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Michael drives x miles due north at arrives at point A. He then heads due east for y miles. Finally, he drives z miles..

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Michael drives x miles due north at arrives at point A. He then heads due east for y miles. Finally, he drives z miles..

by BTGmoderatorLU » Wed Dec 09, 2020 4:23 am

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Source: Magoosh

Michael drives x miles due north at arrives at point A. He then heads due east for y miles. Finally, he drives z miles in a straight line until he reaches his starting point. If x, y, and z are integers, then how many miles did Michael drive if the shortest leg was 5 miles?

A. 5 miles
B. 12 miles
C. 25 miles
D. 30 miles
E. Cannot be determined by the information given

The OA is D
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Re: Michael drives x miles due north at arrives at point A. He then heads due east for y miles. Finally, he drives z mil

by swerve » Wed Dec 09, 2020 4:39 pm
BTGmoderatorLU wrote: ↑
Wed Dec 09, 2020 4:23 am
 Source: Magoosh

Michael drives x miles due north at arrives at point A. He then heads due east for y miles. Finally, he drives z miles in a straight line until he reaches his starting point. If x, y, and z are integers, then how many miles did Michael drive if the shortest leg was 5 miles?

A. 5 miles
B. 12 miles
C. 25 miles
D. 30 miles
E. Cannot be determined by the information given

The OA is D
Here we go!

The shortest leg \(= 5\)
\(z=\)hypotenuse
\(x=3\)rd side

\(5^2+x^2=z^2\)
\(25=z^2-x^2\)
\(25=(z-x)(z+x)\)

Therefore,
\(z-x=25\) or \(z+x=25\)
\(z=25-x\) or \(z=25+x\)

\(30\) is ruled out and cannot be \(5\) as doesn't satisfy Pythagoras theorem. The only option left is \(12\).

Hence, B
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Re: Michael drives x miles due north at arrives at point A. He then heads due east for y miles. Finally, he drives z mil

by Scott@TargetTestPrep » Wed Dec 16, 2020 11:18 am
BTGmoderatorLU wrote: ↑
Wed Dec 09, 2020 4:23 am
 Source: Magoosh

Michael drives x miles due north at arrives at point A. He then heads due east for y miles. Finally, he drives z miles in a straight line until he reaches his starting point. If x, y, and z are integers, then how many miles did Michael drive if the shortest leg was 5 miles?

A. 5 miles
B. 12 miles
C. 25 miles
D. 30 miles
E. Cannot be determined by the information given

The OA is D
Solution:

We see that the distances of x, y and z miles must form a right triangle. Since x, y and z are integers, and the shortest leg is 5, then the triangle must be a 5-12-13 right triangle. So Michael drove 5 + 12 + 13 = 30 miles.

Answer: D

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