Time, Speed and Distance Exercise 2
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- Question 1 of 25
The speed of a car increases by 2 kms after every one hour. If the distance travelled in the first one hour was 35 kms, what was the total distance travelled in 12 hours?
Total distance travelled in 12 hour
= (35 + 37 + 39 + ….. upto 12 terms)
This is an A.P. with first term a = 35,
number of terms n = 12,
common difference d = 2.
∴ Required distance
- Question 2 of 25
Anna left for city A from city B at 5.20 a.m. She travelled at the speed of 80 km/h for 2 hours 15 minutes. After that the speed was reduced to 60 km/h. If the distance between two cities is 350 kms, at what time did Anna reach city A?
Distance covered in 2 hr 15 min,
i.e., 2¼ hr
Time taken to cover remaining distance
= 2 hr 50 min.
Total time taken
= (2 hr 15 min + 2 hr 50 min)
= 5 hr 5 min.
So, Anna reached city A at 10.25 a.m.
- Question 3 of 25
A and B walk around a circular track. They start at 8 a.m. from the same point in the opposite directions. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. How many times shall they cross each other before 9 : 30 a.m.?
= (2 + 3) = 5 rounds per hour.
So, they cross each other 5 times in an hour and 2 times in half an hour.
Hence, they cross each other 7 times before 9 :.30 a.m.
- Question 4 of 25
A circular running path is 726 metres in circumference. Two men start from the same point and walk in opposite directions @ 3.75 km/h and 4.5 km/h, respectively. When will they meet for the first time?
Their relative speeds
= (4.5 + 3.75) = 8.25 km/h
Distance = 726 metres
- Question 5 of 25
A person has to cover a distance of 6 km in 45 minutes. If he covers one-half of the distance in two-thirds of the total time; to cover the remaining distance in the remaining time, his speed (in km/h) must be:
Remaining distance = 3 km
and remaining time
∴ Required speed = (3 × 4) km/h
= 12 km / hr.
- Question 6 of 25
In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is:
Let the duration of the flight be x hour Then,
⇒ x (2x + 1) = 3
⇒ 2x² + x – 3 = 0
⇒ (2x + 3) (x – 1) = 0
⇒ x = 1 hr. [neglecting the –ve value of x].
- Question 7 of 25
A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minutes?
Relative speed of the thief and policeman
= (11 – 10) km/h = 1 km/h.
Distance covered in 6 minutes
∴ Distance between the thief and policeman
= (200 – 100) m = 100 m.
- Question 8 of 25
Two persons living in different towns 32 km apart started at 1 pm to see each other at the rate of 3.5 and 4.5 km per hour respectively. When will they meet?
= (3.5 + 4.5) = 8 kmph.
Time to meet = 32/8 = 4 hour
So when they meet at 5 pm, one will have walked
3.5 × 4 = 14 km and the other will have walked
4.5 × 4 = 18 km.
- Question 9 of 25
A man rides at the rate of 11 miles an hour, but stops for 5 min to change horse at the end of every seventh mile. How long will be take to go a distance of 96 miles? (Approx.)
Time taken to travel 96 miles
= ⁹⁶⁄₁₁ hr = 8 hr 43 minutes
During the journey of 96 miles, he has to stop for 13 times to change the horse.
∴Total stoppage time
= 13 × 5 = 65 mins. = 1 hr. 5 mins.
Hence the total time
= 8 hr 43 mins + 1 hr. 5 mins.
= 9 hr 48 mins.
- Question 10 of 25
Subbu starts from a point O at 10:00 a.m., overtakes Ajay, who is moving in the same direction, at 11:00 a.m. and Bhuvan moving in the opposite direction at 12:00 (noon). If the speed of Bhuvan is one fourth the speed of Subbu, at what time will Ajay and Bhuvan cross each other?
Let the speed of Ajay be V and the speed of Bhuvan and Subbu be 1 and 4 respectively.
Then OA = 4 and OB = 4.
At 12:00 noon.
Let Ajay be at C at 12:00 noon at a distance of V from A (towards B)
∴ Time taken for them to meet from 12:00 noon. =
Since V is not known cannot be determined.
- Question 11 of 25
On a journey across Bombay, a tourist bus averages 10 km/h for 20% of the distance, 30 km/h for 60% of it and 20 km/h for the remainder. The average speed for the whole journey was
Let the average speed be x km/h.
and total distance = y km. then,
- Question 12 of 25
A man walks half of the journey at 4 km/h by cycle does one third of journey at 12 km/h and rides the remainder journey in a horse cart at 9 km/h, thus completing the whole journey in 6 hours and 12 minutes. The length of the journey is
Let the length of the journey =x km.
∴ Journey rides by horse cart
Then, total time taken to complete journey
- Question 13 of 25
A train leaves station X at 5 a.m. and reaches station Y at 9 a.m. Another train leaves station Y at 7 a.m. and reaches station X at 10: 30 a.m. At what time do the two trains cross each other?
Let the distance between X and Y be x km.
Then, the speed of A is x/4 km/h and that of B is 2x/7 km/h.
Relative speeds of the trains
Therefore the distance between the trains at 7 a.m.
Hence, time taken to cross each other
Thus, both of them meet at 7 : 56 a.m.
- Question 14 of 25
A man covers a certain distance on a toy train. If the train moved 4 km/h faster, it would take 30 minutes less. If it moved 2 km/h slower, it would have taken 20 minutes more. Find the distance.
Let the distance be x km. Let speed of train be y km/h. Then by question, we have
On solving (i) and (ii), we get x = 3y
Put x = 3y in (i) we get
⇒ y = 20
Hence, distance = 20 × 3 = 60 km.
- Question 15 of 25
If a man walks to his office at 5/4 of his usual rate, he reaches office 30 minutes early than usual. What is his usual time to reach office.
- Question 16 of 25
Mary jogs 9 km at a speed of 6 km per hour. At what speed would she need to jog during the next 1.5 hours to have an average of 9 km per hour for the entire jogging session?
Let speed of jogging be x km/h.
Total time taken
Total distance covered = (9 + 1.5x) km.
- Question 17 of 25
Mohan travels 760 km to his home, partly by train and partly by car. He takes 8 hours if he travels 160 km by train and the rest by car. He takes 12 minutes more if he travels 240 km by train and the rest by car. The speed of the train and the car, respectively are:
Let speed of the train be x km/h and that of the car be y km/h.
Solving (i) and (ii),
we have x = 80 km/h and
y = 100 km/h.
- Question 18 of 25
It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the speed of the car is :
Let the speed of the train and the car be x km/h and y km/h, respectively.
120y + 480x = 8xy and …(iii)
200y + 400x = …(iv)
From (iii) and (iv),
⇒ 15y + 60x = 24y + 48x
⇒ 12x = 9y
- Question 19 of 25
In a flight of 6000 km, an aircraft was slowed down due to bad weather. The average speed for the trip was reduced by 400 kmph and the time of flight increased by 30 minutes. The original planned duration of the flight was
Let the original planned time of the flight be x hour
∴ The average speed of the flight
If the average speed is
then the time of the flight is
⇒ – 400x +
⇒ – 4x +
⇒ – 4x² + 30 – 2x = 0
⇒ 4x² + 2x – 30 = 0
⇒ 2x² + x – 15 = 0
⇒ x =
- Question 20 of 25
Points A and B are 70 km apart on a highway. One car starts form A and the another one from B at the same time. If they travel in the same direction, they meet in 7 hours But if they travel towards each other, they meet in one hour. The speeds of the two cars are, respectively.
Let the speed of the car be x km/h and y km/h, respectively.
Their relative speeds when they are moving in same direction = (x – y) km/h.
Their relative speeds when they are in opposite directions = (x + y) km/h.
or x + y = 70 … (i)
and or x – y = 10 … (ii)
Solving (i) and (ii), we have
x = 40 km/h and y = 30 km/h.
- Question 21 of 25
A plane left 30 minutes later than the scheduled time and in order to reach the destination 1500 km away in time, it had to increase the speed by 250 km/h from the usual speed. Find its usual speed.
Let the usual speed of the plane be x km/h
∴ Time taken in covering the distance of 1500 km
⇒ 3000 x = 3000 (x + 250) – x(x + 250)
⇒ x² + 250 x – 3000 × 250 = 0
- Question 22 of 25
Walking ⁶⁄₅ of his usual speed, a person takes 10 min less to reach his office. His usual time taken to reach the office is
- Question 23 of 25
A man starts from B to K, another from K to B at the same time. After passing each other they complete their journeys in 3⅓ and 4⅘ hours, respectively. Find the speed of the second man if the speed of the first is 12 km/hr.
= = ⁶⁄₅
∴ = ⁶⁄₅
∴ 2nd man’s speed
= ⁶⁰⁄₆ = 10 km/hr.
- Question 24 of 25
A man travelled a distance of 50 km on his first trip. On a later trip, travelling three times as fast, he covered 300 km. Compare the times he took.
Let speed be x
⇒50/x = time taken
also = time taken.
Hence ratio is 1 : 2.
- Question 25 of 25
The driver of a car driving @ 36 kmph locates a bus 40 meters ahead of him. After 20 seconds the bus is 60 meters behind. The speed of the bus is :
Net distance gained by car over the bus
= 40 + 60 = 100m, in 20 sec.
⇒ S₂ = 5 m/s = 18 kmph.