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During a 40-mile trip, Marla traveled at an average speed of [#permalink]
27 Apr 2010, 08:55

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Difficulty:

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Question Stats:

61% (02:17) correct
39% (01:32) wrong based on 470 sessions

During a 40-mile trip, Marla traveled at an average speed of x miles per hour for the first y miles of the trip and and at an average speed of 1.25x miles per hour for the last 40 - y miles of the trip. The time that Marla took to travel the 40 miles was what percent of the time it would have taken her if she had traveled at an average speed of x miles per hour for the entire trip?

During a 40-mile trip, Marla traveled at an average speed of x miles per hour for the first y miles of the trip and and at an average speed of 1.25x mph for the last 40-y miles of the trip. The time that Marla took to travel the 40 miles was what percent of the time it would have taken her if she has traveled at an average speed of x miles per hour for the entire trip?

Q: \(\frac{t_1}{t_2}=\frac{0.25y+40}{1.25x}*\frac{x}{40}=\frac{0.25y+40}{1.25}*\frac{1}{40}\). So we see that the value of this fraction does not depend on \(x\), only on \(y\).

First read the stem and derive the appropriate equations.

We get,

[(y/x) + (40-y)/1.25x] is total time of the journey done by Marla. By simplifying, we get (y+160)/5x. Now, lets process this, 'what percent of the time it would have taken her if she had traveled at an average speed of x miles per hour for the entire trip' i.e. total time '40/x'

(40/x) P% = (y+160)/5x On simplifying, the x is removed from the equation and we are left with only y. i.e. P% = (y+160)/(5*40)

So we need only y value to obtain value of P. Therefore, we need only B.

Re: During a 40-mile trip, Marla traveled at an average speed of [#permalink]
29 Jul 2013, 10:14

During a 40-mile trip, Marla traveled at an average speed of x miles per hour for the first y miles of the trip and and at an average speed of 1.25x mph for the last 40-y miles of the trip. The time that Marla took to travel the 40 miles was what percent of the time it would have taken her if she has traveled at an average speed of x miles per hour for the entire trip?

(1) x = 48.

Answering the stem requires that we know how many miles she covered at what speed. For example, if someone covers 99 miles at 1 mile/hour and 1 mile at 1000 miles/hour, their average speed (and the time it takes for them to cover point A to B will be far more than if someone covers 99 miles at 1000 miles/hour and 1 mile at 1 mile/hour. This problem is no different. We know the speed she covered for the first portion of the trip and the speed she covered for the second portion of the trip (i.e. 1.25*48) but we don't know how many miles she covered for each speed. INSUFFICIENT

(2) y = 20.

If y = 20 that means she traveled 1/2 of the trip at speed x and 1/2 of the trip at 1.25x. Regardless of the speed of x the ratio of x to 1.25x will be the same because the distance covered by speed x and the distance covered by speed 1.25 x is the same. SUFFICIENT

I understand this problem conceptually (and in my post above this, got it right) but I have difficultly setting up the algebra to solve the problem as I do with many word and rate/distance problems. Here are my questions:

I.) I know why we have to add up y and (40-y) but I don't understand why it's set up as y/x + (40-y)/1.25x. Is it because distance/speed = time and we are looking for time in this question?

II.) Time 2 is x/40 which = speed/distance = time so we are multiplying the two averages (y/x + (40-y)/1.25x) combined by the single average (x/40) to get the percentage difference between the two averages and the one average, but why in your solution do you flip x/40 to 40/x?

Thanks!!!

Bunuel wrote:

During a 40-mile trip, Marla traveled at an average speed of x miles per hour for the first y miles of the trip and and at an average speed of 1.25x mph for the last 40-y miles of the trip. The time that Marla took to travel the 40 miles was what percent of the time it would have taken her if she has traveled at an average speed of x miles per hour for the entire trip?

Q: \(\frac{t_1}{t_2}=\frac{0.25y+40}{1.25x}*\frac{x}{40}=\frac{0.25y+40}{1.25}*\frac{1}{40}\). So we see that the value of this fraction does not depend on \(x\), only on \(y\).

I understand this problem conceptually (and in my post above this, got it right) but I have difficultly setting up the algebra to solve the problem as I do with many word and rate/distance problems. Here are my questions:

I.) I know why we have to add up y and (40-y) but I don't understand why it's set up as y/x + (40-y)/1.25x. Is it because distance/speed = time and we are looking for time in this question?

II.) Time 2 is x/40 which = speed/distance = time so we are multiplying the two averages (y/x + (40-y)/1.25x) combined by the single average (x/40) to get the percentage difference between the two averages and the one average, but why in your solution do you flip x/40 to 40/x?

Thanks!!!

Bunuel wrote:

During a 40-mile trip, Marla traveled at an average speed of x miles per hour for the first y miles of the trip and and at an average speed of 1.25x mph for the last 40-y miles of the trip. The time that Marla took to travel the 40 miles was what percent of the time it would have taken her if she has traveled at an average speed of x miles per hour for the entire trip?

Q: \(\frac{t_1}{t_2}=\frac{0.25y+40}{1.25x}*\frac{x}{40}=\frac{0.25y+40}{1.25}*\frac{1}{40}\). So we see that the value of this fraction does not depend on \(x\), only on \(y\).

(1) x = 48. Not sufficient.

(2) y = 20. Sufficient.

Answer: B.

x mph for the first y miles --> \(time=\frac{distance}{speed}=\frac{y}{x}\) 1.25x mph for the last 40-y miles of the trip --> \(time=\frac{distance}{speed}=\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}\) So, \(t_1=\frac{y}{x}+\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}\);

x miles per hour for the entire trip --> \(t_2=\frac{40}{x}\).

But why when you multiply by x/40 do you flip it to 40/x?

Bunuel wrote:

WholeLottaLove wrote:

Hi Bunuel,

I understand this problem conceptually (and in my post above this, got it right) but I have difficultly setting up the algebra to solve the problem as I do with many word and rate/distance problems. Here are my questions:

I.) I know why we have to add up y and (40-y) but I don't understand why it's set up as y/x + (40-y)/1.25x. Is it because distance/speed = time and we are looking for time in this question?

II.) Time 2 is x/40 which = speed/distance = time so we are multiplying the two averages (y/x + (40-y)/1.25x) combined by the single average (x/40) to get the percentage difference between the two averages and the one average, but why in your solution do you flip x/40 to 40/x?

Thanks!!!

Bunuel wrote:

During a 40-mile trip, Marla traveled at an average speed of x miles per hour for the first y miles of the trip and and at an average speed of 1.25x mph for the last 40-y miles of the trip. The time that Marla took to travel the 40 miles was what percent of the time it would have taken her if she has traveled at an average speed of x miles per hour for the entire trip?

Q: \(\frac{t_1}{t_2}=\frac{0.25y+40}{1.25x}*\frac{x}{40}=\frac{0.25y+40}{1.25}*\frac{1}{40}\). So we see that the value of this fraction does not depend on \(x\), only on \(y\).

(1) x = 48. Not sufficient.

(2) y = 20. Sufficient.

Answer: B.

x mph for the first y miles --> \(time=\frac{distance}{speed}=\frac{y}{x}\) 1.25x mph for the last 40-y miles of the trip --> \(time=\frac{distance}{speed}=\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}\) So, \(t_1=\frac{y}{x}+\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}\);

x miles per hour for the entire trip --> \(t_2=\frac{40}{x}\).

But why when you multiply by x/40 do you flip it to 40/x?

Bunuel wrote:

WholeLottaLove wrote:

Hi Bunuel,

I understand this problem conceptually (and in my post above this, got it right) but I have difficultly setting up the algebra to solve the problem as I do with many word and rate/distance problems. Here are my questions:

I.) I know why we have to add up y and (40-y) but I don't understand why it's set up as y/x + (40-y)/1.25x. Is it because distance/speed = time and we are looking for time in this question?

II.) Time 2 is x/40 which = speed/distance = time so we are multiplying the two averages (y/x + (40-y)/1.25x) combined by the single average (x/40) to get the percentage difference between the two averages and the one average, but why in your solution do you flip x/40 to 40/x?

Thanks!!!

x mph for the first y miles --> \(time=\frac{distance}{speed}=\frac{y}{x}\) 1.25x mph for the last 40-y miles of the trip --> \(time=\frac{distance}{speed}=\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}\) So, \(t_1=\frac{y}{x}+\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}\);

x miles per hour for the entire trip --> \(t_2=\frac{40}{x}\).

Ahh...okay. You divide t1/t2 and you flip the denominator in the process. The problem therefore is looking for the difference between the average of the two times and the average if she traveled for the one figure of time?

Ahh...okay. You divide t1/t2 and you flip the denominator in the process. The problem therefore is looking for the difference between the average of the two times and the average if she traveled for the one figure of time?

Read the question: The time that Marla took to travel the 40 miles was what percent of the time it would have taken her if she has traveled at an average speed of x miles per hour for the entire trip? _________________

Re: During a 40-mile trip, Marla traveled at an average speed of [#permalink]
08 Oct 2013, 06:03

Is it possible to solve the problem by taking the weighted averages of the distance and speed, and equating the same to find the average speed of the trip.

My thinking was that if there was a total line, it would have been total miles/total rate --> 40/(x + 1.25x), which does not equal the above T1 + T2

Why is this the case?

Also, as a broader question: why is it that in distance problems, such as this one, we cannot add the speeds but in work problems, we commonly add, for example, the rates of two machines working individually to find their combined rate?

Re: During a 40-mile trip, Marla traveled at an average speed of [#permalink]
27 Feb 2015, 17:18

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