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Updated on: 03 May 2018, 12:09
Originally posted by MathRevolution on 02 May 2018, 03:03.
Last edited by MathRevolution on 03 May 2018, 12:09, edited 1 time in total.
Last edited by MathRevolution on 03 May 2018, 12:09, edited 1 time in total.
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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[GMAT math practice question]
The distance between Alice’s house and Bob's is 120 miles. Alice and Bob travel in opposite directions along the same road, starting at the same time. They meet at a place that is 40 miles away from Alice’s house. If Alice’s velocity is x miles per hour, what is Bob's velocity in terms of x?
\(A. (\frac{1}{3})x\)
\(B. (\frac{1}{2})x\)
\(C. x\)
\(D. 2x\)
\(E. 3x\)
The distance between Alice’s house and Bob's is 120 miles. Alice and Bob travel in opposite directions along the same road, starting at the same time. They meet at a place that is 40 miles away from Alice’s house. If Alice’s velocity is x miles per hour, what is Bob's velocity in terms of x?
\(A. (\frac{1}{3})x\)
\(B. (\frac{1}{2})x\)
\(C. x\)
\(D. 2x\)
\(E. 3x\)
02 May 2018, 07:15
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This is what i am getting alice's distance = 40mi alice's rate = x mi/hr --> Alice's time which must equal Bob's time = 40/x
Bob's distance must = 80 Mi Bobs rate then must be 80/40/x=2x? I must be missing something.
Bob's distance must = 80 Mi Bobs rate then must be 80/40/x=2x? I must be missing something.
02 May 2018, 07:21
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MathRevolution
Dear MathRevolution,
If they met 40 miles from Alice house, so A walked 40 miles and Bob 80 miles. The time is the same, so
40/x = 80/y
y=2x
Answer (D)
Please check OA. Logicaly Bob's speed can not be twice slower than Alice's, otherwise they would met closer to Bob's house.
