Difference between revisions of "2002 AMC 12P Problems/Problem 4"
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== Problem == | == Problem == | ||
− | + | Let <math>a</math> and <math>b</math> be distinct real numbers for which | |
+ | <cmath>\frac{a}{b} + \frac{a+10b}{b+10a} = 2.</cmath> | ||
− | <math> \ | + | Find <math>\frac{a}{b}</math> |
− | == Solution == | + | <math> |
− | + | \text{(A) }0.4 | |
+ | \qquad | ||
+ | \text{(B) }0.5 | ||
+ | \qquad | ||
+ | \text{(C) }0.6 | ||
+ | \qquad | ||
+ | \text{(D) }0.7 | ||
+ | \qquad | ||
+ | \text{(E) }0.8 | ||
+ | </math> | ||
+ | |||
+ | == Solution 1== | ||
+ | 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> | ||
+ | |||
+ | == 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= | + | {{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
Contents
Problem
Let and be distinct real numbers for which
Find
Solution 1
For sake of speed, WLOG, let . This means that the ratio will simply be because Solving for with some very simple algebra gives us a quadratic which is Factoring the quadratic gives us . Therefore, or However, since and must be distinct, cannot be so the latter option is correct, giving us our answer of
Solution 2
The only tricky part about this equation is the fact that the left-hand side has fractions. Multiplying both sides by gives us Moving everything to the left-hand side and dividing by gives which factors into Because giving us our answer of
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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