Question

A man walks a distance 48 km in a given time. If he walks 2 km an hour faster, then he will perform the journey four hours earlier. Find its normal rate of walking.
A). 4 km/hr
B). 2 km/hr
C). 6 km/hr
D). 5 km/hr

Answer 1

Hint: In this question it is given that a man walks a distance 48 km in a given time. If he walks 2 km/hr faster, then he will perform the journey four hours earlier. So we have to find its normal rate of walking. So to find the solution we need to know that,
$$\text{Speed} =\dfrac{\text{Distance} }{\text{Time} }$$
From the above we can write,
$$\text{Time} =\dfrac{\text{Distance} }{\text{Speed} }$$............(1)
So by using the above formula we will get our required solution.

Complete step-by-step solution:
Let us assume that the man walks at a speed of x km/hr.
Here it is given the man covers 48 km in a certain time $$t_{1}$$,
Then the time taken by the man to complete the journey,
$$\text{Time} =\dfrac{\text{Distance} }{\text{Speed} }$$ [by formula (1)]
$$\Rightarrow t_{1}=\dfrac{48}{x}$$
Now in the second case when speed increased by 2 kmph, i.e when speed = (x+2) kmph then then let us assume the man covers 48 km in a time $$t_{2}$$.
Therefore,
$$ t_{2}=\dfrac{48}{x+2}$$ [by formula (1)]
Now since in the second case the man performed the journey four hours earlier, so by according to the question, we can write,
  $$t_{1}-t_{2}=4$$
$$\Rightarrow \dfrac{48}{x} -\dfrac{48}{x+2} =4$$
$$\Rightarrow 48\left( \dfrac{1}{x} -\dfrac{1}{x+2} \right) =4$$ [by taking 48 common from the left side]
$$\Rightarrow \left( \dfrac{1}{x} -\dfrac{1}{x+2} \right) =\dfrac{4}{48}$$ [dividing both side by 48]
$$\Rightarrow \left( \dfrac{1}{x} -\dfrac{1}{x+2} \right) =\dfrac{1}{12}$$
$$\Rightarrow \dfrac{(x+2)-x}{\left( x+2\right) x} =\dfrac{1}{12}$$
$$\Rightarrow \dfrac{x+2-x}{\left( x+2\right) x} =\dfrac{1}{12}$$
$$\Rightarrow \dfrac{2}{\left( x+2\right) x} =\dfrac{1}{12}$$
$$\Rightarrow \dfrac{2\times 12}{\left( x+2\right) x} =\dfrac{1}{12} \times 12$$ [multiplying both side by 12]
$$\Rightarrow \dfrac{24}{\left( x+2\right) x} =1$$
$$\Rightarrow 24=\left( x+2\right) x$$ [multiplying both side by (x+2)x]
$$\Rightarrow \left( x+2\right) x=24$$
$$\Rightarrow x^{2}+2x=24$$
$$\Rightarrow x^{2}+2x-24=0$$
$$\Rightarrow x^{2}+(6x-4x)-24=0$$
$$\Rightarrow x^{2}+6x-4x-24=0$$
Now taking ‘x’ common from the first two term and (-4) from the last two terms, we get
$$x\left( x+6\right) -4\left( x+6\right) =0$$
$$\Rightarrow \left( x+6\right) \left( x-4\right) =0$$ [taking (x+6) common]
$$\text{Either} ,\ x+6=0\ \text{or} ,\ x-4=0$$
Therefore we get, $x = -6$ or $x = 4$.
Speed cannot be a negative value and hence $x = 4$ kmph
Hence option A is correct.

Note: While solving this type of question you need to keep in mind that if the walking speed of a man increases then time taken by a man decreases, i.e, we can say that time and speed are inversely proportional. Also the normal rate of walking is nothing but the normal walking speed.