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> | ||
− | ==Solution== | + | ==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>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | WLOG, let <math>b=4, a=2, c=12.</math> | ||
+ | Thus, the answer is <math>\frac{4+2}{12+4}= \frac{3}{8}</math> | ||
== See also == | == See also == |
Latest revision as of 17:50, 22 June 2020
Contents
Problem
If and , what is the ratio of to ?
Solution 1
Since we are finding ratios, it would be helpful to put everything in terms of one variable. Since is in both equations, that would be a place to start. We manipulate the equations yielding and . Since we are asked to find the ratio of to , we need to find . We found the and in terms of so that means we can plug them in. We have: . Thus the answer is .
Solution 2
WLOG, let Thus, the answer is
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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