 # Time and Distance Practice Set 4 5 Steps - 3 Clicks

# Time and Distance Practice Set 4

### Introduction

Time and distance as a topic involves a variety of areas which include speed-time-distance concepts, relative speed, moving and stationary bodies, boats and streams, circular motion, and so on. Time and Distance Practice Set 4 article is useful for candidates preparing for different competitive examinations like RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC, IBPS PO Exams and etc.

### Quiz

1. How many minutes does Mayank take to cover a distance of 400 m, If he runs at speed of 20 km/hr?

A. 1$$\frac{1}{5}$$min
B. 1$$\frac{2}{5}$$min
C. 2$$\frac{2}{5}$$min
D. 2$$\frac{3}{5}$$min

Explanation:
Mayank’s speed = 20 km /hr
= (20*$$\frac{5}{18}$$) m/sec =$$\frac{50}{9}$$m/sec
So, time taken to cover 400 m
(400*$$\frac{9}{50}$$)sec= 72 sec= 1$$\frac{1}{5}$$min

2. A cyclist covers a distance of 750 m in 2 min 30 sec. what is the speed in km /hr of the cyclist ?

A. 12 km/hr
B. 15 km/hr
C. 18 km / hr
D. 20 km / hr

Explanation:
By applying a formula,
Speed =$$\frac{Distance}{Time}$$
Speed = $$\frac{750}{150}$$m/sec =5m/sec
=5*$$\frac{18}{5}$$km / hr =18 km / hr

3. A Jackal takes 4 leaps for every 5 leaps of goat but 3 leaps of a Jackal are equal to 4 leaps of the goat. compare their speeds

A. 12 : 10
B. 7 : 5
C. 1 : 4
D. 16 : 15

Explanation:
Let the distance covered in 1 leap of the jackal be x and that covered in 1 leap of the goat be y.
Then, 3x = 4y ⇒ x =$$\frac{4}{3}$$y
4x= $$\frac{16}{3}$$y
So, Ratio speed of jackal and goat = Ratio of distances covered by them in the same time,
⇒ 4x : 5y =$$\frac{16}{3}$$y: 5y
16 : 15

4. While covering a distance of 24 km, a man noticed that after walking for 1 hour and 40 minutes, the distance covered by him was $$\frac{5}{7}$$ of the remaining distance. What was his speed in meters per second?

A. 2$$\frac{4}{3}$$m/sec
B. 2$$\frac{4}{7}$$m/sec
C. 1$$\frac{2}{3}$$m/sec
D. 2$$\frac{4}{7}$$m/sec

Explanation:
Let the speed be x km/hr.
Then, distance covered in 1 hr. 40 min.
i.e , 1$$\frac{2}{3}$$hrs= $$\frac{5x}{3}$$km
Remaining distance
(24-$$\frac{5x}{3}$$)Km.
7x = 72 – 5x
12x = 72
x = 6
Hence, speed = 6 km/hr
(6*$$\frac{5}{18}$$)m/sec
=1$$\frac{2}{3}$$m/sec

5. A boy travelled from the home to the college at the rate of 25 kmph and walked back at the rate of 4 kmph. If the whole journey took 5 hours 48 minutes, Find the distance of the college from the home.

A. 5 km
B. 10 km
C. 15 km
D. 20 km

Explanation:
Average speed =$$\frac{2xy}{x+y}$$km/hr
= $$\frac{ 2 * 25 * 4}{25 + 4}$$km/hr
= $$\frac{ 200}{29}$$km/hr
Distance travelled in 5 hours 48 minutes i.e,5$$\frac{4}{5}$$hrs
($$\frac{200}{29}$$ * $$\frac{29}{5}$$)km = 40 km.
So, Distance of the college from the home
$$\frac{40}{2}$$= 20km

1. A thief is spotted by a policeman from a distance of 100 m. When the policeman starts the chase, the thief also starts running. If the speed of the thief be 8 km/hr and that of the policeman 10 km/hr. How far the thief will have run before he is overtaken?

A. 100 m
B. 150 m
C. 200 m
D. 400 m

Explanation:
Realtive speed of policeman = (10 – 8) km/hr = 2 km/hr.
Time =$$\frac{Distance}{Speed}$$
Time taken by policeman to cover 100 m
= ($$\frac{100}{1000}$$* $$\frac{1}{2}$$)hr= $$\frac{1}{20}$$hr
In $$\frac{1}{20}$$hr the thief covers a distance of
= (8*$$\frac{1}{20}$$)= 400m.

2. I walk a certain distance and ride back taking a total time of 37 minutes. I could walk both ways in 55 minutes. How long would it take me to ride both ways?

A. 5 min.
B. 10 min.
C. 13 min.
D. 19 min.

Explanation:
Let the distance be x, then,
(Time taken to walk x km) + (Time taken to ride x km) = 37min.
⇒ (Time taken to walk 2x km) + (Time taken to ride 2x km) = 74 min.
But, time taken to walk 2x km = 55 min.
So, Time to ride 2x km = (74 – 55) min ⇒ 19 min.
Hence, option D is correct.

3. A motor-cycle covers 40 km with a speed of 20 km/hr. find the speed of the motor-cycle for the next 40 km journey so that the average speed of the whole journey will be 30 km/hr.

A. 70 km/hr
B. 52.5 km/hr
C. 60 km/hr
D. 60.5 km/hr

Explanation:
Average speed of whole journey = 30 km/hr
x = speed of first 40 km = 20 km/hr
30= $$\frac{2 * 20 * y}{20 + y}$$
600 + 30y = 40y ⇒ 10y = 600
y= $$\frac{600}{10}$$= 60 km/hr

4. A man rides at the rate of 18 km/hr, but stops for 6 minutes to change horses at the end of every 7th km. The time that he will take to cover a distance of 90 km is

A. 6 hrs
B. 6 hrs. 12 min.
C. 6 hrs. 18 min.
D. 6 hrs. 24 min.

Explanation:
Number of stoppages
= $$\frac{90}{7}$$= 12.8,
it means there is 12 stoppages.
Distance to be covered = 90 km, Speed = 18,
Time for each stoppage = 6 mins.
By the short trick approach, we get
= $$\frac{90}{18}$$ + 12 * 6
5 hrs + 72 mins
Convert mins to hours then,
= 6 hrs 12 mins.

5. Walking at 3 km/hr . Pintu reaches his school 5 minutes late. If he walks at 4 km per hour he will be 5 minutes early. The distance of Pintu’s from his house is

A. 1$$\frac{1}{2}$$km
B. 2 km
C. 2$$\frac{1}{2}$$km
D. 5km

Explanation:
Speed1 = 3 km/hr, Speed2 = 4 km/hr
Time1 = 5 mins late, Time2 = 5 min early
Reqd. Distance
$$\frac{3 * 4}{4 – 3}$$ * $$\frac{5 + 5}{60}$$= 2 km

1. Two buses start at the same time from Delhi and Agra, which are 300 km. apart, towards each other. After what time will they cross each other if their speeds are 38 km per hour and 37 km per hour?

A. 4 hours
B. 3 hours
C. 5 hours
D. 6 hours

Explanation:
Distance between Agra and Delhi = 300 km
Relative speed = 38 + 37 = 75 km/hr
Time taken to cross each other = $$\frac{Distance}{Speed}$$
= $$\frac{300}{75}$$= 4 hours.

2. Raj and Prem walk in opposite direction at the rate of 3 km and 2 km per hour respectively. How far will they be from each other after 2 hrs?

A. 6 km
B. 8 km
C. 10 km
D. 2 km

Explanation:
Distance covered per hour = Relative speed × Time
= (3 + 2) * 1 = 5 km
Distance covered in 2 hours = 5 × 2 = 10 km.

3. Walking $$\frac{2}{3}$$ times of his usual speed a man gets late by 20 minutes. The usual time taken by him to cover that distance:

A. 1 hour
B. 30 minutes
C. 40 minutes
D. 1 hr 20 min

Explanation:
New Speed =$$\frac{2}{3}$$ of the usual speed
New time taken =$$\frac{3}{2}$$of the usual time
So, ($$\frac{3}{2}$$of the usual time)– (usual time) = 20 min
=$$\frac{1}{2}$$of the usual time = 20 min
usual time = 40 mins.

4. A car travels the first one-third of a certain distance with a speed of 10 km/hr, the next one-third distance with a speed of 20 km/hr and the last one-third distance with a speed of 60 km/hr. The average speed of the car for the whole journey is

A. 18 km/hr
B. 24 km/hr
C. 30 km/hr
D. 36 km/hr

Explanation:
Let the total distance covered = LCM of (10, 20, 60) = 60.
As per the question,
Distance covered by the car with each speed
$$\frac{1}{3}$$*60= 20 km
$$\frac{20}{10}$$+$$\frac{20}{20}$$+$$\frac{20}{60}$$= $$\frac{60}{avg. speed}$$
average speed = 18 km/hr

5. A student goes to school at the rate of 2$$\frac{1}{2}$$km/hr and reaches 6 minutes late. If he travels at the speed of 3 km/hr he is 10 minutes early. What is the distance to the school?

A. 1 km
B. 4 km
c. 3$$\frac{1}{4}$$ km
D. 3$$\frac{1}{2}$$ km

Explanation:
Where,
‘a’ is the extra of time taken by first speed = 6 mins
‘b’ is the less of time taken by second speed = 10 mins
a + b = 10 + 6 = 16 mins = $$\frac{4}{15}$$hrs
By the short trick approach, we get
Reqd. distance = 4km

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