Difference between revisions of "1988 AHSME Problems/Problem 8"
(Created page with "==Problem== If <math>\frac{b}{a} = 2</math> and <math>\frac{c}{b} = 3</math>, what is the ratio of <math>a + b</math> to <math>b + c</math>? <math>\textbf{(A)}\ \frac{1}{3}\qqu...") |
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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. | |
+ | 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.