Piyush covers a certain distance on bike. Had he moved 3 km/h faster, he would have taken 20 min less. If he had moved 2 km/h slower, he would have taken 20 min more. The distance covered (in km) is:

20 KM

The correct answer is Option (2) → 20 KM

To find the distance covered by Piyush, we can set up equations based on the relationship $\text{Distance} = \text{Speed} \times \text{Time}$.

1. Define Variables

Let:

• $s$ = initial speed in km/h
• $t$ = initial time in hours
• $d$ = distance in km (where $d = s \times t$)

2. Set up the Scenarios

Scenario 1: Had he moved 3 km/h faster, he would have taken 20 min (1/3 hour) less.

$d = (s + 3)\left(t - \frac{1}{3}\right)$

Since $d = st$, we can write:

$st = st - \frac{s}{3} + 3t - 1$

$0 = -\frac{s}{3} + 3t - 1 ⇒s = 9t - 3 \quad \text{---(Equation 1)}$

Scenario 2: If he had moved 2 km/h slower, he would have taken 20 min (1/3 hour) more.

$d = (s - 2)\left(t + \frac{1}{3}\right)$

$st = st + \frac{s}{3} - 2t - \frac{2}{3}$

$0 = \frac{s}{3} - 2t - \frac{2}{3} ⇒s = 6t + 2 \quad \text{---(Equation 2)}$

3. Solve for Time ($t$) and Speed ($s$)

Equate the two expressions for $s$:

$9t - 3 = 6t + 2$

$3t = 5$

$t = \frac{5}{3} \text{ hours}$

Now, substitute $t$ back into Equation 2 to find $s$:

$s = 6\left(\frac{5}{3}\right) + 2$

$s = 10 + 2 = 12 \text{ km/h}$

4. Calculate Distance ($d$)

$d = s \times t = 12 \times \frac{5}{3}$

$d = 4 \times 5 = 20 \text{ km}$

The distance covered is: 20 KM