2002 AMC 12P Problems/Problem 4

The following problem is from both the 2002 AMC 12P #4 and 2002 AMC 10P #10, so both problems redirect to this page.

Problem

Let $a$ and $b$ be distinct real numbers for which \[\frac{a}{b} + \frac{a+10b}{b+10a} = 2.\]

Find $\frac{a}{b}$

$\text{(A) }0.4 \qquad \text{(B) }0.5  \qquad \text{(C) }0.6 \qquad \text{(D) }0.7 \qquad \text{(E) }0.8$

Solution 1

For sake of speed, WLOG, let $b=1$. This means that the ratio $\frac{a}{b}$ will simply be $a$ because $\frac{a}{b}=\frac{a}{1}=a.$ Solving for $a$ with some very simple algebra gives us a quadratic which is $5a^2 -9a +4=0.$ Factoring the quadratic gives us $(5a-4)(a-1)=0$. Therefore, $a=1$ or $a=\frac{4}{5}=0.8.$ However, since $a$ and $b$ must be distinct, $a$ cannot be $1$ so the latter option is correct, giving us our answer of $\boxed{\textbf{(E) } 0.8}$ for the AMC 12P and $\boxed{\textbf{(C) } 0.8}$ for the AMC 10P.

Solution 2

The only tricky part about this equation is the fact that the left-hand side has fractions. Multiplying both sides by $b(b+10a)$ gives us $2ab+10a^2+10b^2=2b^2+20ab.$ Moving everything to the left-hand side and dividing by $2$ gives $5a^2-4b^2 -9ab,$ which factors into $(5a-4b)(a-b)=0.$ Because $a \neq b, 5a=4b \implies \frac{a}{b}=0.8$ giving us our answer of $\boxed{\textbf{(E) } 0.8}$ for the AMC 12P and $\boxed{\textbf{(C) } 0.8}$ for the AMC 10P.

Note

For some unknown reason, the answer choices for the 2002 AMC 10P are different from the answer choices for the 2002 AMC 12P, even though the question is exactly the same. Indeed, $\boxed{\textbf{(C) }0.8}$ is the correct answer choice for the 2002 AMC 10P.

See also

2002 AMC 10P (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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 AMC 10 Problems and Solutions
2002 AMC 12P (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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 AMC 12 Problems and Solutions

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