Speed, Time and Distance
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A man with 3 of his usual speed reaches 5
Find his usual time to reach the destination.the destination 2 1 hours late. 2
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3/5 of usual speed will take 5/3 of usual time.
[∵ time & speed are inversely proportional]∴ 5 of usual time 3 = usual time + 5 2 ⇒ 2 of usual time = 5 ⇒ usual time 3 2 = 5 × 3 = 15 = 3 3 hours. 2 2 4 4
Second Method :Here, A = 3, B = 5, t = 2 1 2 Usual time = A × time Diff. of A and B = 3 × 2 1 5 − 3 2 = 3 × 5 2 2 = 15 = 3 3 hours 4 4 Correct Option: C
3/5 of usual speed will take 5/3 of usual time.
[∵ time & speed are inversely proportional]∴ 5 of usual time 3 = usual time + 5 2 ⇒ 2 of usual time = 5 ⇒ usual time 3 2 = 5 × 3 = 15 = 3 3 hours. 2 2 4 4
Second Method :Here, A = 3, B = 5, t = 2 1 2 Usual time = A × time Diff. of A and B = 3 × 2 1 5 − 3 2 = 3 × 5 2 2 = 15 = 3 3 hours 4 4
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in 22 hours. How much time could be saved if the train would run at its own speed?A train running at 7 its own speed reached a place 11
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Since the train runs at 7/11 of its own speed, the time it takes is 11/7 of its usual speed.
Let the usual time taken be t hours.Then we can write, 11 t = 22 7 ∴ t = 22 × 7 = 14 hours 11
Hence, time saved
= 22 – 14 = 8 hoursCorrect Option: C
Since the train runs at 7/11 of its own speed, the time it takes is 11/7 of its usual speed.
Let the usual time taken be t hours.Then we can write, 11 t = 22 7 ∴ t = 22 × 7 = 14 hours 11
Hence, time saved
= 22 – 14 = 8 hours
- Two trains start at the same time from Aligarh and Delhi and proceed towards each other at the rate of 14 km and 21 km per hour respectively. When they meet, it is found that one train has travelled 70 km more than the other. The distance between two stations is
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Let the trains meet each other after t hours.
Distance = Speed × Time
According to the question,
21t – 14t = 70⇒ 7t = 70 ⇒ t = 70 = 10 hours 7
∴ Required distance
= 21t + 14t = 35t
= 35 × 10 = 350 km.
Second Method :
Here, a = 21, b = 14, d = 70Required distance = a + b × d a − b = 21 + 14 × 70 21 − 14 = 35 × 70 = 350 km. 7 Correct Option: A
Let the trains meet each other after t hours.
Distance = Speed × Time
According to the question,
21t – 14t = 70⇒ 7t = 70 ⇒ t = 70 = 10 hours 7
∴ Required distance
= 21t + 14t = 35t
= 35 × 10 = 350 km.
Second Method :
Here, a = 21, b = 14, d = 70Required distance = a + b × d a − b = 21 + 14 × 70 21 − 14 = 35 × 70 = 350 km. 7
- Two trains of lengths 150m and 180m respectively are running in opposite directions on parallel tracks. If their speeds be 50 km/hr and 58 km/hr respectively, in what time will they cross each other?
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Relative speed = (50 + 58) kmph
= 108 × 5 m/sec. 18
= 30 m/sec∴ Required time = Total length of trains Relative Speed = 150 + 180 seconds 30 = 330 seconds 30
= 11 secondsCorrect Option: D
Relative speed = (50 + 58) kmph
= 108 × 5 m/sec. 18
= 30 m/sec∴ Required time = Total length of trains Relative Speed = 150 + 180 seconds 30 = 330 seconds 30
= 11 seconds
- Two trains start at the same time from A and B and proceed toward each other at the speed of 75 km/hr and 50 km/hr respectively. When both meet at a point in between, one train was found to have travelled 175 km more than the other. Find the distance between A and B.
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Let the trains meet after t hours.
Distance = Speed × Time
According to the question,
75t – 50t = 175
⇒ 25t = 175⇒ t = 175 = 7 hours 25
∴ Distance between A and B
= 75t + 50t = 125t
= 125 × 7 = 875 km.
Second Method :
Here, a = 75, b = 50, d = 175Required distance = a + b × d a − b = 75 + 50 × 175 75 − 50 = 125 × 175 25
= 125 × 7 = 875 kmCorrect Option: A
Let the trains meet after t hours.
Distance = Speed × Time
According to the question,
75t – 50t = 175
⇒ 25t = 175⇒ t = 175 = 7 hours 25
∴ Distance between A and B
= 75t + 50t = 125t
= 125 × 7 = 875 km.
Second Method :
Here, a = 75, b = 50, d = 175Required distance = a + b × d a − b = 75 + 50 × 175 75 − 50 = 125 × 175 25
= 125 × 7 = 875 km