Professor Cummings laid this our logically: Consider
distance:
P: a t
Q: (a+b)(t-p)
R: (a+2b)(t-p-x)
but they are all the same distance, so
at = at + bt -ap -bp
so
bt = ap + bp
and
at = (a+2b)(t-p-x)
=at-ap-ax+2bt-2bp-2bx then
(a+2b)x = -ap+2bt-2bp or
x = (-ap+2bt-2bp)/(a+2b) but remembering from above what bt is equal to
= (-ap+2ap +2bp-2bp)/(a+2b)
= ap/(a+2b)
whew !
P, Q and R start from the same place X at (a) kmph, (a+b) kmph and (a+2b) kmph respectively.
If Q starts p hours after P, how many hours after Q should R start, so that both Q and R overtake P at the same time?
Answer is pa/a+2b
Your faculty have posted the answer but I didn't get some part of the answer properly and in that answers how i come to know that we have to take T and t-p and t-p-x and one request is that can you explain question properly in easy way as you understand
1 answer
1 answer