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Marcia took a trip consisting of three segments at three different spe

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mikemcgarry
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Marcia took a trip consisting of three segments at three different spe [#permalink] New post   Updated on: 10 Jun 2019, 01:05
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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:
https://magoosh.com/gmat/2013/how-to-plu ... questions/

Mike :-)
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Originally posted by mikemcgarry on 17 Jul 2013, 09:50.
Last edited by Bunuel on 10 Jun 2019, 01:05, edited 2 times in total.
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Re: Marcia took a trip consisting of three segments at three different spe [#permalink] New post   17 Jul 2013, 10:00
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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:
https://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 different spe [#permalink] New post   17 Jul 2013, 10:49
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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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Re: Marcia took a trip consisting of three segments at three different spe [#permalink] New post   06 Jan 2014, 17:26
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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) 4D/v
(B) 4v/D
(C) (10D)/(11v)
(D) (10v)/(11D)
(E) (11D)/(10v)

For a discussion of picking numbers as a problem solving strategy, as well as a complete explanation of this particular question, see:
https://magoosh.com/gmat/2013/how-to-plu ... questions/

Mike :-)
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Re: Marcia took a trip consisting of three segments at three different spe [#permalink] New post   06 Jan 2014, 18:50
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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) 4D/v
(B) 4v/D
(C) (10D)/(11v)
(D) (10v)/(11D)
(E) (11D)/(10v)

For a discussion of picking numbers as a problem solving strategy, as well as a complete explanation of this particular question, see:
https://magoosh.com/gmat/2013/how-to-plu ... questions/

Mike :-)



Total time can be evaluated by total time in segments.

Time = distance/Speed

T1 = 5D/2V
T2 = 4D/3V
T3= D/6V

Total Time = T1+T2+T3

5D/2V+4D/3V+D/6V = 24D/6V = 4D/V

Hence A is this answer.
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Re: Marcia took a trip consisting of three segments at three different spe [#permalink] New post   07 Jan 2014, 05:19
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) 4D/v
(B) 4v/D
(C) (10D)/(11v)
(D) (10v)/(11D)
(E) (11D)/(10v)

Trip distance: 5D.............4D................D

Velocity :2v...............3v................6v

Time :5D/2v...........4D/3v..........D/6v

So, total time of Marcia's Trip = 5D/2v + 4D/3v + D/6v
= 15D/6v + 8D/6v + D/6v
= 24D/6v
= 4 D/v
Answer (A)

Alternative Solution

Let D = 100 and v= 50.
So, D/v = 2

Trip distance:500...........400...........100

Velocity :100...........150............300

Time [= Trip Distance/Velocity]:5........... 8/3............1/3

So, Total Trip Time = 5 + 8/3 + 1/3
= 24/3
= 8..................(1)

Back solving, starting from (C), plugging (D/v) as 2 and evaluating each choice to see which one equals 8 [as calculated in (1)]

(A) 4D/v ............................. 4* 2 =8.....matches the answer (1) calculated above.
(B) 4v/D . ........................... eliminate as v/D is not time
(C) (10D)/(11v)................... (10/11)*2....definitely not 8, hence eliminate.
(D) (10v)/(11D)
(E) (11D)/(10v)[/color]

Answer (A).
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Re: Marcia took a trip consisting of three segments at three different spe [#permalink] New post   14 Feb 2018, 11:01
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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) 4D/v
(B) 4v/D
(C) (10D)/(11v)
(D) (10v)/(11D)
(E) (11D)/(10v)


The time for her first leg of the trip is 5D/(2v).

The time for her second leg is 4D/(3v).

The time for her last leg is: D/(6v).

Thus the total time is:

5D/(2v) + 4D/(3v) + D/(6v)

15D/(6v) + 8D/(6v) + D/(6v)

24D/(6v)

4D/v

Answer: A
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Re: Marcia took a trip consisting of three segments at three different spe [#permalink]
14 Feb 2018, 11:01
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