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605-655 (Medium)|   Distance and Speed Problems|                                 
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During a trip, Francine traveled x percent of the total distance at an 31 Oct 2016, 06:12
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WilDThiNg
Can we solve it by assuming Francine travelled 100% of the distance at 40 - However, when i am plugging in the assumption, several options are yielding correct result. Where is it going wrong?

Thanks
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Yes, you can plug in values like that. If multiple options give you the correct answer, try some other values on the shortlisted options only.

For example, next you should try x = 0 (all distance travelled at 60 only and hence avg is 60). You get options (C) and (E).

Next try x = 50
Avg Speed = 2*40*60/(40+60) = 48
Out of options (C) and (E), only option (E) gives 48.
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Sorry if this was mentioned in some way in the replies above, but it wasn't immediately clear to me -
If I don't plug in any assumption for d, I get the answer 12000 / (x+2).
It is only when I plug in an assumption for d, say d=100 or d=200, then I get 12000 / (x+200).
Does anyone get this answer as well, without assuming for d?

Fortunately in this question there is no MCQ option for 12000 / x+2, so I would pick (E) which looks the closest to what I got, even though it is not exactly the same.

I'd like to know why it is OK to plug in an assumption for d in this case, as during the actual GMAT I might not think of assuming a value for d.
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suntorytea
Sorry if this was mentioned in some way in the replies above, but it wasn't immediately clear to me -
If I don't plug in any assumption for d, I get the answer 12000 / (x+2).
It is only when I plug in an assumption for d, say d=100 or d=200, then I get 12000 / (x+200).
Does anyone get this answer as well, without assuming for d?

Fortunately in this question there is no MCQ option for 12000 / x+2, so I would pick (E) which looks the closest to what I got, even though it is not exactly the same.

I'd like to know why it is OK to plug in an assumption for d in this case, as during the actual GMAT I might not think of assuming a value for d.
Show more

How do you get 12000/(x + 2)?
You don't need to assume a value for d.

Average Speed = Total Distance / Total Time

Total Distance = d

Time 1 \(= \frac{D1}{S1} = \frac{(\frac{x}{100})*d}{40} = \frac{xd}{4000}\)

Time 2 \(= \frac{D2}{S2} = \frac{(1 - x/100)*d}{60} = \frac{(100 - x)d}{6000}\)

Average Speed \(= \frac{d}{\frac{xd}{4000} + \frac{(100-x)d}{6000}}\)

We take d common from the denominator and it cancels out with the d in the numerator

Average Speed \(= \frac{d}{d * (\frac{x}{4000} + \frac{(100-x)}{6000})}\)

Average Speed \(= \frac{1}{(\frac{x}{4000} + \frac{(100-x)}{6000})}\)

Average Speed \(= \frac{12000}{x + 200}\)
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During a trip, Francine traveled x percent of the total distance at an Updated on: 19 Jan 2020, 13:49
Originally posted by BrentGMATPrepNow on 24 Aug 2017, 12:17.
Last edited by BrentGMATPrepNow on 19 Jan 2020, 13:49, edited 1 time in total.
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vksunder
During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine's average speed for the entire trip?


A. \(\frac{(180-x)}{2}\)

B. \(\frac{(x+60)}{4}\)

C. \(\frac{(300-x)}{5}\)

D. \(\frac{600}{(115-x)}\)

E. \(\frac{12,000}{(x+200)}\)
Show more

I like to begin with a word equation:
Average speed = (total distance)/(total time)
For this question, let's let the total distance = D

Next, observe that: total time = (time spent driving 40 mph) + (time spent driving 60 mph)

time spent driving 40 mph = distance/speed
Aside: distance driven = (x/100)(D)
So, time spent driving 40 mph = (x/100)(D)/40

time spent driving 60 mph = distance/speed
Aside: if x% of the distance was driven at 40 mph, then the distance driven at 60 mph = [(100-x)/100](D)
So, time spent driving 60 mph = [(100-x)/100](D)/60

Here comes the awful algebra ...

Total time = (x/100)(D)/40 + [(100-x)/100](D)/60

Simplify ...
Total time = xD/4000 + [100D-xD]/6000
Total time = 3xD/12000 + [200D-2xD]/12000
Total time = (xD+200D)/12000

And finally,
Average speed = (total distance)/(total time)
= D/[(xD+200D)/12000]
= (12000D)/(xD+200D)
= (12000)/(x+200)

Answer:
Show Spoiler
E

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During a trip, Francine traveled x percent of the total distance at an 09 Sep 2017, 18:56
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vksunder
During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine's average speed for the entire trip?


A. \(\frac{(180-x)}{2}\)

B. \(\frac{(x+60)}{4}\)

C. \(\frac{(300-x)}{5}\)

D. \(\frac{600}{(115-x)}\)

E. \(\frac{12,000}{(x+200)}\)
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Answer: option E

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During a trip, Francine traveled x percent of the total distance at an 13 Sep 2019, 18:20
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Hi All,

We're told that During a trip, Francine traveled X percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. We're asked, in terms of X, what was Francine's AVERAGE SPEED was for the entire trip. This question can be approached in an number of different ways - and can be solved rather easily by TESTing VALUES.

IF....
Total Distance = 100 miles and X = 40....

40% of 100 = 40 miles, so Francine traveled 40 miles at 40 miles/hour --> 1 hour traveled
The rest = 60 miles, so Francine traveled 60 miles at 60 miles/hour --> 1 hour traveled

Total Distance traveled = 100 miles
Total Time traveled = 2 hours
Average Speed = 100/2 = 50 miles/hour

So, we're looking for an answer that equals 50 when X = 40. There's only one answer that matches...

Final Answer:
Show Spoiler
E

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During a trip, Francine traveled x percent of the total distance at an 27 Oct 2019, 15:04
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Can someone help with one aspect:

In other posts, people have been saying that the average speed is 48 m.p.h by showing this math:

(2)(40)(60)/(40+60)

How is 2 X 40 X 60 the total distance? Where do the individual numbers come from?

Thanks
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Hi mborski,

In this question, the Average Speed will depend on the total distance that the car travels at each of the two speeds (40 miles/hour and 60 miles/hour). The ONLY time that the Average Speed will be exactly 48 miles/hour is when the car travels the SAME distance at each of those two speeds. Under all other situations, the Average Speed will be something OTHER than 48 miles/hour.

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Bunuel
During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine’s average speed for the entire trip?

A. (1800 - x) /2
B. (x + 60) /2
C. (300 - x ) / 5
D. 600 / (115 - x )
E. 12,000 / ( x + 200)

Algebraic approach:

\(Average \ speed=\frac{distance}{total \ time}\), let's assume \(distance=40\) (distance \(d\) will cancel out from the equation, so we can assume distance to be some number.) so we should calculate total time.

Francine traveled \(x\) percent of the total distance at an average speed of 40 miles per hour --> time needed for this part of the trip: \(t_1= \frac{distance_1}{speed_1}=\frac{\frac{x}{100}*40}{40}=\frac{x}{100}\);

Timed needed for the rest of the trip: \(t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*40}{60}=\frac{100-x}{150}\);

\(Total \ time=t_1+t_2=\frac{x}{100}+\frac{100-x}{150}=\frac{x+200}{300}\);

\(Average \ speed=\frac{distance}{total \ time}=\frac{40}{\frac{x+200}{300}}=\frac{12000}{x+200}\).

Answer: E.
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Hi Bunuel,

2 Questions:
(1) Can we use the weighted average approach to solve this question? If so, how would we do that?
(2) Why can't we use the p1A1 + p2A2 = A(total) approach for this question?

Thank you! :please:
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During a trip, Francine traveled x percent of the total distance at an 25 Apr 2020, 04:58
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Bunuel
During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine’s average speed for the entire trip?

A. (1800 - x) /2
B. (x + 60) /2
C. (300 - x ) / 5
D. 600 / (115 - x )
E. 12,000 / ( x + 200)

Algebraic approach:

\(Average \ speed=\frac{distance}{total \ time}\), let's assume \(distance=40\) (distance \(d\) will cancel out from the equation, so we can assume distance to be some number.) so we should calculate total time.

Francine traveled \(x\) percent of the total distance at an average speed of 40 miles per hour --> time needed for this part of the trip: \(t_1= \frac{distance_1}{speed_1}=\frac{\frac{x}{100}*40}{40}=\frac{x}{100}\);

Timed needed for the rest of the trip: \(t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*40}{60}=\frac{100-x}{150}\);

\(Total \ time=t_1+t_2=\frac{x}{100}+\frac{100-x}{150}=\frac{x+200}{300}\);

\(Average \ speed=\frac{distance}{total \ time}=\frac{40}{\frac{x+200}{300}}=\frac{12000}{x+200}\).

Answer: E.

Hi Bunuel,

2 Questions:
(1) Can we use the weighted average approach to solve this question? If so, how would we do that?
(2) Why can't we use the p1A1 + p2A2 = A(total) approach for this question?

Thank you! :please:
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Hi pruekv

Please remember: The average Speed is NOT the average of two speeds and it's a function of total distance and total time

For detail please watch the video here.
Video Time - 2:26 onward

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During a trip, Francine traveled x percent of the total distance at an 07 Dec 2020, 16:54
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vksunder
During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine's average speed for the entire trip?

A. (180-x)/2
B. (x+60)/4
C. (300-x)/5
D. 600/(115-x)
E. 12,000/(x+200)

Don't be scared of plugging in numbers. Sometimes it is just the most straightforward way to solve.

For this problem, assume Distance = D = 240
Then assume x = 50

Half of the distance = 120 @ 40mph = 3 hrs
The other half = 120 @ 60mph = 2 hrs

Then avg. speed = total distance / total time = 240 / 5 =48 (this is our target)

Answer choice (E) just replace x with 50.

So we get 240/5 = 48. Bingo.

This is our answer (E)

Gimme Kudos! I need those GMAT Club tests!! :P
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why do we assume x=50?
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Hi Elnur1997,

This is a reference to TESTing VALUES, which is a really useful Tactic in the Quant section of the GMAT. Since "X" refers to the percent of the distance traveled at 40 miles/hour (with the remaining distance traveled at 60 miles/hour), using X = 50 is an easy way to 'split' the distance into 2 pieces.

Based on the way that the overall question is written, there's a really easy way to answer the question with this same strategic approach. Make sure to note what the question is asking for: we're asked, in terms of X, what was Francine's AVERAGE SPEED was for the entire trip.

IF....
Total Distance = 100 miles and X = 40....

40% of 100 = 40 miles, so Francine traveled 40 miles at 40 miles/hour --> 1 hour traveled
The rest = 60 miles, so Francine traveled 60 miles at 60 miles/hour --> 1 hour traveled

Total Distance traveled = 100 miles
Total Time traveled = 2 hours
Average Speed = 100/2 = 50 miles/hour

So, we're looking for an answer that equals 50 when X = 40. There's only one answer that matches...

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vksunder
During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine's average speed for the entire trip?


A. \(\frac{(180-x)}{2}\)

B. \(\frac{(x+60)}{4}\)

C. \(\frac{(300-x)}{5}\)

D. \(\frac{600}{(115-x)}\)

E. \(\frac{12,000}{(x+200)}\)
Show more

Put x = 0, you should get 60 because the entire distance was travelled at 60 mph. Only option (C) and (E) are possible.
Put x = 50, half the distance would be travelled at 40 and half at 60 so avg speed will be 2ab/(a+b) = 2*40*60/(40 + 60) = 48
Out of (C) and (E) only option (E) gives 48.

Answer (E)
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vksunder
During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine's average speed for the entire trip?

A. (180-x)/2
B. (x+60)/4
C. (300-x)/5
D. 600/(115-x)
E. 12,000/(x+200)

Don't be scared of plugging in numbers. Sometimes it is just the most straightforward way to solve.

For this problem, assume Distance = D = 240
Then assume x = 50

Half of the distance = 120 @ 40mph = 3 hrs
The other half = 120 @ 60mph = 2 hrs

Then avg. speed = total distance / total time = 240 / 5 =48 (this is our target)

Answer choice (E) just replace x with 50.

So we get 240/5 = 48. Bingo.

This is our answer (E)

Gimme Kudos! I need those GMAT Club tests!! :P


why do we assume x=50?
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Elnur1997

You use x = 50 because you are assuming that 50% of the distance (half) was travelled at 40 mph and the other 50% at 60 mph.
We know how to calculate avg speed when two equal distances are travelled at speeds of a and b. The avg speed is given by the formula 2ab/(a+b)

Check this post for more: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2015/0 ... -the-gmat/
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Video solution from Quant Reasoning starts at 15:35
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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Generally on the GMAT:
When the same distance is traveled at two different speeds, the average speed for the entire trip will be a little closer to the lower speed than to the higher speed.
The reason:
It takes longer to travel the distance at the lower speed than at the higher speed.
Since more time is spent at the lower speed, the average speed for the whole trip will be closer to the lower speed.

vksunder
During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine's average speed for the entire trip?


A. \(\frac{(180-x)}{2}\)

B. \(\frac{(x+60)}{4}\)

C. \(\frac{(300-x)}{5}\)

D. \(\frac{600}{(115-x)}\)

E. \(\frac{12,000}{(x+200)}\)
Show more

Let x=50%, implying that half the distance is traveled at 40 mph and the other half at 60 mph, with the result that the average speed for the whole trip will be a bit closer to 40 than to 60.
Implication:
When x=50 is plugged into the correct answer, the result will be a bit less than 50.

A. \(\frac{(180-x)}{2} = \frac{180-50}{2} = 65\)
B. \(\frac{(x+60)}{4} = \frac{50+60}{4} = \frac{110}{4} = 27.5\)
C. \(\frac{(300-x)}{5} = \frac{300-50}{5} = \frac{250}{5} = 50\)
D. \(\frac{600}{(115-x)} = \frac{600}{115-50} = \frac{600}{65}\) = less than 10
Eliminate A, B, C and D.

Show Spoiler
E


E. \(\frac{12,000}{(x+200)} = \frac{12,000}{50+200} = \frac{12,000}{250} = \frac{1200}{25} = 48\)
Only E yields a value a bit less than 50.
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During a trip, Francine traveled x percent of the total distance at an 27 Dec 2022, 01:57
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vksunder
During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine's average speed for the entire trip?


A. \(\frac{(180-x)}{2}\)

B. \(\frac{(x+60)}{4}\)

C. \(\frac{(300-x)}{5}\)

D. \(\frac{600}{(115-x)}\)

E. \(\frac{12,000}{(x+200)}\)

I like to begin with a word equation:
Average speed = (total distance)/(total time)
For this question, let's let the total distance = D

Next, observe that: total time = (time spent driving 40 mph) + (time spent driving 60 mph)

time spent driving 40 mph = distance/speed
Aside: distance driven = (x/100)(D)
So, time spent driving 40 mph = (x/100)(D)/40

time spent driving 60 mph = distance/speed
Aside: if x% of the distance was driven at 40 mph, then the distance driven at 60 mph = [(100-x)/100](D)
So, time spent driving 60 mph = [(100-x)/100](D)/60

Here comes the awful algebra ...

Total time = (x/100)(D)/40 + [(100-x)/100](D)/60

Simplify ...
Total time = xD/4000 + [100D-xD]/6000
Total time = 3xD/12000 + [200D-2xD]/12000
Total time = (xD+200D)/12000

And finally,
Average speed = (total distance)/(total time)
= D/[(xD+200D)/12000]
= (12000D)/(xD+200D)
= (12000)/(x+200)

Answer:
Show Spoiler
E

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Hi BrentGMATPrepNow, does total distance can only be 100? As if I use 120 but getting a different answer below and not sure why is that? Thanks Brent

120/ x/40 + 120-x/60
120/ (3x+240-2x)/120
120 * 120/ x+240
14400 / x+240
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During a trip, Francine traveled x percent of the total distance at an 27 Dec 2022, 10:49
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Kimberly77
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During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine's average speed for the entire trip?


A. \(\frac{(180-x)}{2}\)

B. \(\frac{(x+60)}{4}\)

C. \(\frac{(300-x)}{5}\)

D. \(\frac{600}{(115-x)}\)

E. \(\frac{12,000}{(x+200)}\)

I like to begin with a word equation:
Average speed = (total distance)/(total time)
For this question, let's let the total distance = D

Next, observe that: total time = (time spent driving 40 mph) + (time spent driving 60 mph)

time spent driving 40 mph = distance/speed
Aside: distance driven = (x/100)(D)
So, time spent driving 40 mph = (x/100)(D)/40

time spent driving 60 mph = distance/speed
Aside: if x% of the distance was driven at 40 mph, then the distance driven at 60 mph = [(100-x)/100](D)
So, time spent driving 60 mph = [(100-x)/100](D)/60

Here comes the awful algebra ...

Total time = (x/100)(D)/40 + [(100-x)/100](D)/60

Simplify ...
Total time = xD/4000 + [100D-xD]/6000
Total time = 3xD/12000 + [200D-2xD]/12000
Total time = (xD+200D)/12000

And finally,
Average speed = (total distance)/(total time)
= D/[(xD+200D)/12000]
= (12000D)/(xD+200D)
= (12000)/(x+200)

Answer:
Show Spoiler
E

RELATED VIDEO

Hi BrentGMATPrepNow, does total distance can only be 100? As if I use 120 but getting a different answer below and not sure why is that? Thanks Brent

120/ x/40 + 120-x/60
120/ (3x+240-2x)/120
120 * 120/ x+240
14400 / x+240
Show more

Be careful. In my solution, D = the total distance.
The 100 does not represent a distance.
The 100 is the "cent" (100 in French) part of x perCENT, as in "Francine traveled x percent of the total distance at an average speed of 40 miles per hour"
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Greg1234
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During a trip, Francine traveled x percent of the total distance at an 07 May 2024, 05:36
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oldstudent

Easier approach:



Formula

:When an body covers m part of journey at speed p and next n part of the journey at speed q then the Average speed of the total journey is:
(m+n)*pq / (np+mq).

Using above formula:
initial part of journey =x
remaining part 100-x (since x is in percent)
m+n=100

so we have => 100*40*60/x*60+(100-x)40 -> solves to 12000/x+200 -E is the Answer
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­Why is it that we must multiply x or in your case m by 60 in the denominator, when m was done at 40mph. I would assume that the denominator must be x * 40 and then 100-x * 60 since that is what the question tells us. I know this is wrong, I would like to understand why. 
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During a trip, Francine traveled x percent of the total distance at an 29 Oct 2024, 03:48
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Thanks for the clever, quick method.
How is the formula derived?
oldstudent

Easier approach:



Formula

:When an body covers m part of journey at speed p and next n part of the journey at speed q then the Average speed of the total journey is:
(m+n)*pq / (np+mq).

Using above formula:
initial part of journey =x
remaining part 100-x (since x is in percent)
m+n=100

so we have => 100*40*60/x*60+(100-x)40 -> solves to 12000/x+200 -E is the Answer
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