Question

Nathan is out rafting. He rafts 16 miles with the river current. At the end of 16 miles, he turns around and rafts the same distance against the river current. The journey takes him 4 hours overall. If he can raft at a speed of 9 mph in still water, what is the speed of the current of the river he is in?
A 5 mph
B 4 mph
C 6 mph
D 3 mph

Answer 1

S = d/t

st = d

t = d/s

The time going is t1.
The time returning is t2.
The total time is 4 hours, so we have t1 + t2 = 4

The speed of the current is c.
The speed going is 9 + c.
The speed returning is 9 - c.

t1 = 16/(9 + c)

t2 = 16/(9 - c)

t1 + t2 = 16/(9 + c) + 16/(9 - c)

4 = 16/(9 - c) + 16/(9 + c)

1 = 4/(9 - c) + 4/(9 + c)

(9 + c)(9 - c) = 4(9 - c) + 4(9 + c)

81 - c^2 = 36 - 4c + 36 + 4c

81 - c^2 = 72

c^2 = 9

c^2 - 9 = 0

(c + 3)(c - 3) = 0

c + 3 = 0   or   c - 3 = 0

c = -3   or   c = 3

We discard the negative answer, and we get c = 3.

The speed of the current is 3 mph.

Answer 2

Answer: D. 3 miles per hour.

Explanation:

Since, the speed of the boat in still water = 9 mph

Let, x be the speed of the current of the river.

Then, the speed of boat with the river current = 9+x mph

And, speed of boat against the river current = 9-x mph

According to the question,

Total distance with river=total distance against river = 16 miles.

Thus, Time taken by boat with the current= $\frac{16}{9+x}$ ( because time=distance/speed)

And, Time taken by boat against the current= $\frac{16}{9-x}$

But, total time taken by boat in both with and against the current= 4 hours.

⇒$\frac{16}{9+x}+\frac{16}{9-x}= 4$

⇒$16(\frac{1}{9+x}+\frac{1}{9-x})= 4$

⇒$4(\frac{1}{9+x}+\frac{1}{9-x})= 1$

⇒$4(\frac{1}{81-x^2})= 1$

⇒$\frac{18}{81-x^2}= 1/4$

⇒$x^2=9$⇒$x=\sqrt{9}=3$ ( we will not take -3 value because speed can not be negative)

Therefore, speed of the current of the river= 3mph