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Average Speed of Bus

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Average Speed of Bus

by hardikm » Tue May 31, 2011 9:36 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.


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by Frankenstein » Tue May 31, 2011 9:51 am
Hi,
Let the distance between A and B be 'd'. so, avg speed is d/4
From(1): bus traveled 100 miles in first 2 hours. We don not know the distance it traveled in the next 2 hours.
Not sufficient
From(2): (d/2)/t1 = 2.(d/2)/t2 =>t2 = 2t1. But, t1+t2=4 =>t1=4/3, t2=8/3.We know nothing about 'd'
Not sufficient
From (1) & (2) we know nothing about 'd'
Not sufficient

Hence, E

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by cans » Tue May 31, 2011 11:46 pm
time=4 hours. to find speed
a)avg speed for first 2 hours = 100/2 = 50 miles/hr
but no information about next 2 hours, thus insufficient.
b)let total distance=d
for first d/2 miles, speed = 2s and for next d/2 miles, speed=s
avg speed = d/4 (total distance/total time)
also total time = d/2*(1/2s + 1/s) = 3d/(4s)=4
insufficient.
both together, 2s=50 or s=25
thus 3d/100=4 or d=400/3
thus avg speed = d/4 = 100/3 miles/hr
IMO C
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by sjmit4 » Wed Jun 01, 2011 12:03 am
cans wrote:time=4 hours. to find speed
a)avg speed for first 2 hours = 100/2 = 50 miles/hr
but no information about next 2 hours, thus insufficient.
b)let total distance=d
for first d/2 miles, speed = 2s and for next d/2 miles, speed=s
avg speed = d/4 (total distance/total time)
also total time = d/2*(1/2s + 1/s) = 3d/(4s)=4
insufficient.
both together, 2s=50 or s=25
thus 3d/100=4 or d=400/3
thus avg speed = d/4 = 100/3 miles/hr
IMO C
Hey cans
Both together 2s= 50 ?
I think it is not the case as 2 hrs does not mean the distance traveled is half.
So s is not equal to 50 (you actually can not get value of s )
Hence option E
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by Frankenstein » Wed Jun 01, 2011 12:04 am
cans wrote:time=4 hours. to find speed
a)avg speed for first 2 hours = 100/2 = 50 miles/hr
but no information about next 2 hours, thus insufficient.
b)let total distance=d
for first d/2 miles, speed = 2s and for next d/2 miles, speed=s
avg speed = d/4 (total distance/total time)
also total time = d/2*(1/2s + 1/s) = 3d/(4s)=4
insufficient.
both together, 2s=50 or s=25
thus 3d/100=4 or d=400/3
thus avg speed = d/4 = 100/3 miles/hr
IMO C
Hi,
In (1), avg speed is calculated for first 2 hours but in (2), '2s' is average speed for half distance and this half distance is not traveled in 2 hours. So, you can't equate them.

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by cans » Wed Jun 01, 2011 12:16 am
oops! thanks for correction :)
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by phanideepak » Mon Jun 06, 2011 8:47 pm
Hi guys what the OA??

Please correct me if my approach is wrong. I agree that both A and B are not sufficient but yeah when we combine both

eq 2 : d/2v + d/4v = 4 so from this d/2v = 8/3 and d/4v = 4/3

from the first eqn we know that 4/3(2v) + (2-4/3)v = 100 So from this we can solve for v

v = 60 2v = 120 so we have no unknowns isn't the answer C ?
Last edited by phanideepak on Wed Jun 08, 2011 7:51 pm, edited 1 time in total.
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by aftableo2006 » Mon Jun 06, 2011 8:54 pm
i think the answer is E
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by phanideepak » Mon Jun 06, 2011 9:02 pm
Hi could you tell me what is wrong with the approach that I have used?
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by sjmit4 » Mon Jun 06, 2011 11:33 pm
phanideepak wrote:Hi guys what the OA??

Please correct me if my approach is wrong. I agree that both A and B are not sufficient but yeah when we combine both

eq 2 : d/2v + d/4v = 4 so from this d/2v = 8/3 and d/4v = 4/3

from the first eqn we know that 4/3(2v) + (2-4/3)v = 200 So from this we can solve for v

v = 60 2v = 120 so we have no unknowns isn't the answer C ?
Hey phanideepak there is problem in your first equation 4/3(2v) + (2-4/3)v = 200.
Problem here is 200 is a value which is not correct.
In first 2 hours the bus covered 100 miles does not mean in 4 hours it covered 200 miles as speed of the bus changed in between.Yes this change occurred when bus had traveled half the distance but it does not mean the time was also half used. So we can not get value 200 or in fact any value using 1st line
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by phanideepak » Tue Jun 07, 2011 2:31 am
sjmit4 wrote:
phanideepak wrote:Hi guys what the OA??

Please correct me if my approach is wrong. I agree that both A and B are not sufficient but yeah when we combine both

eq 2 : d/2v + d/4v = 4 so from this d/2v = 8/3 and d/4v = 4/3

from the first eqn we know that 4/3(2v) + (2-4/3)v = 200 So from this we can solve for v

v = 60 2v = 120 so we have no unknowns isn't the answer C ?
Hey phanideepak there is problem in your first equation 4/3(2v) + (2-4/3)v = 200.
Problem here is 200 is a value which is not correct.
In first 2 hours the bus covered 100 miles does not mean in 4 hours it covered 200 miles as speed of the bus changed in between.Yes this change occurred when bus had traveled half the distance but it does not mean the time was also half used. So we can not get value 200 or in fact any value using 1st line
@sjmit4

Hey that was a typo.

We do know that in the first two hours the bus travelled 100 kms and we know that the bus has travelled at 2v for 4/3 and v for 8/3

so I used this in the first eqn.

2v(4/3) + v(2-4/3) = 100

from this we can solve for v and also get total d all are known now so the answer can be C. Please correct me if I'm wrong.
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by krishnasty » Tue Jun 07, 2011 7:20 am
IMO E
phanideepak wrote:
sjmit4 wrote:
phanideepak wrote:Hi guys what the OA??

Please correct me if my approach is wrong. I agree that both A and B are not sufficient but yeah when we combine both

eq 2 : d/2v + d/4v = 4 so from this d/2v = 8/3 and d/4v = 4/3

from the first eqn we know that 4/3(2v) + (2-4/3)v = 200 So from this we can solve for v

v = 60 2v = 120 so we have no unknowns isn't the answer C ?
Hey phanideepak there is problem in your first equation 4/3(2v) + (2-4/3)v = 200.
Problem here is 200 is a value which is not correct.
In first 2 hours the bus covered 100 miles does not mean in 4 hours it covered 200 miles as speed of the bus changed in between.Yes this change occurred when bus had traveled half the distance but it does not mean the time was also half used. So we can not get value 200 or in fact any value using 1st line
@sjmit4

Hey that was a typo.

We do know that in the first two hours the bus travelled 100 kms and we know that the bus has travelled at 2v for 4/3 and v for 8/3

so I used this in the first eqn.

2v(4/3) + v(2-4/3) = 100

from this we can solve for v and also get total d all are known now so the answer can be C. Please correct me if I'm wrong.
Its getting confused by every post.
we can create 2 equations here.
1st - 2hours - 100 miles. speed of the second half not known. Hence, insufficient
2nd - first half - 2x mph, 2nd half - x mph - not sufficient

Combining both the equations,
(D1/2x) + (D2/x) = 4
D1 = 100
hence,
(50 + D2) = 4x

One equation, 2 unknown. Hence, not sufficient.

Hope it's clear now :)
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by phanideepak » Tue Jun 07, 2011 7:52 am
Krish i think you have understood the question wrong. There is no question of D1 and D2 as he says that first half and second half of the distance. so d1 = d2 = d/2.
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by krishnasty » Tue Jun 07, 2011 7:56 am
phanideepak wrote:Krish i think you have understood the question wrong. There is no question of D1 and D2 as he says that first half and second half of the distance. so d1 = d2 = d/2.
sorry, my bad..missed one of d most imp factor in the question.. :P
In this case, can v even put these equations together?
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by phanideepak » Tue Jun 07, 2011 7:59 am
Yep please look at my previous post. from the second question we can calculate the times for which the average velocities are v and 2v respectively and accordingly we can substitute in equation 1 and solve for v and d. Please verify my approach.
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