Question

Two cities x and y are 400km apart. Q leaves x 8 hours after p. Both p and q arrive simultaneously. Find the time the slower person spent on the trip if the speed of one of them was 15kmph higher than that of other. Solution:

Answer 1

Suppose, the speed of  P= s kmph and takes t time to reach y

Then, the speed of Q= s+15 kmph and takes (t-8) time to reach y

According to the question,

st=400 and st= (s+15)(t-8)

st= st-8s+15t-120

Further, solving the above equation, we get:

15t-8s-120=0

Putting the value of s from st=400, we get:

15t-8*400/t-120=0

15t^2-120t-3200=0

Solving the above equation for t, we get:

t=19.14 hours, the second value of t is irrelevant.

Therefore, time taken by the slower person=19.14 hours