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Janine travelled 1,320 miles on her road trip at an average

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pranab223
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Janine travelled 1,320 miles on her road trip at an average [#permalink] New post   Updated on: 01 Oct 2017, 20:32
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Janine travelled 1,320 miles on her road trip at an average speed of 45 miles per hour.

Quantity A
Quantity B
The length of time, in hours, Janine spent driving
The length of time, in minutes, Janine spent driving to cover the first 2% of her total travel distance



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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D

Originally posted by pranab223 on 30 Sep 2017, 11:59.
Last edited by pranab223 on 01 Oct 2017, 20:32, edited 3 times in total.
Renamed the topic and edited the question.
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Re: Janine travelled 1,320 miles on her road trip at an average [#permalink] New post   01 Oct 2017, 05:45
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Total distance = 1320 miles.

Average speed is = 45 miles/hour.

Time taken for travel = \(\frac{1320}{45}=29.3333\)hours. (Quantity A)


2% of total distance = \(\frac{2}{100} \times 1320\) miles = 26.4 miles.

Time taken to travel this @ average speed = \(\frac{26.4}{45}=0.58666666666\)hours or 35.2 minutes. (Quantity B)


So numerical value of Quantity B is greater!
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Re: Janine travelled 1,320 miles on her road trip at an average [#permalink] New post   01 Oct 2017, 20:42
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sandy wrote:
Total distance = 1320 miles.

Average speed is = 45 miles/hour.

Time taken for travel = \(\frac{1320}{45}=29.3333\)hours. (Quantity A)


2% of total distance = \(\frac{2}{100} \times 1320\) miles = 26.4 miles.

Time taken to travel this @ average speed = \(\frac{26.4}{45}=0.58666666666\)hours or 35.2 minutes. (Quantity B)


So numerical value of Quantity B is greater!


This is not the right answer,

The option indeed is D.

Since her average speed is given and we do not know what is her actual speed at first 2% of trip so it could be any value for example, she might have driven 60 miles per hour for the first hour and then later traveled slower than 45 miles per hour (suppose 30 m/hr)for a period of time. Because it is not known how much Janine’s speed varied on the trip, the relationship cannot be determined from the information given.
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Re: Janine travelled 1,320 miles on her road trip at an average [#permalink] New post   27 May 2021, 03:16
pranab223 wrote:
sandy wrote:
Total distance = 1320 miles.

Average speed is = 45 miles/hour.

Time taken for travel = \(\frac{1320}{45}=29.3333\)hours. (Quantity A)


2% of total distance = \(\frac{2}{100} \times 1320\) miles = 26.4 miles.

Time taken to travel this @ average speed = \(\frac{26.4}{45}=0.58666666666\)hours or 35.2 minutes. (Quantity B)


So numerical value of Quantity B is greater!


This is not the right answer,

The option indeed is D.

Since her average speed is given and we do not know what is her actual speed at first 2% of trip so it could be any value for example, she might have driven 60 miles per hour for the first hour and then later traveled slower than 45 miles per hour (suppose 30 m/hr)for a period of time. Because it is not known how much Janine’s speed varied on the trip, the relationship cannot be determined from the information given.

Is not it enough providing average speed sir?
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Re: Janine travelled 1,320 miles on her road trip at an average [#permalink] New post   27 May 2021, 04:09
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No because the average is just......the average. We do not have the EXACT value

Please refer to our book to brush your math skills

https://gre.myprepclub.com/forum/gre-math- ... -2609.html

Regards
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Re: Janine travelled 1,320 miles on her road trip at an average [#permalink] New post   13 Oct 2025, 05:00
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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Re: Janine travelled 1,320 miles on her road trip at an average [#permalink]
13 Oct 2025, 05:00
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