# 1995 AHSME Problems/Problem 4

## Problem

If $M$ is $30 \%$ of $Q$, $Q$ is $20 \%$ of $P$, and $N$ is $50 \%$ of $P$, then $\frac {M}{N} =$ $\mathrm{(A) \ \frac {3}{250} } \qquad \mathrm{(B) \ \frac {3}{25} } \qquad \mathrm{(C) \ 1 } \qquad \mathrm{(D) \ \frac {6}{5} } \qquad \mathrm{(E) \ \frac {4}{3} }$

## Solution 1

We are given: $M=\frac{3Q}{10}$, $Q=\frac{P}{5}$, $N=\frac{P}{2}$. We want M in terms of N, so we substitute N into everything: $\frac{2}{5}N=\frac{P}{5}=Q$ $M=\frac{3N}{25}$ $\frac{M}{N}=\frac{3}{25} \Rightarrow \mathrm{(B)}$

## Solution 2

Alternatively, picking an arbitrary value for $Q$ of $100$, we find that $M = 30\% \cdot 100 = 0.30 \cdot 100 = 30$.

We find that $Q = 20\% \cdot P$, meaning $100 = 0.2\cdot P$, giving $P = \frac{100}{0.2} = \frac{1000}{2} = 500$.

Finally, since $N$ is $50\%$ of $P$, we have $N = 50\% \cdot 500 = 0.5 \cdot 500 = 250$.

Thus, $M = 30$ and $N = 250$, so their ratio $\frac{M}{N} = \frac{30}{250} = \frac{3}{25} \Rightarrow \mathrm{(B)}$

This method does not prove that the answer must be constant, but it proves that if the answer is a constant, it must be $B$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 