Question

Ben's boat will take 1 1/2 hours longer to go 12 miles up a river than to return. What is the rate of his boat if the rate of the current is 2 miles an hour? 4 mph 5 mph 6 mph

Answer 1

Answer:

6 mph

Step-by-step explanation:

Let x mph be the rate of the boat. Going up the river boat has the rate x-2 mph (the current decreases the rate of the boat) and going down the river boat has the rate x+2 mph (the current increases the rate of the boat). Then it will take $\dfrac{12}{x-2}$ hours to go up the river and $\dfrac{12}{x+2}$ hours to go down the river. Thus,

$\dfrac{12}{x-2}-\dfrac{12}{x+2}=1\dfrac{1}{2},\\ \\\dfrac{12x+24-12x+24}{(x-2)(x+2)}=\dfrac{3}{2},\\ \\96=3(x^2-4),\\ \\x^2-4=32,\\ \\x^2=36,\\ \\x=\pm6\ mph.$

The rate of the boat cannot be negative, hence, x=6 mph.

Answer 2

Answer:

The rate of his boat is:

                    6 mph

Step-by-step explanation:

It is given that:

Ben's boat will take 1 1/2 hours longer to go 12 miles up a river than to return.

Let u denote  the speed of the boat in still water.

and v denote the speed of the current.

Then the speed of boat upstream= u-v km/h

and speed of boat downstream=u+v km/h

Let t denote the time taken by the boat downstream.

Then the time taken by boat upstream is: t+(3/2) hours

Distance traveled by boat each way is: 12 miles.

Hence, we have:

Speed of boat upstream is:

$\dfrac{12}{t+\dfrac{3}{2}}$

i.e.

$u-v=\dfrac{12}{t+\dfrac{3}{2}}----------(1)$

and

speed of boat downstream i.e.

$u+v=\dfrac{12}{t}------------(2)$

On subtracting equation (1) from equation (2) we have:

$2v=\dfrac{12}{t}-\dfrac{12}{t+\dfrac{3}{2}}$

Also, we are given :

$v=2\ mph$

i.e.

$2\times 2=\dfrac{12}{t}-\dfrac{12}{t+\dfrac{3}{2}}$

i.e.

$4=\dfrac{12\times (t+\dfrac{3}{2})-12\times t}{t(t+\dfrac{3}{2}}\\\\i.e.\\\\4=\dfrac{12t+18-12t}{t(t+\dfrac{3}{2})}\\\\i.e.\\\\4(t(t+\dfrac{3}{2}))=18\\\\i.e.\\\\2(t(t+\dfrac{3}{2}))=9\\\\i.e.\\\\2t^2+3t=9\\\\i.e.\\\\2t^2+3t-9=0$

i.e.

$2t^2+6t-3t-9=0\\\\i.e.\\\\2t(t+3)-3(t+3)=0\\\\i.e.\\\\(2t-3)(t+3)=0\\\\i.e.\\\\t=\dfrac{3}{2}\ or\ t=-3$

But t can't be negative.

Hence, we have:

$t=\dfrac{3}{2}$

Hence, from equation (2) we have:

$u+v=\dfrac{12}{\dfrac{3}{2}}\\\\i.e.\\\\u+v=\dfrac{12\times 2}{3}\\\\i.e.\\\\u+v=8\\\\i.e.\\\\u+2=8\\\\i.e.\\\\u=8-2\\\\i.e.\\\\u=6\ mph$

              Hence, the answer is: 6 mph