A motorboat travels 244 km in 4 hours going upstream and 891km in 9 hours going downstream. What is the rate of the boat in still water and what is the rate of the current?

Question

Rate of the boat in still water: ___km/h

Rate of the current: ___km/h

Answer 1

Suppose, V_r be the rate of current of the river.

V_b be the rate of boat.

For Upstream, boat travels so 144 km in 4 hours so,

(244/4) =61= V_r - V_b

For Downstream, boat travels 891 km in 9 hours so,

(891/9) = 99=V_r + Vb

Adding above equations we get,

61 + 99 = V_r - Vb + V_r + Vb

160 = 2V_r

V_r= (160/2) = 80 km/hr

Hence, rate of current is 80 km/hr.

Substituting the value of V_r in 1st eqaution then,

61= 80-V_b

V_b = 80-61= 19 km/hr

Hence, rate of the boat is 19km/hr.

Answer 2

Let

u = the speed of the water (the rate of the current), km/h

v = the speed of the boat (rate) in still water, km/h

When traveling upstream, the net speed of the boat is diminished by the speed of the river to (v-u) km/h.

The distance traveled is 244 km and the time taken is 4 h.

Because time = distance/speed, therefore

244/(v-u) = 4

Multiply each side by (v-u).

4(v-u) = 244

Divide each side by 4.

v - u = 61 (1)

When traveling downstream, the net speed of the boat is increased by the speed of the river to (v+u) km/h.

The distance traveled is 891 km and the time taken is 9 h. Therefore

891/(v+u) = 9

Multiply each side by (v+u).

9(v+u) = 891

Divide each side by 9.

v + u = 99 (2)

Add equations (1) and (2).

v - u + v + u = 61 + 99

2v = 160

v = 160/2 = 80

From (2), obtain

80 + u = 99

u = 99 - 80 = 19

Answer:

v = 80 km/h, the speed (rate) of the boat in still water

u = 19 km/h, the speed (rate) of the current.