Time and Work Questions for SSC CPO Set-2 PDF

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Time and Work Questions for SSC CPO Set-2 PDF
Time and Work Questions for SSC CPO Set-2 PDF

Time and Work Questions for SSC CPO Set-2 PDF

Download SSC CPO Time and work Questions with answers set-2 PDF based on previous papers very useful for SSC CPO exams. Very important time and work Questions for SSC exams.

Download Time and Work Questions for SSC CPO Set-2 PDF

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Question 1: If A and B together can complete a piece of work in 15 days and B alone in 20 days, in how many days can A alone complete the work ?

a) 60

b) 45

c) 40

d) 30

Question 2: By walking at 3/4 of his usual speed, a man reaches his office 20 minutes later than his usual time. The usual time taken by him to reach his office is

a) 75 minutes

b) 60 minutes

c) 40 minutes

d) 30 minutes

Question 3: A can complete a piece of work in 18 days, B in 20 days and C in 30 days. B and C together start the work and are forced to leave after 2 days. The time taken by A alone to complete the remaining work is

a) 10 days

b) 12 days

c) 15 days

d) 16 days

Question 4: A can complete 1/3 of a work in 5 days and B, 2/5 of the work in 10 days. In how many days both A and B together can complete the work ?

a) 10

b) 9$\frac{3}{8}$

c) 8$\frac{4}{5}$

d) 7$\frac{1}{2}$

Question 5: 7 men can complete a piece of work in 12 days. How many additional men will be required to complete double the work in 8 days ?

a) 28

b) 21

c) 14

d) 7

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Question 6: One pipe fills a water tank three times faster than another pipe. If the two pipes together can fill the empty tank in 36 minutes, then how much time will the slower pipe alone take to fill the tank ?

a) 1 hour 21 minutes

b) 1 hour 48 minutes

c) 2 hours

d) 2 hour 24 minutes

Question 7: Water flows into a tank which is 200m long and 150m wide, through a pipe of cross-section $0.3m \times 0.2m$ at 20 km/hour. Then the time (in minutes) for the water level in the tank to reach 8cm is

a) 50

b) 120

c) 150

d) 200

Question 8: A and B together can do a works in 12 days. B and C together do it in 15 days. If A’s efficiency is twice that of C, then the days required for B alone to finish the work is

a) 60

b) 30

c) 20

d) 15

Question 9: A and B can do a work in 12 days, B and C can do the same work in 15 days, C and A can do the same time work in 20 days. The time taken by A, B and C to do the same work is

a) 5 days

b) 10 days

c) 15 days

d) 20 days

Question 10: A sum of money becomes eight times in 3 years, if the rate is compounded annually. In how much time will the same amount at the same compound rate become sixteen times?

a) 6 years

b) 4 years

c) 8 years

d) 5 years

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Answers & Solutions:

1) Answer (A)

B does complete work in 20 days
hence in unit time amount of work done will be $\frac{1}{20}$
suppose in unit time amount of work done by A will be $x$
hence when they are working together amount of work done in unit time will be
$x$ + $\frac{1}{20}$ = $\frac{1}{15}$ (given)
$x$= $\frac{1}{60}$
it means A does $\frac{1}{60}$ amount of work in unit time
now unit work will be done ,when A alone is working, in 60 days

2) Answer (B)

As distance is constant and we know s = \frac{d}{t} (where s is distance and t is time)
hence st = constant
or $s_{1}t_{1}$= $s_{2}t_{2}$
or $s_{1}t_{1}$=$ \frac{3}{4}s_{1}(t_{1}+ 20)$
hence $t_{1}$ = 60

3) Answer (C)

Amount of work done by B and C together in unit time will be $\frac{1}{20}+\frac{1}{30}$=$\frac{5}{60}$

hence in two days amount of work done by B and C together = $\frac{10}{60}$
=$\frac{1}{6}$
remaining work = $\frac{5}{6}$

A does $\frac{1}{18}$ amount of work in unit time

hence  $\frac{5}{6}$ amount of work will be done in 15 days.

4) Answer (B)

A does 1/3 of work in 5 days
hence in 1 day it will do 1/15 amount of work

similarly B will do 2/50 amount of work in one day

Together they will do $\frac{1}{15} + \frac{2}{50}$ = $\frac{16}{150}$ amount of work in one day

hence unit work will be done in $\frac{150}{16}$ days or 9$\frac{3}{8}$ days

5) Answer (C)

In 12 days unit work is done by 7 men
hence in 1 day $\frac{1}{12}$ amount of work is done by 7 men
so 1 man does $\frac{1}{12\times7}$ amount of work in one day.
we have to find no of man to do double the work in 8 days with above efficiency
let’s say we have $x$ men
now total work done in single day will be $\frac{1}{12\times7} \times x $
hence in 8 days = $\frac{1}{12\times7} \times 8x $ = 2 ( as work is doubled)
$x$ = 21
additional man required = 14

6) Answer (D)

let’s say slower pipe fills tank in $x$min. so in 1 min. it fills $\frac{1}{x}$
and faster one will fill in $\frac{x}{3}$ min. so in 1 min. it fills $\frac{3}{x}$
together in 1 min. they will fill ($\frac{1}{x}$+$\frac{3}{x}$) of tank
as it is given that together they will fill the tank in 36 min.
so
$(\frac{1}{x}+\frac{3}{x})\times36$ = 1
or $x$ = 144 min.

7) Answer (B)

Volume of water in time t to reach at 8 cm. height will be equal to volume of tank till height 8 cm.
hence volume of water in time t to reach at height 8 cm. =$ .3\times .2\times \frac{2000}{6}\times$( spped in per minute)
which is equal to volume of tank =$ 200\times150\times \frac{8}{100}$

equating both and after solving we will get t = 120 min.

8) Answer (C)

A and B will do in 1 day, $\frac{1}{12}$ amount of work
B and C will do in 1 day, $\frac{1}{15}$ amount of work
Hence adding up above two equations
A+2B+C will do in 1 day, $\frac{9}{60}$ amount of work
as it is mentioned A=2C (i.e. efficiency of A is twice of C)
So 2C+2B+C will do in 1 day $\frac{9}{60}$ amount of work
or 2(B+C) + C = $\frac{9}{60}$
or C = $\frac{1}{60}$ (as B+C = $\frac{1}{15}$ )
or B = $\frac{1}{15}$ – $\frac{1}{60}$ = $\frac{1}{20}$
So B alone will do work in 20 days.

9) Answer (B)

A+B will do $\frac{1}{12}$ amount of work in a day.
B+C will do $\frac{1}{15}$ amount of work.
C+A will do $\frac{1}{20}$ amount of work.
Adding above 3 equation 2(A+B+C) = $\frac{1}{5}$
or (A+B+C) = $\frac{1}{10}$
Hence together A,B and C will do complete work in 10 days.

10) Answer (B)

$P(1+ \frac{r}{100})^3 = 8P$ (Where P is principal amount , r is rate)
So $(1+ \frac{r}{100}) = 2$
Now for 16 times
$P(1+ \frac{r}{100})^t = 16P$
or $2^t = 16$
or $t = 4$

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