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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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 T1 hours
Speed=Distance/Time
S=400/T
S+20=400/(T1)
(replace S in the second equation with 400/T)
(400/T)+20=400/(T1)
This gives us 1 equation with 1 variable, which should be sufficient.
2) Travelled 400KM in T hours.
Had it been 20kph faster in would have travelled 400km in T1 hours
Speed=Distance/Time
S=400/T
S+20=400/(T1)
(replace S in the second equation with 400/T)
(400/T)+20=400/(T1)
This gives us 1 equation with 1 variable, which should be sufficient.

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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.
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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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.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 T1 hours
Speed=Distance/Time
S=400/T
S+20=400/(T1)
(replace S in the second equation with 400/T)
(400/T)+20=400/(T1)
This gives us 1 equation with 1 variable, which should be sufficient.
Any help would be greatly appreciated.
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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.saritalr wrote: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.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 T1 hours
Speed=Distance/Time
S=400/T
S+20=400/(T1)
(replace S in the second equation with 400/T)
(400/T)+20=400/(T1)
This gives us 1 equation with 1 variable, which should be sufficient.
Any help would be greatly appreciated.
Multiply by t(t1):
400(t1) + 20(t(t1) = 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)(t5)=0, or t1=4 and t2=5.

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Hi Geva 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(t1):
400(t1) + 20(t(t1) = 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)(t5)=0, or t1=4 and t2=5.
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!
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"
t1=400/(s+20), if t=400/s we have
400/s1=400/(s+20), where s=80
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"
t1=400/(s+20), if t=400/s we have
400/s1=400/(s+20), where s=80