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

Revision as of 21:14, 9 January 2008

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

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:

$\frac{2}{5}N=\frac{p}{5}=Q$

$M=\frac{3N}{25}$

$\frac{M}{N}=\frac{3}{25} \Rightarrow \mathrm{(B)}$

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

Revision as of 21:14, 9 January 2008

Problem

Solution

See also

Revision as of 08:21, 9 January 2008 (view source) 1=2 (talk \| contribs) (→‎See also) ← Older edit	Revision as of 21:14, 9 January 2008 (view source) Azjps (talk \| contribs) m (1995 AMC 12 Problems/Problem 4 moved to 1995 AHSME Problems/Problem 4) Newer edit →
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