Question

A boat can travel 1010 miles against a current in the same time that it can travel 110110 miles with the current. The rate of the current is 55 mph. Find the rate of the boat in still water.

Answer 1

Answer:

56.02 mph

Explanation:

Let the velocity of boat in still water is, $v$

Given that velocity of the current is 55 mph

Now, velocity against current is, v-55

velocity along the current is v+55.

And according to question, time taken by boat to travel 1010 miles against the current is equivalent to time take to travel 110110 miles with the current.

Therefore,

Time taken against the current is, $t=\frac{1010}{v-55}$

Time taken with the current is, $t=\frac{110110}{v+55}$

Now this time is equal.

Therefore,

$\frac{1010}{v-55}=\frac{110110}{v+55}\\\frac{{v+55}}{{v-55}} =109.02\\v+55=109.02(v-55)\\v=56.02mph$

Therefore, this is the velocity in still water.