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To arrive at its destination on time the bus should have [#permalink]
29 Oct 2007, 16:02
2
This post was BOOKMARKED
00:00
A
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Difficulty:
95% (hard)
Question Stats:
43% (02:54) correct
57% (02:09) wrong based on 86 sessions
To arrive at its destination on time, a bus should have maintained a speed of \(V\) kmh throughout its 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. If, as a result, the bus arrived at its destination \(X\) minutes earlier than planned, what was the actual duration of the trip?
Re: To arrive at its destination on time the bus should have [#permalink]
29 Oct 2007, 19:52
beckee529 wrote:
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
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!!!
Re: To arrive at its destination on time the bus should have [#permalink]
29 Oct 2007, 20:49
1
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beckee529 wrote:
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
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
Re: To arrive at its destination on time the bus should have [#permalink]
30 Oct 2007, 02:49
bkk145 wrote:
beckee529 wrote:
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
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
Re: To arrive at its destination on time the bus should have [#permalink]
01 Nov 2011, 14:52
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:
Re: To arrive at its destination on time the bus should have [#permalink]
01 Nov 2011, 22:34
Expert's post
beckee529 wrote:
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
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. _________________
Re: To arrive at its destination on time the bus should have [#permalink]
12 Sep 2014, 03:33
Expert's post
1
This post was BOOKMARKED
m10 q08
To arrive at its destination on time the bus should have maintained a speed of \(v\) kilometers per hour throughout the journey. Instead, after going the first third of the distance at \(v\) kilometers per hour, the bus increased its speed and went the rest of the distance at \(1.2v\) kilometers per hour. As a result, the bus arrived at its destination \(x\) minutes earlier than planned. What was the actual duration of the trip?
A bus covered 1/3 of the distance at \(v\) kilometers per hour and the remaining 2/3 of the distance at \(1.2v\) kilometers per hour.
Say the actual duration of the trip is \(t\) and the distance is \(d\);
Then \(t=\frac{(\frac{d}{3})}{v}+\frac{(\frac{d2}{3})}{1.2v}\) --> \(t=\frac{d}{v}*(\frac{1}{3}+\frac{2}{3.6})\) --> \(t=\frac{d}{v}*\frac{8}{9}\)
Also we know that if the speed throughout the journey had been \(v\) kilometers per hour the bus would need \(\frac{x}{60}\) hours more time to cover the same distance: \(t+\frac{x}{60}=\frac{d}{v}\);
Substitute \(\frac{d}{v}\) in the first equation: \(t=(t+\frac{x}{60})*\frac{8}{9}\). So, to get \(t\) we need to know the value of \(x\).
(1) v = 60. Not sufficient. (2) x = 20. Sufficient.
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