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During a 40-mile trip, Marla traveled at an average speed of 27 Apr 2010, 09:55
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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?

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


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During a 40-mile trip, Marla traveled at an average speed of 27 Apr 2010, 11:06
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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 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?

\(t_1=\frac{y}{x}+\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}\);

\(t_2=\frac{40}{x}\);

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.
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During a 40-mile trip, Marla traveled at an average speed of 29 Jul 2013, 11:14
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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 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

(B)
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During a 40-mile trip, Marla traveled at an average speed of 18 Aug 2010, 21:38
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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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During a 40-mile trip, Marla traveled at an average speed of 22 Nov 2011, 15:03
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jimmyjamesdonkey
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 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 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=2
Show more

Marla spent \(\frac{y}{x}\) hours for the first part of the trip and \(\frac{4*(40-y)}{5*x}\) hours for the second.

Total time spent= \(\frac{y}{x}+\frac{4*(40-y)}{5*x}=\frac{y+160}{5*x}\)

If Marla drove for 40 miles with x miles/hour then time taken would be \(\frac{40}{x}\)

Question is: \(\frac{y+160}{5*x}\) divided by \(\frac{40}{x}=\frac{y+160}{200}\) ?? So we only need y to calculate the answer.

So the correct answer is B.
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During a 40-mile trip, Marla traveled at an average speed of 29 Jul 2013, 11:31
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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
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?

\(t_1=\frac{y}{x}+\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}\);

\(t_2=\frac{40}{x}\);

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.
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During a 40-mile trip, Marla traveled at an average speed of 29 Jul 2013, 11:41
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WholeLottaLove
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
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?

\(t_1=\frac{y}{x}+\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}\);

\(t_2=\frac{40}{x}\);

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.
Show more

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

Hope it's clear.
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During a 40-mile trip, Marla traveled at an average speed of 29 Jul 2013, 12:29
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Thanks!

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

Bunuel
WholeLottaLove
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
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?

\(t_1=\frac{y}{x}+\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}\);

\(t_2=\frac{40}{x}\);

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

Hope it's clear.
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During a 40-mile trip, Marla traveled at an average speed of 29 Jul 2013, 12:35
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WholeLottaLove
Thanks!

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

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

Hope it's clear.
Show more

Do you mean this:

\(\frac{t_1}{t_2}=\frac{(\frac{0.25y+40}{1.25x})}{(\frac{40}{x})}=(\frac{0.25y+40}{1.25x})*(\frac{x}{40})\).
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During a 40-mile trip, Marla traveled at an average speed of 29 Jul 2013, 14:46
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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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WholeLottaLove
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?
Show more

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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During a 40-mile trip, Marla traveled at an average speed of 29 Jan 2017, 01:30
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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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During a 40-mile trip, Marla traveled at an average speed of 13 Feb 2017, 09:56
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Hi GMATters,

Here's my video explanation of this question:



Enjoy!
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During a 40-mile trip, Marla traveled at an average speed of 14 Feb 2017, 17:01
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LM
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.
Show more

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
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During a 40-mile trip, Marla traveled at an average speed of 16 Oct 2020, 05:35
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Bunuel
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?

\(t_1=\frac{y}{x}+\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}\);

\(t_2=\frac{40}{x}\);

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.
Show more

Hi Bunuel,

Just want to confirm that we can't use the "If time is x to do 1 job, speed is \(\frac{1}{x}\)" rule in this question, because we already know that the speed is x, right? It's true that we can't find time from speed alone, right?
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During a 40-mile trip, Marla traveled at an average speed of 10 Oct 2021, 21:48
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Bunuel
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?

\(t_1=\frac{y}{x}+\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}\);

\(t_2=\frac{40}{x}\);

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.

Hi Bunuel,

Just want to confirm that we can't use the "If time is x to do 1 job, speed is \(\frac{1}{x}\)" rule in this question, because we already know that the speed is x, right? It's true that we can't find time from speed alone, right?
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Hey, tagheueraquaracer,
I see that it is old question; just for clarity, we cannot figure out time taking into consideration only ratio because Time = Distance/Ration. Knowing how long Marla traveled give as insight how much time she spent.
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During a 40-mile trip, Marla traveled at an average speed of 19 Oct 2021, 11:03
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Hi Bunuel please someone help me put in a different approach.

when distance travelled is same, ratio of time is inverse of ratio of rates

from 2) we know that distance travelled is same -->> ratio of rates x/1.25x -->> 4/5 , hencce ratio of time -->> 5/4
-->> adding an unknown multiplier to the ratios -->> rate ratio 4h/5h & time ratio 5h/4h
-->> x = 4h now
-->> the question basically says that J% of 40/4h = 9h
J% = 9h * 4h / 40 }-->> here we still need a value of h to get to the final NUMERICAL ANS

please let me know whwre i am going wrong using this method as i know that the variable cancel approach is workling fine but i wanna know where am i going wrong in this ??
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Nipunh
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Schools: HBS '26 ISB '26 IIMB
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During a 40-mile trip, Marla traveled at an average speed of 14 Feb 2025, 01:23
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Your first line is an assumption " when distance travelled is same, ratio of time is inverse of ratio of rates ". Acc to the question we dont know
about this, hence, we cant use the proportionality approach here.
parth2424
Hi Bunuel please someone help me put in a different approach.

when distance travelled is same, ratio of time is inverse of ratio of rates

from 2) we know that distance travelled is same -->> ratio of rates x/1.25x -->> 4/5 , hencce ratio of time -->> 5/4
-->> adding an unknown multiplier to the ratios -->> rate ratio 4h/5h & time ratio 5h/4h
-->> x = 4h now
-->> the question basically says that J% of 40/4h = 9h
J% = 9h * 4h / 40 }-->> here we still need a value of h to get to the final NUMERICAL ANS

please let me know whwre i am going wrong using this method as i know that the variable cancel approach is workling fine but i wanna know where am i going wrong in this ??
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