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

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

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30 Dec 2017, 13:23
Thanks SO much niks18 for a fantastic explanation! Everything is clear! Highly appreciated !
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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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08 Jan 2018, 18:54
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.
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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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18 Apr 2018, 14:17
Bunuel wrote:
hardnstrong wrote:
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

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

$$Total \ time=t_1+t_2=\frac{dx}{40*100}+\frac{(100-x)d}{60*100}=\frac{d(x+200)}{12000}$$;

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

Hope it's clear.

Hi Bunuel

can you expain step by step(with comments) how from adding up two TIMES you get this $$\frac{d(x+200)}{12000}$$

$$Total \ time=t_1+t_2=\frac{dx}{40*100}+\frac{(100-x)d}{60*100}=\frac{d(x+200)}{12000}$$;

thank you

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

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20 Jun 2018, 21:57
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.

Hello - anyone can help me explain the rationale of the formula in bold ?
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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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22 Jul 2018, 01:30
Top Contributor
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. $$\frac{(180-x)}{2}$$

B. $$\frac{(x+60)}{4}$$

C. $$\frac{(300-x)}{5}$$

D. $$\frac{600}{(115-x)}$$

E. $$\frac{12,000}{(x+200)}$$

Assume total distance = 100 miles
so, x percent = 100*(x/100) = x, the rest is 100-x
Time 1 = x/40 [Time * speed = Total distance, i.e, TS = D, T = D/S]
Time 2 = (100-x)/60

Now, average speed = Total Distance/ Total time

100/[(x/40)+(100-x/60)]
= 12000/(x+200)
Ans. E
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Re: During a trip, Francine traveled x percent of the total distance at an  [#permalink]

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02 Sep 2018, 00:56
i just assumed total distance to be 100. We can assume distance here because we are only bothered about x% and remaining distance. The total distance can be assumed here in this case as distance will cancel out

So x% of 100 at 40 mph . Distance covered will be x
Time taken will be x/40

Remaining distance : 100-x since (100-x)% of 100 will be 100-x
Speed = 60mph
Time taken = (100-x)/60

Avg speed= total distance/ total time
So Total distance = 100-x+x =100
Avg speed = 100/[(x+40) +(100-x/60)]

Simplify and you will get the answer
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Re: During a trip, Francine traveled x percent of the total distance at an &nbs [#permalink] 02 Sep 2018, 00:56

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

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