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## GMATPrep Work question

##### This topic has expert replies
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### GMATPrep Work question

by aimhigh715 » Sun Dec 09, 2007 12:32 pm
How much time did it take a certain car to travel 400 kilometers?

1) The car travelled the first 200 kilometer in 2.5 hours

2) If the car's AVG speed had been 20 kilometers per hour greater than it was, it would have travelled the 400 kilometers in 1 hour time less than it did.

OA: B

I chose A thinking that distance / time = speed

Given that we have distance and time, find speed. 400 km / speed = time it takes to travel, right?

In 2) we're missing speed (x + 20) but ALSO the actual travel time to get y - 1 (minus one hour).

Am I missing smth here?

Thanks!
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by Tud » Sun Dec 09, 2007 3:00 pm
1) seems insufficient because there is no indication of how long it took to do the remaining distance, or how fast they travelled. They're left with 200km and could have taken 2 hours or 2 days. Insuff.

2) Travelled 400KM in T hours.
Had it been 20kph faster in would have travelled 400km in T-1 hours
Speed=Distance/Time

S=400/T
S+20=400/(T-1)
(replace S in the second equation with 400/T)

(400/T)+20=400/(T-1)

This gives us 1 equation with 1 variable, which should be sufficient.

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by aimhigh715 » Sun Dec 09, 2007 6:24 pm
Thanks Tud!

I missed the 400km total distance from the question which made 2) sufficient it seems!

1) does seem awkward saying the initial 200km, so NS due to lack of knowledge about rest of the journey seems to make sense.
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by richardwang6430 » Tue Apr 15, 2008 6:08 pm

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by saritalr » Tue Nov 02, 2010 12:00 pm
Tud wrote:1) seems insufficient because there is no indication of how long it took to do the remaining distance, or how fast they travelled. They're left with 200km and could have taken 2 hours or 2 days. Insuff.

2) Travelled 400KM in T hours.
Had it been 20kph faster in would have travelled 400km in T-1 hours
Speed=Distance/Time

S=400/T
S+20=400/(T-1)
(replace S in the second equation with 400/T)

(400/T)+20=400/(T-1)

This gives us 1 equation with 1 variable, which should be sufficient.
Hi - I know I'm reviving an old thread. But I'm hoping someone can help me work through the algebra in the second statement as described above. I keep ending up with an equation with exponents - although I do see in theory why having only one variable would work.

Any help would be greatly appreciated.

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by Geva@EconomistGMAT » Wed Nov 03, 2010 1:56 am
saritalr wrote:
Tud wrote:1) seems insufficient because there is no indication of how long it took to do the remaining distance, or how fast they travelled. They're left with 200km and could have taken 2 hours or 2 days. Insuff.

2) Travelled 400KM in T hours.
Had it been 20kph faster in would have travelled 400km in T-1 hours
Speed=Distance/Time

S=400/T
S+20=400/(T-1)
(replace S in the second equation with 400/T)

(400/T)+20=400/(T-1)

This gives us 1 equation with 1 variable, which should be sufficient.
Hi - I know I'm reviving an old thread. But I'm hoping someone can help me work through the algebra in the second statement as described above. I keep ending up with an equation with exponents - although I do see in theory why having only one variable would work.

Any help would be greatly appreciated.
If you factor the equation (see below), you will get one positive and one negative root. Since time cannot logically be negative (you cannot travel 400 kilometers in -4 hours), you are left with only one possible values for time.

Multiply by t(t-1):

400(t-1) + 20(t(t-1) = 400t
400t - 400 + 20t^2 - 20t = 400t
20t^2 - 20t - 400 = 0
t^2 - t - 20=0
Factoring this equation will get you (t+4)(t-5)=0, or t1=-4 and t2=5.
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by saritalr » Wed Nov 03, 2010 6:52 am
Geva@MasterGMAT wrote:
If you factor the equation (see below), you will get one positive and one negative root. Since time cannot logically be negative (you cannot travel 400 kilometers in -4 hours), you are left with only one possible values for time.

Multiply by t(t-1):

400(t-1) + 20(t(t-1) = 400t
400t - 400 + 20t^2 - 20t = 400t
20t^2 - 20t - 400 = 0
t^2 - t - 20=0
Factoring this equation will get you (t+4)(t-5)=0, or t1=-4 and t2=5.
Hi Geva -

Thanks for working that out. Believe it or not, I actually got the same thing - then tried plugging 5 back in to check and thats where I made my mistake (again and again). One question I have, when doing the question originally, I figured I'd have enough information with just the second statement so I picked B. But in a perfect world (without timers), I'd have time to work the equation and the check. But obviously this isn't possible given the 2 minute time frame (at least for me).

I know that sometimes in DS questions that deal with quadratic equations, you need to be careful about picking what appears to be a viable solution because maybe the equation yields two possible answers (or no possible answers). Do you think the same thing applies to a problem like this -- you couldn't be absolutely sure that the second statement is sufficient unless you checked it?

Thanks again!

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by crisro » Sat Jun 18, 2011 10:21 am
t(time)=distance/speed

t=400/s

"if the car's average speed had been 20km/h greater than it was, it would have traveled the 400km in 1 hr less than it did"

t-1=400/(s+20), if t=400/s we have

400/s-1=400/(s+20), where s=80

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