Question

A motor boat whose speed is 24 km/hr in still water takes 1 hour more to go32 km upstream than to return downstream to the same spot. Find the speed of the stream

Answer 1

:
let c = the rate of the stream current
then
(24+c) = the effective speed of the boat downstream
and
(24-c) = the effective speed up stream
:
Write a time equation; time = dist/rate
:
Time up - time down = 1 hr
Multiply by (24-c)(24+c)
(24-c)(24+c)* - (24-c)(24+c)* = 1(24-x)(24+c)
Cancel the denominators


32(24+c) - 32(24-c) = (24-c)(24+c)
768 + 32c - 768 + 32c = 576 + 24c - 24c - c^2
Combine like terms
64c = 576 - c^2
Arrange as a quadratic equation
c^2 + 64c - 576 = 0
You can use the quadratic formula a=1, b=64, c=-576, but this will factor to:
(x+72)(x-8) = 0
The positive solution is all we want here
x = 8 km/hr is the rate of the current
:
:
See if that checks out, find the actual time each way
Effective speeds: 16 upstream; 32 down stream
32/16 = 2 hrs
32/32 = 1 hr
-------------
differ: 1 hr

Answer 2

Answer:

Let the speed of stream be x.

Then,

Speed of boat in upstream is 24 ‒ x

In downstream, speed of boat is 24 + x

According to question,

Time taken in the upstream journey ‒ Time taken in the downstream journey = 1 hour

$\implies\tt \dfrac{32}{24 - x} - \dfrac{32}{24 + x} = 1 \\\\\\\implies\tt\dfrac{24 + x - 24 + x}{{24}^{2} -{x}^{2}} = \dfrac{1}{32} \\\\\\\implies\tt \dfrac{2x}{576 -{x}^{2}} = \dfrac{1}{32}\\\\\\\implies\tt 2x \times 32 = 576 -{x}^{2}\\\\\\\implies\tt {x}^{2} + 64x - 576 = 0\\\\\\\implies\tt {x}^{2} + 72x - 8x - 576 = 0\\\\\\\implies\tt x(x + 72) - 8(x + 72) = 0\\\\\\\implies\tt (x - 8)(x + 72) = 0\\\\\\\implies\tt \green{x = 8} \quad or \quad \red{x =- 72}$

⠀

∴ Speed of the Stream will be 8 km/hr.