At an ordinary rate, a man can row the distance from A to B, about 15 km, in 5 hours less time than it takes him to return. Could he double his rate, his time to B would only be one hour less than his time to A. What is his usual rate in still water? What is the rate of the stream?
How were you able to obtain the answer? My equation is similar to what steve formulated but im stuck in the problem thanks for the help.
Since time=distance/speed, if his speed is s and the current's is c, then
15/(s+c) = 15/(s-c)-5
15/(2s+c) = 15/(2s-c)-1
Now just solve for s and c.
My working equations were 15/(s+c) = 5 - 15/(s-c) and 15/(2s+c) = 15/(2s-c) - 1
Whichever I use, I can't seem to get the right values of s and c. My answers are not in the choices given.
I don't think Steve is on line right now, so I will try.
Did you notice the the right side of both of Steve's equations are the opposite of yours?
Steve's solution would result in
s = 4
c = 2
Yours give inadmissible results of
s = appr -14 and c = 16.7
Your translation of the given data into equations is incorrect.
Oh yeah, I see that.
Well, I guess I'm a bit confused with the less thingy. Anyway, could you please explain how you got 4&2?
Thank you.
Nevermind, I got it. Thanks for all the help!
To solve this problem, let's break it down into two parts:
1. Determine the man's usual rate in still water.
2. Determine the rate of the stream.
Let's start with the first part:
1. Determine the man's usual rate in still water:
Let's assume the man's usual rate in still water is "R" km/h.
Now, let's consider the first part of the problem statement: "At an ordinary rate, a man can row the distance from A to B, about 15 km, in 5 hours less time than it takes him to return."
So, the time it takes for the man to row from A to B at his usual rate (R km/h) is:
Time for A to B = Distance / Rate = 15 / R
And the time it takes for the man to row back from B to A is:
Time for B to A = Distance / (Rate + Stream rate)
According to the problem, Time for B to A is 5 hours more than Time for A to B. Therefore:
Time for B to A = Time for A to B + 5
Substituting the expressions we derived above for the times:
Distance / (Rate + Stream rate) = 15 / R + 5
Now, let's move on to the second part of the problem:
2. Determine the rate of the stream:
Let's assume the rate of the stream is "S" km/h.
The rate of the stream affects the man's effective speed as he rows back from B to A. So, his effective rate will be his usual rate in still water (R km/h) plus the rate of the stream (S km/h), resulting in (R + S) km/h.
Now, we can rewrite the equation we derived earlier using the expressions for the times and incorporating the rate of the stream:
Distance / (R + S) = 15 / R + 5
To solve these two equations simultaneously, we can apply the method of substitution:
1. Rearrange the first equation to solve for Distance / R:
Distance / R = 15 / R + 5 - 5
Simplifying:
Distance / R = 15 / R
2. Substitute this expression for Distance / R in the second equation:
(15 / R) / (R + S) = 15 / R + 5
Now, we can cancel out the common factor of 15 / R:
1 / (R + S) = 1 + 5
Simplifying:
1 / (R + S) = 6
Cross-multiplying:
R + S = 1 / 6
R + S = 1/6
Now, we can use the second part of the problem statement: "Could he double his rate, his time to B would only be one hour less than his time to A."
When the man doubles his rate, his new rate would be 2R km/h. Therefore:
Time for A to B = 15 / R
Time for B to A = 15 / (2R + S)
According to the problem, Time for B to A is one hour less than Time for A to B. Therefore:
Time for B to A = Time for A to B - 1
Substituting the expressions we derived above for the times:
15 / (2R + S) = 15 / R - 1
Now, let's solve this equation for R + S:
15 / (2R + S) = 15 / R - 1
Cross-multiplying:
15R - R(2R + S) = 15(2R + S)
Simplifying:
15R - 2R^2 - RS = 30R + 15S
Moving all terms to one side:
2R^2 + (R + S) + 15R - 15S = 0
But we know that R + S = 1/6 from the previous equation we derived. Substituting this value in:
2R^2 + 1/6 + 15R - 15S = 0
Since our goal is to solve for R, let's solve this equation using the quadratic formula:
2R^2 + 15R + (1/6 - 15S) = 0
Using the quadratic formula:
R = (-b ± √(b^2 - 4ac)) / 2a
Substituting the values:
R = (-(15) ± √((15)^2 - 4(2)(1/6 - 15S))) / (2(2))
Simplifying:
R = (-15 ± √(225 - 4(2)(1/6 - 15S))) / 4
Simplifying further:
R = (-15 ± √(225 - (8/6 - 120S))) / 4
Simplifying the expression inside the square root:
R = (-15 ± √(225 - 16/6 + 240S)) / 4
R = (-15 ± √((1350 - 16 + 1440S) / 6)) / 4
R = (-15 ± √(1420 + 1440S) / 6) / 4
R = (-15 ± √(10(142 + 144S)) / 6) / 4
R = (-15 ± √(10(71 + 72S)) / 6) / 4
R = (-15 ± √(10(71 + 72S)) / 24
Therefore, the usual rate of the man in still water is (-15 ± √(10(71 + 72S)) / 24.
To solve for the rate of the stream (S), we would need additional information or equations.