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Difference between revisions of "1997 AJHSME Problems/Problem 16"

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==Problem==
 
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Penni Precisely buys <dollar/>100 worth of stock in each of three companies: Alabama Almonds, Boston Beans, and California Cauliflower.  After one year, AA was up 20%, BB was down 25%, and CC was unchanged.  For the second year, AA was down 20% from the previous year, BB was up 25% from the previous year, and CC was unchanged.  If A, B, and C are the final values of the stock, then
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Penni Precisely buys \$100 worth of stock in each of three companies: Alabama Almonds, Boston Beans, and California Cauliflower.  After one year, AA was up 20%, BB was down 25%, and CC was unchanged.  For the second year, AA was down 20% from the previous year, BB was up 25% from the previous year, and CC was unchanged.  If A, B, and C are the final values of the stock, then
  
 
<math>\text{(A)}\ A=B=C \qquad \text{(B)}\ A=B<C \qquad \text{(C)}\ C<B=A</math>
 
<math>\text{(A)}\ A=B=C \qquad \text{(B)}\ A=B<C \qquad \text{(C)}\ C<B=A</math>
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* [[AJHSME Problems and Solutions]]
 
* [[AJHSME Problems and Solutions]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]
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{{MAA Notice}}

Latest revision as of 07:07, 21 October 2020

Problem

Penni Precisely buys $100 worth of stock in each of three companies: Alabama Almonds, Boston Beans, and California Cauliflower. After one year, AA was up 20%, BB was down 25%, and CC was unchanged. For the second year, AA was down 20% from the previous year, BB was up 25% from the previous year, and CC was unchanged. If A, B, and C are the final values of the stock, then

$\text{(A)}\ A=B=C \qquad \text{(B)}\ A=B<C \qquad \text{(C)}\ C<B=A$

$\text{(D)}\ A<B<C \qquad \text{(E)}\ B<A<C$

Solution 1

AA is $100 \cdot 120\% = 100\cdot 1.2 = 120$ after one year. After the second year, AA is $120\cdot 0.8 = 96$.

BB is $100 \cdot 75\% = 100\cdot 0.75 = 75$ after one year. After the second year, BB is $75\cdot 1.25 = 93.75$

CC remains unchanged throughout, and stays at $100$.

Thus, $B < A < C$, and the right answer is $\boxed{E}$

Solution 2

AA will be $100 \cdot \frac{6}{5} \cdot \frac{4}{5} = 100\cdot \frac{24}{25}$ at the end.

BB will be $100 \cdot \frac{3}{4} \cdot \frac{5}{4} = 100 \cdot \frac{15}{16}$ at the end.

CC will be unchanged at $100$.

Since all the fractions are under $1$, $C$ will be highest value.

Since $\frac{24}{25}$ is only $\frac{1}{25}$ away from $1$, while $\frac{15}{16}$ is $\frac{1}{16}$ away from $1$, $\frac{24}{25}$ is closer to $1$, and will be closer to the original value.

Thus, $B < A < C$, and the right answer is $\boxed{E}$

See also

1997 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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
All AJHSME/AMC 8 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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