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Difference between revisions of "1981 AHSME Problems/Problem 12"

(Created page with "==Problem== If <math>p</math>, <math>q</math>, and <math>M</math> are positive numbers and <math>q<100</math>, then the number obtained by increasing <math>M</math> by <math>p...")
 
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==Problem==
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==Problem 12==
 
If <math>p</math>, <math>q</math>, and <math>M</math> are positive numbers and <math>q<100</math>, then the number obtained by increasing <math>M</math> by <math>p\%</math> and decreasing the result by <math>q\%</math> exceeds <math>M</math> if and only if
 
If <math>p</math>, <math>q</math>, and <math>M</math> are positive numbers and <math>q<100</math>, then the number obtained by increasing <math>M</math> by <math>p\%</math> and decreasing the result by <math>q\%</math> exceeds <math>M</math> if and only if
  
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<math>\textbf{(A)}\ p>q \qquad\textbf{(B)}\ p>\dfrac{q}{100-q}\qquad\textbf{(C)}\ p>\dfrac{q}{1-q}\qquad \textbf{(D)}\ p>\dfrac{100q}{100+q}\qquad \\ \textbf{(E)}\ p>\dfrac{100q}{100-q}</math>
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<math>\textbf{(A)}\ p>q \qquad\textbf{(B)}\ p>\dfrac{q}{100-q}\qquad\textbf{(C)}\ p>\dfrac{q}{1-q}\qquad \textbf{(D)}\ p>\dfrac{100q}{100+q}\qquad\textbf{(E)}\ p>\dfrac{100q}{100-q}</math>
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==Solution 1==
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The problem asks when <math>M\left(1+\frac{p}{100}\right)\left(1-\frac{q}{100}\right) > M</math>.  We can simplify this inequality and isolate <math>p</math> as follows:
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<math>\left(1+\frac{p}{100}\right)\left(1-\frac{q}{100}\right) > 1</math>
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<math>\left(\frac{100+p}{100}\right)\left(\frac{100-q}{100}\right) > 1</math>
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<math>\frac{(100+p)(100-q)}{10000} > 1</math>
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<math>100+p > \frac{10000}{100-q}</math>
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<math>p > \frac{10000}{100-q} - 100 = \frac{100q}{100-q} \fbox{(E)}</math>
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-j314andrews
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==Solution 2 (Answer Choices)==
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Answer Choice <math>A</math>: It is obviously incorrect because if <math>M</math> is <math>50</math> and we increase by <math>50</math>% and then decrease <math>49</math>%, <math>M</math> will be around <math>37</math>.
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Answer Choice <math>B</math>: If <math>p</math> is <math>100</math> and <math>q</math> is <math>50</math>, it should be equal but instead we get <math>100</math> is more than <math>1</math>. This is therefore also incorrect.
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Answer Choice <math>C</math>: Obviously incorrect since if <math>q</math> is larger than <math>1</math>, this is always valid since <math>\frac {1}{1-q}</math> is less than <math>0</math> which is obviously false.
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Answer Choice <math>D</math>: If <math>p</math> is <math>100</math> and <math>q</math> is <math>50</math>, it should be equal but instead we get <math>100</math> is less than <math>\frac {5000}{150}</math>. Therefore, <math>D</math> is also incorrect.
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Answer Choice <math>E</math>: If <math>p</math> is <math>100</math> and <math>q</math> is <math>50</math>, it should be equal, and if we check our equation, we get <math>\frac {5000}{50}</math> = <math>100</math>. Therefore, our answer is <math>\boxed {(E)\dfrac{100q}{100-q}}</math>
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~Arcticturn
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==See also==
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{{AHSME box|year=1981|num-b=11|num-a=13}}
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{{MAA Notice}}

Latest revision as of 13:54, 28 June 2025

Problem 12

If $p$, $q$, and $M$ are positive numbers and $q<100$, then the number obtained by increasing $M$ by $p\%$ and decreasing the result by $q\%$ exceeds $M$ if and only if

$\textbf{(A)}\ p>q \qquad\textbf{(B)}\ p>\dfrac{q}{100-q}\qquad\textbf{(C)}\ p>\dfrac{q}{1-q}\qquad \textbf{(D)}\ p>\dfrac{100q}{100+q}\qquad\textbf{(E)}\ p>\dfrac{100q}{100-q}$

Solution 1

The problem asks when $M\left(1+\frac{p}{100}\right)\left(1-\frac{q}{100}\right) > M$. We can simplify this inequality and isolate $p$ as follows:

$\left(1+\frac{p}{100}\right)\left(1-\frac{q}{100}\right) > 1$

$\left(\frac{100+p}{100}\right)\left(\frac{100-q}{100}\right) > 1$

$\frac{(100+p)(100-q)}{10000} > 1$

$100+p > \frac{10000}{100-q}$

$p > \frac{10000}{100-q} - 100 = \frac{100q}{100-q} \fbox{(E)}$

-j314andrews

Solution 2 (Answer Choices)

Answer Choice $A$: It is obviously incorrect because if $M$ is $50$ and we increase by $50$% and then decrease $49$%, $M$ will be around $37$.

Answer Choice $B$: If $p$ is $100$ and $q$ is $50$, it should be equal but instead we get $100$ is more than $1$. This is therefore also incorrect.

Answer Choice $C$: Obviously incorrect since if $q$ is larger than $1$, this is always valid since $\frac {1}{1-q}$ is less than $0$ which is obviously false.

Answer Choice $D$: If $p$ is $100$ and $q$ is $50$, it should be equal but instead we get $100$ is less than $\frac {5000}{150}$. Therefore, $D$ is also incorrect.

Answer Choice $E$: If $p$ is $100$ and $q$ is $50$, it should be equal, and if we check our equation, we get $\frac {5000}{50}$ = $100$. Therefore, our answer is $\boxed {(E)\dfrac{100q}{100-q}}$

~Arcticturn

See also

1981 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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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