Question

A boat downstream from A to B takes $7hours$ , upstream from B to A $9hours$ . How much time does a log take from A to B? (Driftwood along the water velocity).

Answer 1

It is known that the speed of the downstream is the sum of speed of boat and speed of the water and for the speed of the upstream subtract the speed of the water from the speed of boat.
Then calculate the speed of the water stream using the given conditions.

Complete step by step answer:
We are given here, the time taken to cover the distance in the upstream and the downstream.
Let us assume that the speed of water is $xkm{h^{ - 1}}$ the speed of the boat is $ykm{h^{ - 1}}$ and the distance between the shores is $dkm$ .
We know that the speed of the boat in downstream is the sum of speed of boat and speed of the water. Hence, the speed of the boat when moving downstream from A to B is $(x + y)km{h^{ - 1}}$ .
Now for the speed of the boat in upstream, we have to subtract the speed of the water from the speed of the boat. Hence, speed of the boat upstream from B to A is $(y - x)km{h^{ - 1}}$
Now we know time travelled by the boat is the ratio of total distance travelled and the speed. So we can write the mathematical expression for time taken by boat to cover the distance in the upstream and downstream as,
$\dfrac{d}{{(x + y)}} = 7$ …when moving downstream.
$\dfrac{d}{{(y - x)}} = 9$ …when moving upstream.
Dividing the equations we get,
$\dfrac{{\dfrac{d}{{(x + y)}}}}{{\dfrac{d}{{(y - x)}}}} = \dfrac{7}{9}$
$\dfrac{{x + y}}{{y - x}} = \dfrac{7}{9}$
Using Componendo-Dividendo method we get,
Or, $\dfrac{y}{x} = \dfrac{{7 + 9}}{{7 - 9}}$
Or, $\dfrac{y}{x} = \dfrac{{16}}{{ - 2}}$
Or, $x = \dfrac{y}{8}$
Hence, the speed of the stream is one eighth of the speed of the boat. Now, the upstream boat takes 7 hours.
So, in terms of the speed of the stream we can write, $\dfrac{d}{{(\dfrac{y}{8} + y)}} = 7$
Or, $\dfrac{d}{{\dfrac{{9y}}{8}}} = 7$
Or, $\dfrac{d}{{\dfrac{y}{8}}} = 7 \times 9 = 63$
Now, if a log is moving in the water the time taken by log to reach the other shore (A to B) will be,
$\dfrac{d}{x}hours$ . Since, log itself has no velocity the stream carries it.
So, earlier we have found that, $x = \dfrac{y}{8}$
So, time taken by the log is, $\dfrac{d}{{\dfrac{y}{8}}}hours$
That will be equal to, $\dfrac{d}{{\dfrac{y}{8}}} = 63hours$
So, the log will take $63hours$ to reach from A to B.

Note:
It is observed that to find the time taken by log to reach from A to B we do not have to find the speed of the boat or the distance between the shores. We just need a relation between the speeds to find the time taken when time taken in upstream and downstream is given.