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> | ||
− | ==Solution== | + | ==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: | ||
− | <math>\frac{2}{5}N=\frac{ | + | <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> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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== | ||
+ | {{AHSME box|year=1995|num-b=3|num-a=5}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 21:30, 16 February 2018
Contents
[hide]Problem
If is of , is of , and is of , then
Solution 1
We are given: , , . We want M in terms of N, so we substitute N into everything:
Solution 2
Alternatively, picking an arbitrary value for of , we find that .
We find that , meaning , giving .
Finally, since is of , we have .
Thus, and , so their ratio
This method does not prove that the answer must be constant, but it proves that if the answer is a constant, it must be .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.