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bus speed!

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bus speed!

by vscid » Sun Apr 18, 2010 10:28 am
If it took a bus 4 hours to get from town A to town B, what was the average speed of the bus for the trip?

1. In the first 2 hours the bus covered 100 miles.
2. The average speed of the bus for the first half of the distance was twice its speed for the second half.
The GMAT is indeed adaptable. Whenever I answer RC, it proficiently 'adapts' itself to mark my 'right' answer 'wrong'.
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by chrsrook » Sun Apr 18, 2010 12:46 pm
IMO C
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by anoopprasad » Sun Apr 18, 2010 5:44 pm
C

From 1 - speed for first 2 hrs = 50 mph
From 2 - speed for next 2 hrs = 25 mph
distance covered = 100 + 25 * 2

avg speed = tot dist/tot time
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by iamtensai » Mon Apr 19, 2010 1:21 pm
E ... Statement 2 talks about the first half of the DISTANCE, which is different from the first half of the time.
while we know total time = 4 hrs, total dist is unknown
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by nisha.menon294 » Mon Apr 19, 2010 9:11 pm
IMO E , what is the OA
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by liferocks » Mon Apr 19, 2010 9:22 pm
vscid wrote:If it took a bus 4 hours to get from town A to town B, what was the average speed of the bus for the trip?

1. In the first 2 hours the bus covered 100 miles.
2. The average speed of the bus for the first half of the distance was twice its speed for the second half.
IMO ans is C

clearly either statement alone is not sufficient.

now from statement 2 we get average speed of the bus for the first half say s1 was twice of average speed of the bus for the second half,say s2
s1:s2=2:1
if distance is same time and speed are inversely proportional so t1:t2=1:2 (t1 time to cover first half of distance t2 time to cover second half)
total time 4 hr
hence t1=4/3
and t2=8/3
so in first 2 hours it has covered 4s1/3+2s2/3 miles=8s2/3+2s2/3=10s2/3 miles
from statement 1 this is 100 miles hence s2= 30miles/hour and s1=60miles/hour

so distance Between town A and B =60*4/3 + 30*8/3 =80+80=160 miles
hence average speed for journey is 160/4=40 miles per hour

And is C
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by bhooshang » Tue Apr 20, 2010 1:14 pm
Hi All!

Doesn't 'iamtensai' have a point here?
He says, "Statement 2 talks about the first half of the DISTANCE, which is different from the first half of the time. "

If we consider speed in first half as '2s' and second half as 's' then we cannot assume that the speed in the first half will be 2s=100/2. Lets say we assumed that and got s=25 and subsequently a total distance d=150. Then, as per statement 2, the speed in the first half of the distance i.e. 150/2=75 is 2s. Doesn't this contradict our assumption of 2s=100/2 ?

Please advise.
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by tpr-becky » Wed Apr 21, 2010 1:48 pm
In order to get the average speed we use the rate formula - D=R(T)

as stated before neither of the statements work on their own but

We know that T1 + T2 = 4 hours and we know if you go twice as fast you get there in 1/2 the time so that T1=1/2(T2) Which means that T2+1/2(T2)=4 which means you cover the second 1/2 of the distance in 8/3 hours and the first half of the distance in 4/3 hours

The key is noting that 2 hours in the first statement is morethan the 4/3 hours - so we know that in 2 hours he went R1 for 4/3 hours and then R2 for 2/3 hours(for a total of 2 hours travel) . since R1=2(R2) we also know that 4/3(r1) + 2/3(1/2(r1)=100 - we don't have to sovle but we are left with one variable and thus the problem is solveable

Answer C
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