(\frac{3}{4})^3xy^2$

$\frac{27}{64}xy^2$

$xy^2-\frac{27}{64}xy^2 = \frac{37}{64}xy^2 \implies \textbf{(C)}$

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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