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

Revision as of 21:47, 30 December 2023

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=

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 17: / Line 17: @@
 </math>
-== Solution ==
+== Solution 1==
-If <math>\log_{b} 729 = n</math>, then <math>b^n = 729</math>. Since <math>729 = 3^6</math>, <math>b</math> must be <math>3</math> to some [[factor]] of 6. Thus, there are four (3, 9, 27, 729) possible values of <math>b \Longrightarrow \boxed{\mathrm{E}}</math>.
+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, $a=
 == See also ==
 {{AMC12 box|year=2002|ab=P|num-b=3|num-a=5}}
 {{MAA Notice}}

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

Revision as of 21:47, 30 December 2023

Problem

Solution 1

See also