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Re: During a trip, Francine traveled x percent of the total [#permalink]
02 Mar 2013, 09:50

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 _________________

Happy are those who dream dreams and are ready to pay the price to make them come true

Re: During a trip, Francine traveled x percent of the total [#permalink]
03 Oct 2013, 15:23

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)

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)

Re: Aren't E and C both answers ? [#permalink]
30 Oct 2013, 08:03

dimitri92 wrote:

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]
07 Nov 2013, 09:37

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]
27 May 2014, 12:14

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 _________________

Please consider giving 'kudos' if you like my post and want to thank

Re: During a trip, Francine traveled x percent of the total [#permalink]
16 Jun 2014, 16:42

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 _________________

During a trip, Francine traveled x percent of the total [#permalink]
31 Aug 2014, 05:38

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};

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.

How do i get (100-x)/150 ? I don't know how to solve the fraction.

Re: During a trip, Francine traveled x percent of the total [#permalink]
01 Sep 2014, 01:11

Expert's post

lou34 wrote:

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};

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.

How do i get (100-x)/150 ? I don't know how to solve the fraction.

Re: During a trip, Francine traveled x percent of the total [#permalink]
12 Sep 2014, 21:36

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]
07 Oct 2014, 03:32

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 [#permalink]
08 Oct 2014, 03:00

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.

Kudos if I helped. thanks!

gmatclubot

Re: During a trip, Francine traveled x percent of the total
[#permalink]
08 Oct 2014, 03:00