We are
aware that this chapter is important for competitive exam from work and time
chapter question are asked in various
exam like ssc, upsc, banking etc. that work and time are time consuming
chapter.
If a
person A can do a piece of work in 'a' days, then working at the same uniform
speed A will do 1/a fraction of the work in one day.
If B
can do a work in 'b' days, then work done by him in 1 day is 1/b.
Then,
in one day, if A and B work together, then work done by them in 1 day, = (1/a +
1/b);
For example,
Days
taken to complete 'a' work
|
15
|
10
|
1/3
|
Fraction
of work done in a day
|
1/15th
|
1/10
|
3
|
Relationship between Men and Work
More men ------- can do
------->
More work
Less
men ------- can do
------->
Less work
Relationship
between Work and Time
More work -------- takes------>
More
Time
Less
work --------
takes------>
Less Time
Relationship
between Men and Time
More men ------- can do in
-------> Less Time
Less men -------
can do in ------->
More Time
· If A can do a work in ‘x’
days and B can do the same work in ‘y’ days, then the number of days required
to complete the work if A and B work together is
(X x Y) / (X + Y)
·
If A is thrice
as good a workman as B, then:
Ratio of work done by A and B = 3 :
1.
Ratio of times taken by A and B to
finish a work = 1 : 3
· If
M1 persons can do W1 work in D1 days
and M2 persons can do W2 work in D2 days,
then
M1D1 / W1 = M2D2 /
W2
· If
M1 persons can do W1 work in D1 days
for h1 hours and M2 persons can do W2 work
in D2 days for h2 hours, then
(M1D1h1) / W1 =
(M2D2h2) / W2
Note: If
works are same, then M1D1h1 = M2D2h2
Type 1: On Basic Concept
1. 1.A piece of work can be
done by Ram and Shyam and Hari in 15 days, by Shyam and Hari in 15 days and by
Hari and Ram in 20 days. Ram lone will complete the work in
(CGL-2013)
(a)
30 days
(b) 32 days
(c)36
days
(d) 42 days
Solution: (a) (Ram’s + Shyam’s)
1 day’s work = 1/12
(Shyam’s
+ Hari’s) 1 day’s work = 1/15
(Hari’s
+ Ram’s) 1 day’s work = 1/20
Adding
all three,
2
(Ram’s + Shyam’s + Hari’s) 1 day’s work = 1/12+1/15+1/20=(5+4+3)/60=1/5
Ram’s
1 day’s work = 1/10-1/15=(3-2)/30=1/30
Ram
alone will do the work in 30 days.
2. A and B can separately do a
piece of work in 6 days and 12 days respectively. How long will they together
take to do the work?
(CGL-2012)
(a)
9 days
(b) 18 days
(c)6
days
(d) 4 days
Solution: (d) (A+B)’s 1 day’s
work = 1/6+1/12=(2+1)/2=1/4
A
and B together will complete the work in 4 days.
3. A can do a piece of work in
10 days, B can do it in 15 days. In how many days both of these can together
complete the job?
(a)
9 days
(b) 18 days
(c)6
days
(d) 4 days
Solution:: A's per day work
= 1/10; B's per day work = 1/15
So
per day work of these people together = 1/10+1/15 ⇒(3+2)/30⇒5/30⇒1/6
So
A and B together 1/6 th of the total work perday. So total work can be
completed in 6 days.
Alternate
Method:
Ask
yourself that why the same work done by A in 10 days will be done by B in 15
days?
The
answer is due to their efficiencies. We know that number of days are
inversely proportional to their efficiencies. So we can solve this problem by
converting days into efficiencies.
Step
1: Estimate
the total work. This can be done by taking LCM of the days given. LCM
(10,15) = 30. Assume there is 30 meters length of wall to be
constructed.
Step
2: Calculate
the efficiencies of A and B: This can be done by dividing total work by their
days Efficiency = (Total work)/days
So
we get A's per day efficiency = 30 meters/10days=3mts
We
get B's per day efficiency = 30meters/15days=2mts
So
total work done by A and B together = 3 + 2 = 5 meters
Total
work of 30 meters can be done in 30/5 = 6 Days
Type II – They Work Together For n days and A/B leaves.
1. A and B can do a job in 6
and 12 days respectively. They began the work together but A leaves after 3
days. Then the total number of days needed for the completion of the work is: (CGL-2000)
(a)
4
(b) 5
(c)6
(d) 9
Solution: (c) A’s one day’s
work = 1/6
B’s
one day’s work = 1/12
(A+B)’s
one day’s work = 1/6+1/12=(2+1)/12=1/4
(A+B)’s
three day’s work = ¾
Remaining
work = 1- 3/4=1/4
Required
number of days = 1/4×12/1+3=3+3 days
= 6
days.
2. A and B can together finish
a work in 30 days. They worked together for 20 days and then B left. After
another 20 days. A finished the remaining work. In how many days A alone can
finish the job?
(CGL-2003)
(a)
50
(b) 60
(c)
48
(d) 54
Solution:: (b) (A+B)’s 1 day’s
work = 1/30
(A+B)’s
20 day’s work = 20/30=2/3
Remaining
work = 1- 2/3=1/3
Now,
1/3 part of work is done by A in 20 days.
Whole
work will be done by A alone in 20 x 3 = 60 days.
Type-III Based on A Man,
B Woman
1. 3 men and 4 boys can
complete a piece of work in 12 days. 4 men and 3 boys can do the same work in
10 days. Then 2 men and 3 boys can finish the work in number of days is (CGL-2012)
(a)
17 1/2 (b)
5 5/11
(c)
8
(d) 22
Solution (a) 12 (3 men + 4
boys) = 10 (4 men + 4 boys)
36
men + 48 boys = 40 men + 30 boys
4
men = 18 boys , 2 men = 9 boys
4
men + 3 boys = 21 boys who do the work in 10 days
and,
2 men + 3 boys = 12 boys
M1D1
= M2D2
D2=(21×10)/12=35/2=171/2
days
2. If 3 men or 6 women can do
a piece of wok in 16 days, in how many days can 12 men and 8 women do the same
piece of work?
(CGL-2000)
(a)
4 days
(b) 5 days
(c)3
days
(d) 2 days
Solution:(c) 3 m = 6w
1m
= 2w
12m
+ 8w = (12 x 2w) + 8w = 32w
6
women can do the work in 16 days.
32
women can do the work in (16×6)/32=3 days.
Type-IV Based on
Fraction
1. A can cultivate 2/5th of a
land in 6 days and B can cultivate 1/3rd of the same land in 10 days working
together A and B can cultivate 4/5th of the land in: (CGL-2002)
(a)
4 days
(b) 5 days
(c)8
days
(d) 10 days
Solution: (c) The part of field
cultivated by A in 1 day = 2/(5×6)=1/15
The
part of field cultivated by B in 1 day = 1/(3×10)=1/30
The
part of field cultivated by A and B together = 1/15+1/30=3/30=1/10
4/5
part of field cultivated by A and B together in = (4/5)/(1/10) days =
(4×10)/5=8 days
2. A does 4/3 of a piece of
work in 20 days: He then calls in B and they finish the remaining work in 3
days. How long B alone will take to do whole work? (CGL-2002)
(a)
37 1/2 days
(b) 37 days
(c)40
days
(d) 23 days
(a)
A can do the whole work in (20×5)/4=25 days
Remaining
work = 1 -4/5=1/5
(A+B)’s
1 day’s work = 1/15 and A’s 1 day’s work = 1/25
B’s
1 day’s work = 1/15-1/25=(5-3)/75=2/75
B
can finish the work in 75/2 days i.e., 37 1/2 days
TypeV- Efficient Worker
1. X is 3 times as fast as Y
and is able to complete the work in 40 days less than Y. Then the time in which
they can complete the work together is
(CGL
– 2011)
(a)
15 days
(b) 10 days
(c)
7 1/2 days (d) 5 days
Solution: (a) If X completes a
work in x days, Y will do the same in 3x days.
3x-
x = 40
x
= 20
Y
will finish the work in 60 days.
(X
+Y)’s 1 days work = 1/20+1/60=(3+1)/60=1/15
Both
together will complete the work in 15 days.
2. Kamal can do a work in 15
days. Bimal is 50 per cent more efficient than Kamal in doing the work. In how
many days will Bimal do that wok?
(CGL-2002)
(a)
14 days
(b) 12 days
(c)10
days (d) 10 1/2 days
Solution: (c) Efficiency and
time taken are inversely proportional Bimal : Kamal = 150 : 100 (work) 200 :
150 (Time) = 2 : 3
∵ 3 units = 15 days
2
units ----- 15/3×2 days = 10 days
Type VI Based on
Formula
1. If p men working p hours
per day for p days produce p units of work, then the units of work produced by
n men working n hours a day for n days is
(CGL-2008)
(a)
p^2/n^2
(b) p^3/n^2
(c)
n^2/p^2
(d) n^3/p^2
Solution: (d) ∵ P
men working P hours /day for P days produce P units of work.
1
man working 1 hour /day for 1 day produce
P/P^3
=1/P^2 units of work n men working n hours a day for n day’s
produce n^3/P^2 units of work
2. One man, 3 women and 4 boys
can do a piece of work in 96 hours, 2 men and 8 boys can do it in 80 hours, 2
men and 3 women can do it in 120 hours. 5 men and 12 boys can do it in
(CGL-2012)
(a)
39 1/11 hours
(b) 42 7/11 hours
(c)43
7/11 hours
(d) 44 hours
Solution: (c) 1 hour’s work of
1 man and 4 boys = 1/160
1
hour’s work of 1 man and 3 women = 1/96
1
hour’s work of 3 women = 1/90-1/160=(10-6)/960=1/240
1
hour’s work of 2 men = 1/120-1/240=1/240
1
hour’s work of 4 boys =1/160-1/480
=
(3-1)/480=1/240
2
men = 3 women = 4 boys
2
men + 8 boys = 12 boys
5men
+ 12 boys = 22 boys
By
M1D1 = M2D2
D2
= (12×80)/22
=
480/11=437/11 hours