# Time and Work Problems for SSC CHSL

1
1983

Time, speed, distance and work problems appear in SSC exams (CGL and CHSL). We have given some of the SSC CHSL time and work questions asked in previous year question papers.

Time and Work Problems for SSC CHSL:

Download Time and work problems with solutions for SSC CGL and CHSL PDF

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Question 1:

Rakesh goes on a trip on his motor-cycle and rides for 368 kms. If he rides for 5 hours at a speed of 49 km/hr, at what speed he travels for the remaining 3 hours of the journey?

a) 58 km/hr
b) 41 km/hr
c) 54 km/hr
d) 46 km/hr

Question 2:

To cover a distance of 333 km in 2 hours by a car, what should be the average speed of the car (in meter/second)?

a) 166.5
b) 46.25
c) 83.25
d) 92.5

Question 3:

A thief is stopped by a policeman from a distance of 150 metres. When the policeman starts the chase, the thief also starts running. Assuming the speed of the thief as 7 km/hr and that of policeman as 9 km/hr, how far the thief would have run, before he is over-taken by the policeman?

a) 420 metres
b) 630 metres
c) 315 metres
d) 525 metres

Question 4:

Two cars travel from city A to city B at a speed of 60 and 108 km/hr respectively. If one car takes 2 hours lesser time than the other car for the journey, then the distance between City A and City B?

a) 270 km
b) 324 km
c) 405 km
d) 216 km

Question 5:

Two cars travel from city A to city B at a speed of 30 and 45 km/hr respectively. If one car takes 2.5 hours lesser time than the other car for the journey, then the distance between City A and City B is:

a) 270 km
b) 338 km
c) 180 km
d) 225 km

Question 6:

To cover a distance of 216 km in 3.2 hours, what should be the average speed of the car in meters/second?

a) 67.5 m/s
b) 33.75 m/s
c) 37.5 m/s
d) 18.75 m/s

Question 7:

A mason can build a wall in 70 hours. After 7 hours he takes a break. What fraction of the wall is yet to be built?

a) 0.9
b) 0.8
c) 0.5
d) 0.75

Solutions: (1 to 7)

Total distance covered = 368 km
He rides for 5 hours at a speed of 49 km/hr
=> Distance covered = $$5 \times 49 = 245$$ km
Distance left = 368 – 245 = 123 km
$$\therefore$$ Speed he should travel for the remaining 3 hours of the journey = $$\frac{123}{3} = 41$$ km/hr
=> Ans – (B)

The car covers 333 km in 2 hours
Speed of car (in km/hr) = $$\frac{333}{2} = 166.5$$ km/hr
=> Speed in m/s = $$166.5 \times \frac{5}{18}$$
= $$5 \times 9.25 = 46.25$$ m/s
$$\therefore$$ In 1 second, it travels 46.25 metres
=> Ans – (B)

Since the thief is escaping from the police man, thus they both are running in same direction.
Speed of thief = 7 km/hr and speed of policeman = 9 km/hr
=> Relative speed = 9 – 7 = 2 km/hr
Distance between them = 150 metres = 0.15 km
=> Time taken = $$\frac{distance}{speed}$$
= $$\frac{0.15}{2} = \frac{3}{40}$$ hr
$$\therefore$$ Distance covered by thief before he was caught = $$7 \times \frac{3}{40}$$
= 0.525 km = 525 metres
=> Ans – (D)

Let the distance between City A and City B = $$d$$ km
Speed of first car = 60 km/hr and speed of second car = 108 km/hr
Let time taken by first car = $$t$$ hrs and time taken by second car = $$(t – 2)$$ hrs
Using, speed = distance/time for first car :
=> $$\frac{d}{t} = 60$$
=> $$d = 60t$$ ————–(i)
For second car, => $$\frac{d}{t – 2} = 108$$
Substituting value of $$d$$ from equation (i), we get :
=> $$60t = 108t – 216$$
=> $$108t – 60t = 48t = 216$$
=> $$t = \frac{216}{48} = 4.5$$ hrs
From equation (i), => $$d = 60 \times 4.5 = 270$$ km
=> Ans – (A)

Let the distance between City A and City B = d km
Speed of first car = 30 km/hr and speed of second car = 45 km/hr
Let time taken by first car = t hrs and time taken by second car = $$(t – 2.5)$$ hrs
Using, speed = distance/time for first car :
=> $$\frac{d}{t} = 30$$
=> $$d = 30t$$ ————–(i)
For second car, => $$\frac{d}{t – 2.5} = 45$$
Substituting value of $$d$$ from equation (i), we get :
=> $$30t = 45t – 112.5$$
=> $$45t – 30t = 15t = 112.5$$
=> $$t = \frac{112.5}{15} = 7.5$$ hrs
From equation (i), => $$d = 30 \times 7.5 = 225$$ km
=> Ans – (D)

The car covers 216 km in 3.2 hours
Speed of car (in km/hr) = $$\frac{216}{3.2} = 67.5$$ km/hr
=> Speed in m/s = $$67.5 \times \frac{5}{18}$$
= $$5 \times 3.75 = 18.75$$ m/s
$$\therefore$$ In 1 second, it travels 18.75 metres
=> Ans – (D)

=> Fraction of the wall yet to be built = $$\frac{(70 – 7)}{70} = \frac{63}{70}$$
= $$\frac{9}{10} = 0.9$$