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
Answer
Verified
483.3k+ views
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.
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.
Recently Updated Pages
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE
The highest possible oxidation states of Uranium and class 11 chemistry CBSE
Find the value of x if the mode of the following data class 11 maths CBSE
Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE
A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE
Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE
Trending doubts
10 examples of friction in our daily life
The correct order of melting point of 14th group elements class 11 chemistry CBSE
Difference Between Prokaryotic Cells and Eukaryotic Cells
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
State and prove Bernoullis theorem class 11 physics CBSE
What organs are located on the left side of your body class 11 biology CBSE