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 $\fbox{C}$ .

1970 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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1970 AHSME Problems/Problem 9

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