Mr A started half an hour later than usual for the marketplace. But by increasing his speed to $$\frac{3}{2}$$ times his usual speed he reached 10 minutes earlier than usual. What is his usual time for this journey?

Solution (By Examveda Team)

Let the usual speed be S.
So, increased speed = $$\frac{{3{\text{S}}}}{2}$$
Let usual time taken be T
Thus, S × T = D, and also $$\frac{{3{\text{S}}}}{2}$$ × (T - 40) = D. (T - 40, Because he started 30 minutes later and reached 10 minutes earlier, means he saved 40 minutes)
$$\frac{{3{\text{S}}}}{2}$$ × (T - 40) = ST
$$\frac{{3 \times {\text{ST}}}}{2}$$  - 60S = ST
$$\frac{1}{2}$$ × ST = 60S
T = 120 minutes = 2 hours

Alternatively,
As, ST = D [S ∝ $$\frac{1}{{\text{T}}}$$ , speed is inversely proportional to time] Speed increase to $$\frac{{3}}{{2}}$$ times so time decreases to $$\frac{{2}}{{3}}$$ times .
Net decrease = $$\frac{{1}}{{3}}$$ = 40 minutes.
Hence actual time taken
= 3 × 40 = 120 mnts.
= 2 hours