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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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New post Updated on: 04 Jul 2014, 05:40
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Question Stats:

59% (02:02) correct 41% (02:13) wrong based on 488 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

Originally posted by GMATT73 on 14 May 2005, 06:43.
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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New post 14 May 2005, 06:59
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2
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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New post 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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New post 10 Nov 2011, 16:03
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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

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

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New post Updated on: 11 Nov 2011, 12:37
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The correct answer is 300.

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.

Originally posted by kostyan5 on 10 Nov 2011, 20:30.
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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New post 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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New post 11 Nov 2011, 12:35
1
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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New post 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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New post 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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New post 25 Aug 2017, 08:11
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Assuming x=50, i.e. 50% road with speed of 60mph and remaining 50% with speed of 50mph.
Now knowing formula for average speed (when distance traveled is same) = \(\frac{2ab}{a+b}\), (a,b are speeds)
average speed in our case = \(\frac{2*50*60}{50+60} = \frac{2*50*60}{110} = \frac{2*300}{11}\)
Question asking in terms of x,i.e. 50% = \(\frac{1}{2}\)
thus \(\frac{1}{2}*\frac{2*300}{11}*100\)
Numerator = 30000
Thus option E
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Re: Martha takes a road trip from point A to point B. She drives [#permalink]

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New post 18 Apr 2018, 05:49
HolaMaven wrote:
Assuming x=50, i.e. 50% road with speed of 60mph and remaining 50% with speed of 50mph.
Now knowing formula for average speed (when distance traveled is same) = \(\frac{2ab}{a+b}\), (a,b are speeds)
average speed in our case = \(\frac{2*50*60}{50+60} = \frac{2*50*60}{110} = \frac{2*300}{11}\)
Question asking in terms of x,i.e. 50% = \(\frac{1}{2}\)
thus \(\frac{1}{2}*\frac{2*300}{11}*100\)
Numerator = 30000
Thus option E



an someone explain why this is being multiplies with 100 ??

Bunuel VeritasPrepKarishma
Re: Martha takes a road trip from point A to point B. She drives   [#permalink] 18 Apr 2018, 05:49
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Martha takes a road trip from point A to point B. She drives

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