A person goes from P to Q at a speed of 20 km/h. Then he goes from Q to R at a speed of q km/h. Finally the person goes from R to S at a speed of r km/h. The distances from P to Q, Q to R and R to S are equal. If the average speed from P to R is $$\frac{280}{11}$$ km/h, and the average speed from Q to S is $$\frac{112}{3}$$ km/h, then what is the value of r?

The distances from P to Q, Q to R and R to S are equal.

Let's assume the distances from P to Q = distances from Q to R = distances from R to S = y.

A person goes from P to Q at a speed of 20 km/h. Then he goes from Q to R at a speed of q km/h.

If the average speed from P to R is $$\frac{280}{11}$$ km/h

average speed = $$\frac{total\ distance}{total\ time}$$

$$\frac{280}{11}=\frac{y+y}{\frac{y}{20}+\frac{y}{q}}$$

11q-7q = 140

4q = 140

q = 35 km/h

the person goes from R to S at a speed of r km/h. the average speed from Q to S is $$\frac{112}{3}$$ km/h.

average speed = $$\frac{total\ distance}{total\ time}$$

$$\frac{112}{3}=\frac{y+y}{\frac{y}{q}+\frac{y}{r}}$$

$$\frac{112}{3}=\frac{2qr}{\left(r+q\right)}$$

Put the value of 'q' in the above equation.

$$\frac{112}{3}=\frac{2\times\ 35r}{\left(r+35\right)}$$

$$\frac{56}{3}=\frac{35r}{\left(r+35\right)}$$

$$\frac{8}{3}=\frac{5r}{\left(r+35\right)}$$

$$8r+280=15r$$

15r-8r = 280

7r = 280
r = 40

Value of r = 40