Difference between revisions of "1970 AHSME Problems/Problem 9"
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== Solution == | == Solution == | ||
− | <math>\ | + | 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: | ||
+ | |||
+ | <math>\frac{\frac{2}{3}(2 + y) + 2}{y} = \frac{3}{4}</math>. | ||
+ | |||
+ | <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 and are on line segment , and both points are on the same side of the midpoint of . Point divides in the ratio , and divides in the ratio . If =2, then the length of segment is
Solution
In order, the points from left to right are . Let the lengths between successive points be , respectively.
Since , we have .
Since , we have .
The first equation gives . Plugging that in to the second equation gives:
.
\frac{1}{12}y$$ (Error compiling LaTeX. ! Missing $ inserted.)y = 40x = \frac{2}{3}(2 + y)x = 28x + 2 + y70\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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