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

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<math> \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} }  </math>
 
<math> \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} }  </math>
  
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==Solution==
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==Solution 1==
 
We are given: <math>M=\frac{3Q}{10}</math>, <math>Q=\frac{P}{5}</math>, <math>N=\frac{P}{2}</math>. We want M in terms of N, so we substitute N into everything:
 
We are given: <math>M=\frac{3Q}{10}</math>, <math>Q=\frac{P}{5}</math>, <math>N=\frac{P}{2}</math>. We want M in terms of N, so we substitute N into everything:
  
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<math>\frac{2}{5}N=\frac{p}{5}=Q</math>
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<math>\frac{2}{5}N=\frac{P}{5}=Q</math>
  
 
<math>M=\frac{3N}{25}</math>
 
<math>M=\frac{3N}{25}</math>
  
 
<math>\frac{M}{N}=\frac{3}{25} \Rightarrow \mathrm{(B)}</math>
 
<math>\frac{M}{N}=\frac{3}{25} \Rightarrow \mathrm{(B)}</math>
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==Solution 2==
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Alternatively, picking an arbitrary value for <math>Q</math> of <math>100</math>, we find that <math>M = 30\% \cdot 100 = 0.30 \cdot 100 = 30</math>.
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We find that <math>Q  = 20\% \cdot P</math>, meaning <math>100 = 0.2\cdot P</math>, giving <math>P = \frac{100}{0.2} = \frac{1000}{2} = 500</math>.
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Finally, since <math>N</math> is <math>50\%</math> of <math>P</math>, we have <math>N = 50\% \cdot 500 = 0.5 \cdot 500 = 250</math>.
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Thus, <math>M = 30</math> and <math>N = 250</math>, so their ratio <math>\frac{M}{N} = \frac{30}{250} = \frac{3}{25} \Rightarrow \mathrm{(B)}</math>
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This method does not prove that the answer must be constant, but it proves that if the answer is a constant, it must be <math>B</math>.
  
 
==See also==
 
==See also==
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{{Old AMC12 box|year=1995|num-b=3|num-a=5}}
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{{AHSME box|year=1995|num-b=3|num-a=5}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 22:30, 16 February 2018

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 $B$.

See also

1995 AHSME (Problems • Answer Key • Resources)
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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