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# Davis drove from Amityville to Beteltown at 50 miles per

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Davis drove from Amityville to Beteltown at 50 miles per [#permalink]  15 Jul 2018, 02:55
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Question Stats:

80% (00:48) correct 20% (00:04) wrong based on 15 sessions
Davis drove from Amityville to Beteltown at 50 miles per hour, and returned by the same route at 60 miles per hour.

 Quantity A Quantity B Davis’s average speed for the round trip, in miles per hour $$55$$

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Sandy
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GRE Prep Club Legend
Joined: 07 Jun 2014
Posts: 4857
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 105

Kudos [?]: 1776 [0], given: 397

Re: Davis drove from Amityville to Beteltown at 50 miles per [#permalink]  15 Jul 2018, 03:05
Expert's post
Explanation

Never take an average speed by just averaging the two speeds (50 mph and 60 mph). Instead, use the formula Average Speed = Total Distance ÷ Total Time. Fortunately, for Quantitative Comparisons, you can often sidestep actual calculations.

Davis’s average speed can be thought of as an average of the speed he was traveling at every single moment during his journey—for instance, imagine that Davis wrote down the speed he was going during every second he was driving, then he averaged all the seconds. Since Davis spent more time going 50 mph than going 60 mph, the average speed will be closer to 50 than 60, and Quantity B is greater.

If the distances are the same, average speed is always weighted towards the slower speed.

To actually do the math, pick a convenient number for the distance between Amityville and Beteltown —for instance, 300 miles (divisible by both 50 and 60). If the distance is 300 miles, it took Davis 6 hours to drive there at 50 mph, and 5 hours to drive back at 60 mph. Using Average Speed = Total Distance ÷ Total Time (and a total distance of 600 miles, for both parts of the journey), you get the following:

Average Speed =$$\frac{600}{11}\frac{miles}{hours}$$

Average Speed = 54.54 … (which is less than 55)

The result will be the same for any value chosen. Quantity B is greater.
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Re: Davis drove from Amityville to Beteltown at 50 miles per [#permalink]  05 Aug 2018, 00:18
Its good to keep in mind that the average speed will be close to slower speed. hecne B
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Re: Davis drove from Amityville to Beteltown at 50 miles per [#permalink]  05 Aug 2018, 05:59
Expert's post
sandy wrote:
Davis drove from Amityville to Beteltown at 50 miles per hour, and returned by the same route at 60 miles per hour.

 Quantity A Quantity B Davis’s average speed for the round trip, in miles per hour $$55$$

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

GIVEN: Davis drove from Amityville to Beteltown at 50 miles per hour, and returned by the same route at 60 miles per hour.

Let's determine Davis’s average speed for the round trip, in miles per hour (Quantity A)
Average speed = (total distance)/(total time)

Let d = distance from Amityville to Beteltown

So, 2d = total distance

Total time = time spent driving 50 mph + time spent driving 60 mph
time = distance/speed
So, Total time = d/50 + d/60
= 6d/300 + 5d/300
= 11d/300

So, Average speed = (2d)/(11d/300)
= 600d/11d
= 600/11
= 54 6/11

We get:
Quantity A: 54 6/11
Quantity B: 55

Cheers,
Brent
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Re: Davis drove from Amityville to Beteltown at 50 miles per   [#permalink] 05 Aug 2018, 05:59
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