Question

If a person travels x% faster than normal, he reaches y minutes earlier than normal. what is his normal time of travel?

Answer 1

The person’s normal time of travel for the same distance at a constant speed is (100/x + 1)y minutes.

Explanation:

If distance is D, speed is S and time is T,

The person travels same distance at different speed and time.

D = ST ----- (1)

When he travels x% faster and reaches y minutes earlier then,

D = (xS/100 + S) (T-y) -------- (2)

Equating distance from equations (1) and (2)

ST = (100S + xS)/100 (T-y)

100T = (100+x)(T-y)

Tx = 100y+xy

T = (100/x + 1)y

Answer 2

Answer:

Y(100+x)/x or (100/x+1)y