Answer
425.4k+ 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
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)