Speed, Time and Distance
 A train covers a distance between station A and station B in 45 minutes. If the speed of the train is reduced by 5 km/hr, then the same distance is covered in 48 minutes. The distance between station A and B is

 60 km
 64 km
 80 km
 55 km

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Let the distance between stations be x km, then speed of train
= 2x = (4x/3) kmph 45 60 ∴ x = (48/60) 4x − 5 3 ⇒ 3x = 4 4x − 15 5
⇒ 16x – 60 = 15x
⇒ x = 60 kmCorrect Option: A
Let the distance between stations be x km, then speed of train
= 2x = (4x/3) kmph 45 60 ∴ x = (48/60) 4x − 5 3 ⇒ 3x = 4 4x − 15 5
⇒ 16x – 60 = 15x
⇒ x = 60 km
 When a person cycled at 10 km per hour he arrived at his office 6 minutes late. He arrived 6 minutes early, when he increased his speed by 2 km per hour. The distance of his office from the starting place is

 6 km
 7 km
 12 km
 16 km

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Let the distance be x km.
∴ x − x = 12 10 12 60 ⇒ 6x − 5x = 1 60 5 ⇒ x = 1 × 60 = 12 km 5
Second Methdod :
Here, S_{1} = 10, t_{1} = 6
S_{2} = 12, t_{2} = 6Distance = (S_{1} ×S_{2})(t_{1} + t_{2}) S_{2} −S_{1} = (10 × 12)(6 + 6) 12 − 10 = 120 × 12 2 = 60 × 12 km = 12 km 60 Correct Option: C
Let the distance be x km.
∴ x − x = 12 10 12 60 ⇒ 6x − 5x = 1 60 5 ⇒ x = 1 × 60 = 12 km 5
Second Methdod :
Here, S_{1} = 10, t_{1} = 6
S_{2} = 12, t_{2} = 6Distance = (S_{1} ×S_{2})(t_{1} + t_{2}) S_{2} −S_{1} = (10 × 12)(6 + 6) 12 − 10 = 120 × 12 2 = 60 × 12 km = 12 km 60
 A student goes to school at the rate of
If he travels at the speed of 3 km/h. he is 10 minutes early. The distance (in km) between the school and his house is2 1 km/h and reaches 6 minutes late. 2

 5
 4
 3
 1

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Let the required distance be x km.
x − x = 16 5/2 3 60 ⇒ 2x − x = 4 5 3 15 ⇒ 6x − 5x = 4 15 15
⇒ x = 4 km.
Second Methdod :Here, S_{1} = 2 1 , t_{1} = 6 2
S_{2} = 3, t_{2} = 10Distance = (S_{1} ×S_{2})(t_{1} + t_{2}) S_{2} −S_{1} = 5 × 3(6 + 10) 2 3 − 5 2 = 15 × 16 km = 4 km 60 Correct Option: B
Let the required distance be x km.
x − x = 16 5/2 3 60 ⇒ 2x − x = 4 5 3 15 ⇒ 6x − 5x = 4 15 15
⇒ x = 4 km.
Second Methdod :Here, S_{1} = 2 1 , t_{1} = 6 2
S_{2} = 3, t_{2} = 10Distance = (S_{1} ×S_{2})(t_{1} + t_{2}) S_{2} −S_{1} = 5 × 3(6 + 10) 2 3 − 5 2 = 15 × 16 km = 4 km 60
 Walking at 5 km/hr a student reaches his school from his house 15 minutes early and walking at 3 km/hr he is late by 9 minutes. What is the distance between his school and his house ?

 5 km
 8 km
 3 km
 2 km

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Let the required distance be x km.
Then, x − x = 24 3 5 60 ⇒ 5x − 3x = 2 15 5 ⇒ 2x = 2 3
⇒ 2x = 2 × 3 ⇒ x = 3 km
Second Methdod :
Here, S_{1} = 3, t_{1} = 9
S_{2} = 5, t_{2} = 15Distance = (S_{1} ×S_{2})(t_{1} + t_{2}) S_{2} −S_{1} = (3 × 5)(9 + 15) 5 − 3 = 15 × 24 2 60
= 3 kmCorrect Option: C
Let the required distance be x km.
Then, x − x = 24 3 5 60 ⇒ 5x − 3x = 2 15 5 ⇒ 2x = 2 3
⇒ 2x = 2 × 3 ⇒ x = 3 km
Second Methdod :
Here, S_{1} = 3, t_{1} = 9
S_{2} = 5, t_{2} = 15Distance = (S_{1} ×S_{2})(t_{1} + t_{2}) S_{2} −S_{1} = (3 × 5)(9 + 15) 5 − 3 = 15 × 24 2 60
= 3 km
 Shri X goes to his office by scooter at a speed of 30km/h and reaches 6 minutes earlier. If he goes at a speed of 24 km/h, he reaches 5 minutes late. The distance of his office is

 20 km
 21 km
 22 km
 24 km

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Let the distance of office be x km.
∴ x − x = 11 24 30 60 ⇒ 5x − 4x = 11 120 60 ⇒ x = 11 120 60 ⇒ x = 11 × 120 = 22 km. 60
Second Methdod :
Here, S_{1} = 24, t_{1} = 5
S_{2} = 30, t_{2} = 6Distance = (S_{1} ×S_{2})(t_{1} + t_{2}) S_{2} −S_{1} = 24 × 30(5 + 6) 30 − 24 = 720 × 11 = 22 km 6 × 60 Correct Option: C
Let the distance of office be x km.
∴ x − x = 11 24 30 60 ⇒ 5x − 4x = 11 120 60 ⇒ x = 11 120 60 ⇒ x = 11 × 120 = 22 km. 60
Second Methdod :
Here, S_{1} = 24, t_{1} = 5
S_{2} = 30, t_{2} = 6Distance = (S_{1} ×S_{2})(t_{1} + t_{2}) S_{2} −S_{1} = 24 × 30(5 + 6) 30 − 24 = 720 × 11 = 22 km 6 × 60