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

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Manager
Joined: 03 Jan 2017
Posts: 201

Kudos [?]: 4 [0], given: 4

Re: During a 40-mile trip, Marla traveled at an average speed of [#permalink]

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28 Mar 2017, 10:49
let's write down what the question asks and them make it up: we will get at (0,25y+40)/1,25x : 40/x or (0,25y+40)/1,25*40
So we don't need x, but y
1) NS
2) S

Kudos [?]: 4 [0], given: 4

Intern
Joined: 19 May 2016
Posts: 24

Kudos [?]: 6 [0], given: 86

Location: United States
Concentration: General Management, Strategy
GMAT 1: 710 Q46 V41
GMAT 2: 730 Q49 V41
GMAT 3: 680 Q46 V37
WE: Operations (Manufacturing)
Re: During a 40-mile trip, Marla traveled at an average speed of [#permalink]

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16 Apr 2017, 16:08
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.

Thanks, Banuel!

This explanation is perfectly clear. However, when I got this question, I started by adding the rates, so the actual time became 40/(9/4x) and the time had x been used all along would be 40/x. I ended up only needing x to solve the problem, which is obviously wrong. When is it okay to add the rates?

Thank you!

Kudos [?]: 6 [0], given: 86

Manager
Joined: 26 Dec 2015
Posts: 143

Kudos [?]: 36 [0], given: 1

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) = $$\frac{X}{100}$$ * ($$\frac{40}{X}$$)

Kudos [?]: 36 [0], given: 1

Re: During a 40-mile trip, Marla traveled at an average speed of   [#permalink] 06 Jul 2017, 19:40

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