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?

A. 1 hours

B. 2 hours

C. 3 hours

D. 45 minutes

E. None of these

Answer: Option B

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