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

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02 Mar 2013, 10:50

let the whole distance be 100, and x=40

then we got that the 1st distance took 1 h ( distance =40%*100=40 . time = distance/speed =40/40=1) the 2nd distance also took 1 h (distance =100-40=60 ; time = 60/60=1)

so, average speed = total distance/total time =100/2 =50

lets plug in answer choice E 12000/ (40+200)=50 bingo
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Re: During a trip, Francine traveled x percent of the total [#permalink]

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03 Oct 2013, 16:23

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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)

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)

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.

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

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07 Nov 2013, 10:37

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?

Average speed = total distance/total time.

shortcut: avg speed = 1/(d1/v1 + d2/v2)

d1 = (percentage traveled on first leg of journey)*d/100(where 100 represents total distance) d2 = (100%-percentage traveled on first leg of journey)/100(where 100 represents total distance)*d

v1=average speed for first portion of journey v2=average speed for second portion of journey

In this problem, what is d?

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 [#permalink]

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27 May 2014, 13:14

I think the easiest way to do this problem is to assume distance is 100.

So X= distance traveled at 40mph and (100-x)=distance traveled at 60mph

Let A = average speed

Time required to travel total distance = time required to travel x + time required to travel (100-x) 100/A = x/40 + (100-x)/60 ===> A = 12,000 / ( x + 200). Answer E
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Re: During a trip, Francine traveled x percent of the total [#permalink]

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16 Jun 2014, 17:42

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)

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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During a trip, Francine traveled x percent of the total [#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 [#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...

During a trip, Francine traveled x percent of the total [#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)

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Re: During a trip, Francine traveled x percent of the total [#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.
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Re: During a trip, Francine traveled x percent of the total [#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)

During a trip, Francine traveled x percent of the total [#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.

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

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10 May 2016, 14:23

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)

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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Re: During a trip, Francine traveled x percent of the total
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10 May 2016, 14:23

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