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Difference between revisions of "1996 AJHSME Problems/Problem 7"

(Created page with "==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...")
 
 
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<math>\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8</math>
 
<math>\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8</math>
  
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==Solution==
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==Solution 1==
  
 
Call this month "Month 0".  Make a table of the fish that Brent and Gretel have each month.
 
Call this month "Month 0".  Make a table of the fish that Brent and Gretel have each month.
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<math>\text{5 / 4096 / 4096}</math>
 
<math>\text{5 / 4096 / 4096}</math>
  
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You could create a similar table without doing all of the calculations by converting all the goldfish into powers of <math>2</math>.  In this table, you could increase Brent's goldfish by two powers of <math>2</math>, while increasing Greta's fish by one power of <math>2</math>.
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You could create a similar table without doing all of the calculations by converting all the goldfish into powers of <math>2</math>.  In this table, you could increase Brent's goldfish by two powers of <math>2</math>, while increasing Gretel's fish by one power of <math>2</math>.
  
 
Either way, in <math>5</math> months they will have the same number of fish, giving an answer of <math>\boxed{B}</math>
 
Either way, in <math>5</math> months they will have the same number of fish, giving an answer of <math>\boxed{B}</math>
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 +
==Solution 2==
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Create two equations.
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 +
Brent starts with <math>4</math> goldfish and they quadruple each month, giving us <math>y=4\cdot 4^{x}</math> or <math>y=2^{2x+2}</math>.
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Gretel starts with <math>128</math> goldfish and they double each month, giving us <math>y=128 \cdot 2^{x}</math> or <math>y=2^{x+7}</math>.
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Using substitution, <math>2^{2x+2}=2^{x+7}</math>. Thus <math>2x+2=x+7</math> giving us <math>x=5</math>, which is <math>\boxed{B}</math>.
  
 
== See also ==
 
== See also ==
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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 10:26, 27 June 2023

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 $\boxed{B}$.

See also

1996 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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

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

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