1961 AHSME Problems/Problem 23

Problem

Points $P$ and $Q$ are both in the line segment $AB$ and on the same side of its midpoint. $P$ divides $AB$ in the ratio $2:3$, and $Q$ divides $AB$ in the ratio $3:4$. If $PQ=2$, then the length of $AB$ is:

$\textbf{(A)}\ 60\qquad \textbf{(B)}\ 70\qquad \textbf{(C)}\ 75\qquad \textbf{(D)}\ 80\qquad \textbf{(E)}\ 85$

Solution

[asy] draw((0,0)--(70,0)); dot((0,0)); dot((28,0)); dot((30,0)); dot((70,0)); label("2",(29,0),N); label("x",(14,0),N); label("y",(50,0),N); label("A",(0,0),S); label("P",(28,0),SW); label("Q",(30,0),SE); label("B",(70,0),S); [/asy]

Draw diagram as shown, where $P$ and $Q$ are on the same side. Let $AP = x$ and $QB = y$.

Since $P$ divides $AB$ in the ratio $2:3$, $\frac{x}{y+2} = \frac{2}{3}$. Since $Q$ divides $AB$ in the ratio $3:4$, $\frac{x+2}{y} = \frac{3}{4}$. Cross multiply to get a system of equations. \[3x = 2y+4\] \[4x+8=3y\] Solve the system to get $x = 28$ and $y = 40$. Thus, $AB = 40 + 28 + 2 = 70$, so the answer is $\boxed{\textbf{(B)}}$.

See Also

1961 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
All AHSME Problems and Solutions


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

Invalid username
Login to AoPS