To arrive at its destination on time the bus should have

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To arrive at its destination on time the bus should have [#permalink]

29 Oct 2007, 17:02

To arrive at its destination on time the bus should have maintained a speed of V kmh throughout the journey. Instead, after going the first third of the distance at V kmh, the bus increased its speed and went the rest of the distance at (1.2)*V kmh. As a result, the bus arrived at its destination X minutes earlier than planned. What was the actual duration of the trip?

1. V = 60
2. X = 20

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29 Oct 2007, 18:29

If distance = d,

then time taken travelling at constant speed v kmh = d/v h
time taken travelling at v kmh for x/3km and 6v/5 kmh for 2x/3 km = d/3v + 5d/9v = 8d/9v

8d/9v = d/v - x/60 -- [1]

St1:
v = 60 ---> Reduces [1] to d = 9x. Insufficient.

St2:
x = 20 --> Reduce [1] to v = 3d. Insufficient.

St1 and St2:
d = 9x --[a]
d = v/3 --[b]

[a] = [b]
9x = v/3
x = v/27 = 2.22 kmh. Sufficient.

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29 Oct 2007, 19:09

I think B should do it

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Re: distance [#permalink]

29 Oct 2007, 20:52

beckee529 wrote:

We have 3 rates: Original rate, 1/3 original rate, and 2/3 new rate.

Let: t be the time for 1/3 rate
T be the time for 2/3 rate
Z be the time for the original rate.
d be the distance

We really want to know what T+t is. But the problem isn't that easy so well need to do some rewording. So here are a few equations well need

:Z- (T+t)=X

riginal time: Z=d/V

:1/3 time: t=(d/3)/V --> (6d/18)/V

:2/3 time: T= (2d/3)/1.2V --> (10d/18)/V

Now we can reword our first equation to: d/V - ((10d/18)/V+(6d/18)/V)

(18d/18)/V - (16d/18)/V --> (2d/18)/V = X --> (d/9)/V=X

Now we can finally look at the statements:

S1: V=60. This doesn't help us. We cannot find out what T+t is.

All we can get is that (16d/18)/60= T+t We still have d in there and there is no way to elim the d without adding another variable. Insuff.

S2: x=20 well now we can refer back to our original Z-(T+t)=X

Convieniently reworked to: (d/9)/V=20 180V=d. Plug this into

T+t--> T+t= (16(180V)/18)/V ---> 160V^2=T+t...

ughhh NOOOOOOO on paper I was able to cancel out the V's now i cant. Arghhh!!!

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Re: distance [#permalink]

29 Oct 2007, 21:49

beckee529 wrote:

I got B.

d = total distance
d/v is the original total time
d/(3v) + 2d/(3*(1.2v)) is the new total time
Therefore,
d/v - d/(3v) - 2d/(3.6v) = x
d/v*(1 - 1/3 - 1/1.8) = x

Looking for d/v = ?

(1) Given v, you can find d in term of x. This doesn't help us. Imagine the distance is infinite, there is no way we can find the actual time. INSUFFICIENT

(2) Given x, we can plug in
d/v*(1 - 1/3 - 1/1.8) = x
to obtain d/v
SUFFICIENT

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Re: distance [#permalink]

30 Oct 2007, 03:49

bkk145 wrote:

beckee529 wrote:

wow!!! what a fundoo way.. great bkk..

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Re: distance [#permalink]

01 Nov 2011, 15:52

Just showing some more of the steps a little different way....

velocity * time = distance
vt = d
t = d/v

Thus, if we can find d/v, we can find t.

Now, Let:
t = original planned time (hours)
x = minutes arrived early; Must divide by 60 to convert to hours
v= original velocity (km/h)
d = total distance

Then, actual time of the trip (what we are solving for) is (t - x/60) which equals (from substitution from above):
(d/v) - x/60; So, if we know d/r and x, we can find the actual time of the trip

For the later 2/3rds of the trip, what is given in the problem is:

(1.2v)(2/3t - x/60)= (2/3)d

For B, plug in x = 20

(1.2v)(2/3t - 20/60) = (2/3)d

Substitute t= d/v
(1.2v)((2/3)(d/v)) - 1/3) = (2/3)d

Multiple out, .8d - 0.4v = (2/3)d
(2/15)d = 0.4v
d= (0.4*15/2) v
d = 3v
d/v =3

now, we know d/v and we know x (which is given), so we can find the actual time for the trip.

The answer is B.

Choice A doesn't work for reason explained in earlier posts...

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Re: distance [#permalink]

01 Nov 2011, 23:34

beckee529 wrote:

Here is my take.
Let the actual duration of the trip be t.
Distance is same in both the cases.
So, V*t = Increased Average Speed*(t - X/60)

What is the Increased Average Speed?
Average Speed = Total Distance/Total Time = \frac{1}{[\frac{\frac{1}{3}}{V} + \frac{\frac{2}{3}}{1.2V}]}
We will get this increased average speed in terms of V.

When we put it in the equation above, V will get canceled leaving us with t in terms of X. So to get t, we only need X.
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Re: distance [#permalink] 01 Nov 2011, 23:34

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