Speed, Time and Distance
- A car travels a distance of 300 kms at uniform speed. If the speed of the car is 5 km per hr more it takes two hours less to cover the same distance. Find the original speed of the car.
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Let the original speed of the car = x km per hr.
When it is increased by 5 km
per hr, the speed = x + 5 km per hr.
As per the given information in the question,300 − 300 = 2 x x + 5 ⇒ 300(x + 5) − 300x = 2 x(x + 5) ⇒ 300x + 1500 − 300x = 2 x2 + 5x ⇒ 1500 = 2 x2 + 5x ⇒ 750 = 2 x2 + 5x
⇒ x2 + 5x = 750
⇒ x2 + 5x – 750 = 0
⇒ x2 + 30x – 25x – 750 = 0
⇒ x (x + 30) – 25 (x + 30) = 0
⇒ (x + 30) (x – 25) = 0
⇒ x = – 30 or 25
The negative value of speed is inadmissible.
Hence, the required speed = 25 km per hr.Correct Option: A
Let the original speed of the car = x km per hr.
When it is increased by 5 km
per hr, the speed = x + 5 km per hr.
As per the given information in the question,300 − 300 = 2 x x + 5 ⇒ 300(x + 5) − 300x = 2 x(x + 5) ⇒ 300x + 1500 − 300x = 2 x2 + 5x ⇒ 1500 = 2 x2 + 5x ⇒ 750 = 2 x2 + 5x
⇒ x2 + 5x = 750
⇒ x2 + 5x – 750 = 0
⇒ x2 + 30x – 25x – 750 = 0
⇒ x (x + 30) – 25 (x + 30) = 0
⇒ (x + 30) (x – 25) = 0
⇒ x = – 30 or 25
The negative value of speed is inadmissible.
Hence, the required speed = 25 km per hr.
- A car can finish a certain journey in 10 hours at a speed of 48 km per hr. In order to cover the same distance in 8 hours, how much the speed be increased by ?
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Time = 10 hours,
Speed = 48 km per hr.
∴ Distance = Speed × Time
= 48 × 10 = 480 km
Now, this distance of 480 kms is to be covered in 8 hours.
Hence, the required Speed= Distance = 480 New time 8
= 60 km per hr.
∴ Increase in speed
= 60 – 48 = 12 km per hr.Correct Option: B
Time = 10 hours,
Speed = 48 km per hr.
∴ Distance = Speed × Time
= 48 × 10 = 480 km
Now, this distance of 480 kms is to be covered in 8 hours.
Hence, the required Speed= Distance = 480 New time 8
= 60 km per hr.
∴ Increase in speed
= 60 – 48 = 12 km per hr.
- If a boy walks from his house to school at the rate of 4 km per hr, he reaches the school 10 minutes earlier than the scheduled time. However if he walks at the rate of 3 km per hr, he reaches 10 minutes late. Find the distance of his school from his house.
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Let the distance be x kms.
∴ Time taken at 4 km per hr. t1= x hrs. 4
Time taken at 3 km per hr.t2= x hrs. 3
Difference in timings
= 10 + 10 = 20 minutesor 20 = 1 hour 60 3 ∴ x − x = 1 3 4 3 ⇒ 4x − 3x = 1 12 3 ⇒ x = 1 12 3
∴ x = 4 km.
Hence the required distance
= 4 kms.Correct Option: C
Let the distance be x kms.
∴ Time taken at 4 km per hr. t1= x hrs. 4
Time taken at 3 km per hr.t2= x hrs. 3
Difference in timings
= 10 + 10 = 20 minutesor 20 = 1 hour 60 3 ∴ x − x = 1 3 4 3 ⇒ 4x − 3x = 1 12 3 ⇒ x = 1 12 3
∴ x = 4 km.
Hence the required distance
= 4 kms.
- A man has to reach a place 40 kms away. He walks at the rate of 4 km per hr. for the first 16 kms and then he hires a rickshaw for the rest of the journey. However if he had travelled by the rickshaw for the first 16 kms and the remaining distance on foot at 4 km per hr, he would have taken an hour longer to complete the journey. Find the speed of rickshaw.
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Let the speed of Rickshaw be 'x ' .
Then, time taken to cover 16 km on foot and 24 km on Rikshaw= 16 + 24 hrs 4 x
and time taken to travel 24 km on foot & 16 km on Rikshaw= 16 + 24 hrs x 4
According to question,= 16 + 24 + 1 = 16 + 24 4 x x 4 ⇒ 5 + 24 = 16 + 6 x x ⇒ 24 − 16 = 1 x
⇒ x = 8 km/hrCorrect Option: D
Let the speed of Rickshaw be 'x ' .
Then, time taken to cover 16 km on foot and 24 km on Rikshaw= 16 + 24 hrs 4 x
and time taken to travel 24 km on foot & 16 km on Rikshaw= 16 + 24 hrs x 4
According to question,= 16 + 24 + 1 = 16 + 24 4 x x 4 ⇒ 5 + 24 = 16 + 6 x x ⇒ 24 − 16 = 1 x
⇒ x = 8 km/hr
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is marked on my cards by 10 minutes. Find my usual time.Walking 3 of my usual speed, a late 4
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Since I walk at 3/4 of my usual speed the time taken is 4/3 of my usual time.
[∵ the speed and time are in the inverse ratio]∴ 4 of usual time 3
= Usual time + Time I reach late∴ 1 of usual time 3
= 10 minutes
∴ Usual time
= 10 × 3 = 30 minutes.Correct Option: A
Since I walk at 3/4 of my usual speed the time taken is 4/3 of my usual time.
[∵ the speed and time are in the inverse ratio]∴ 4 of usual time 3
= Usual time + Time I reach late∴ 1 of usual time 3
= 10 minutes
∴ Usual time
= 10 × 3 = 30 minutes.
