# What is the ratio between the speed of the boat and speed of the water current?

A boat running upstream takes 8 hours 48 minutes to cover a certain distance, while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and speed of the water current respectively?

A. $2 : 1$

B. $3 : 2$

C. $8 : 3$

D. Cannot be determined

E. None of these

What I think can be useful, they travel the same distance, so we can equate them taking up down speeds as x and y, but then what? Also, how are these types of question solved quickly?

• Up speed is boat speed minus river speed, and down speed is boat speed plus river speed. – Arthur Oct 14 '14 at 12:22

Denote with $x$ the speed of the boat and with $y$ the speed of the water current. Then it is given that when the boat travels upstream with a speed of $x-y$ it covers a certain distance, say $D$, in $8$h and $48$m or equivalently in $528$ minutes. Accordingly, when the boat travels downstream with a speed of $x+y$ it covers the same distance $D$ in $4$h or equivalently in $240$ minutes. Thus $$D=528(x-y)$$ as well as $$D=240(x+y)$$ Equating the right sides you have $$528(x-y)=240(x+y)$$ which gives $$288x=768y$$ By dividing both sides with $96$ the last equation reduces to $$3x=8y$$ so that the correct answer is C.

• Thanks for the clear explanation! – user1502 Oct 14 '14 at 12:32

Time taken to travel upstream is 528 min, while time taken to travel downstream is 240 min. Let speeds of boat and current be x and y respectively; then we have

(x+y)/(x-y) = (a/240)/(a/528) = 528/240 (letting distance travelled be a).

and from this we can solve for y in terms of x. x : y = 384 : 144 = 8 : 3 and hence answer is C.

Say that the boat has a standard velocity $v_0$ when there is no current at all. Say the water runs with a constant speed $v_w$. Let $\Delta x$ be the distance they travel. We now know, specifically from what is given: $t_{upstream}=\frac{\Delta x}{v_0-v_w}=8.8 h$ and $t_{downstream}=\frac{\Delta x}{v_0+v_w}$. Now just fix $\Delta x$, say to $1m$. Then we see $\frac{1}{v_0-v_w}=8.8$ and $\frac{1}{v_0+v_w}=4$.

Now we find $v_0-v_w=\frac{1}{8.8}$ and $v_0+v_w=\frac{1}{4}$, so $v_0=\frac{1}{2}(\frac{1}{8.8}+\frac{1}{4})=\frac{2}{11}$, from which we can calculate $v_w$ and find the ratio $\frac{8}{3}$