unuel', ), ); ?> Was Rylon’s average speed greater than 55 miles per hour? (1) Rylon dr : Data Sufficiency (DS)
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# Was Rylon’s average speed greater than 55 miles per hour? (1) Rylon dr

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Was Rylon’s average speed greater than 55 miles per hour? (1) Rylon dr  [#permalink]

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17 Apr 2018, 05:30
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25% (medium)

Question Stats:

87% (01:17) correct 13% (01:36) wrong based on 43 sessions

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Was Rylon’s average speed greater than 55 miles per hour?

(1) Rylon drove 100 miles in 1 hour 48 minutes.
(2) Rylon drove 60 miles at an average speed of 60 miles per hour, and then drove 40 miles at an average speed of 50 miles per hour.

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Re: Was Rylon’s average speed greater than 55 miles per hour? (1) Rylon dr  [#permalink]

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18 Apr 2018, 00:25
Bunuel wrote:
Was Rylon’s average speed greater than 55 miles per hour?

(1) Rylon drove 100 miles in 1 hour 48 minutes.
(2) Rylon drove 60 miles at an average speed of 60 miles per hour, and then drove 40 miles at an average speed of 50 miles per hour.

Average Speed = Total distance/Total time

St 1: A.S. = 100 /1.8

A.S. = 55.555 m/h Sufficient

St 2: Time 1 = 60/60 = 1hr

Time 2 = 40/50 = 0.8 hr

A.S. = (60 + 40)/1.8

A.S = 55.555 m/h Sufficient

(D)
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Re: Was Rylon’s average speed greater than 55 miles per hour? (1) Rylon dr  [#permalink]

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18 Apr 2018, 08:34
The answer is (D). We need to know whether the average speed is over 50 mph. Therefore, we will work with the average speed formula, of which we have 2. When we just have the distance and time, we work with the $$\frac{total distance}{total time}$$ formula. But when we have distance and speed, we can calculate the time and then work out the average speed.

Lets get to the good stuff now.

Statement (1) states that Rylon drove 100 miles in 1 hr 48 mins. Speed = $$\frac{distance}{time}$$
=> Speed = $$100/1\frac{48}{60}$$
=> 100/1.8=55.5555. Therefore, statement (1) is sufficient.

Statement (2) states that Rylon drove 60 miles at 60mph and 40 miles at 50mph. We can use a weighted average here or we can do a manual calculation of distance travelled in time. I prefer the following calculation.
=> travelling 60 miles at 60 mph will take 1 hour. Travelling 40 miles at 50 mph will take 4/5th of an hour (40/50) so 60 * \frac{4}{5} is 48 minutes.
Thus, 1 hour + 48 minutes to travel a distance of 60 miles + 40 miles. Thats 1 hr 48 mins to travel 100 miles. Plug this into our first formula and we get the average speed.
Therefore, statement (2) is sufficient as well.

Thus, the correct answer is option (D).
Re: Was Rylon’s average speed greater than 55 miles per hour? (1) Rylon dr &nbs [#permalink] 18 Apr 2018, 08:34
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