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# Time and Work Questions for SSC-CPO Set-3 PDF

Download SSC CPO Time and Work questions with answers PDF based on previous papers very useful for SSC CPO exams. Very important Time and Work Questions for SSC exams

Question 1: A drum of kerosene is $\frac{3}{4}$ full. When 30 litres of kerosene is drawn from it, it remains $\frac{1}{2}$ full. The capacity of the drum is

a) 120 litres

b) 135 litres

c) 150 litres

d) 180 litres

Question 2: By what least number should 675 be multiplied so as to obtain a perfect cube number ?

a) 3

b) 5

c) 24

d) 40

Question 3: $0.\overline{001}$ is equal to

a) 1/1000

b) 1/999

c) 1/99

d) 1/9

Question 4: $\frac{4.41 \times 0.16}{2.1 \times 1.6 \times 0.21}$ is equal to

a) 1

b) 0.1

c) 0.01

d) 10

Question 5: If a and b are two odd positive integers, by which of the following integers is $(a^4-b^4)$ always divisible ?

a) 3

b) 6

c) 8

d) 12

Question 6: $\frac{256 \times 256 – 144 \times 144}{112}$ is equal to

a) 420

b) 400

c) 360

d) 320

Question 7: Buses start from a bus terminal with a speed of 20 km/hr at intervals of 10 minutes. What is the speed of a man coming from the opposite direction towards the bus terminal if he meets the buses at intervals of 8 minutes ?

a) 3 km/hr

b) 4 km/hr

c) 5 km/hr

d) 7 km/hr

Question 8: A and B together can do a piece of work in 9 days. If A does thrice the work of B in a given time, the time A alone will take to finish the work is

a) 4 days

b) 6 days

c) 8 days

d) 12 days

Question 9: If $a^3+\frac{1}{a^3}=2$, then value of $\frac{a^2+1}{a}$ is (a is a positive number)

a) 1

b) 2

c) 3

d) 4

Question 10: The mean of 100 observations was calculated as 40. It was found later on that one of the observations was misread as 83 instead of 53. The correct mean is:

a) 39

b) 39.7

c) 40.3

d) 42.7

Question 11: The time taken by a train 160 m long, running at 72 km/hr, in crossing an electric pole is

a) 8 sec

b) 9 sec

c) 6 sec

d) 4 sec

Question 12: A and B start running at the same time and from the same point around a circle. If A can complete one round in 40 seconds and B in50 seconds, how many seconds will they take to reach the starting point simultaneously?

a) 10

b) 200

c) 90

d) 2000

Question 13: In what time will a 100 metre long train running with a speed of 50 km/hr cross a pillar

a) 7.0 sec

b) 72 sec

c) 7.2 sec

d) 70 sec

Question 14: A gun is fired at a distance of 6.64 km away from Ram. He hears the sound 20 seconds later. Then the speed of sound is

a) 664 m/s

b) 664 km/s

c) 332 m/s

d) 332 km/s

Question 15: A train leaves a station A at 7 am and reaches another station B at 11 am. Another train leaves B at 8 am and reaches A at 11.30 am. The two trains cross one another at

a) 8:36 am

b) 8:56 am

c) 9:00 am

d) 9:24 am

Question 16: A can do a piece of work in 4 days and B can do it in 12 days. In how many days will they finish the work, both working together ?

a) 4 days

b) 6 days

c) 2 days

d) 3 days

Question 17: A can do 1/4 of a work in 10 days. B can do 1/3 of the work in 20 days. In how many days can both A and B together do the work ?

a) 30 days

b) 32 days

c) 24 days

d) 25 days

Question 18: A and B can together finish a work in 30 days. They worked at it for 20 days and then B left. The remaining work was done by A alone in 20 more days.A alone can finish the work in

a) 60 days

b) 54 days

c) 48 days

d) 50 days

Question 19: P can do a piece of work in 9 days. Q is 50% more efficient than P. The number of days it takes for Q to do the same piece of work is

a) $3$

b) $13\frac{1}{2}$

c) $4\frac{1}{2}$

d) $6$

Question 20: Sixteen men can complete a work in fifteen days, twenty-four children can do the same work in twenty days. In how many days will eight men and eight children, complete the same work ?

a) $18$ days

b) $16 days$

c) $13\frac{1}{3}$ days

d) $20$ days

Drum is $\frac{3}{4}$ full.

When 30 liters are drawn out of it, it becomes $\frac{1}{2}$ full.
Therefore $\frac{3}{4}$ – $\frac{1}{2}$ of drum = 30
$\frac{1}{4}$ of drum = 30
Total capacity of drum = 30 $\times$ 4 = 120 litres

Factorising 675 we get
675 = 5 $\times$ 5 $\times$ 3 $\times$ 3 $\times$ 3
There in order to make 675 a perfect cube, it has to be multiplied by 5.

Let $x =0.\overline{001}$ be the first equation

$1000x = 1000 \times 0.\overline001$

= $1.\overline{001}$

Let $1000x = 1.\overline{001}$ be second equation

Subtracting equation 1 from 2

$999x = 1$

$x = \frac{1}{999}$

$\frac{4.41 \times 0.16}{2.1 \times 1.6 \times 0.21}$

= $\frac{{2.1}^2 \times 0.16}{2.1 \times 0.16 \times 10 \times 0.21}$

=$\frac{{2.1}^21 \times 0.16}{2.1 \times 0.16 \times 2.1}$

=$\frac{{2.1}^2 \times 0.16}{{2.1}^2 \times 0.16}$

=1

$(a^4-b^4) = (a^2+b^2)\times(a^2-b^2)$
= $(a^2+b^2)\times(a+b)\times(a-b)$

Let a = 5 , b= 3
54 – 34 = (52 + 32)(5 + 3)(5 – 3) = 34 x 8 x 2 which is divisible by 8
Let a  = 7 and b = 5
74 – 5= (72 + 52)(7 + 5)(7 – 5) = 74 x 12 x 2 which is also divisible by 8

$\frac{256 \times 256 – 144 \times 144}{112} = \frac{256^2 – 144^2}{112} = \frac{(256 – 144)(256 + 144)}{112} = \frac{(112)(400)}{112} = 400$

Distance between buses will $20\times\frac{10}{60}$ = $\frac{10}{3}$ km.
Now man is travelling this distance in 8 min. with the relative speed of (20+$x$) (let’s assume speed of man is $x$ km/hr )
hence (20+$x$) = $\frac{\frac{10}{3}}{\frac{8}{60}}$
$x$= 5

A does thrice the work of B in a given time.

Let B’s efficiency = $x$ units/day

=> A’s efficiency = $3x$ units/day

Thus, (A+B)’s 1 day’s work = $x+3x=4x$ units/day

=> Total work done by them in 9 days = $4x \times 9=36x$ units

$\therefore$ Time taken by A alone to finish the work = $\frac{36x}{3x}=12$ days

=> Ans – (D)

Given : $a^3+\frac{1}{a^3}=2$

To find : $\frac{a^2+1}{a}=(a+\frac{1}{a}) = x = ?$

We know that, $(a+\frac{1}{a})^3=a^3+\frac{1}{a^3}+3(a)(\frac{1}{a})(a+\frac{1}{a})$

=> $(a+\frac{1}{a})^3=2+3(a+\frac{1}{a})$

=> $x^3=2+3x$

=> $x^3-3x=2$

=> $x(x^2-3)=2 \times 1$

Thus, the only value that satisfy above equation is $x=2$

=> Ans – (B)

Mean of 100 observations = 40

=> Sum of 100 observations = $100 \times 40=4000$

Now, replacing 53 instead of 83

=> Correct sum = $4000-83+53=3970$

$\therefore$ Correct mean = $\frac{3970}{100}=39.7$

=> Ans – (B)

Speed of train = 72 km/hr

= $(72 \times \frac{5}{18})$ m/s = $20$ m/s

Length of train = 160 m

Using, time = distance/speed

=> Time taken = $\frac{160}{20}=8$ sec

=> Ans – (A)

Time taken by A to complete 1 round = 40 seconds and by B = 50 seconds

=> Time taken by them to reach the starting point simultaneously = L.C.M.(40,50)

= 200 seconds

=> Ans – (B)

Speed of train = 50 km/hr

= $(50 \times \frac{5}{18})$ m/s = $\frac{125}{9}$ m/s

Length of train = 100 m

Using, time = distance/speed

=> Time taken = $100\div \frac{125}{9}$

= $100 \times \frac{9}{125}$

= $\frac{36}{5}=7.2$ sec

=> Ans – (C)

Distance of gun from Ram = 6.64 km = 6640 m

Time = 20 seconds

=> Speed of sound = distance/time

= $\frac{6640}{20}=332$ m/s

=> Ans – (C)

Time taken by 1st train to travel from A to B = 11-7 = 4 hours

Time taken by 2nd train to travel from B to A = 11:30-8 = 3.5 hours

=> ratio of time taken by 1st train to 2nd train = $4 : \frac{7}{2}$ = 8 : 7

Since, speed is inversely proportion to time

=> Ratio of speeds of 1st train to 2nd train = 7 : 8

Let the speed of 1st train = $7x$ and 2nd train = $8x$ km/hr

Distance between the two stations = time * speed = $7x * 4 = 28x$ km

We know that, 1st train starts one hour early, thus it will cover $7x$ distance till the time 2nd train starts.

So, at 8.00 a.m., remaining distance between two trains = $28x-7x = 21x$ km

Also, the two trains are moving in opposite directions, =>relative speed of two trains = $8x+7x = 15$ km/hr

Now, time taken to meet = $\frac{21x}{15x} = \frac{7}{5}$ hours

=> $\frac{7}{5} * 60$ = 84 minutes after 8.00 a.m.

=> Time when they meet = 8.00 a.m. + 84 min = 9:24 a.m.

A’s 1 day’s work=$\frac{1}{4}$
B’s 1 day’s work=$\frac{1}{12}$
(A+B)’s 1 day’s work=$\frac{1}{4}+\frac{1}{12}$
$=\frac{3+1}{12}$
$=\frac{4}{12}$
$=\frac{1}{3}$
$\therefore$ A and B together does the work in $3$ days.
Hence,Option D is correct.

A does$\frac{1}{4}$ work in 10 days.
$\therefore$A will complete the work in $10\times4=40$ days.
Similarly, B will complete the work in $20\times3=60$ days.
(A+B)’s 1 day’s work=$\frac{1}{40}+\frac{1}{60}$
$=\frac{3+2}{120}$
$=\frac{1}{24}$
$\therefore$ time taken by A and B together to complete the work=$24$ days.
Hence, Correct option is C.

(A+B)together do the work in 30 days.
$\therefore$ (A+B)’s 1 day’s work=$\frac{1}{30}$
$\therefore$ (A+B)’s 20 days’ work=$\frac{20}{30}$$=\frac{2}{3}$
Remaining Work=$1-\frac{2}{3}$
=$\frac{1}{3}$
$\because$ Time taken by A in doing $\frac{1}{3}$ work=20 days
$\therefore$ Time taken by A to complete the work=$20\times3$
=$60$days
Hence, Option A is correct.

P can do a piece of work in 9 days.

let efficiency of p is 100%

then efficiency of Q will be 150%

let Q can do the same piece of work in x days.

work is constant. so

efficiency*no of days = constant

100*9 = 150*x

x = 6.

16 men can complete a work in 15 days

1 man can complete the work in (16*15) days

24 children can do the same work in 20 days

1 child can complete the work in (24*20) days

let x be the no.of days taken by 8 men and 8 children to complete the work, then

1/x = 8/(16*15) + 8/(24*20)

1/x = 1/30 + 1/60

1/x = 3/60

x = 20