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# Easy algebra 1 math problem?

A boat traveled 40 miles downstream and back. The trip downstream took 2 hours. The trip back took 10 hours. What is the speed of the boat in still water. What is the speed of the current?

Can't figure this out! explain please and thank you

---the answer is boat= 12 mph current= 8mph

??

### 5 Answers

- TomVLv 79 years agoFavorite Answer
b = speed of boat in still water

c = speed of current

time = distance/speed

Upstream against the current:

10 =40/(b-c)

Downstream with the current

2 = 40/(b+c)

10b-10c = 40

2b +2c = 40 : multiplied by 5 => 10b + 10c = 200

add the two equations:

10b + 10b - 10c + 10c = 40 + 200

20b = 240

b = 12 mph

2(12) + 2c = 40

2c = 40-24 = 16

c = 8

Speed of current c = 8 mph

Speed of boat b = 12 mph

- peabodyLv 79 years ago
Let x = boat speed & y = current speed

upstream speed = x -y

downstream speed = x+y

use d =vt

Upstream

20 = (x-y)10

2 = x -y (1)

Downstream

20 = (x +y)2

10 = x +y (2)

(1) + (2) gives 12 = 2x

so x = 6 mph

sub. x=6 into (2)

10 = 6 +y

so y = 4 mph

Boat speed is 6 mph & current speed is 4 mph

- Anonymous9 years ago
I'll give it a go.

Trip down stream is 40mi takes 2hours - Speed = distance/time

s=d/t

s=40/2

s=20mi/hr

Trip back is 40mi takes 10 hours - Speed = distance/time

s=d/t

s=40/10

s=4mi/hr

Therefore the current would be 20mi/hr - 4 mi/hr = 16mi/hr so 8mi/hr for each trip

so the boat speed down stream is 20mi/hr less current 8mi/hr leaves 12mi/hr

and the trip back is 4mi/hr plus currnt 8mi/hr makes it also 12mi/hr.

That's how I would work it out anyway.

- Mike GLv 79 years ago
40/[B+C] = 2 where B = boat speed and C = current

40 = 2B + 2C ..........(1)

200 = 10B+10C....(2)

40/[B-C] = 10

40 = 10B-!0C

Add equation (2)

240 = 20B

B= 12 mph

40 = 24 + 2C

C = 8 mph

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