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Question

A motor boat whose speed is 18 Km/h in still water takes 1 hour more to go 24 Km upstream than to return downstream to the same spot. Find the speed of the stream.


Solution

Given:- 
Speed of boat $$= 18 \; {km}/{hr}$$
Distance $$= 24 \; km$$

Let $$x$$ be the speed of stream.
Let $${t}_{1}$$ and $${t}_{2}$$ be the time for upstream and downstream.
As we know that,

$$\text{speed} = \cfrac{\text{distance}}{\text{time}}$$
$$\Rightarrow \text{time} = \cfrac{\text{distance}}{\text{speed}}$$

For upstream,
Speed $$= \left( 18 - x \right) \; {km}/{hr}$$
Distance $$= 24 \; km$$
Time $$= {t}_{1}$$
Therefore,

$${t}_{1} = \cfrac{24}{18 - x}$$

For downstream,
Speed $$= \left( 18 + x \right) \; {km}/{hr}$$
Distance $$= 24 \; km$$
Time $$= {t}_{2}$$
Therefore,

$${t}_{2} = \cfrac{24}{18 + x}$$

Now according to the question-

$${t}_{1} = {t}_{2} + 1$$

$$\cfrac{24}{18 - x} = \cfrac{24}{18 + x} + 1$$

$$\Rightarrow \cfrac{1}{18 - x} - \cfrac{1}{18 + x} = \cfrac{1}{24}$$

$$\Rightarrow \cfrac{\left( 18 + x \right) - \left( 18 - x \right)}{\left( 18 - x \right) \left( 18 + x \right)} = \cfrac{1}{24}$$

$$\Rightarrow 48x = \left( 18 - x \right) \left( 18 + x \right)$$

$$\Rightarrow 48x = 324 + 18x - 18x - {x}^{2}$$

$$\Rightarrow {x}^{2} + 48x - 324 = 0$$
$$\Rightarrow {x}^{2} + 54x - 6x - 324 = 0$$
$$\Rightarrow x \left( x + 54 \right) - 6 \left( x + 54 \right) = 0$$
$$\Rightarrow \left( x + 54 \right) \left( x - 6 \right) = 0$$

$$\Rightarrow x = -54 \text{ or } x = 6$$

Since speed cannot be negative.

$$\Rightarrow x \ne -54$$

$$\therefore x = 6$$

Thus the speed of stream is $$6 \; {km}/{hr}$$

Hence the correct answer is $$6 \; {km}/{hr}$$.

Maths

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