Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution 2)
 
(13 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2002 AMC 12P Problems|2002 AMC 12P #4]] and [[2002 AMC 10P Problems|2002 AMC 10P #10]]}}
 +
 
== Problem ==
 
== Problem ==
 
Let <math>a</math> and <math>b</math> be distinct real numbers for which
 
Let <math>a</math> and <math>b</math> be distinct real numbers for which
Line 18: Line 20:
  
 
== Solution 1==
 
== 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, $a=
+
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> for the AMC 12P and <math>\boxed{\textbf{(C) } 0.8}</math> 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 <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> for the AMC 12P and <math>\boxed{\textbf{(C) } 0.8}</math> for the AMC 10P.
 +
 
 +
=== Note ===
 +
For some unknown reason, the answer choices for the [[2002 AMC 10P|2002 AMC 10P]] are different from the answer choices for the [[2002 AMC 12P|2002 AMC 12P]], even though the question is exactly the same. Indeed, <math>\boxed{\textbf{(C) }0.8}</math> is the correct answer choice for the 2002 AMC 10P.
  
 
== See also ==
 
== See also ==
 +
{{AMC10 box|year=2002|ab=P|num-b=9|num-a=11}}
 
{{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 17:40, 15 July 2024

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 (Problems • Answer Key • Resources)
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 (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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12P_Problems/Problem_4&oldid=222650"