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Alice travels 120 km, the first 40 km at x km/h and the remaining 80 k

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MathRevolution
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Alice travels 120 km, the first 40 km at x km/h and the remaining 80 k [#permalink] New post   20 Jul 2018, 01:29
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[GMAT math practice question]

Alice travels 120 km, the first 40 km at x km/h and the remaining 80 km at y km/h. What is her average speed for the entire trip?

\(A. \frac{2xy}{(2x+y)}\)
\(B. \frac{2xy}{(3x+y)}\)
\(C. \frac{3xy}{(2x+y)}\)
\(D. \frac{xy}{(2x+y)}\)
\(E. \frac{xy}{(x+2y)}\)
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C
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Re: Alice travels 120 km, the first 40 km at x km/h and the remaining 80 k [#permalink] New post   20 Jul 2018, 01:43
MathRevolution wrote:
[GMAT math practice question]

Alice travels 120 km, the first 40 km at x km/h and the remaining 80 km at y km/h. What is her average speed for the entire trip?

\(A. \frac{2xy}{(2x+y)}\)
\(B. \frac{2xy}{(3x+y)}\)
\(C. \frac{3xy}{(2x+y)}\)
\(D. \frac{xy}{(2x+y)}\)
\(E. \frac{xy}{(x+2y)}\)


Avg speed =Total Distance/Total Time Taken
= \(\frac{120}{(40/x+80/y)}\)
=\(\frac{3xy}{(2x+y)}\)
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LevanKhukhunashvili
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Alice travels 120 km, the first 40 km at x km/h and the remaining 80 k [#permalink] New post   20 Jul 2018, 04:41
For me the easiest way is to draw a diagram

..............Distance.....Rate........Time
..............40 km........x km/h....\(\frac{40}{x}\)
..............80 km........y km/h....\(\frac{80}{y}\)
--------------------------------------------------------------------
....Total...120 km.......{--}........\(\frac{40y+80x}{xy}\)

to calculate rate: 120*\(\frac{xy}{40y+80x}\)=\(\frac{3xy}{y+2x}\)

IMO
Ans: C
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Re: Alice travels 120 km, the first 40 km at x km/h and the remaining 80 k [#permalink] New post   20 Jul 2018, 06:19
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MathRevolution wrote:
[GMAT math practice question]

Alice travels 120 km, the first 40 km at x km/h and the remaining 80 km at y km/h. What is her average speed for the entire trip?

\(A. \frac{2xy}{(2x+y)}\)
\(B. \frac{2xy}{(3x+y)}\)
\(C. \frac{3xy}{(2x+y)}\)
\(D. \frac{xy}{(2x+y)}\)
\(E. \frac{xy}{(x+2y)}\)


This is a simple Distance-Time problem!

We know that \(Speed = \frac{Total Distance}{Total Time}\) ----------(1)

Total Distance here is 40 km + 80 km = 120 kms ----------(2)

Time taken for the first 40 km at x km/h = \(\frac{40}{x}\). Similarly, Time taken for the remaining 80 km at y km/h = \(\frac{80}{y}\)

Hence, Total Time =\(\frac{40}{x}\) + \(\frac{80}{y}\) ---------- (3)

Substitute (2) and (3) in (1)

We will then get, \(Speed = \frac{120}{40/x + 80/y}\)

\(Speed = \frac{120 *x*y}{40*y+80*x}\)

Simplifying the above, we get, \(Speed = \frac{3xy}{2x+y}\)

Answer is C!



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Re: Alice travels 120 km, the first 40 km at x km/h and the remaining 80 k [#permalink] New post   22 Jul 2018, 18:25
Expert Reply
=>

The average speed is (Total Distance) / (Total Time).
The time for the first 40 km is 40 / x and the time for the remaining 80 km is 80 / y. Thus, the total time for the trip is ( 40/x + 80/y ).
Her average speed for the trip is
120 / ( 40/x + 80/y ) = 120xy / ( 40y + 80x ) = 3xy / ( y + 2x ) = 3xy /
(2x +y).

Therefore, the answer is C.
Answer : C
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Re: Alice travels 120 km, the first 40 km at x km/h and the remaining 80 k [#permalink] New post   23 Jul 2018, 18:21
Expert Reply
MathRevolution wrote:
[GMAT math practice question]

Alice travels 120 km, the first 40 km at x km/h and the remaining 80 km at y km/h. What is her average speed for the entire trip?

\(A. \frac{2xy}{(2x+y)}\)
\(B. \frac{2xy}{(3x+y)}\)
\(C. \frac{3xy}{(2x+y)}\)
\(D. \frac{xy}{(2x+y)}\)
\(E. \frac{xy}{(x+2y)}\)


We can use the average rate formula: average rate = total distance/total time

Average rate = 120/(40/x + 80/y)

Average rate = 120/(40y/xy + 80x/xy)

Average rate = 120/[(40y + 80x)/xy]

Average rate = 120xy/(40y + 80x)

Average rate = 120xy/40(y + 2x) = 3xy/(y + 2x)

Answer: C
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Re: Alice travels 120 km, the first 40 km at x km/h and the remaining 80 k [#permalink]
23 Jul 2018, 18:21
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