A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours it goes 40 km upstream and 55 km downstream. Determine the speed of the stream and that of boat in still water

Answer 1

let the boat have speed x and the river have speed y.

Since time = distance/speed,

30/(x-y) + 44/(x+y) = 10
40/(x-y) + 55/(x+y) = 13

Just looking at the equations, I expect we will find that (x+y) = 11 and (x-y) is a multiple of 5. Doing the actual solutions, we find that indeed,

x=8 and y=3

I expect you will check to be sure that those numbers fit the problem's conditions.

Answer 2

speed of boat --- x km/h
speed of current -- y km/h

case1: 30/(x-y) + 44/(x+y) = 10
times (x+y)(x-y) ....
30(x+y) + 44(x-y) = 10(x+y)(x-y)
30x + 30y + 44x - 44y = 10x^2 - 10y^2

10x^2 - 10y^2 = 74x - 14y ----> #1

case2: 40/(x-y) + 55(x + y) = 13
40(x+y) + 55(x-y) = 13(x+y)(x-y)
13x^2 - 13y^2 = 95x - 15y -----> #2

two hyperbolas !
messy to solve

with the help of Wolfram ....
http://www.wolframalpha.com/input/?i=solve+10x%5E2+-+10y%5E2+%3D+74x+-+14y+%2C+13x%5E2+-+13y%5E2+%3D+95x+-+15y+

(8,3) works in both

So the speed of the boat in still water is 8 km/h and the speed of the current is 3 km/h

actual work:
#1 times 13 --> 130x^2 - 130y^2 = 962x - 182y
#2 times 10 --> 130x^2 - 130y^2 = 950x - 150y

subtract them:
0 = 12x - 32y
12x = 32y
y = 3x/8

sub into the 1st
10x^2 - 10y^2 = 74x - 14y
10x^2 - 10(9x^2/64) = 74x - 14(3x/8)

times 64:
640x^2 - 90x^2 = 4736x - 336x
550x^2 - 4400x = 0
x^2 - 8x = 0
x(x-8) = 0
x = 0 or x = 8

if x = 0 , -----> y = 0, makes no sense

if x = 8, then y = 3

Answer 3

argggghhh!

go with Steve's method, I'm going back to bed