# 1970 AHSME Problems/Problem 9

## Problem

Points $P$ and $Q$ are on line segment $AB$, and both points are on the same side of the midpoint of $AB$. Point $P$ divides $AB$ in the ratio $2:3$, and $Q$ divides $AB$ in the ratio $3:4$. If $PQ$=2, then the length of segment $AB$ is $\text{(A) } 12\quad \text{(B) } 28\quad \text{(C) } 70\quad \text{(D) } 75\quad \text{(E) } 105$

## Solution

In order, the points from left to right are $A, P, Q, B$. Let the lengths between successive points be $x, 2, y$, respectively.

Since $\frac{AP}{PB} = \frac{2}{3}$, we have $\frac{x}{2 + y} = \frac{2}{3}$.

Since $\frac{AQ}{QB} = \frac{3}{4}$, we have $\frac{x + 2}{y} = \frac{3}{4}$.

The first equation gives $x = \frac{2}{3}(2 + y)$. Plugging that in to the second equation gives: $\frac{\frac{2}{3}(2 + y) + 2}{y} = \frac{3}{4}$. $\frac{2}{3}(2 + y) + 2 = \frac{3}{4}y$ $\frac{4}{3} + \frac{2}{3}y + 2 = \frac{3}{4}y$ $\frac{10}{3} = \frac{1}{12}y$ $y = 40$

Since $x = \frac{2}{3}(2 + y)$, we have $x = 28$. The length of the entire segment is $x + 2 + y$, which is $70$, or option $\fbox{C}$.

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