Speed, Time and Distance

If the speed is increased by 5 km/hour, it would takeA car can cover a certain distance in 4 1 hours. 2 1 hour less to cover the same distance. 2
Find the slower speed of the car.

 50 km/hour
 40 km/hour
 45 km/hour
 60 km/hour

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Let the initial speed of the car be x kmph and the distance be y km.
Then, y = 9 x ...(i) 2
and, y = 4 (x + 5) ...(ii)∴ 9x = 4 (x + 5) 2
⇒ 9x = 8x + 40
⇒ x = 40 kmphCorrect Option: B
Let the initial speed of the car be x kmph and the distance be y km.
Then, y = 9 x ...(i) 2
and, y = 4 (x + 5) ...(ii)∴ 9x = 4 (x + 5) 2
⇒ 9x = 8x + 40
⇒ x = 40 kmph
 A boy is late by 9 minutes if he walks to school at a speed of 4 km/hour. If he walks at the rate of 5 km/hour, he arrives 9 minutes early. The distance to his school is

 9 km
 5 km
 4 km
 6 km

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Let the required distance be x km.
According to the question,x − x = 18 4 5 60 ⇒ 5x − 4x − 3 20 10 ⇒ x = 3 × 20 = 6 km 10
Second Methdod :
Here, S_{1} = 4, t_{1} = 9
S_{2} = 5, t_{2} = 9Distance = (S_{1} ×S_{2})(t_{1} + t_{2}) S_{2} −S_{1} = (4 × 5)(9 + 9) 5 − 4 = 20 × 18 = 6 km 60 Correct Option: D
Let the required distance be x km.
According to the question,x − x = 18 4 5 60 ⇒ 5x − 4x − 3 20 10 ⇒ x = 3 × 20 = 6 km 10
Second Methdod :
Here, S_{1} = 4, t_{1} = 9
S_{2} = 5, t_{2} = 9Distance = (S_{1} ×S_{2})(t_{1} + t_{2}) S_{2} −S_{1} = (4 × 5)(9 + 9) 5 − 4 = 20 × 18 = 6 km 60
 A student walks from his house at
6 minutes late. The next day he increases his speed by 1 km per hour and reaches 6 minutes before school time. How far is the school from his house ?a speed of 2 1 km per hour and reaches his school 2


5 km 4 
7 km 4 
9 km 4 
11 km 4


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Let the required distance be x km.
Difference of time= 6 + 6 = 12 minutes = 1 hr. 5
According to the question,⇒ x − x = 1 ⇒ 2x − 2x = 1 5/2 7/2 5 5 7 5 ⇒ 14x − 10x − 1 35 5 ⇒ 4x = 1 ⇒ x = 35 = 7 km. 35 5 20 4
Second Methdod :Here, S_{1} = 2 1 , t_{1} = 6 2 Here, S_{2} = 3 1 , t_{2} = 6 2 Distance = (S_{1} ×S_{2})(t_{1} + t_{2}) S_{2} −S_{1} = 5 × 7 × (6 + 6) 2 2 5 − 7 2 2 = 35 × 12 = 7 km 4 60 4 Correct Option: B
Let the required distance be x km.
Difference of time= 6 + 6 = 12 minutes = 1 hr. 5
According to the question,⇒ x − x = 1 ⇒ 2x − 2x = 1 5/2 7/2 5 5 7 5 ⇒ 14x − 10x − 1 35 5 ⇒ 4x = 1 ⇒ x = 35 = 7 km. 35 5 20 4
Second Methdod :Here, S_{1} = 2 1 , t_{1} = 6 2 Here, S_{2} = 3 1 , t_{2} = 6 2 Distance = (S_{1} ×S_{2})(t_{1} + t_{2}) S_{2} −S_{1} = 5 × 7 × (6 + 6) 2 2 5 − 7 2 2 = 35 × 12 = 7 km 4 60 4
 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

 13
 15
 19
 21

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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 ?


5 km 8  8 km
 5 km
 10 km


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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, S_{1} = 8, t_{1} = 2.5
S_{2} = 10, t_{2} = 5Distance = (S_{1} ×S_{2})(t_{1} + t_{2}) S_{2} −S_{1} = (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, S_{1} = 8, t_{1} = 2.5
S_{2} = 10, t_{2} = 5Distance = (S_{1} ×S_{2})(t_{1} + t_{2}) S_{2} −S_{1} = (8 × 10)(2.5 + 5) 10 − 8 = 40 × 7.5 = 5 km 60