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

10

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B

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E

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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?

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27 Apr 2010, 10:06

40

38

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 \(y\).

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18 Aug 2010, 20:38

4

The answer is B.

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.

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29 Jul 2013, 10:14

2

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

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29 Jul 2013, 10:31

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 \(y\).

Re: During a 40-mile trip, Marla traveled at an average speed of
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29 Jul 2013, 10:41

2

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 --> \(t_2=\frac{40}{x}\).

Re: During a 40-mile trip, Marla traveled at an average speed of
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29 Jul 2013, 11:29

Thanks!

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 --> \(t_2=\frac{40}{x}\).

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29 Jul 2013, 11:35

WholeLottaLove wrote:

Thanks!

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 --> \(t_2=\frac{40}{x}\).

Re: During a 40-mile trip, Marla traveled at an average speed of
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29 Jul 2013, 13:46

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?

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29 Jul 2013, 13:58

2

WholeLottaLove wrote:

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?
_________________

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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?

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19 Nov 2015, 10:50

Bunuel,

My approach was:

The question stem deals with a ratio (total time in cenario 1 / total time in cenario 2).

We know by definition that ratios are not affected if both a/b are increased or decreased by the same percentage. Statement 1 stablishes a steady 25% difference between average speeds, so that alone could not give us a unique ratio. Eliminate Statement 1.

Statement 2 gives a value for distance, which is not subject to any proportion rule. X could be anything, as long as the 25% increase in speed still holds. Bingo! That alone could affect our final ratio. Statement 2 is enough.

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29 Jan 2017, 00:30

From the question, I see that there are two situations we are comparing: (1) The time for traveling x mph for the first y miles and 1.25x mph for the last 40-y miles. We'll call this T1. (2) The time for traveling x mph the entire 40 miles. We'll call this T2.

We want to find T1 / T2 * 100%

Let's rephrase T2 first. Time = Distance / Rate, so T2 = 40 miles / x mph = 40/x hours.

Now let's take a look at T1. Using an RTD Chart: ----------Rate * Time = Distance----- 1st Part x mph * ?? = y miles 2nd Part 1.25x mph * ?? = 40-y miles

The time for the first part is y/x hours, and the time for the second part is (40-y)/(1.25x) hours.

Now let's plug these values for T1 and T2 into the the formula. T1 / T2 * 100% = [(40-y)/(1.25x)] / (40/x) * 100% = [(40-y)/(1.25x)][x / 40] * 100% = [(100)(40-y)]/[(40)(1.25)]

Our rephrased question is "what is y?"

Clearly (2) alone is clearly sufficient, so (B) is the answer. Hope that helps.
_________________

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14 Feb 2017, 16:01

LM 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 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?

(1) x = 48. (2) y = 20.

We are given that 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 at an average speed of 1.25x miles per hour for the last 40 - y miles of the trip. We are asked to determine what percentage the travel time for the 40-mile trip was of the time it would have taken her, had she traveled the entire trip at an average speed of x miles per hour.

Since time = distance/rate, the time it would have taken her to travel the entire trip at an average speed of x miles per hour is 40/x, and the actual time it took her to travel the entire trip at an average speed of x miles per hour for the first y miles of the trip and at an average speed of 1.25x miles per hour for the last 40 - y miles of the trip is y/x + (40 – y)/(1.25x). So the question becomes:

y/x + (40 – y)/(1.25x) is what percent of 40/x ?

[y/x + (40 – y)/(1.25x)]/(40/x) * 100 = ?

[y/x + (40 – y)/(1.25x)]/(40/x) * 100 = ?

[y/x + (40 – y)/(1.25x)] * (x/40) * 100 = ?

[y/40 + (40 – y)/50] * 100 = ?

2.5y + 2(40 – y) = ?

Thus, knowing the value of y will allow us to determine the above percentage.

Statement One Alone:

x = 48

Since we do not have any information about y, statement one alone is not sufficient. We can eliminate answer choices A and D.

Statement Two Alone:

y = 20

Since we have a value for y, statement two alone is sufficient. For practice, we can calculate the percentage as follows:

2.5(20) + 2(40 – 20) = 50 + 40 = 90

Answer: B
_________________

Jeffery Miller Head of GMAT Instruction

GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

Re: During a 40-mile trip, Marla traveled at an average speed of
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28 Mar 2017, 09:49

let's write down what the question asks and them make it up: we will get at (0,25y+40)/1,25x : 40/x or (0,25y+40)/1,25*40 So we don't need x, but y 1) NS 2) S

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16 Apr 2017, 15:08

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.

Thanks, Banuel!

This explanation is perfectly clear. However, when I got this question, I started by adding the rates, so the actual time became 40/(9/4x) and the time had x been used all along would be 40/x. I ended up only needing x to solve the problem, which is obviously wrong. When is it okay to add the rates?

Thank you!

gmatclubot

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16 Apr 2017, 15:08