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# During a 40-mile trip, Marla traveled at an average speed of

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Senior Manager
Joined: 26 Dec 2015
Posts: 286
Location: United States (CA)
Concentration: Finance, Strategy
WE: Investment Banking (Venture Capital)
Re: During a 40-mile trip, Marla traveled at an average speed of [#permalink]

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06 Jul 2017, 19:40
help here would be greatly appreciated. i'm having a hard time setting up this equation algebraically.

this is what i understand:

- the entire trip consists of 2 parts (below)
1) traveling distance of Y at X mph
2) traveling distance of 40-Y @ 1.25X mph
*Note: Y must be < X
- D=RT. BUT you want to know T, so T = $$\frac{D}{R}$$

JeffTargetTestPrep, can you please explain why you did the below?

y/x + (40 – y)/(1.25x) is what percent of 40/x ?

[y/x + (40 – y)/(1.25x)]/(40/x) * 100 = ?

The way I translated the first statement is: y/x + (40 – y)/(1.25x) = $$\frac{X}{100}$$ * ($$\frac{40}{X}$$)
Manager
Joined: 19 Aug 2016
Posts: 73
Re: During a 40-mile trip, Marla traveled at an average speed of [#permalink]

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26 Nov 2017, 20:41
Bunuel wrote:
WholeLottaLove wrote:
Hi Bunuel,

I understand this problem conceptually (and in my post above this, got it right) but I have difficultly setting up the algebra to solve the problem as I do with many word and rate/distance problems. Here are my questions:

I.) I know why we have to add up y and (40-y) but I don't understand why it's set up as y/x + (40-y)/1.25x. Is it because distance/speed = time and we are looking for time in this question?

II.) Time 2 is x/40 which = speed/distance = time so we are multiplying the two averages (y/x + (40-y)/1.25x) combined by the single average (x/40) to get the percentage difference between the two averages and the one average, but why in your solution do you flip x/40 to 40/x?

Thanks!!!

Bunuel wrote:
During a 40-mile trip, Marla traveled at an average speed of x miles per hour for the first y miles of the trip and and at an average speed of 1.25x mph for the last 40-y miles of the trip. The time that Marla took to travel the 40 miles was what percent of the time it would have taken her if she has traveled at an average speed of x miles per hour for the entire trip?

$$t_1=\frac{y}{x}+\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}$$;

$$t_2=\frac{40}{x}$$;

Q: $$\frac{t_1}{t_2}=\frac{0.25y+40}{1.25x}*\frac{x}{40}=\frac{0.25y+40}{1.25}*\frac{1}{40}$$. So we see that the value of this fraction does not depend on $$x$$, only on $$y$$.

(1) x = 48. Not sufficient.

(2) y = 20. Sufficient.

x mph for the first y miles --> $$time=\frac{distance}{speed}=\frac{y}{x}$$
1.25x mph for the last 40-y miles of the trip --> $$time=\frac{distance}{speed}=\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}$$
So, $$t_1=\frac{y}{x}+\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}$$;

x miles per hour for the entire trip --> $$t_2=\frac{40}{x}$$.

Hope it's clear.

When u multiplied 0.25 with y what about the y in (40-y)?

Math Expert
Joined: 02 Sep 2009
Posts: 44351
Re: During a 40-mile trip, Marla traveled at an average speed of [#permalink]

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26 Nov 2017, 20:58
zanaik89 wrote:
Bunuel wrote:

Bunuel wrote:
During a 40-mile trip, Marla traveled at an average speed of x miles per hour for the first y miles of the trip and and at an average speed of 1.25x mph for the last 40-y miles of the trip. The time that Marla took to travel the 40 miles was what percent of the time it would have taken her if she has traveled at an average speed of x miles per hour for the entire trip?

$$t_1=\frac{y}{x}+\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}$$;

$$t_2=\frac{40}{x}$$;

Q: $$\frac{t_1}{t_2}=\frac{0.25y+40}{1.25x}*\frac{x}{40}=\frac{0.25y+40}{1.25}*\frac{1}{40}$$. So we see that the value of this fraction does not depend on $$x$$, only on $$y$$.

(1) x = 48. Not sufficient.

(2) y = 20. Sufficient.

x mph for the first y miles --> $$time=\frac{distance}{speed}=\frac{y}{x}$$
1.25x mph for the last 40-y miles of the trip --> $$time=\frac{distance}{speed}=\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}$$
So, $$t_1=\frac{y}{x}+\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}$$;

x miles per hour for the entire trip --> $$t_2=\frac{40}{x}$$.

Hope it's clear.

When u multiplied 0.25 with y what about the y in (40-y)?

Are you talking about the highlighted part?

There we are NOT multiplying by 0.25. We are adding two fraction by finding the common denominator.

$$t_1=\frac{y}{x}+\frac{40-y}{1.25x}=\frac{1.25y}{1.25x}+\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}$$
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Re: During a 40-mile trip, Marla traveled at an average speed of   [#permalink] 26 Nov 2017, 20:58

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