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
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!!!