Difference between revisions of "1995 AHSME Problems/Problem 4"

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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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Revision as of 12:58, 5 July 2013

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

Altenrately, picking an arbitrarly 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 $B$.

See also

1995 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

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