Question 1145608
basic formula is r * t = d
r is the rate
t is the time
d is the distance.


when t = 2, the formula becomes 2 * r = d
let b = the rate of the boat.
let w = the rate of the water.


with the current, the formula becomes 2 * (b + w) = d
against the current, the formula becomes 2 * (b - w) = d - 16


this conforms to the requirement of the problem that that the return trip comes 16 miles short of the distance traveled when going with the current.


simplify these two equations to get:
2b + 2w = d
2b - 2w = d - 16


subtract the second equation from the first to get:
4w = 16


solve for w to get w = 4


you are given that the rate of the water is 1/3 the rate of the boat.
the equation for that is 1/3 * b = w
since w = 4, this becomes:
1/3 * b = 4
solve for b to get:
b = 12


you now have the rate of the boat is 12 mph and the rate of the water is 4 mph.


the rate of the boat is 12 mph is your answer.


you can confirm by replacing b and w in the original equations to get:


2b + 2w = d becomes 24 + 8 = d which gets you d = 32.


2b - 2w = d - 16 becomes 24 - 8 = d - 16 which gets you 16 = d - 16
solve for d to get d = 32.


replacing all variables with their values in the original equations gets you.


2b + 2w = d becomes 24 + 8 = 32 which becomes 32 = 32, which is true.


2b - 2w = d - 16 becomes 24 - 8 = 32 - 16 which becomes 16 = 16, which is true.


the values of b and w and d are confirmed to be true.
your solution is that the speed of the boat is 12 mph.