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

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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Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

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31 Aug 2014, 06:38

Bunuel wrote:

hardnstrong wrote:

Is there any clear way of getting correct answer without plugging in different numbers ????????????????

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: \(t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*40}{60}=\frac{100-x}{150}\);

Is there any clear way of getting correct answer without plugging in different numbers ????????????????

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: \(t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*40}{60}=\frac{100-x}{150}\);

Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

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12 Sep 2014, 22:36

vksunder wrote:

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)

I have used plug in method.. let the total distance be 100 & X=10

time for travelling x = 1/4 ; time for travelling remaining = 90/60 = 3/2

t= t1+t2

1/4 +3/2 = 7/4

D=ST ==> 100 = (7/4)s

s= 400/7

apply x= 10 in options .. close call between C & E but E Wins...

Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

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07 Oct 2014, 04:32

The approach I followed is to assume total distance as 100 miles. Then x% of total distance is x miles, at the speed of 40 miles per hour and the remaining distance is 100-x.

Formula is Speed = Total Distance/Total Time

T1 = X/40 T2 = 100-X/50 T1 + T2 = X+200/120

Thus, Speed = 100/(X+200/120) and that leads to the answer 12,000/(x+200)

_________________ press kudos, if you like the explanation, appreciate the effort or encourage people to respond.

Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

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08 Oct 2014, 04:00

I assumed x=20. so remaining distance =80. first part= s=d/t. put in speed as 40 (given in the question), distance= 20. find out time. It comes out to be 1/2.

Second part distance= 80. speed =60. Time will now be t=4/3

average speed for the whole journey is total distance / total time.

That's 100/(11/6)

simplifies to 600/11, which is equal to option E. Start with option C first and see if your target answer (in this case, 600/11) is less than or greater than C. Then decide whether you need to go higher or lower.

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)

Responding to a pm:

Quote:

is this a good formula for different distances? so if i learn just these two (first one given in my pm above), then I can pretty much solve anything...is that right? how will this formula change for three different average speeds?

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.
_________________

Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

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17 May 2015, 00:00

xyz21 wrote:

vksunder wrote:

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)

Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

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29 Nov 2015, 13:14

Given Info: Average Speed for some distance (x% of total distance) in a trip is given, and the average speed for the remaining distance is also given. We have to find average speed for entire trip

Interpreting the Problem: We are given average speed for some distances. Based on that we can calculate the total time taken for these distances, and then divide the total distance traveled by the total time calculated we will be able to calculate the average speed for the entire trip.

Solution: Let the total distance be a

Distance travelled at a speed of 40 miles per hour = x%*a

Time taken to travel x% of total distance=\(\frac{xa}{100*40}\)

Time taken will be \(\frac{xa}{4000}\)

Distance traveled at a speed of 60 miles per hour = (100-x%)a.

Time taken to travel (100-x)% of total distance=\(\frac{(100-x)a}{100*60}\)

Time taken will be \(\frac{((100-x)a)}{6000}\)

Average speed= \(\frac{TotalDistanceTraveled}{TotalTimeTaken}\)

Total distance traveled =a

Total time taken = \(\frac{xa}{4000} + \frac{((100-x)a)}{6000}\)

Total time taken = \(\frac{xa+200a}{12000} = \frac{a(x+200)}{12000}\)

Average speed for entire trip = \(\frac{a}{(a(x+200)/12000)}\)

Average speed =\(\frac{12000}{(x+200)}\)

Hence, The answer is E

Key Takeaways:

Always remember that the average speed is total distance traveled/total time taken. It is never the average of the speeds for various distances.

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)

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.
_________________

Jeffery Miller Head of GMAT Instruction

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

Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

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06 Jun 2016, 05:58

vksunder wrote:

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)

Hello, I found assuming total distance to be 100 much easier.

D1= x and D2= 100-x

Average speed = Total distance /total time taken = 100/(T1 + T2) where T1= D1/S1=x/40 and T2= D2/S2=(100-x)/60

or, Average Speed = 100/[(x/40)+(100-x)/60] or 12000/(x+200) _________________

It is not who I am underneath but what I do that defines me.

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!
_________________

Chelsey Cooley | Manhattan Prep Instructor | Seattle and Online

Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

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24 Aug 2016, 11:17

ccooley wrote:

Mariwa wrote:

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!

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!

I got it thanks!

This is applicable to all problems concerning rates? No only the ones about speed right?

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!

I got it thanks!

This is applicable to all problems concerning rates? No only the ones about speed right?

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.
_________________

Chelsey Cooley | Manhattan Prep Instructor | Seattle and Online

Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

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29 Sep 2016, 18:12

step1: make sure that you isolate the variable you are looking for and at the end double check for it. in this question you are being asked to find x => x is the percentage at which car drove at 40 miles per hour. step2: decide whether you want to use the variables given or smart numbers. I think the easiest way to solve this problem is by using smart numbers yet after this I am going to do this problem using the variables. step3: assumptions: x=50% and the total distance is 240. why 240? because 50% of 240 divides easily with rates 40 and 60. step4: "draw" the equations => R1T1=D1 = 40(t1)=120 R2T2=D2 = 60(t2)=120 total RT = 240 ... T= T1+T2 step5: you are asked the total rate it took to complete, so you use the total formula for R = 240 / t1+t2 step6: rewrite t1 and t2 => 40(t1)=120 => t1=3 => 60(t2)=120 => t2=20 => 240 / t1+t2 => 240 / 2+5 => 48 step7: AT THIS POINT STOP AND THINK. you are being told to find the rate (48) in terms of X, what is X? 50! remember in 3 we decided x = 50. step8: replace x for 50 in the answer choices and look for answer to be 48. E is the answer.

Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

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31 Oct 2016, 03:04

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?

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

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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