Speed, Time and Distance
 A car travels a distance of 300 kms at uniform speed. If the speed of the car is 5 km per hr more it takes two hours less to cover the same distance. Find the original speed of the car.

 25 kmph
 20 kmph
 24 kmph
 28 kmph

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Let the original speed of the car = x km per hr.
When it is increased by 5 km
per hr, the speed = x + 5 km per hr.
As per the given information in the question,300 − 300 = 2 x x + 5 ⇒ 300(x + 5) − 300x = 2 x(x + 5) ⇒ 300x + 1500 − 300x = 2 x^{2} + 5x ⇒ 1500 = 2 x^{2} + 5x ⇒ 750 = 2 x^{2} + 5x
⇒ x^{2} + 5x = 750
⇒ x^{2} + 5x – 750 = 0
⇒ x^{2} + 30x – 25x – 750 = 0
⇒ x (x + 30) – 25 (x + 30) = 0
⇒ (x + 30) (x – 25) = 0
⇒ x = – 30 or 25
The negative value of speed is inadmissible.
Hence, the required speed = 25 km per hr.Correct Option: A
Let the original speed of the car = x km per hr.
When it is increased by 5 km
per hr, the speed = x + 5 km per hr.
As per the given information in the question,300 − 300 = 2 x x + 5 ⇒ 300(x + 5) − 300x = 2 x(x + 5) ⇒ 300x + 1500 − 300x = 2 x^{2} + 5x ⇒ 1500 = 2 x^{2} + 5x ⇒ 750 = 2 x^{2} + 5x
⇒ x^{2} + 5x = 750
⇒ x^{2} + 5x – 750 = 0
⇒ x^{2} + 30x – 25x – 750 = 0
⇒ x (x + 30) – 25 (x + 30) = 0
⇒ (x + 30) (x – 25) = 0
⇒ x = – 30 or 25
The negative value of speed is inadmissible.
Hence, the required speed = 25 km per hr.
 A boy walks from his house at 4 km per hr. and reaches his school 9 minutes late. If his
speed had been 5 km per hr. he would have reached his school 6 minutes earlier. How far his school from house?

 6.5 km.
 5.5 km.
 6 km.
 5 km.

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Let time taken to reach school at 4 kmph be x hrs.
Then time taken to reach schoolat 5 kmph = x + 15 hrs 60
Since, distance is equal.∴ 4x = 5 x + 15 60 x = 5 hrs. 4
Hence, distance between school & house= 4 × 5 km = 5 km 4 Correct Option: D
Let time taken to reach school at 4 kmph be x hrs.
Then time taken to reach schoolat 5 kmph = x + 15 hrs 60
Since, distance is equal.∴ 4x = 5 x + 15 60 x = 5 hrs. 4
Hence, distance between school & house= 4 × 5 km = 5 km 4
 When a person covers the distance between his house and office at 50 km per hr. he is late by 20 minutes. But when he travels at 60 km per hr. he reaches 10 minutes early. What is the distance between his office and his house?

 140 km.
 160 km.
 150 km.
 120 km

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Let time taken to reach office at 50 kmph be x hrs
Then time taken to reach officeat 60 kmph = x + 30 hrs 60
As, distance covered is same,∴ x × 50 = 60 x + 30 60
50x = 60x + 30
⇒ x = 3 hrs
Hence, distance = 3 × 50
= 150 kmCorrect Option: C
Let time taken to reach office at 50 kmph be x hrs
Then time taken to reach officeat 60 kmph = x + 30 hrs 60
As, distance covered is same,∴ x × 50 = 60 x + 30 60
50x = 60x + 30
⇒ x = 3 hrs
Hence, distance = 3 × 50
= 150 km
 A man covers onethird of his journey at 30 km per hr. and the remaining twothird at 45 km per hr. If the total journey is of 150 kms, what is his average speed for the whole journey?

 38 kmph

38 4 kmph 7  64 kmph

39 4 kmph 7

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Length of journey = 150 kms
1 rd of journey = 150 = 50 kms 3 3 Remaining 2 journey 3
= 150 – 50 = 100 kmsTime taken in 1 rd journey at 30 km per hr. 3 t_{1} = 50 = 5 hrs. 30 3 Time taken in 2 rd journey at 45 km per hr. 3 t_{2} = 100 = 20 hrs. 45 9
Total time taken in whole journey = t_{1} + t_{2}= 5 + 20 = 15 + 20 = 35 hrs. 3 9 9 9 Average Speed = 150 = 150 × 9 = 270 35/9 35 7 = 38 4 km per hr. 7 Correct Option: B
Length of journey = 150 kms
1 rd of journey = 150 = 50 kms 3 3 Remaining 2 journey 3
= 150 – 50 = 100 kmsTime taken in 1 rd journey at 30 km per hr. 3 t_{1} = 50 = 5 hrs. 30 3 Time taken in 2 rd journey at 45 km per hr. 3 t_{2} = 100 = 20 hrs. 45 9
Total time taken in whole journey = t_{1} + t_{2}= 5 + 20 = 15 + 20 = 35 hrs. 3 9 9 9 Average Speed = 150 = 150 × 9 = 270 35/9 35 7 = 38 4 km per hr. 7
 An aeroplane travels a distance in the form of a square with the speed of 400 km per hr, 600 km per hr, 800 km per hr. and 1200 km per hr respectively. Find the average speed for the whole distance along the four sides of the square.

 640 kmph
 620 kmph
 630 kmph
 650 kmph

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As distance is covered along four sides (equal) of a square at different speeds, the average speed of the aeroplane
= 4 1 + 1 + 1 + 1 400 600 800 1200
[∵ All the sides of square are equal, so distance between them is same]= 4 30 + 20 + 15 + 10 12000 = 48000 = 640 km per hr. 75 Correct Option: A
As distance is covered along four sides (equal) of a square at different speeds, the average speed of the aeroplane
= 4 1 + 1 + 1 + 1 400 600 800 1200
[∵ All the sides of square are equal, so distance between them is same]= 4 30 + 20 + 15 + 10 12000 = 48000 = 640 km per hr. 75