Difference between revisions of "1970 AHSME Problems/Problem 9"

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== Solution ==
 
== Solution ==
<math>\fbox{C}</math>
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In order, the points from left to right are <math>A, P, Q, B</math>.  Let the lengths between successive points be <math>x, 2, y</math>, respectively. 
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Since <math>\frac{AP}{PB} = \frac{2}{3}</math>, we have <math>\frac{x}{2 + y} = \frac{2}{3}</math>.
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Since <math>\frac{AQ}{QB} = \frac{3}{4}</math>, we have <math>\frac{x + 2}{y} = \frac{3}{4}</math>.
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The first equation gives <math>x = \frac{2}{3}(2 + y)</math>.  Plugging that in to the second equation gives:
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<math>\frac{\frac{2}{3}(2 + y) + 2}{y} = \frac{3}{4}</math>.
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<math>\frac{2}{3}(2 + y) + 2 = \frac{3}{4}y</math>
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<math>\frac{4}{3} + \frac{2}{3}y + 2 = \frac{3}{4}y</math>
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<math>\frac{10}{3} = </math>\frac{1}{12}y<math>
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</math>y = 40<math>
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Since </math>x = \frac{2}{3}(2 + y)<math>, we have </math>x = 28<math>.  The length of the entire segment is </math>x + 2 + y<math>, which is </math>70<math>, or option </math>\fbox{C}$.
  
 
== See also ==
 
== See also ==

Revision as of 20:10, 13 July 2019

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$$ (Error compiling LaTeX. ! Missing $ inserted.)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$\fbox{C}$.

See also

1970 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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
All AHSME Problems and Solutions

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