Running at $$\frac{5}{4}$$ of his usual speed a person improveshis timing by 30 minutes. What was his usual time in hours?

Solution:

Let old speed and time be S1 and T1.

Let new speed and time be S2 and T2.

A/q,

S2=(5/4)S1 = 1.25 S1 .....Eqn (i)

T1 - T2 = 30minutes = 0.5 hr.....Eqn (ii)

As Distance remains constant ; S x T = constant;

S1 x T1 = S2 x T2

S1 x T1 = (1.25 S1) x T2......from Eqn.(i)

T1/T2 = 1.25

T1= 1.25 T2

and T1- T2 = 0.5 hr .........from Eqn.(ii)

1.25T2 - T2 = 0.5Hr

0.25 T2 = 0.5 Hr.

T2 = 2 Hr.

T1 = old time = T2 + 0.5 Hr. = 2.5 Hr.Answer