1996 AJHSME Problems/Problem 7

Problem

Brent has goldfish that quadruple (become four times as many) every month, and Gretel has goldfish that double every month. If Brent has 4 goldfish at the same time that Gretel has 128 goldfish, then in how many months from that time will they have the same number of goldfish?

$\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$

Solution 1

Call this month "Month 0". Make a table of the fish that Brent and Gretel have each month.

$\text{Month / Brent / Gretel}$

$\text{0 / 4 / 128}$

$\text{1 / 16 / 256}$

$\text{2 / 64 / 512}$

$\text{3 / 256 / 1024}$

$\text{4 / 1024 / 2048}$

$\text{5 / 4096 / 4096}$

You could create a similar table without doing all of the calculations by converting all the goldfish into powers of $2$ . In this table, you could increase Brent's goldfish by two powers of $2$ , while increasing Gretel's fish by one power of $2$ .

Either way, in $5$ months they will have the same number of fish, giving an answer of $\boxed{B}$

Solution 2

Create two equations.

Brent starts with $4$ goldfish and they quadruple each month, giving us $y=4\cdot 4^{x}$ or $y=2^{2x+2}$ .

Gretel starts with $128$ goldfish and they double each month, giving us $y=128 \cdot 2^{x}$ or $y=2^{x+7}$ .

Using substitution, $2^{2x+2}=2^{x+7}$ . Thus $2x+2=x+7$ giving us $x=5$ , which is $\boxed{B}$ .

1996 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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1996 AJHSME Problems/Problem 7

Contents

Problem

Solution 1

Solution 2

See also