# MATH PROBLEM DOWN THERE! :)?

A motorboat travels 306 km in 6 hours going upstream and 873 km in 9 hours going downstream. What is the rate of the boat in still water and what is the rate of the current?

Rate of the boat in still water: __km/h

Rate of the current: __km/h

Relevance
• 9 years ago

Let b = boat's rate in still water and c = current's rate.

6(b - c) = 306

9(b + c) = 873

solve the system of equations:

6(b - c) = 306

b - c = 51

b = 51 + c

9(51 + c + c) = 873

51 + c + c = 97

2c = 46

c = 23

b = 51 + 23 = 74

• TBT
Lv 7
9 years ago

Rate of the boat in still water: u km/h

Rate of the current: v km/h

306 / (u-v) = 6

51= u - v ---- (1)

873 / (u +v) = 9

97 = u + v ---- (2)

(1) +(2)

148 = 2 u

51 = 74 - v

Source(s): my brain (Prof TBT)
• 9 years ago

Let the speed of boat in still waters be x km per hour

And rate of current be y km /hr

UPSTREAM

306/6 = 51 = x -y (A)

Downstream

873/9 = 97 = x+y (B)

A +B = 2x = 148

x = 74 km/ hour

• mayi o
Lv 4
9 years ago

going with the current: v + vc

going against the current: v-vc

going through still water: v

306/6 = 51 km/hr

873/9 = 97 km/hr

v+vc = 97

v-vc = 51

2v = 148

v = 74 km/hr

rate of current, vc = 23 km/hr

• Mike G
Lv 7
9 years ago

306 = 6(V-C) so 153 = 3V-3C

873 = 9(V+C) so 291 = 3V+3C subtract equations

138 = 6C

C = 23 km/h = current

V = 74 km/h = boat