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# Martha takes a road trip from point A to point B. She drives

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Martha takes a road trip from point A to point B. She drives [#permalink]

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14 May 2005, 06:43
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85% (hard)

Question Stats:

57% (02:59) correct 43% (02:13) wrong based on 416 sessions

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Martha takes a road trip from point A to point B. She drives x percent of the distance at 60 miles per hour and the remainder at 50 miles per hour. If Martha's average speed for the entire trip is represented as a fraction in its reduced form, in terms of x, which of the following is the numerator?

(A) 110
(B) 300
(C) 1,100
(D) 3,000
(E) 30,000
[Reveal] Spoiler: OA

Last edited by Bunuel on 04 Jul 2014, 05:40, edited 1 time in total.
Edited the question and added the OA.
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Re: Martha takes a road trip from point A to point B. She drives [#permalink]

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14 May 2005, 06:59
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total distance = d

total time taken = x/(100*60) + (100-x)/(100*50)

speed = distance / time

gives numerator = 30000
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Re: Martha takes a road trip from point A to point B. She drives [#permalink]

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14 May 2005, 11:12
total distance/total time=>d/((xd/100)/60+(d-(xd/100))/50)=>solve it and you will see that the numerator is 30000.
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Martha's Average Speed - Question 13 of Guide 2 [#permalink]

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10 Nov 2011, 16:03
1
KUDOS
Martha takes a road trip from point A to point B. She drives x percent of the distance at 60 miles per hour and
the remainder at 50 miles per hour. If Martha's average speed for the entire trip is represented as a fraction
in its reduced form, in terms of x, which of the following is the numerator?

A)110
B) 300
C)1,100
D)3,000
E) 30,000

Guys - the answer for this question is not provided. I have calculated and got 30000 i.e. E. Can someone please tell me whether my answer is correct or incorrect?
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GMAT ==> 730

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Re: Martha's Average Speed - Question 13 of Guide 2 [#permalink]

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10 Nov 2011, 20:30
2
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Let total distance = D.

d1 = xD/100 v1 = 60 t1 = d1/v1 = xD/6000
d2 = (1-x)D/100 v2 = 50 t2 = d2/v2 = (1-x)D/5000

For the total trip:

V = D/T = (d1+d2) / (t1+t2)

Note that (d1+d2) = D

$$V = \frac{D}{t1+t2}= \frac{D}{\frac{xD}{6000}+\frac{(1-x)D}{5000}}$$

For this it's easy to see that the numerator in reduced form will be the least common multiple of 5000 and 6000, which is 30,000.

Let me know if you need help on how to find the least common multiple.

EDIT: fixed expressing x in percent.

Last edited by kostyan5 on 11 Nov 2011, 12:37, edited 1 time in total.
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Re: Martha's Average Speed - Question 13 of Guide 2 [#permalink]

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11 Nov 2011, 11:01
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i keep getting 30,000.

@ kostyan5: x% 0f D is xD/100, so t1 is xD/6000 isn't it?
what am i doing wrong?
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Re: Martha's Average Speed - Question 13 of Guide 2 [#permalink]

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11 Nov 2011, 12:35
1
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BDSunDevil wrote:
i keep getting 30,000.

@ kostyan5: x% 0f D is xD/100, so t1 is xD/6000 isn't it?
what am i doing wrong?

Yes, you are right. I didn't realize that percent (/100) doesn't cancel out at the end because there's on x in D. So we do need to express d1 and d2 in terms of x/100. I'm going to modify my solution above.

Sorry for the confusion. The answer is 30,000.
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Re: Martha takes a road trip from point A to point B. She drives [#permalink]

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04 Jul 2014, 03:34
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Re: Martha takes a road trip from point A to point B. She drives [#permalink]

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17 Aug 2014, 06:59
Question asks us for the fraction to be in its reduced form..

So if we solve, numerator will be 30 and not 30000....Where am I going wrong
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Re: Martha takes a road trip from point A to point B. She drives [#permalink]

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20 Feb 2016, 19:13
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Re: Martha takes a road trip from point A to point B. She drives [#permalink]

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21 Feb 2016, 15:28
let d=total distance
let x=fraction of distance driven at 60 mph
d/[(xd/60)+(d-xd/50)]=300/(6-x) average speed
numerator=(300)(100)=30,000
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Re: Martha takes a road trip from point A to point B. She drives [#permalink]

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22 Feb 2017, 12:44
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Re: Martha takes a road trip from point A to point B. She drives   [#permalink] 22 Feb 2017, 12:44
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