2002 AMC 12P Problems/Problem 4

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 $\boxed{\textbf{(E) } 0.8}.$

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 $\boxed{\textbf{(E) } 0.8}.$

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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All AMC 12 Problems and Solutions

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2002 AMC 12P Problems/Problem 4

Contents

Problem

Solution 1

Solution 2

See also