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
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.
argggghhh!
go with Steve's method, I'm going back to bed
To determine the speed of the stream and the speed of the boat in still water, we can use a system of equations. Let's assume the speed of the boat in still water is "b" km/h and the speed of the stream is "s" km/h.
First, let's analyze the first scenario: the boat goes 30 km upstream and 44 km downstream in 10 hours.
When going upstream (against the current), the effective speed of the boat is reduced by the speed of the stream. Therefore, the relative speed for the upstream journey is (b - s) km/h.
When going downstream (with the current), the effective speed of the boat is increased by the speed of the stream. Therefore, the relative speed for the downstream journey is (b + s) km/h.
Using the formula distance = speed * time, we can write two equations:
30 = (b - s) * t1 ----(1)
44 = (b + s) * t2 ----(2)
where t1 and t2 are the times taken for the upstream and downstream journeys respectively.
Next, let's analyze the second scenario: the boat goes 40 km upstream and 55 km downstream in 13 hours.
Using the same logic, we can write two more equations:
40 = (b - s) * t3 ----(3)
55 = (b + s) * t4 ----(4)
where t3 and t4 are the times taken for the upstream and downstream journeys in the second scenario.
Now we have a system of four equations (equations 1, 2, 3, and 4) with four unknowns (b, s, t1, and t2).
To solve this system of equations, we can use a variety of methods such as substitution or elimination. However, since our ultimate goal is to determine the speed of the stream and the speed of the boat, we can solve the system of equations by elimination method. By subtracting equation 1 from equation 3 and equation 2 from equation 4, we eliminate t1 and t2:
40 - 30 = (b - s) * t3 - (b - s) * t1
55 - 44 = (b + s) * t4 - (b + s) * t2
Simplifying these expressions, we get:
10 = (b - s) * (t3 - t1) ----(5)
11 = (b + s) * (t4 - t2) ----(6)
Now we have two equations (equations 5 and 6) and two unknowns (b - s and b + s).
To solve this system of equations, we can divide equation 6 by equation 5:
11 / 10 = (b + s) * (t4 - t2) / (b - s) * (t3 - t1)
Simplifying further, we get:
11 / 10 = (b + s) / (b - s)
To solve this equation, we can cross-multiply:
11 * (b - s) = 10 * (b + s)
Expanding both sides, we get:
11b - 11s = 10b + 10s
Rearranging terms, we get:
11b - 10b = 11s + 10s
Simplifying further, we get:
b = 21s
Now we have the relationship between the speed of the boat in still water (b) and the speed of the stream (s).
To solve for the specific values of b and s, we need to substitute this relationship into one of the original equations. Let's substitute it into equation 1:
30 = (21s - s) * t1
Simplifying further, we get:
30 = 20s * t1
Dividing both sides by 20t1, we get:
s = 30 / (20t1)
Now we have the value of s in terms of t1.
Similarly, we can substitute the relationship between b and s into equation 2 to get the value of t2, and substitute it into equations 3 and 4 to get the values of t3 and t4, respectively.
Now that we have the values of s, t1, t2, t3, and t4, we can substitute them back into the original equations to find the specific values of b and s.
By following this method, we can determine the speed of the stream (s) and the speed of the boat in still water (b).
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