Difference between revisions of "2015 AMC 10A Problems/Problem 6"

Revision as of 15:24, 24 April 2022

The following problem is from both the 2015 AMC 12A #4 and 2015 AMC 10A #6, so both problems redirect to this page.

Problem

The sum of two positive numbers is $5$ times their difference. What is the ratio of the larger number to the smaller number?

$\textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \frac{5}{2}$

Solution

Let $a$ be the bigger number and $b$ be the smaller.

$a + b = 5(a - b)$ .

Multiplying out gives $a + b = 5a - 5b$ and rearranging gives $4a = 6b$ and then solving gives

$\frac{a}{b} = \frac32$ , so the answer is $\boxed{\textbf{(B) }\frac32}$ .

Solution 2

Without loss of generality, let the two numbers be $3$ and $2$ , as this pair clearly satisfies the condition of the problem. The ratio is $\boxed{\textbf{(B) }\frac32}$ .

Video Solution

https://youtu.be/JLKoh-Nb0Os

~savannahsolver

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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

2015 AMC 12A (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.

@@ Line 15: / Line 15: @@
 <math>\frac{a}{b} = \frac32</math>, so the answer is <math>\boxed{\textbf{(B) }\frac32}</math>.
+==Solution 2==
+Without loss of generality, let the two numbers be <math>3</math> and <math>2</math>, as this pair clearly satisfies the condition of the problem. The ratio is <math>\boxed{\textbf{(B) }\frac32}</math>.
 ==Video Solution==

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