Question

The current of a stream runs at the rate of 4 kmph. A boat goes 6 km and back to the starting point in 2 hours, then find the speed of the boat in still water​

Answer 1

Answer:

This is the answer

Step-by-step explanation:

Let the speed of boat in still water be x km/hr.

Let the speed of boat in still water be x km/hr.\therefore \frac{6}{x+4}+\frac{6}{x-4} = 2

Let the speed of boat in still water be x km/hr.\therefore \frac{6}{x+4}+\frac{6}{x-4} = 2=> 6\left ( \frac{x-4+x+4}{(x+4)(x-4)} \right ) = 2

Let the speed of boat in still water be x km/hr.\therefore \frac{6}{x+4}+\frac{6}{x-4} = 2=> 6\left ( \frac{x-4+x+4}{(x+4)(x-4)} \right ) = 2=> 6x = x^{2}-16

Let the speed of boat in still water be x km/hr.\therefore \frac{6}{x+4}+\frac{6}{x-4} = 2=> 6\left ( \frac{x-4+x+4}{(x+4)(x-4)} \right ) = 2=> 6x = x^{2}-16=> x^{2}-6x-16 = 0

Let the speed of boat in still water be x km/hr.\therefore \frac{6}{x+4}+\frac{6}{x-4} = 2=> 6\left ( \frac{x-4+x+4}{(x+4)(x-4)} \right ) = 2=> 6x = x^{2}-16=> x^{2}-6x-16 = 0=> x^{2}-8x+2x-16 = 0

Let the speed of boat in still water be x km/hr.\therefore \frac{6}{x+4}+\frac{6}{x-4} = 2=> 6\left ( \frac{x-4+x+4}{(x+4)(x-4)} \right ) = 2=> 6x = x^{2}-16=> x^{2}-6x-16 = 0=> x^{2}-8x+2x-16 = 0=> x(x-8)+2(x-8) = 0

Let the speed of boat in still water be x km/hr.\therefore \frac{6}{x+4}+\frac{6}{x-4} = 2=> 6\left ( \frac{x-4+x+4}{(x+4)(x-4)} \right ) = 2=> 6x = x^{2}-16=> x^{2}-6x-16 = 0=> x^{2}-8x+2x-16 = 0=> x(x-8)+2(x-8) = 0=> (x+2)(x-8) = 0

Let the speed of boat in still water be x km/hr.\therefore \frac{6}{x+4}+\frac{6}{x-4} = 2=> 6\left ( \frac{x-4+x+4}{(x+4)(x-4)} \right ) = 2=> 6x = x^{2}-16=> x^{2}-6x-16 = 0=> x^{2}-8x+2x-16 = 0=> x(x-8)+2(x-8) = 0=> (x+2)(x-8) = 0=> x = 8km/hr

Answer 2

I think of the answer to be so complicated.

But i look at the question again.

The answer is 6km/hr