Marcia took a trip consisting of three segments at three : GMAT Problem Solving (PS)
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# Marcia took a trip consisting of three segments at three

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Magoosh GMAT Instructor
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Marcia took a trip consisting of three segments at three [#permalink]

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17 Jul 2013, 08:50
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72% (02:19) correct 28% (01:09) wrong based on 121 sessions

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Marcia took a trip consisting of three segments at three different speeds: she drove a distance of (5D) at a speed of (2v), then a distance of (4D) at a speed of (3v), then a distance of D at a speed of (6v). In terms of D & v, what was the total time of Marcia’s trip?
(A) $$\frac{4D}{v}$$

(B) $$\frac{4v}{D}$$

(C) $$\frac{10D}{11v}$$

(D) $$\frac{10v}{11D}$$

(E) $$\frac{11D}{10v}$$

For help with plugging in numbers as a strategy, as well as a complete explanation of this problem, see:
http://magoosh.com/gmat/2013/how-to-plu ... questions/

Mike
[Reveal] Spoiler: OA

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Mike McGarry
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Last edited by mikemcgarry on 17 Jul 2013, 09:47, edited 1 time in total.
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Re: Marcia took a trip consisting of three segments at three [#permalink]

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17 Jul 2013, 09:00
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mikemcgarry wrote:
Marcia took a trip consisting of three segments at three different speeds: she drove a distance of (5D) at a speed of (2v), then a distance of (4D) at a speed of (3v), then a distance of D at a speed of (6v). In terms of D & v, what was the total time of Marcia’s trip?
(A) $$\frac{2D}{v}$$

(B) $$\frac{2v}{D}$$

(C) $$\frac{10D}{11v}$$

(D) $$\frac{10v}{11D}$$

(E) $$\frac{11D}{10v}$$

For help with plugging in numbers as a strategy, as well as a complete explanation of this problem, see:
http://magoosh.com/gmat/2013/how-to-plu ... questions/

Mike

$$speedXtime=space$$ so $$time=\frac{space}{speed}$$

$$time=time_1+time_2+time_3=\frac{5D}{2v}+\frac{4D}{3v}+\frac{D}{6v}=\frac{4D}{v}$$

Am I right? Is there a typo Mike?
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Re: Marcia took a trip consisting of three segments at three [#permalink]

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17 Jul 2013, 09:49
Zarrolou wrote:

$$speedXtime=space$$ so $$time=\frac{space}{speed}$$

$$time=time_1+time_2+time_3=\frac{5D}{2v}+\frac{4D}{3v}+\frac{D}{6v}=\frac{4D}{v}$$

Am I right? Is there a typo Mike?

Odds Bodkins! I'm sorry! Indeed, that is a typo! Sir, you are of course quite correct. Thank you very much for pointing this out! I changed it here, and now I have to change it on the blog.
Thanks again,
Mike
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Mike McGarry
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Re: Marcia took a trip consisting of three segments at three [#permalink]

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17 Jul 2013, 10:01
d=rt....therefore t = d/r

1st trip t = 5d/2v
2nd trip t = 4d/3v
3rd trip t = d/6v

Add all the t's up and you get 24D/6v = 4D/v. Answer (A)
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Re: Marcia took a trip consisting of three segments at three [#permalink]

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03 Aug 2013, 11:21
Total time = total distance/total rate:

We have three different distances and three different rates:

5d/2v
4d/3v
1d/6v

5d/2v + 4d/3v + 1d/6v
15d/6v + 8d/6v + 1d/6v
24d/6v
4d/v

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Re: Marcia took a trip consisting of three segments at three [#permalink]

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05 Dec 2014, 09:04
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Re: Marcia took a trip consisting of three segments at three   [#permalink] 05 Dec 2014, 09:04
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