# If a man walks 20 Km at 5 km/hr. he will be late by 40 minutes. If he walks at 8 km per hour, how early from the fixed time will he reach?

$\left( a \right)$ 15 minutes

$\left( b \right)$ 25 minutes

$\left( c \right)$ 50 minutes

$\left( d \right)$ 1.5 hours

$\left( e \right)$ None of these

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**Hint:**In this particular question assume any variable in hours be the actual time taken by the man to cover 20 km at the rate of 5km/hr. but he will be late by 40 minutes i.e. (2/3 hr.) so the time taken by the man (t + $\dfrac{2}{3}$) hr. and use the relation between the speed, distance and time which is given as, speed = (distance/time), so use these concepts to reach the solution of the question.

__Complete step-by-step answer__:Given data:

Distance walked by the man = 20Km.

Let the distance be denoted by D, so D = 20Km.

Now he walked at the rate of 5 km/hr.

So the walking speed of the man = 5 km/hr.

Let it be denoted by v, so v = 5 km/hr.

Now it is given that he will be late by 40 minutes.

As we all know that, 60 min = 1 hour.

So 40 min = $\dfrac{{40}}{{60}} = \dfrac{2}{3}$ hour.

Let the original time taken by the man be t hours.

But he will be late by $\dfrac{2}{3}$ hours.

So the time taken by the man to walk 20 km at the speed of 5 km/hr. is, T = (t + $\dfrac{2}{3}$) hours.

Now as we know that the speed, distance and time is related as, speed = (distance/time).

Therefore, v = (D/T)

Now substitute the values we have,

$ \Rightarrow 5 = \dfrac{{20}}{{t + \dfrac{2}{3}}}$

Now simplify we have,

$ \Rightarrow t + \dfrac{2}{3} = \dfrac{{20}}{5} = 4$

$ \Rightarrow t = 4 - \dfrac{2}{3} = \dfrac{{10}}{3}$ Hours.

$ \Rightarrow t = \dfrac{{10}}{3} \times 60 = 200$ Minutes

Now if he walks 20 km at the speed of 8 km/hr.

Let the time taken by the man is t’ hours.

Therefore, $8 = \dfrac{{20}}{{t'}}$

$ \Rightarrow t' = \dfrac{{20}}{8} = \dfrac{5}{2}$ Hours.

$ \Rightarrow t' = \dfrac{5}{2} \times 60 = 150$ Minutes.

So the difference between the fixed time and the new time is = t – t’ = 200 – 150 = 50 minutes early.

So he reached 50 minutes early.

So this is the required answer.

**Hence option (c) is the correct answer.**

**Note**: Whenever we face such types of questions the key concept we have to remember is the relation between the speed, distance and time which is stated above, so simplify substitute the values in this formula and calculate the original time taken by the man, then again use this formula and calculate the time taken by the man if he increases his speed, then subtract both the time if it is come positive then he will arrive early, if negative then he will arrive late.