
A train moves at a constant speed of $ 75km/hr $ . How far will it travel in 20 minutes? Find the time required to cover a distance of 250 km.
(A) 200 min
(B) 300 min
(C) 100 min
(D) 50 min
Answer
535.3k+ views
Hint: For a train moving with a constant speed. Distance travelled is the product of speed and time. And since the train will cover equal distance in equal intervals of time. If we multiply the distance by some constant, then the time required to cover that distance will also be multiplied by the same constant.
Complete step-by-step answer:
It is given that the train is moving with a constant speed $ v = 75km/hr $
Since, the speed of the train is constant, we can use the formula
$ s = vt $ . . . (1)
Where,
$ s $ is the distance travelled
$ v $ is the constant speed
$ t $ is the time taken
Now, we have to calculate the distance travelled by the train in $ t = 20\min $
i.e. in $ t = \dfrac{{20}}{{60}} = \dfrac{1}{3}hr $
By substituting the values of speed and time in equation (1), we get
$ s = 75 \times \dfrac{1}{3} $
$ \Rightarrow s = 25\;km $
Therefore, the train travels a distance of $ 25\;km $ in 20 minutes.
Now, since the speed of the train is constant. The time required to travel ten times the distance travelled will be ten times the time.
i.e. time required to travel 250 km will be equal to 200 minutes.
Therefore, from the above explanation, the correct answer is, option (A) 200 min
So, the correct answer is “Option A”.
Note: Here it was important to know that we could use the formula of distance equal to the product of speed and time because the speed was constant. We cannot use the same formula for a variable speed. Second important point was to understand that the speed is constant. So if the distance is increased, then the time will increase in the same ratio. That is why, we did not need to calculate the time taken to travel 250 km, separately.
Complete step-by-step answer:
It is given that the train is moving with a constant speed $ v = 75km/hr $
Since, the speed of the train is constant, we can use the formula
$ s = vt $ . . . (1)
Where,
$ s $ is the distance travelled
$ v $ is the constant speed
$ t $ is the time taken
Now, we have to calculate the distance travelled by the train in $ t = 20\min $
i.e. in $ t = \dfrac{{20}}{{60}} = \dfrac{1}{3}hr $
By substituting the values of speed and time in equation (1), we get
$ s = 75 \times \dfrac{1}{3} $
$ \Rightarrow s = 25\;km $
Therefore, the train travels a distance of $ 25\;km $ in 20 minutes.
Now, since the speed of the train is constant. The time required to travel ten times the distance travelled will be ten times the time.
i.e. time required to travel 250 km will be equal to 200 minutes.
Therefore, from the above explanation, the correct answer is, option (A) 200 min
So, the correct answer is “Option A”.
Note: Here it was important to know that we could use the formula of distance equal to the product of speed and time because the speed was constant. We cannot use the same formula for a variable speed. Second important point was to understand that the speed is constant. So if the distance is increased, then the time will increase in the same ratio. That is why, we did not need to calculate the time taken to travel 250 km, separately.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Trending doubts
What is BLO What is the full form of BLO class 8 social science CBSE

Citizens of India can vote at the age of A 18 years class 8 social science CBSE

The number of corners in a cube are A 4 B 6 C 8 D class 8 maths CBSE

The pH of the gastric juices released during digestion class 8 biology CBSE

What are the methods of reducing friction. Explain

Advantages and disadvantages of science


