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# During a trip, Francine traveled x percent of the total distance at an

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PS Forum Moderator
Joined: 25 Feb 2013
Posts: 810

Kudos [?]: 379 [0], given: 42

Location: India
GPA: 3.82
Re: During a trip, Francine traveled x percent of the total distance at an [#permalink]

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30 Dec 2017, 11:02
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dave13 wrote:
niks18 wrote:
dave13 wrote:
Hi niks18
thanks for your comments, why so? i used 4 variables in two "t"s and two "x"s i just didnt denote x and t as x1 and x2 - but it can clearly be seen that 40t and 60t is the same as 40t1 and 60t2

Hi dave13

what you are assuming here is t1=t2=t. is this really the case?

do you know the values of t1, t2 or t. no

as per your equation 40t1+60t2/x1/40+x2/60, if you solve this you should get

(40t1+60t2)/[(3x1+4x2)/120]=120*20(2t1+3t2)/(3x1+4x2)

how are you going to proceed further from here?

Hi niks18, and Bunuel

(40t1+60t2)/[(3x1+4x2)/120]=120*20(2t1+3t2)/(3x1+4x2) a minor correcton

I solved denominator seperatly but the point is that i didnt indicate x as x1 and x2 and t as t1 and t2 - was it a must ?

so i did this 3x+2x/120 = 5x/120 is total total time

than numerotor calculated 40t+60t = 100t is a total distance

average speed = total distance / total time

100t / (5x/120) = 100t/1 *120/5x = 12000t/5x this is how i solved it

you are saying i need additional equation for linking variables, but i seperatly expressed time and distance for one part and seperatly expressed time and distance for second part

so i i linked all these variables through formula of finding average speed = total distance divided by total time.

if so which additional equation am i missing and why ?

many thanks!

hi dave13
what you are doing is really confusing . Nonetheless i'll take a final shot in explaining the calculation to you using the variables you used.

distance traveled in the first part $$=x_1$$

so time taken to travel first part $$t_1=\frac{x_1}{40}$$

distance traveled in the second part $$=x_2$$

so time taken to travel second part $$t_2=\frac{x_2}{60}$$

Hence total time taken to complete the journey $$= t_1+t_2=\frac{x_1}{40}+\frac{x_2}{60}=\frac{3x_1+2x_2}{120}$$------------(1)

Now let the total distance traveled be $$d$$.

given distance traveled in first part $$x_1= x$$% of $$d=\frac{xd}{100}$$

so distance traveled during the second part $$x_2= d-\frac{xd}{100}=\frac{d(100-x)}{100}$$

Now substitute the value of $$x_1$$ & $$x_2$$ in equation (1) to get total time $$= \frac{\frac{3xd}{100}+\frac{2d(100-x)}{100}}{120}$$

=> total time $$= \frac{d(x+200)}{12000}$$

therefore average speed $$= \frac{total distance}{total time}$$ $$=\frac{d}{\frac{d(x+200)}{12000}}=\frac{12000}{(x+200)}$$

Kudos [?]: 379 [0], given: 42

Manager
Joined: 09 Mar 2016
Posts: 134

Kudos [?]: 23 [0], given: 184

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

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30 Dec 2017, 12:23
Thanks SO much niks18 for a fantastic explanation! Everything is clear! Highly appreciated !

Kudos [?]: 23 [0], given: 184

Intern
Joined: 16 Jul 2016
Posts: 33

Kudos [?]: 10 [0], given: 2

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

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08 Jan 2018, 17:54
rate * time = distance

40_____ x/40_______________ = x

60_____ (100-x)/60_________ = 100-x

(100)/(x/40+[100-x]/60)

after simplifying we get choice E.

A good idea is to assume the distance is 100.

Kudos [?]: 10 [0], given: 2

Re: During a trip, Francine traveled x percent of the total distance at an   [#permalink] 08 Jan 2018, 17:54

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