Work &Time,
Work
and Time
In most of the problems based on Work and Time, either the amount of time taken to finish a given job or
the amount of work done is to be calculated. Unless otherwise
specified, the amount of work done is generally taken as unity (1). Also, if it is
given that a person X can finish a job in D days, then it implies that X alone can do the job in D
days.
·
If a man can do a piece of work in N days (or hours or
any other unit of time), then the work done
by him in one day will be 1/N of the total work.
El. Animesh
can finish a piece of work by himself in 14 days. Then calculate the amount of work done by him in 1 day and 11
days.
Sol. The amount of work done by Animesh, working alone
in 14 days = 1 unit of work.
So, the amount of work done
by Animesh, working alone in 1 day = 1/14 unit of work and in 11 days = 11/14 units
of work.
·
If A
is twice as good a workman as B, then A will take half the time B takes to
finish a piece of work.
E2. Suresh can finish a piece of work by himself
in 42 days. Mahesh, who is 1/5 times more efficient as Suresh, requires X days to finish the work by working all by himself. Then what is the value of X?
Sol. Suresh, working alone 42 days = 1 unit of work.
Mahesh is 1/5 times more
efficient than Suresh. So Mahesh is
6/5 times as efficient as Suresh. Hence Mahesh should
require 5/6th of the time, the time taken by Suresh.
Therefore
time taken by Mahesh = 5/6 x 42 = 35 days.
Measurement of work
Work can be measured by
many units. Some of them are
Man-days
A certain number of men working for a
certain number of days.
For example, 5 men can finish a piece of work in
13 days, This implies that the work is of 5
x 13 = 65 Man-days, i.e., if one man alone
works to finish the given work then he is going to take 65 days to
finish it and if 65 men work together, then the work would be finished in 1
day.
Man-hours
A certain number
of men working for a certain number of hours.
For example, 3 men can
finish a piece of work in 13 hours. This implies that the work is of 3 x 13 = 39 Man-hours, i.e.,
if one man alone
works to finish the given work then he is going to take 39 hours to finish it and if 39 men work together,
then the work would be finished in 1 hour.
Some
others are, Man-minutes, Machine-hours etc.
E3. 4 men of equal
efficiencies can finish a piece of work by working 10 hrs a day in 7 days. How many men will be required to finish the same work if they work 3.5 hrs a day and
the work is to be completed in 10 days?
Sol. The total work to be
done = 4 x 10 x 7 = 280 man hours Let X men be required to finish this
job, then
280 = X x 3.5 x 10 .so, X =
8. So, 8 men are required.
If A and B
can do a piece of work in X & Y days respectively while working alone, they will
together take X x Y/X + Y days to complete it.
It can
be explained as follows:
A's one day
work = 1/X , B's one day work = 1/Y
(A + B)'s
total one day work = 1/X + 1/Y of the total.
Let the
total work be 1 (unity). Now1/ X+1/Y of 1 can be
finished in one day.
Total work can be finished in 1 / 1/X + 1/Y = X x Y / X+Ydays.
E4. A can finish a piece of work by
working alone in 12 days. B while working alone, can finish the same work in 24
days. If both of them work together, then in
how many days, the work will be finished?
Sol. Applying formula, 12 x 24/ (12 + 24) = 288/ 36 =
8.
The work will be finished in 8 days.
If A, B, C can do a piece of
work in X, Y, Z days respectively while
working alone, they will together take XxYxZ / XxY+YxZ+ZxX days to finish it.
It
can be explained as follows:
A's one day work =1/X , B's
one day work = 1/Y and C's one day work = 1/Z .
Their total
one day work = 1/X + 1/Y + 1/Z of the total work.
Let the total work be 1
(unity).
Now 1/X +1/ Y +1/ Z of 1 can be
finished in one day
Total work can be finished in
1 = XxYxZ
days.
1/X+1/Y+1/Z XxY+YxZ+ZxX X+Y+Z
E5. Ganpat , yogesh , bhagwat can do job
in 15, 20 and 30 days respectively. In how
many days can the job be finished if they work together?
Sol. Ganpat's 1 day work = 1/15 of the
total. Yogesh's 1 day work = 1/20 of the total. Bhagwat's 1 day work = 1/30 of
the total.
They can do (1/15+1/20+1/30) of the
total work in 1 day. Total
work can be finished in
= 20/3 days.
E6. A and B require 10 days to complete a
job. B and C require 12 days to complete the same job. A and C require 15 days to complete the same job. Find the number of days
required, if all are at work,
to complete the same job.
Sol.
A+B= 1/10 ,
B+C= 1/12 and C+A= 1/15
2(A+B+C)= 1/10+1/12+1/15 6 + 5 + 4 = 15/60
= 1/4
60
A + B + C = 1/8.
Hence if A, B and C all work together, they will need total 8
days to complete the work.
If A, B and C can finish a
piece of work in X, Y and Z days respectively, while working alone and together
they require M days to finish the work, then the amount of work done by A is
M/X, B is M/Y and C is M/Z.
Applying the concept of variation
Most of the questions given in
this chapter can be solved using the concept of variation. The following points
should be kept in mind while solving the questions by this method.
→Number of persons employed to
do the work is directly proportional to the amount of work done. (More the
number of persons employed, more the work done)
E7. 4 men take 3 days to paint a wall. If
20 men with same efficiencies are working,
then how many walls can be painted in 3 days?
Sol. Since 4 men can paint the wall in three days, so 20
men can paint 20/4 i.e. 5 such walls in 3 days.
The number of days is directly
proportional to the work done. (More the number of days for which a work was
done, more shall be the total amount of
work done)
E8. A man takes 3 days to paint a wall. If
he works for 12 days, then how many walls can be painted by him?
Sol. Since in 3 days he can paint the
wall, so in 12 days, he can paint 12/3 i.e. 4 such walls. The number of persons employed is
inversely proportional to the number of days reciuir~d to finish a work. (More
the number of persons employed, less will be the time required to finish the
work)
E9. A man takes 5 days to complete a job.
If he is assisted by 4 of his friends of
same efficiency, then in how many days can the job be completed?
Sol. Since the man
takes 5 days to complete the job, the job is of
5 man — days. When he is assisted by 4 of his friends, they can do the job (5
man — days) in one day.
E10.If 12 men can build a wall 100 m long,
3 m high and 0.5 m thick in 25 days. In how many days will 20 men build a wall
60 m long, 4 m high and 0;25 m thick?
Sol. Man Days Length Height Thickness
12 25 100 3 0.5
20 X
60 4
0.25
25 = 20 x 100 x 3 x
0.50
X 12
60 4 0.25
X=6.
The
concepts used to solve this question are
Men and days are inversely proportional
to each other.
Length,
height and thickness of the wall are directly proportional to the number of days required.
E11.A
contractor undertook to do a certain work in 75 days and employed 60 men to do
it. After 25 days he found that only 1/4th of the work was done. How many more men must be employed so that the work may be finished in time?
Sol. 60 men in 25 days
can do 1/4 of work. For 3/4th of work in 50 days,
men required=60 x (3/2) =
90.
Additional men = 90 —
60 = 30.
Some important formulae
ü
Two persons A and B, working together, can finish a piece of
work in M days. If A can finish the job in X days by working alone, then B will finish the job by working alone in
M x X / X - M days.
E12.
Ram and Shyam working together can finish
a work in 12 days. If Ram alone can do the
same in 20 days, then in how days, Shyam alone can complete the work?
Sol.
Let Shyam take x days to complete the work.
Work
done by both in 1 day = 1/20 + 1/X
Number of days required = 20 x X =
12 (given)
20+X
=20x-12x= 240=x=30.
So Shyam
alone can do the work in 30 days.
Short-cut: Using formula we can get the answer as
12 x 20/ 20-12= 30 days
If A can complete a/b part of a
work in X days, then c/d part of the
work will be done in
b x c x
X /a x d days.
E13.If A can
complete a work in 16 days, then in how many days can he complete 3/4 th of the work?
Sol. Using formula we get 1/1 x
3/4 x 16 = 12 days.
If A can finish a work in X days and B is K
times as efficient as A, then the time
required by both A and B working together to finish the job will be X/1+k days.
E14.Mayank can finish a work in 10 days.
Mahesh is 2 times as efficient as Mayank. If
they work together, in how many days will the work be finished?
Sol. Time
Tken by mayank to do the work = 10days
.•.
Time taken by Mahesh to do the work = 5 days. (He is twice as efficient)
1 day's
work = 1/10 +1/5 = 3/10
Time required to complete the job = 10/3 days.
Short-cut: Using
formula, we can get the answer as
10
= 10 days.
1+2
3
A is K times as good a worker as
B and takes X days less than B to finish the work. Then the amount of time
required
by A and B working together is k
x X / k2 - 1days.
E15. Ram is three times as good a worker as Rahim and takes 8
days less than Rahim to finish the work. Find the amount of time required by Ram and Rahim
working together, to do the work.
Sol. Let Rahim take x days to complete the
work and so Ram takes x — 8 days =x/3 days (Ram is 3 times efficient).
So, x = 12 and so x - 8 = 4.
Time
required to complete the work, working together
= 1
1/12 + ¼ =
12/4 = 3days
Short-cut:
Using
formula we get = 3 X 8 / 32 - 1= 3 days.
Concept of WORK EQUIVALENCE
In order to solve certain types of questions,
there is a very logical method that can be used. This is the
WORK EQUIVALENCE method which works on a simple premise — make
the LHS equal to the RHS on the basis of SAME UNITS in terms of
which work can be measured.
Let's take some examples.
E16.A group of soldiers can completely destroy an enemy bunker
in 7 days. However 12
soldiers fell ill. The remaining now can do
the job in 10 days. Find the original group strength.
Sol. Here, first of all, let us see how WORK can be defined.
It is obvious that work
can be measured as "destruction of the enemy bunkers."
In the first case, let us
say that there were S number of soldiers in
the group. So they had to work for 7 days for the work which we call W.
→ Sx7=W ...(1)
Now 12
fell ill and the remaining did the work in 10 days. Hence the new equation is
(S-12)x10=W ...(2)
Just compare
the two equations to get the answer
S x
7 = (S - 12) x 10 =7S=10S-120
=120=3S
= S = 40 soldiers.
Hence, there were 40 soldiers in the
group initially.
E17.A
group of 20 cows can
graze a field 3 acres in size in 10 days. How many cows can graze a field twice
as large in 8 days?
Sol.
Here,
first of all, let us see how WORK can be defined. It is obvious that
work can be measured as "acres grazed." In the first case, there were 20 cows
in the group.
They had to work
for 10 days to do the work which we call W (which =
3).
→20x10=3 ...(1)
Do
not be worried about the numerical values on either side. The point is that logically this
equation is consistent as the LHS indicates
"Cowdays" and the RHS indicates "Acres'; both of which
are correct ways of measuring work done.
Now the
field is twice as large. Hence the new equation is
= Cx8=6 ...(2)
Just divide (2) by (1) to get the answer.
8C / 200
= 6 / 3
8C=2x200
C = 400/ 8 = 50 cows.
Hence, there were 50 cows in the second group.
3/4 1/2
The expression for emptying pipe which
can empty el part of
cistern in tl min and e2 part
in t2 min is tl
/el = t2/e2.
E25.An empty pipe can empty 5/6 of the
cistern in 20 min. What part of the cistern will be emptied in 9 min?
Sol. Using the formula,
t1
= t2 = 20 = 9 = e2 = 3
e1 e2 5/6
e2 8
in 9 min, 3/8 part of the cistern will be
emptied
If two pipes A and B are opened for different time periods, then to
solve the questions we use
A's amount of work + B's amount
of work = 1.
A's open time + B's open time =1
Alone fill time of A Alone fill time
for B
E26.Two pipes A and B
can fill a cistern in 10 and 15 min respectively. Both pipes are opened together but at the end of 3 min, 'B' is turned off. How much
time will the cistern take to fill?
Sol. Let the cistern be filled in T min.
.•. Pipe A is opened for T min and pipe B
is opened for 3 min. Using the formula, for filling the
cistern,
A's amount of work + B's amount of work = 1.
T/10
+ 3/15 = 1 , T = 10 (1 – 1/5) = 8.
The cistern is filled in 8 min.
E27.Two fill pipes A and B can fill a cistern in
12 and 16 min respectively. Both 'fill
pipes' are opened together but 4 min before the cistern is full, one
pipe A is closed. How much time will the cistern take to fill?
Sol. Let
the cistern be filled in T min.
.•. Pipe B is opened for T min and pipe A
is opened for (T- 4) min.
Using the formula, for filling the
cistern
A's
amount of work + B's amount of work = 1.
A's open time + B's open time =1
Alone fill time of A Alone fill time for B
Alone fill time of A Alone fill time for B
The cistern is filled in 91/7 min.
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