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Tavi drives 113 miles at 50 miles per hour and returns

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Tavi drives 113 miles at 50 miles per hour and returns [#permalink] New post   20 Oct 2017, 16:05
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Tavi drives 113 miles at 50 miles per hour and returns via the same route at 60 miles per hour.

Quantity A
Quantity B
Tavi's average speed for the entire
round trip
55 mph


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.
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B
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IlCreatore
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Re: Tavi drives 113 miles at 50 miles per hour and returns [#permalink] New post   21 Oct 2017, 07:11
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Here is the trick, the average speed is not computed as a usual average, which would have lead to a wrong answer (C).

It is computed as \(\bar{speed} = \frac{distance_1+distance_2}{time_1+time_2}\). In our case, the distances are equal since the trip is one way and the other, i.e. 113 + 113 = 226 miles, while the time spent is computed as 113/50 = 2.6 hours and 113/60 = 1.883 hours. Substituting in the formula the average speed is 50.41.

Answer B
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Re: Tavi drives 113 miles at 50 miles per hour and returns [#permalink] New post   11 Mar 2018, 07:17
the right answer is B
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Re: Tavi drives 113 miles at 50 miles per hour and returns [#permalink] New post   12 Mar 2018, 19:32
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1st leg: t1 = 113/50
2nd leg: t2 = 113/60

Total time = t1 + t2 = 113/50 + 113/60

Average speed for the round trip = Total distance/Total time = (113 + 113)/(113/50 + 113/60) = 54.5 mph < 55 mph.

So Ans: B
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Re: Tavi drives 113 miles at 50 miles per hour and returns [#permalink] New post   13 Mar 2018, 15:49
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correct: B
route = 113 miles (50 miles / hour) 113/50 hours
return_route = 113 miles (60 miles /hour) 113 / 60 hours
his average speed is: total miles passed / total time passed = (113 + 113)/ (113/50 + 113/60) = 226 * 300/ 1243 = 54.4
average_speed = 54.4 miles per hour
so B is bigger than A
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Re: Tavi drives 113 miles at 50 miles per hour and returns [#permalink] New post   17 Apr 2019, 21:40
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Quick Solution.

Since Tavi spent more time driving at 50 mph than he did at 60 mph. His average speed must be closer to 50 than to 60. His average speed cannot be 55 since the amount of time he spent driving at each speed is different so. if he had spent the same amount of time at each speed then his average for the entire trip would have been 55 mph. So, here Qty B is greater than A.
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Re: Tavi drives 113 miles at 50 miles per hour and returns [#permalink] New post   08 Jun 2020, 06:26
IlCreatore wrote:
Here is the trick, the average speed is not computed as a usual average, which would have lead to a wrong answer (C).

It is computed as \(\bar{speed} = \frac{distance_1+distance_2}{time_1+time_2}\). In our case, the distances are equal since the trip is one way and the other, i.e. 113 + 113 = 226 miles, while the time spent is computed as 113/50 = 2.6 hours and 113/60 = 1.883 hours. Substituting in the formula the average speed is 50.41.

Answer B


113/50=2.26 not 2.6.
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Re: Tavi drives 113 miles at 50 miles per hour and returns [#permalink] New post   09 Jul 2020, 02:57
If the distance is same and speeds are different then you can use this easy formula to compute avg speed:
Avg Speed = 2ab/a+b
where a and b are speeds.
2(50)(60)/50+60
Answer is 54.5 so Quantity B is greater.
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Re: Tavi drives 113 miles at 50 miles per hour and returns [#permalink] New post   12 Jul 2020, 06:44
We essentially need average time to get average speed

113/50 = 2.26 (t1)
113/60 = 1.8 (t2)

=1.8 + 2.26 = 4.14
Aveg time = 4.14/2 = 2.07

Aveg Speed = 113/2.07 = 54.5
Thus its lesser than 55, ans B

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Re: Tavi drives 113 miles at 50 miles per hour and returns [#permalink] New post   21 Dec 2022, 02:28
Is the OA correct?
113/50= 2.26
113/60= 1.88

Total time is 4.14
226 divided by 4.14 is 54.58 which is less than 55

But if I take only 2.2 and 1.8 as the times the total time is 4
which when dividing 226 gives 56.5 which is more than 55
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Re: Tavi drives 113 miles at 50 miles per hour and returns [#permalink] New post   21 Dec 2022, 02:41
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The OA is correct

His speed is between 50 and 60. If you average 50+60/2=55 and the answer would be C. But we cannot simply average the two because the return is faster than the first stretch, which is at 50 mph.

we can use the weighted average concept.

Being the speed between 50 and 60 but because the first stretch is slower we do know that his speed average is NOT 55 but smaller. It does not matter if it is 54.9 or 22.3.

It MUST be < 55.

B is the answer
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Re: Tavi drives 113 miles at 50 miles per hour and returns [#permalink] New post   07 Apr 2024, 10:39
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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