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Difference between revisions of "2002 AMC 12P Problems/Problem 4"

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== Solution 1==
 
== Solution 1==
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For sake of speed, WLOG, let <math>b=1</math>. This means that the ratio <math>\frac{a}{b}</math> will simply be <math>a</math> because <math>\frac{a}{b}=\frac{a}{1}=a.</math> Solving for <math>a</math> with some very simple algebra gives us a quadratic which is <math>5a^2 -9a +4=0</math>. Factoring the quadratic gives us <math>(5a-4)(a-1)=0</math>. Therefore, <math>a=1</math> or <math>a=\frac{4}{5}=0.8</math>. However, since <math>a</math> and <math>b</math> must be distinct, the latter option is correct, giving us our answer of <math>\boxed{/textbf{(E) } 0.8}.</math>
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For sake of speed, WLOG, let <math>b=1</math>. This means that the ratio <math>\frac{a}{b}</math> will simply be <math>a</math> because <math>\frac{a}{b}=\frac{a}{1}=a.</math> Solving for <math>a</math> with some very simple algebra gives us a quadratic which is <math>5a^2 -9a +4=0.</math> Factoring the quadratic gives us <math>(5a-4)(a-1)=0</math>. Therefore, <math>a=1</math> or <math>a=\frac{4}{5}=0.8.</math> However, since <math>a</math> and <math>b</math> must be distinct, <math>a</math> cannot be <math>1</math> so the latter option is correct, giving us our answer of <math>\boxed{\textbf{(E) } 0.8}.</math>
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 +
== Solution 2==
 +
The only tricky part about this equation is the fact that the left-hand side has fractions. Multiplying both sides by <math>b(b+10a)</math> gives us <math>2ab+10a^2+10b^2=2b^2+20ab.</math> Moving everything to the left-hand side and dividing by <math>2</math> gives <math>5a^2-4b^2 -9ab,</math> which factors into <math>(5a-4b)(a-b)=0.</math> Because <math>a \neq b, 5a=4b \implies \frac{a}{b}=0.8</math> giving us our answer of <math>\boxed{\textbf{(E) } 0.8}.</math>
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2002|ab=P|num-b=3|num-a=5}}
 
{{AMC12 box|year=2002|ab=P|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:31, 14 January 2024

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

See also

2002 AMC 12P (Problems • Answer Key • Resources)
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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