Speed, Time and Distance
 If a train runs at 70 km/hour, it reaches its destination late by 12 minutes. But if it runs at 80 km/ hour, it is late by 3 minutes. The correct time to cover the journey is

 58 minutes
 2 hours
 1 hour
 59 minutes

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Distance of journey = x km
Difference of time = 12 – 3 = 9 minutes= 9 hour = 3 hour 60 20 ∴ x − x = 3 70 80 20 ⇒ x − x = 3 7 8 2 ⇒ 8x − 7x = 3 56 2 ⇒ x = 3 56 2 ⇒ x = 3 × 56 = 84 km 2
∴ Required correct time= 84 hours – 12 minutes 70 = 84 × 60 − 12 minutes 70
= 72 – 12 = 60 minutes
= 1 hourCorrect Option: C
Distance of journey = x km
Difference of time = 12 – 3 = 9 minutes= 9 hour = 3 hour 60 20 ∴ x − x = 3 70 80 20 ⇒ x − x = 3 7 8 2 ⇒ 8x − 7x = 3 56 2 ⇒ x = 3 56 2 ⇒ x = 3 × 56 = 84 km 2
∴ Required correct time= 84 hours – 12 minutes 70 = 84 × 60 − 12 minutes 70
= 72 – 12 = 60 minutes
= 1 hour
 A train travelling at a speed of 55 km/hr travels from place X to place Y in 4 hours. If its speed is increased by 5 km/hr., then the time of journey is reduced by

 25 minutes
 35 minutes
 20 minutes
 30 minutes

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Distance between stations X and Y = Speed × Time
= 55 × 4 = 220 km.
New speed = 55 + 5 = 60 kmph∴ Required time = 220 60 = 11 hours 3
= 3 hours 40 minutes.
∴ Required answer
= 4 hours – 3 hours 40 minutes
= 20 minutesCorrect Option: C
Distance between stations X and Y = Speed × Time
= 55 × 4 = 220 km.
New speed = 55 + 5 = 60 kmph∴ Required time = 220 60 = 11 hours 3
= 3 hours 40 minutes.
∴ Required answer
= 4 hours – 3 hours 40 minutes
= 20 minutes
 If a boy walks from his house to school at the rate of 4 km per hour, he reaches the school 10 minutes earlier than the scheduled time. However, if he walks at the rate of 3 km per hour, he reaches 10 minutes late. Find the distance of his school from his house.

 5 km
 4 km
 6 km
 4.5 km

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Let the distance of school be x km, then
x − x = 20 3 4 60 ⇒ x = 1 ⇒ x = 12 = 4 km 12 3 3
Second Methdod :
Here, S_{1} = 3, t_{1} = 10
S_{2} = 4, t_{2} = 10Distance = (S_{1} ×S_{2})(t_{1} + t_{2}) S_{2} −S_{1} = (3 × 4)(10 + 10) 4 − 3 = 12 × 20 = 4 km 60 Correct Option: B
Let the distance of school be x km, then
x − x = 20 3 4 60 ⇒ x = 1 ⇒ x = 12 = 4 km 12 3 3
Second Methdod :
Here, S_{1} = 3, t_{1} = 10
S_{2} = 4, t_{2} = 10Distance = (S_{1} ×S_{2})(t_{1} + t_{2}) S_{2} −S_{1} = (3 × 4)(10 + 10) 4 − 3 = 12 × 20 = 4 km 60
 Walking at a speed of 5 km/hr, a man reaches his office 6 minutes late. Walking at 6 km/hr, he reaches there 2 minutes early. The distance of his office is

 3 km
 4 km
 3.5 km
 2 km

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Let the distance of the office be x km, then
x − x = 8 5 6 60 ⇒ 6x − 5x = 2 30 15
⇒ x = 2 × 2 = 4 km
Second Methdod :
Here, S_{1} = 5, t_{1} = 6
S_{2} = 6, t_{2} = 2Distance = (S_{1} ×S_{2})(t_{1} + t_{2}) S_{2} −S_{1} = (6 × 5)(6 + 2) 6 − 5 = 30 × 8 = 4 km 60 Correct Option: B
Let the distance of the office be x km, then
x − x = 8 5 6 60 ⇒ 6x − 5x = 2 30 15
⇒ x = 2 × 2 = 4 km
Second Methdod :
Here, S_{1} = 5, t_{1} = 6
S_{2} = 6, t_{2} = 2Distance = (S_{1} ×S_{2})(t_{1} + t_{2}) S_{2} −S_{1} = (6 × 5)(6 + 2) 6 − 5 = 30 × 8 = 4 km 60
 A train covers a distance of 10 km in 12 minutes. If its speed is decreased by 5 km/hr, the time taken by it to cover the same distance will be :

 10 minutes
 13 minutes 20 sec
 13 minutes
 11 minutes 20 sec

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Speed of train = Distance Time = 10 kmph 12 60 = 10 × 60 = 50 kmph 12
New speed = 45 kmph∴ Required time = 10 hour 45 = 2 × 60 minutes 9 = 40 minutes 3
or 13 minutes 20 secondsCorrect Option: B
Speed of train = Distance Time = 10 kmph 12 60 = 10 × 60 = 50 kmph 12
New speed = 45 kmph∴ Required time = 10 hour 45 = 2 × 60 minutes 9 = 40 minutes 3
or 13 minutes 20 seconds