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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. \(\frac{12,000}{(x+200)}\)
PS08570
A. \(\frac{(180-x)}{2}\)
B. \(\frac{(x+60)}{4}\)
C. \(\frac{(300-x)}{5}\)
D. \(\frac{600}{(115-x)}\)
E. \(\frac{12,000}{(x+200)}\)
PS08570
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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. (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.
_________________
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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oldstudent
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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
PareshGmat
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Look at the diagram below:
Setting up the equation (We require to find value of s)
\(\frac{x}{40} + \frac{100-x}{60} = \frac{100}{s}\)
\(s = \frac{12,000}{x+200}\)
Answer = E
Setting up the equation (We require to find value of s)
\(\frac{x}{40} + \frac{100-x}{60} = \frac{100}{s}\)
\(s = \frac{12,000}{x+200}\)
Answer = E
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hardnstrong
Nowhere it's said that the total distance is 40 miles. I should have written this more clearly: distance \(d\) will cancel out from the equation (edited the earlier post to clear this). So we can assume distance to be some number. I chose 40 as it's easy for calculation.
The solution with \(d\):
\(Average \ speed=\frac{distance}{total \ time}\).
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: \(t_1= \frac{distance_1}{speed_1}=\frac{\frac{x}{100}*d}{40}=\frac{dx}{40*100}\);
Timed needed for the rest of the trip: \(t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*d}{60}=\frac{(100-x)d}{60*100}\);
\(Total \ time=t_1+t_2=\frac{dx}{40*100}+\frac{(100-x)d}{60*100}=\frac{d(x+200)}{12000}\);
\(Average \ speed=\frac{distance}{total \ time}=\frac{d}{\frac{d(x+200)}{12000}}=\frac{12000}{x+200}\).
Answer: E.
Hope it's clear.
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vksunder
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)
vksunder
If x = 0, the answer should give 60, as this would mean that Francine traveled the whole distance with the average speed of 60.
We should choose between C and E.
Similarly, for x = 100, the answer should give 40. We are still left with choices C and E.
For x = 50, let's say Francine traveled 2 * 120 miles, 120 with 40 mph and the other 120 miles with 60 mph.
The average speed would be 240/(120/40 + 120/60) = 240/(3+2) = 48.
Only E gives the correct answer (12,000/250 = 48, while (300 - 50)/5 = 50).
Answer E.
After I posted my reply, I saw that dimitri92 used a similar approach.
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General Discussion
vksunder
Alt method:
shortcut: avg speed = 1/(d1/v1 + d2/v2)
d1 = x*d/100; d2 = (1 - x/100)*d; v1 = 40, v2 = 60 --> E
dimitri92
Let's say the total mileage = 100 and x = 40%
If x = 40 then it takes 1 hour at 40MPH
Therefore, at 60MPH it takes another hour to go the rest of the distance
Add the numbers:
2r = 100
r = 50MPH
c. 300-40/5 = 52
e. 12000/240 = 50
of course B could be the answer as well
So if we were to reverse the numbers and set x = 60 then the answer is E. Sometimes you have to try different sets of numbers because initially two answers can be correct.
3/2r + 2/3r = 100
13/6 r = 100
r = 600/13
only E would win
Posts: 85
Where does it say that total distance is 40 
Distance is given as x % of total distance
we only have speed, which is 40m/ph and 60m/ph
Distance is given as x % of total distance
we only have speed, which is 40m/ph and 60m/ph
a nice quick way of solving this question in under a min.
First, we should assume x = 50, both distances are the same.
To find the average speed over the same distance, the equation is: 2*s1*s2/(s1+s2).
In this case, that's 2*40*60/100 = 48.
So, plug 50 back into the choices for x, and look for 48... E works.
First, we should assume x = 50, both distances are the same.
To find the average speed over the same distance, the equation is: 2*s1*s2/(s1+s2).
In this case, that's 2*40*60/100 = 48.
So, plug 50 back into the choices for x, and look for 48... E works.
Posts: 577
vksunder
Plug in is the best approach when you find variables in the choices.
1. The average speed is given by (d1+d2) / (d1/40 + d2/60). We see if we substitute 40 for d1 and 60 for d2 we get 50 as the average speed
2. Choice E gives the same value of average speed for the above assumed value of d1 i.e x
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Srinivasan Vaidyaraman
lou34
\(\frac{(1-\frac{x}{100})*40}{60}=\frac{(\frac{100-x}{100})*40}{60}=\frac{100-x}{100*60}*40=\frac{100-x}{150}\)
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vksunder
Responding to a pm:
Quote:
Again, I do not encourage the use of formulas. You will need to learn many formulas to cover various different scenarios and even then you can not cover all.
Say, overall distance is 100. So, distance covered at speed 40 is x. So distance covered at speed 60 will be 100-x
Avg Speed = Total Distance/Total Time \(= \frac{100}{\frac{x}{40} + \frac{100-x}{60}}= \frac{100*40*60}{60x + 40(100-x)}\) (same as given formula)
You might have to take one step extra here but it makes much more sense than learning up every formula you come across and then getting confused whether the formula will work in a particular situation or not.
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Expert reply
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vksunder
This problem can be solved by using a Rate-Time-Distance table. We are given that Francine traveled x percent of the distance at a rate of 40 mph.
Since we are working with percents, and 100% is the total distance percentage, we can say that (100 – x) percent = the percentage of the remaining distance. Thus we know that Francine traveled (100 – x) percent of the distance traveled, at a rate of 60 mph.
Since we are working with percents, we can choose a convenient number for the total distance driven; we'll use 100 miles.
Let’s fill in the table.
Remember, time = distance/rate, so we use the entries from the chart to set up the times:
Time for x percent of the distance = x/40
Time for (100 – x) percent of the distance = (100 – x)/60
Finally, we must remember that average rate = total distance/total time. Our total distance is 100. The total time is the sum of the two expressions that we developed in the previous steps. Here is the initial setup:
100/[(x/40 + (100 – x)/60)]
Now work with the fractions in the denominator, getting a common denominator so that they can be added:
100/[(3x/120 + (200 – 2x)/120)]
100/[(200 + x)/120)]
This fraction division step requires that we invert and multiply:
100 x 120/(200 + x)
12,000/(200 + x)
Answer is E.
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Posts: 18
Why can't I apply the simple average method here?
Average Speed=40*X/100+60*(100-X)/100
That leads to answer C. It makes sense to me.
Please help!
Average Speed=40*X/100+60*(100-X)/100
That leads to answer C. It makes sense to me.
Please help!
Mariwa
The average speed is always the total distance, divided by the total time.
When you multiplied 40*x and 60*(100-x), that actually doesn't make sense, mathematically. You're multiplying a rate (40 miles) by a percent of distance (x% of the total distance). That doesn't give you anything meaningful. Always check before you do the math: what am I taking a percent of?
Here's the version of your formula that would work:
total distance = d
total time for first half of trip = 40/(x/100 * d)
total time for second half of trip = 60/((100-x)/100*d)
average speed = total distance/(total time for first half + total time for second half)
If you try writing that out, though, you'll see why picking numbers is actually the right choice here!
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ccooley
I got it thanks!
This is applicable to all problems concerning rates? No only the ones about speed right?
Mariwa
Any time a problem says 'average speed' or 'average rate', start by writing out 'total work' (or 'total distance') / 'total time' on your paper. Then, just fill in those two values as you work them out! That said, I can't remember ever seeing an official GMAT problem that asked for an average rate, rather than an average speed. But there's always a chance, and in that case, you'd do it exactly like this problem.
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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
Thanks
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