Difference between revisions of "1970 AHSME Problems/Problem 9"
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<math>\frac{4}{3} + \frac{2}{3}y + 2 = \frac{3}{4}y</math> | <math>\frac{4}{3} + \frac{2}{3}y + 2 = \frac{3}{4}y</math> | ||
− | <math>\frac{10}{3} = | + | <math>\frac{10}{3} = \frac{1}{12}y</math> |
− | < | + | <math>y = 40</math> |
− | Since < | + | 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 == |
Latest revision as of 20:11, 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:
.
Since , we have . The length of the entire segment is , which is , or option .
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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