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Source: — Data Sufficiency |
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by eagleeye » Tue May 15, 2012 12:58 pm
Hi Massi:

The OA should be D.

We are given speed from a-b = Sab = 80; speed from b-c = Sbc =60

1) Average speed = (total distance)/(total time);
let distance from b-c = d; then distance from a-b = 4d;
then time from b-c = distance/speed = d/60 ; time from a-b = 4d/80.

Then average speed = (d+4d)/(d/60+4d/80) = 5/(1/60+4/80) , Therefore sufficient.

2) In the same way, now let time from b-c = t ; then time from a-b = 3t
then distance from b-c = time*speed = 80t ; a-b = 60t.

then average speed = (80t+60t)/(3t+t) = 140/4 ; SUFFICIENT
Hence D

Let me know if this helps :)
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by GMATGuruNY » Wed May 16, 2012 3:45 am
What was the average speed at which the train traveled from Station A to Station C?
Statement 2: The amount of time that it took the train to travel from Station A to Station B was 3 times that amount of time that it took the train to travel from Station B to Station C.
Thus, of every 4 hours traveled, 3 hours are traveled at 80 miles per hour and 1 hour is traveled at 60 miles per hour, implying that the average speed every 4 hours = (3*80 + 1*60)/4 = 300/4 = 75 miles per hour.
SUFFICIENT.

Statement 1: The distance that the train traveled from Station A to Station B was 4 times the distance that the train traveled from Station B to Station C.
Learn from Statement 2:
3 hours from A to B at 80 miles per hour = 240 miles.
1 hour from B to C at 60 miles per hour = 60 miles.
When the train travels for the times given in statement 2, the distance from A to B is 4 times the distance from B to C.
Thus, the two statements tell the SAME INFORMATION.
Thus, since statement 2 is sufficient, statement 1 must also be SUFFICIENT.

The correct answer is D.
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by lazarogb » Wed Nov 21, 2012 11:18 am
eagleeye wrote:Hi Massi:

The OA should be D.

We are given speed from a-b = Sab = 80; speed from b-c = Sbc =60

1) Average speed = (total distance)/(total time);
let distance from b-c = d; then distance from a-b = 4d;
then time from b-c = distance/speed = d/60 ; time from a-b = 4d/80.

Then average speed = (d+4d)/(d/60+4d/80) = 5/(1/60+4/80) , Therefore sufficient.

2) In the same way, now let time from b-c = t ; then time from a-b = 3t
then distance from b-c = time*speed = 80t ; a-b = 60t.

then average speed = (80t+60t)/(3t+t) = 140/4 ; SUFFICIENT
Hence D

Let me know if this helps :)
Hello Eagle eye,

I lke your algebraic approach which was my first approach also, but is it correct to use the same variable for the two different distances?

I tried using 2 different variables for distance ab and bc, but maybe this was a mistake from my part.

Any thoughts?
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by Bill@VeritasPrep » Wed Nov 21, 2012 4:54 pm
lazarogb wrote:
eagleeye wrote:Hi Massi:

The OA should be D.

We are given speed from a-b = Sab = 80; speed from b-c = Sbc =60

1) Average speed = (total distance)/(total time);
let distance from b-c = d; then distance from a-b = 4d;
then time from b-c = distance/speed = d/60 ; time from a-b = 4d/80.

Then average speed = (d+4d)/(d/60+4d/80) = 5/(1/60+4/80) , Therefore sufficient.

2) In the same way, now let time from b-c = t ; then time from a-b = 3t
then distance from b-c = time*speed = 80t ; a-b = 60t.

then average speed = (80t+60t)/(3t+t) = 140/4 ; SUFFICIENT
Hence D

Let me know if this helps :)
Hello Eagle eye,

I lke your algebraic approach which was my first approach also, but is it correct to use the same variable for the two different distances?

I tried using 2 different variables for distance ab and bc, but maybe this was a mistake from my part.

Any thoughts?
Since we have a relationship between the distances, using one variable (d and 4d) works better than assigning two variables.
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by mcdesty » Sat Jul 19, 2014 9:29 pm
My scratch paper looked like this when I tackled this one.
Attachments
Gmat_63_resized.jpg
I have made your mistakes before.
I am experienced - I have tutored calculus and linear algebra for over two years.
For a very modest fee, I will ensure that your GMAT journey is a smooth one: Daily assignments and careful micro management.
PM me so we can get started.
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