# 2015 AMC 10A Problems/Problem 6

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 1

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 factorised into $2a = 3b$ 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 they clearly satisfy the condition of the problem. The ratio of the larger to the smaller is $\boxed{\textbf{(B) }\frac32}$.

## Video Solution

~savannahsolver

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