Problem Solving

Hi,

My answer was A, but the correct answer is C) (so, obviously my calculation was wrong) Can you help?? Thanks!


A bus trip of 450 miles would have taken 1 hour less if the average speed S for the trip had been greater by 5 miles per hour. What was the average speed S, in miles per hour, for the trip?
(A) 10
(B) 40
(C) 45
(D) 50
(E) 55

The difference in time (1 hr) can be represented by following equation

450/S - 450/(S+5) = 1

Solving for S, we get 45, -50. Hence C

It might be easier to substitute values from the answer to arrive at
the answer quickly. Choose the middle answer for starters.
We have 450/45 - 450/50 = 1. Which is what we are looking for !

Thanks Jay- it's very clear and good idea to try from C).

The total distance covered during the bus trip is 450 miles.
and, average speed, S, =Total distance covered/ total time taken
S=450/ total time taken
total time taken= 450/S .hour

from the question, the time taken could have been reduced by 1 hour if the average speed S have been increased by 5miles/hr
i.e if Avg. speed (S+5) miles/hr
$$then\ time=\left(\frac{450}{S}d-\frac{1}{ }\right)\ hr$$
$$Therefore,S+5=\frac{total\ dis\tan ce}{\frac{450}{S}-1}$$
Since distance is still the same,
$$Therefore,S+5=\frac{450}{\frac{450}{S}-1}$$
$$\left(S+5\right)\left(\frac{450}{S}-1\right)=450$$
By expanding the expression on the left, we have,
$$450-S+\frac{2250}{S}-5=450$$
$$S+5-\frac{2250}{S}=0$$
$$S^2+5S-2250=0$$
solving this quadratically, we obtain
(S+50)(S-45)=0
S=-50 or S=45 miles/hr

since average speed cannot be negative, we will discard the negative value obtained.
Average speed, S= 45 miles/hr