Speed, Time and Distance
 A car can finish a certain journey in 10 hours at the speed of 42 kmph. In order to cover the same distance in 7 hours, the speed of the car (km/h) must be increased by :

 12
 15
 18
 24

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Distance covered by car = 42 × 10 = 420 km.
New time = 7 hours∴ Required speed = 420 = 60 kmph. 7
∴ Required increase
= (60 – 42) kmph
= 18 kmphCorrect Option: C
Distance covered by car = 42 × 10 = 420 km.
New time = 7 hours∴ Required speed = 420 = 60 kmph. 7
∴ Required increase
= (60 – 42) kmph
= 18 kmph
 A man travels for 5 hours 15 minutes. If he covers the first half of the journey at 60 km/h and rest at 45 km/h. Find the total distance travelled by him.


1028 6 km. 7  189 km.
 378 km.
 270 km.


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Let the distance covered be 2x km.
Time = Distance Speed
According to the question,x + x = 5 15 = 1 60 45 60 4 ⇒ 3x + 4x = 21 180 4 ⇒ 7x = 21 × 180 4 ⇒ x = 21 × 180 = 135 km. 4 × 7
∴ Length of total journey
= 2 × 135 = 270 km.Correct Option: D
Let the distance covered be 2x km.
Time = Distance Speed
According to the question,x + x = 5 15 = 1 60 45 60 4 ⇒ 3x + 4x = 21 180 4 ⇒ 7x = 21 × 180 4 ⇒ x = 21 × 180 = 135 km. 4 × 7
∴ Length of total journey
= 2 × 135 = 270 km.
 A student goes to school at the
If he travels at the speed of 3 km/hr. he is 10 minutes early. What is the distance to the school?rate of 2 1 km/hr and reaches 6 minutes late. 2

 4 km

3 1 km 2  1 km

3 1 km 4

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Let the distance of school be x km.
Difference of time = 6 + 10= 16 minutes = 16 hour 60 = 4 hour 15 Time = Distance Speed ∴ x − x = 4 5/2 3 15 ⇒ 2x − x = 4 5 3 15 ⇒ 6x − 5x − 4 15 15
⇒ x = 4 km.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 60 = 16 = 4 km 4 Correct Option: A
Let the distance of school be x km.
Difference of time = 6 + 10= 16 minutes = 16 hour 60 = 4 hour 15 Time = Distance Speed ∴ x − x = 4 5/2 3 15 ⇒ 2x − x = 4 5 3 15 ⇒ 6x − 5x − 4 15 15
⇒ x = 4 km.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 60 = 16 = 4 km 4
 A man starts from a place P and reaches the place Q in 7 hours.
and the remaining distance at 12 km/hour. The distance between P and Q isHe travels 1 th of the distance at 10 km/hour 4

 72 km
 90 km
 80 km
 70 km

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Time = Distance Speed
According to the question,x + 3x = 7 10 12 ⇒ x + x = 7 10 4 ⇒ 2x + 5x = 7 20
⇒ 7x = 7 × 20∴ x = 7 × 20 = 20 km. 7
∴ PQ = 4x = 4 × 20 = 80 km.Correct Option: C
Time = Distance Speed
According to the question,x + 3x = 7 10 12 ⇒ x + x = 7 10 4 ⇒ 2x + 5x = 7 20
⇒ 7x = 7 × 20∴ x = 7 × 20 = 20 km. 7
∴ PQ = 4x = 4 × 20 = 80 km.
 A student starting from his house walks at a
school 6 minutes late. Next day starting atspeed of 2 1 km/hour and reaches his 2
the same time he increases his speed by 1 km/hour and reaches 6 minutes early. The distance between the school and his house is

 4 km

3 1 km 2 
1 3 km 4  6 km

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Distance between school and house = x km (let)
Time = Distance Speed
According to the question,x − x = 6 + 6 = 1 5/2 7/2 60 5
(Difference of time = 6 + 6 =12 minutes)⇒ 2x − 2x = 1 5 7 5 ⇒ 14x − 10x = 1 ⇒ 4x = 1 35 5 35 5 ⇒ 4x = 35 = 7 5 ⇒ x = 7 = 1 3 km. 4 4
Second Method :Here, S_{1} = 2 1 , t_{1} = 6 2 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 7  5 2 2 = 35 × 12 4 60 = 7 = 1 3 km 4 4 Correct Option: C
Distance between school and house = x km (let)
Time = Distance Speed
According to the question,x − x = 6 + 6 = 1 5/2 7/2 60 5
(Difference of time = 6 + 6 =12 minutes)⇒ 2x − 2x = 1 5 7 5 ⇒ 14x − 10x = 1 ⇒ 4x = 1 35 5 35 5 ⇒ 4x = 35 = 7 5 ⇒ x = 7 = 1 3 km. 4 4
Second Method :Here, S_{1} = 2 1 , t_{1} = 6 2 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 7  5 2 2 = 35 × 12 4 60 = 7 = 1 3 km 4 4