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:

Piyush travels some distance on his bike. If he moves 3 km/h faster, he takes 20 minutes less. If he moves 2 km/h slower, he takes 20 minutes more.

Let Distance = $D$ km, Normal Speed = $S$ km/h, and Normal Time = $T$ hours.

Distance = Speed × Time, so $D = ST$

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Converting 20 minutes to hours: 20 minutes = $\frac{20}{60} = \frac{1}{3}$ hour

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When speed is 3 km/h faster:

New speed = $(S + 3)$ km/h

New time = $(T - \frac{1}{3})$ hours

Distance remains the same:

$D = (S + 3)(T - \frac{1}{3})$

Since $D = ST$:

$ST = (S + 3)(T - \frac{1}{3})$

$ST = ST - \frac{S}{3} + 3T - 1$

$0 = -\frac{S}{3} + 3T - 1$

$\frac{S}{3} = 3T - 1$

$S = 9T - 3$ ... (Equation A)

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When speed is 2 km/h slower:

New speed = $(S - 2)$ km/h

New time = $(T + \frac{1}{3})$ hours

$ST = (S - 2)(T + \frac{1}{3})$

$ST = ST + \frac{S}{3} - 2T - \frac{2}{3}$

$0 = \frac{S}{3} - 2T - \frac{2}{3}$

$\frac{S}{3} = 2T + \frac{2}{3}$

$S = 6T + 2$ ... (Equation B)

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From Equations A and B:

$9T - 3 = 6T + 2$

$3T = 5$

$T = \frac{5}{3}$ hours

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Substituting $T = \frac{5}{3}$ into Equation B:

$S = 6(\frac{5}{3}) + 2$

$S = 10 + 2$

$S = 12$ km/h

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Distance = Speed × Time:

$D = 12 \times \frac{5}{3}$

$D = \frac{60}{3}$

$D = 20$ km

Therefore, the distance covered is 20 km.