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Difference between revisions of "1988 AHSME Problems/Problem 8"

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\textbf{(E)}\ \frac{3}{4}  </math>   
 
\textbf{(E)}\ \frac{3}{4}  </math>   
  
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==Solution==
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==Solution 1 ==
 
Since we are finding ratios, it would be helpful to put everything in terms of one variable. Since <math>b</math> is in both equations, that would be a place to start.  
 
Since we are finding ratios, it would be helpful to put everything in terms of one variable. Since <math>b</math> is in both equations, that would be a place to start.  
 
We manipulate the equations yielding <math>\frac{b}{2}=a</math> and <math>c=3b</math>. Since we are asked to find the ratio of <math>a+b</math> to <math>b+c</math>, we need to find <math>\frac{a+b}{b+c}</math>. We found the <math>a</math> and <math>c</math> in terms of <math>b</math> so that means we can plug them in. We have: <math>\frac{\frac{b}{2}+b}{b+3b}=\frac{\frac{3}{2}b}{4b}=\frac{3}{8}</math>. Thus the answer is <math>\frac{3}{8} \implies \boxed{\text{B}}</math>.
 
We manipulate the equations yielding <math>\frac{b}{2}=a</math> and <math>c=3b</math>. Since we are asked to find the ratio of <math>a+b</math> to <math>b+c</math>, we need to find <math>\frac{a+b}{b+c}</math>. We found the <math>a</math> and <math>c</math> in terms of <math>b</math> so that means we can plug them in. We have: <math>\frac{\frac{b}{2}+b}{b+3b}=\frac{\frac{3}{2}b}{4b}=\frac{3}{8}</math>. Thus the answer is <math>\frac{3}{8} \implies \boxed{\text{B}}</math>.
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== Solution 2 ==
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WLOG, let <math>b=4, a=2, c=12.</math>
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Thus, the answer is <math>\frac{4+2}{12+4}= \frac{3}{8}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 16:50, 22 June 2020

Problem

If $\frac{b}{a} = 2$ and $\frac{c}{b} = 3$, what is the ratio of $a + b$ to $b + c$?

$\textbf{(A)}\ \frac{1}{3}\qquad \textbf{(B)}\ \frac{3}{8}\qquad \textbf{(C)}\ \frac{3}{5}\qquad \textbf{(D)}\ \frac{2}{3}\qquad \textbf{(E)}\ \frac{3}{4}$

Solution 1

Since we are finding ratios, it would be helpful to put everything in terms of one variable. Since $b$ is in both equations, that would be a place to start. We manipulate the equations yielding $\frac{b}{2}=a$ and $c=3b$. Since we are asked to find the ratio of $a+b$ to $b+c$, we need to find $\frac{a+b}{b+c}$. We found the $a$ and $c$ in terms of $b$ so that means we can plug them in. We have: $\frac{\frac{b}{2}+b}{b+3b}=\frac{\frac{3}{2}b}{4b}=\frac{3}{8}$. Thus the answer is $\frac{3}{8} \implies \boxed{\text{B}}$.

Solution 2

WLOG, let $b=4, a=2, c=12.$ Thus, the answer is $\frac{4+2}{12+4}= \frac{3}{8}$

See also

1988 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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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