As part of its plan to upgrade the rail system, a progressive State has bought some new, high speed trains. To help defray the cost of these trains, the government has decided to change the fare structure. Instead of charging according to the distance a passenger wants to travel – as was past practice – the government has decided to set the fares according to the scheduled speed of a train, with a passenger on board a faster train paying more than a passenger on a slower train. In order to get to work, I have to change trains for the last quarter of my journey, and, alas, that leg is by means of an old rail-motor that is soon to be taken out of service! That leg takes, in fact, twice as long as the first leg [if, that is, both legs are according to schedule]. If I pay a $1.20 for the first leg of my journey, how much should I pay for the second leg under the government's new pricing policy?
+36923 20 cents.
rate x time = distance
First distance = 3/4 Second Distance = 1/4 (the two legs of your commute)
r (t) = 3/4 r(2t) = 1/4 (the second leg takes twice as long)
r (1) = 3/4 r(2) = 1/4
r = 3/4 = 6/8 r = 1/8
So you can see the first-leg train is SIX times faster than the second-leg train
so the fair for the second one should be 1/6th the first one
1/6 x 1.20 = 20 cents
+36923 20 cents.
rate x time = distance
First distance = 3/4 Second Distance = 1/4 (the two legs of your commute)
r (t) = 3/4 r(2t) = 1/4 (the second leg takes twice as long)
r (1) = 3/4 r(2) = 1/4
r = 3/4 = 6/8 r = 1/8
So you can see the first-leg train is SIX times faster than the second-leg train
so the fair for the second one should be 1/6th the first one
1/6 x 1.20 = 20 cents
+118629 EP and i have interpreted this question differently.
I think that he has interpreted that the second leg goes half the speed of the first leg and he has found his answer.
This may well be the intended interpretation
BUT
I interpret it that the second leg takes twice as long and how far it is is irrelevant.
If the first leg cost $1.20
and the second leg takes twice as long then I think the cost will be 60c
The rail-motor is 1/6 slower than the first train, for which the person is charged $1.20. To see that rail-motor is 1/6 slower, try and solve the following equations based on the clues given: T=3/4.D/A, and:
2T=1/4.D/B, where T is the time taken for the first leg of the journey, D is the distance of the whole journey, A is the speed of the first train and B the speed of the secod. The result is:6B = A. Therefore, the price of the second leg is:$1.20/6 =.20 cents.
