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

(Created page with "== Problem == Points <math>P</math> and <math>Q</math> are on line segment <math>AB</math>, and both points are on the same side of the midpoint of <math>AB</math>. Point <math>...")
 
(→‎Solution)
 
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== Solution ==
 
== Solution ==
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<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} = \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}</math>.
  
 
== See also ==
 
== See also ==
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{{AHSME box|year=1970|num-b=8|num-a=10}}   
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{{AHSME 35p box|year=1970|num-b=8|num-a=10}}   
  
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[[Category: Introductory Geometry Problems]]
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[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:11, 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:

$\frac{\frac{2}{3}(2 + y) + 2}{y} = \frac{3}{4}$.

$\frac{2}{3}(2 + y) + 2 = \frac{3}{4}y$

$\frac{4}{3} + \frac{2}{3}y + 2 = \frac{3}{4}y$

$\frac{10}{3} = \frac{1}{12}y$

$y = 40$

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 (Problems • Answer Key • Resources)
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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