Speed, Time and Distance
- A man travels for 14 hours 40 minutes. He covers half of the journey by train at the rate of 60 km/hr and rest half by road at the rate of 50 km/hr. The distance travelled by him is :
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Let total distance covered be 2x km.
Total time = 14 hours 40 minutes= 14 40 hours = 14 2 hours 60 3 = 44 hours 3 Time = Distance Speed
According to the question,x + x = 44 60 50 3 ∴ 5x + 6x = 44 300 3 ⇒ 11x = 44 300 3 ⇒ x = 44 × 300 = 400 3 11
∴ Total distance
= 2x = 2 × 400 = 800 kmCorrect Option: B
Let total distance covered be 2x km.
Total time = 14 hours 40 minutes= 14 40 hours = 14 2 hours 60 3 = 44 hours 3 Time = Distance Speed
According to the question,x + x = 44 60 50 3 ∴ 5x + 6x = 44 300 3 ⇒ 11x = 44 300 3 ⇒ x = 44 × 300 = 400 3 11
∴ Total distance
= 2x = 2 × 400 = 800 km
- Two donkeys are standing 400 metres apart. First donkey can run at a speed of 3 m/sec and the second can run at 2 m/sec. If two donkeys run towards each other after how much time (in seconds) will they bump into each other?
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Distance between both donkeys = 400 metre.
Relative speed = (3 + 2) m./sec. = 5 m./sec.∴ Required time = Distance Relative speed = 400 = 80 seconds 5 Correct Option: B
Distance between both donkeys = 400 metre.
Relative speed = (3 + 2) m./sec. = 5 m./sec.∴ Required time = Distance Relative speed = 400 = 80 seconds 5
- A and B are 15 kms apart and when travelling towards each other meet after half an hour whereas they meet two and a half hours later if they travel in the same direction. The faster of the two travels at the speed of
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A’s speed = x kmph.
B’s speed = y kmph.
When A and B move in opposite
directions they meet at C and
when they move in the same direction, they meet at D.
Case I,
AC + CB = ABx + y = 15 2 2
⇒ x + y = 30 ..... (i)
Case II,
AD – BD = AB⇒ x × 5 − y × 5 = 15 2 2 5 (x – y) = 15 2 ⇒ x – y = 15 × 2 = 6 .....(ii) 5
∴ On adding equations (i) and (ii),
x + y + x – y = 30 + 6
⇒ 2x = 36⇒ x = 36 = 18 kmph. 2 Correct Option: B
A’s speed = x kmph.
B’s speed = y kmph.
When A and B move in opposite
directions they meet at C and
when they move in the same direction, they meet at D.
Case I,
AC + CB = ABx + y = 15 2 2
⇒ x + y = 30 ..... (i)
Case II,
AD – BD = AB⇒ x × 5 − y × 5 = 15 2 2 5 (x – y) = 15 2 ⇒ x – y = 15 × 2 = 6 .....(ii) 5
∴ On adding equations (i) and (ii),
x + y + x – y = 30 + 6
⇒ 2x = 36⇒ x = 36 = 18 kmph. 2
- A man walking at 3 km/hour crosses a square field diagonally in 2 minutes. The area of the field (in square metre) is
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Speed of person = 3 kmph
= 
3000 
m./min. 60
= 50 m./min.
∴ Length of the diagonal of square field
= 50 × 2 = 100 metre∴ Required time = 1 × (100)2 2
= 5000 sq. metreCorrect Option: B
Speed of person = 3 kmph
= 3000 m./min. 60
= 50 m./min.
∴ Length of the diagonal of square field
= 50 × 2 = 100 metre∴ Required time = 1 × (100)2 2
= 5000 sq. metre
- A hare sees a dog 100 metres away from her and scuds off in the opposite direction at a speed of 12 km per hr. A minute later the dog perceives her and chases her at a speed of 16 km per hr. How soon will the dog overtake the hare and at what distance from the spot when the hare took flight?
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Let the hare at B sees that dog is at A.
∴ AB = 100 metres
Again, let C be the position of the hare when the dog sees her.
∴ AB = 100 metres
Again, let C be the position of the hare when the dog sees her.
∴ BC= the distance covered by the hare in 1 minute= 12 × 1000 × 1 = 200 metres 60
∴ AC = AB + BC
= 100 + 200 = 300 metres
Thus, hare has a start of 300 metres.
Now, the dog gains 16 – 12 = 4 kms
4000 metres in 1 hour i.e. 60 minutes
∴ The distance gained by dog in 1 minute= 4000 = 200 metres 60 3 ∵ 200 metres is covered in 1 minute 3
∴ 300 metres is covered in300 × 3 = 9 minutes 200 2 Again the distance walked by hare in 9 minutes 2 = 12000 × 9 = 900 metres 60 2
∴ Total distance from
B = 200 + 900 = 1100 metres.Correct Option: D
Let the hare at B sees that dog is at A.
∴ AB = 100 metres
Again, let C be the position of the hare when the dog sees her.
∴ AB = 100 metres
Again, let C be the position of the hare when the dog sees her.
∴ BC= the distance covered by the hare in 1 minute= 12 × 1000 × 1 = 200 metres 60
∴ AC = AB + BC
= 100 + 200 = 300 metres
Thus, hare has a start of 300 metres.
Now, the dog gains 16 – 12 = 4 kms
4000 metres in 1 hour i.e. 60 minutes
∴ The distance gained by dog in 1 minute= 4000 = 200 metres 60 3 ∵ 200 metres is covered in 1 minute 3
∴ 300 metres is covered in300 × 3 = 9 minutes 200 2 Again the distance walked by hare in 9 minutes 2 = 12000 × 9 = 900 metres 60 2
∴ Total distance from
B = 200 + 900 = 1100 metres.
