Speed, Time and Distance
- P and Q are 27 km away. Two trains with speed of 24 km/hr and 18 km/hr respectively start simultaneously from P and Q and travel in the same direction. They meet at a point R beyond Q. Distance QR is
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Let the trains meet after t hours, then
24t – 18t = 27
⇒ 6t = 27⇒ t = 27 6 = 9 hours 2 ∴ QR = 18t = 18 × 9 = 81 km 2 Correct Option: B
Let the trains meet after t hours, then
24t – 18t = 27
⇒ 6t = 27⇒ t = 27 6 = 9 hours 2 ∴ QR = 18t = 18 × 9 = 81 km 2
- Ravi and Ajay start simultaneously from a place A towards B, 60 km apart. Ravi’s speed is 4 km/hr less than that of Ajay. Ajay, after reaching B, turns back and meets Ravi at a place 12 km away from B. Ravi’s speed is
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Let the speed of Ravi be x kmph
then, Ajay’s speed = (x + 4) kmph
Distance covered by Ajay
= 60 + 12 = 72 km
Distance covered by Ravi
= 60 – 12 = 48 km.
According to the question,72 = 48 x + 4 x ⇒ 3 = 2 x + 4 x
⇒ 3x = 2x + 8
⇒ x = 8 kmphCorrect Option: C
Let the speed of Ravi be x kmph
then, Ajay’s speed = (x + 4) kmph
Distance covered by Ajay
= 60 + 12 = 72 km
Distance covered by Ravi
= 60 – 12 = 48 km.
According to the question,72 = 48 x + 4 x ⇒ 3 = 2 x + 4 x
⇒ 3x = 2x + 8
⇒ x = 8 kmph
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Walking at 3 of his usual speed a man is late by 4 2 1 hours. The usual time would have been what? 2
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New speed is 3 of the usual speed 4 ∴ New time taken = 4 of the usual time 3 ∴ 4 of the usual time – Usual time = 5 3 2 ⇒ 1 of the usual time = 5 3 2 ∴ Usual time = 5 × 3 2 = 15 hours or 7.5 hrs 2 Correct Option: B
New speed is 3 of the usual speed 4 ∴ New time taken = 4 of the usual time 3 ∴ 4 of the usual time – Usual time = 5 3 2 ⇒ 1 of the usual time = 5 3 2 ∴ Usual time = 5 × 3 2 = 15 hours or 7.5 hrs 2
- 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?
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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 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?
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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
