a boat averages 6 miles per hour, taking the same amount of time to travel 14 miles downstream and 7 miles upstream. What is the currents rate?

Question

How might I start going about this problem?

 

Answer 1

Hi Megan,

 

Let's begin by assigning letters to represent our unknowns:

    b = rate of the boat in still water

    c = rate of the current

 

Next, let's write expressions for the boat's time downstream with the current and its upstream time against the current (time = distance / rate):

    14 / (b + c)  [boat's downstream time]

    7 / (b - c)     [boat's upstream time]

 

Now let's set these expressions equal to each other (as per the problem) and then express one variable in terms of the other:

    14 / (b + c) = 7 / (b - c)

    14(b - c) = 7(b + c)

    14(b - c)/7 = 7(b + c)/7

    2(b - c) = b + c

    2b - 2c = b + c

    2b - b = c + 2c

    b = 3c (rate of the boat in still water is three times

               the rate of the current)

 

Next, let's write an equation expressing the boat's upstream and downstream average distance traveled in 1 hour (rate times time) given in the problem:

    [(b + c)(1) + (b - c)(1)] / 2 = 6

 

Next, let's substitute "3c" for "b" (as per above) into the equation for c:

    [(3c + c)(1) + (3c - c)(1)] / 2 = 6

    4c + 2c = 6(2)

    6c = 12

    c = 12/6

    c = 2 mph (rate of the current)

    b = 3c

    b = 3(2)

    b = 6 mph (rate of the boat in still water) 

    b + c = 6 + 2

    b + c = 8 mph (the boat's downstream rate)

    b - c = 6 - 2

    b - c = 4 mph (the boat's upstream rate)

 

Finally, we can verify our answer for the rate of the current by plugging in our answers for the boat's downstream and upstream rates in our original time equation to see if the times are equal:

    14 / (b + c) = 7 / (b - c)  [original time equation]

    14/8 = 7/4 [rate plug-ins]

    7/4 = 7/4

    1.75 h = 1.75 h (times are equal)

 

Since the times are equal using the boat's downstream and upstream rates based upon the rate of the current, we are confident that our answer for the rate of the current is correct.

 

Whew - this was a complex problem for sure !!

 

Thanks for submitting this problem and glad to help.

 

God bless, Jordan.