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 $B$ .

1995 AHSME (Problems • Answer Key • Resources)
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1995 AHSME Problems/Problem 4

Contents

Problem

Solution 1

Solution 2

See also