Speed, Time and Distance
- A moving train passes a platform 50 metres long in 14 seconds and a lamp-post in 10 seconds. The speed of the train is
-
View Hint View Answer Discuss in Forum
Suppose length of train be x
According to questionx + 50 = x 14 10
⇒ 14x = 10x + 500
⇒ 4x = 500⇒ x = 500 = 125 m 4
Therefore, speed= 125 × 18 = 45 kmph 10 5 Correct Option: D
Suppose length of train be x
According to questionx + 50 = x 14 10
⇒ 14x = 10x + 500
⇒ 4x = 500⇒ x = 500 = 125 m 4
Therefore, speed= 125 × 18 = 45 kmph 10 5
- If a train runs at 40 km/hour, it reaches its destination late by 11 minutes. But if it runs at 50 km/hour, it is late by 5 minutes only. The correct time (in minutes) for the train to complete the journey is
-
View Hint View Answer Discuss in Forum
If the distance be x km, then
x − x = 6 40 50 60 ⇒ x − x = 1 4 5
⇒ x = 20 km.∴ Required time = 20 hour – 11 minutes 40 = 1 × 60 − 11 minutes 2
= 19 minutesCorrect Option: C
If the distance be x km, then
x − x = 6 40 50 60 ⇒ x − x = 1 4 5
⇒ x = 20 km.∴ Required time = 20 hour – 11 minutes 40 = 1 × 60 − 11 minutes 2
= 19 minutes
- A student rides on bicycle at 8 km/hour and reaches his school 2.5 minutes late. The next day he increases his speed to 10 km/ hour and reaches school 5 minutes early. How far is the school from his house ?
-
View Hint View Answer Discuss in Forum
Let x km. be the required distance.
Difference in time = 2.5 + 5 = 7.5 minutes= 7.5 hrs. = 1 hrs. 60 8 Now, x − x = 1 8 10 8 ⇒ 5x − 4x = 1 40 8 ⇒ x = 40 = 5 km. 8
Second Methdod :
Here, S1 = 8, t1 = 2.5
S2 = 10, t2 = 5Distance = (S1 ×S2)(t1 + t2) S2 −S1 = (8 × 10)(2.5 + 5) 10 − 8 = 40 × 7.5 = 5 km 60 Correct Option: C
Let x km. be the required distance.
Difference in time = 2.5 + 5 = 7.5 minutes= 7.5 hrs. = 1 hrs. 60 8 Now, x − x = 1 8 10 8 ⇒ 5x − 4x = 1 40 8 ⇒ x = 40 = 5 km. 8
Second Methdod :
Here, S1 = 8, t1 = 2.5
S2 = 10, t2 = 5Distance = (S1 ×S2)(t1 + t2) S2 −S1 = (8 × 10)(2.5 + 5) 10 − 8 = 40 × 7.5 = 5 km 60
- If a man reduces his speed to 2/3, he takes 1 hour more in walking a certain distance. The time (in hours) to cover the distance with his normal speed is :
-
View Hint View Answer Discuss in Forum
Since man walks at 2/3 of usual speed, time taken will be 3/2 of usual time.
∴ 3 of usual time = usual time + 1 hour. 2 ⇒ 3 − 1 of usual time = 1 2
⇒ usual time = 2 hours.Correct Option: A
Since man walks at 2/3 of usual speed, time taken will be 3/2 of usual time.
∴ 3 of usual time = usual time + 1 hour. 2 ⇒ 3 − 1 of usual time = 1 2
⇒ usual time = 2 hours.
- Two trains of equal length are running on parallel lines in the same direction at 46 km/hour and 36 km/hour. The faster train passes the slower train in 36 seconds. The length of each train is
-
View Hint View Answer Discuss in Forum
Let the length of each train be x metre.
Relative speed = (46 – 36) kmph = 10 kmph= 10 × 5 m./sec. 18 = 25 m./sec. 9 ∴ 2x = 36 25 9 ∴ 2x = 36 × 25 = 100 9 ⇒ x = 100 = 50 metre 2 Correct Option: D
Let the length of each train be x metre.
Relative speed = (46 – 36) kmph = 10 kmph= 10 × 5 m./sec. 18 = 25 m./sec. 9 ∴ 2x = 36 25 9 ∴ 2x = 36 × 25 = 100 9 ⇒ x = 100 = 50 metre 2