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

Answer: Option B

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