2007 AMC 10B Problem 15

Below is the professionally curated solution for Problem 15 of the 2007 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 10B solutions, or check the answer key.

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Concepts:angle sum ratio and proportion

Difficulty rating: 1370

15.

The angles of quadrilateral $ABCD$ satisfy $\angle A = 2\angle B = 3\angle C = 4\angle D.$ What is the degree measure of $\angle A,$ rounded to the nearest whole number?

$125$

$144$

$153$

$173$

$180$

Solution:

Let $x=\angle A.$ Then $\angle B=\dfrac{x}{2},$ $\angle C=\dfrac{x}{3},$ and $\angle D=\dfrac{x}{4}.$

The angles sum to $360^\circ,$ so $x+\dfrac{x}{2}+\dfrac{x}{3}+\dfrac{x}{4}=\dfrac{25x}{12}=360.$

Thus $x=\dfrac{12\cdot 360}{25}=172.8\approx 173.$

Thus, the correct answer is D.

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