This material
prepared and compiled by Akula. Praveen Kumar, SPM, Papannapet Sub Office-502
303, Medak Division, Andhra Pradesh (9849636361, 8019549939)
Disclaimer:-
All questions/Information provided in this post are Compiled by A. Praveen
Kumar for in good faith of Departmental Employees. The types of questions,
number of questions and standard of questions may be vary in actual
examination. This is my predictions only. Author of blog does not accepts any
responsibility in relation to the accuracy, completeness, usefulness or
otherwise, of the contents.
Time and
Work
1. Work from Days:
|
If A can do a piece of work in n
days, then A's 1 day's work =
|
1
|
.
|
|
n
|
2. Days from Work:
|
If A's 1 day's work =
|
1
|
,
|
then A can finish the work in n
days.
|
|
n
|
3. Ratio:
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.
4. No. of days
= total work / work done in 1 day
5. Relationship
between Men and Work
More men
------- can do ------->
More work
Less men
------- can do
------->
Less work
6. Relationship
between Work and Time
More work
-------- takes------>
More Time
Less work
-------- takes------>
Less Time
7. Relationship
between Men and Time
More men
------- can do in
-------> Less Time
Less men
------- can do in
-------> More Time
8. If M1 persons can do W1
work in D1 days and M2 persons can do W2 work
in D2 days, then
9. 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
Note: If works are same, then M1D1h1 = M2D2h2
10. 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
11. If A can do a work in ‘x’ days and A
+ 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
I think the single most useful formula for the topic Time and Work is
N1H1D1E1W2 = N2H2D2E2W1
Where:
N1 and N2 = number of
person
H1 and H2 = Hours
worked by per person per day (assumed constant)
D1 and D2 = days
E1 and E2 = Efficiency
W1 and W2= Amount of work done
1. A piece of work can be done by 16
men in 8 days working 12 hours a day. How many men are needed to complete
another work, which is three times the first one,in 24 days working 8 hours a
day. The efficiency of the second group is half that of the first group?
Solution –
N1H1D1E1W2 = N2H2D2E2W1
16*12*8*1*3 = N2*8*24*0.5*1
N2 = (16*12*8*1*3)/ (8*24*0.5*1) = 48
So number of men required is 48.
Note – you can remove anything from
formula is not given in the question. For example if the question would have
been –
2.“A piece of work can be done by 16 men
in 8 days working 12 hours a day.How many men are needed to complete another
work, which is three times the first one,in 24 days working 8 hours a day.”
The applicable formula would have
been –
N1H1D1W2 = N2H2D2W1
Since nothing is mentioned about
efficiency, we remove it from both sides.
1. Time taken to do a certain job by a certain number of workers
2. The change in the number of hours required to do a job if the number
of workers is changed.
3. The number of hours required to do the job by different workers
if their speeds are different.
4. Problems on wages earned by various workers executing a certain work
in proportion with the amount of work done by each worker
5. Work done by various people in alternate intervals
6. Time taken by pipes/taps/leakages to fill/empty tanks/cisterns.
There are certain points that are to be kept in mind while doing
problems. They are:-
1. A person does the same amount of work every day(unless specified in
the problem).
Note
a. If A can do a piece of work in ‘n’ days, then the amount of work done
by A in one day is 1/n.
b. Conversely, if A’s 1 day’s work=1/n, then A can finish the work in n
days.
2. If there are more than one person involved carrying out the work
collectively, it is assumed that the working capacity of each person is the
same unless specified in the problem.
In general we can say that if w1 work is done by m1 men working h1 hours
per day in d1 days and w2 work is done by m2 men working h2 hours per day
in d2 days then
m1d1h1w2=m2d2h2w1
Generally, the following types of question are asked in examinations and
here are few time work problems shortcuts
Types of Questions and usage of formulas
METHOD 1
M1 persons do a work in D1 days and M2 persons do the
same work in D2 days then
we have the equation
M1*d1= M2* D2
METHOD 2
M1 persons can do W1 work in D1 days and M2
persons can do W2 work in D2 days then
we have (M1 D1)/W1 = (M2 D2)/W2
METHOD 3
if A can do a work in x days, B can do
a work in y days then
A and B together (A+B) can do the same work in ( xy
)/(x+y) days
METHOD 4
A alone can do a work in x days , A+B can do a work in y days then .
B alone can do the work in (xy)/(x-y)
METHOD 5
here A+B means A and B working together
if A working alone takes x days more than (A+B) and B
working alone takes y days more than (A+B) then
the number of days taken by A and B together is given by
root of (xy)
METHOD 6
here A+B+C means A B C working together
A can do a work in x days , B can do
the same work in y days , C can also do the same work
in z days then
A+B+C can do the work in (xyz)/(xy+yz+zx) days
METHOD 7
here A+B+C means A B C working together
A+B can do a work in x days , B+C can do the work
in y days, C+A can do the work in z days then
A+B+C can do the work in (2xyz
)/xy+yz+zx days
METHOD 8
if x1 men or y1 women can reap the field in D days
,then
x2 men and y2 women take to reap it in D(x1y1)/(x2
y1+x1 y2) days
METHOD 9
a certain men can do a work in D days if there
are x men less it could be finished in d days more, then
the no of men orginaly
are x(D+d)/d
METHOD 10
a certain men can do a work in D days if there
are x men more it could be finished in d days more,then
the no of men originally
are x(D-d)/d
METHOD 11
if x1 men or x2 women or x3
boys can do a work in D days then
1 men + 1 women + 1 boy
can do the same work in d(x1x2x3)/(x1x2+x2x3+x3x1)
1. A can do a
work in 10 days. What is A’s 1 day work?
Sol: 1/10th of work
2. Two people can do a work in 10 days. What
is 1 day 1 man work?
Sol: Since it is given as 2
people can do a work in 10 days, total work is 10 * 2 = 20 units of work.
Therefore,
one day one man work = 1/total work = 1/20 units of work
3. Three men
can do a work in 15 days. One man can do a work in how many days?
Sol: Since it is given as 3
people can do a work in 15 days, total work is 15 * 3 = 45 units of work.
Therefore,
one man can complete the work in 45 days
4. One man
one day work is 1/25. In how many days, can he finish the work?
Sol: Number of days required = 1/(1/25) = 25 days
5. A can work on 1km railway track in 1 day. In how many days, will he
able to complete the work on 12km railway track?
Sol: no. of days = total work / work done in 1 day
Therefore,
no. of days taken = 12/1 = 12 days
6. A can
complete the work in 15 days. What fraction of work will be completed in 1 day?
Sol: Let the total work is 1 unit.
Work in 1day = total work/no.
of days to complete
= 1/15th of work
7.A can do a
piece of work in 3 days and B can do a piece of work in 5 days. In how many
days will the work be completed if both A and B work together?
Sol:
Work done by A in 1 day = 1/3
Work done by B in 1 day = 1/5
Total work done by A and B in 1 day = 1/3 + 1/5 = 8/15
Therefore, no. of days to complete work by A and B
together = 1/(Total work) = 1/(8/15) = 15/8 days which is less than 3 and 5
Using shortcut
Let us consider the total work be 15
units (LCM of 3 and 5)
So work done by A in 1 day = 15/3 = 5 units
Similarly work done by B in 1 day = 15/5 = 3 units
So total work done by A and B in 1 day = 5 + 3 = 8 units
Therefore, no. of days to
complete total work i.e. 15 units = total work/work done in 1 day = 15/8 days
Note:
a. Work done by A and B in 1 day will
always be greater than that of A and B individually
b.
No. of days taken by A and B together will
always be less than that of A and B individually
8.A can do a
piece of work in 6 days, B can do a piece of work in 4 days and C can do a
piece of work in 12 days. Find the no. of days to complete the work if A, B and
C work together?
Sol:
Work done by A in 1 day = 1/6
Work done by B in 1 day = ¼
Work done by C in 1 day = 1/12
Total work done by A, B and C in 1 day = 1/6 + ¼ + 1/12 =
12/24 = 1/2
Therefore, no. of days to complete work by A, B and C
together = 1/(Total work) = 1/(1/2) = 2 days which is less than 4, 6, 12
Using
shortcut
Let us consider the total work be 24
units (LCM of 4, 6, 12)
So work done by A in 1 day = 24/4 = 6 units
work done by B in 1 day = 24/6 = 4 units
work done by C in 1 day = 24/12 = 2 units
So total work done by A, B and C in 1 day = 6 + 4 + 2 =
12 units
Therefore, no. of days to
complete total work i.e. 24 units = total work/work done in 1 day = 24/12 = 2
days
9.A can do a piece of work in 6 days
and B can do a piece of work in 12. Find the no. of days to complete the work
if A and B work alternatively?
Sol:
Work done by A in 1 day = 1/6
Work done by B in 1 day = 1/12
Total work done by A and B working 1 day each = 1/6 +
1/12 = 3/12 = ¼
Therefore, 1/4th of work is done in 2days.
No. of days to complete total
work if A and B work alternatively = 1/((1/4)/2) = 8 days
Using
shortcut
Let us consider the total work as 12
units (LCM of 6, 12)
So work done by A in 1 day = 12/6 = 2 units
work done by B in 1 day = 12/12 = 1 unit
Total work done by A and B working 1 day each = 2 + 1 = 3
units in 2 days
Therefore, work done in 1 day = work/no. of days = 3/2
units
No. of
days to complete work = total work/work in 1 day = 12/(3/2) = 8 days
10.30 men can complete a job in 40 days.
Then 25 men can complete the same job in how many days?
Sol: As per the formula M1D1 =
M2D2
30 * 40 = 25 * X = 30 * 40/25 = 48 days
11.30 men can complete 1500 units in 24
days working 6hrs a day. In how many days can 18 men can complete 1800 units
working 8 hrs a day?
Sol: As per the formula
M1D1h1/W1 = M2D2h2/W2
=> 30*24*6/1500 = 18*x*8/1800
=> x = 36 days
12.
A and B can do a work in 10 and 15 days respectively. Then combined A
& B, in how many days the work will be completed?
Sol:
As per the formula x*y/(x + y)
A and B together can complete the work in 10 * 15/(10 + 15) = 6 days
A and B together can complete the work in 10 * 15/(10 + 15) = 6 days
13.
A can do a work in 10 and, A and B together can do a work in 6 days. In how
many days B can complete the same work?
Sol:
As per the formula x*y/(x - y)
B alone can complete the work in 10 * 6/(10 - 6) = 15 days
B alone can complete the work in 10 * 6/(10 - 6) = 15 days
14. A is twice faster than B and B can complete in 12 days alone.
Find the number of days to complete if A and B together work?
Sol: Given B works in 12 days
A is twice faster than B => A takes 2 times less time
than B
Therefore, A completes work in 12/2 = 6 days
A and B
together can complete in 12 * 6/(12 + 6) = 4 days
15. If 10 men or 18 boys can do a piece of work in 15 days, then
25 men and 15 boys together will do twice the work in:
Sol:10 men = 18boy hence 1 man = 18/10
boys
25 men + 15 boys = (25 * 18/10) + 15 = 60
now more work more days
more boys less days1 * 60 * x = 2*18*15 or x = (2*18*15)/60 = 9 days
25 men + 15 boys = (25 * 18/10) + 15 = 60
now more work more days
more boys less days1 * 60 * x = 2*18*15 or x = (2*18*15)/60 = 9 days
16.A is thrice as good a workman as B and takes 10 days less to
do a piece of work than B takes. B alone can do the whole work in
Sol: Ratio of times taken by A and B =
1:3.
If difference of time is 2 days, B takes 3 days.
If difference of time is 10 days, B takes (3/2) * 10 =15 days.
If difference of time is 2 days, B takes 3 days.
If difference of time is 10 days, B takes (3/2) * 10 =15 days.
Sol: Let 1 mans 1 days work = x and 1
womens a days work = y
then 10x = 1/20 and 20y = 1/15 so x = 1/200 and y = 1/300
(5 men + 10 women)s 1 days work = (5/200) + 10/300 = 7/120.
So, they will finish the work in 120/7 = 17 and 1/7 days.
then 10x = 1/20 and 20y = 1/15 so x = 1/200 and y = 1/300
(5 men + 10 women)s 1 days work = (5/200) + 10/300 = 7/120.
So, they will finish the work in 120/7 = 17 and 1/7 days.
17.A is twice as good a workman as B and ‘together they complete a
work in 15 days. In how many days can the work be completed by B alone
(As 1 days work):(Bs 1 days work) = 2:
1.
(A+B)s 1 day’s work= 1/15.
Divide 1/15 in the ratio 2 : 1
Bs 1 days work = (1/15) * (1/3) = 1/45
B alone can finish the work in 45 days.
(A+B)s 1 day’s work= 1/15.
Divide 1/15 in the ratio 2 : 1
Bs 1 days work = (1/15) * (1/3) = 1/45
B alone can finish the work in 45 days.
18. If Ramesh, Suresh and Harish can do a piece of work in
15 days, 10 days and 6 days respectiVelY how long will they take to do it, if
all the three work at it together ?
(Ramesh + Suresh + Harish)S 1 days work =
[(1/15) + (1/10) + (1/6)] = 10/3 =1/3
So, all the three will finish the work in 3 days.
So, all the three will finish the work in 3 days.
19.A and B can do a piece of
work in 72 days, B and C can do it in 120 days. A and C can do it in 90 days.
In what time can A alone do it?.
Sol: (A+B)s 1 days work = 1/72,
(B+C)s 1 days work = 1/120,
(A+C)s 1 days work= 1/90.
Adding 2(A+B+C)s 1 days work [(1/72) + (1/120) + (1/90)] = 12/360 = 1/30
(A+B+c)s 1 day’s work = 1/60.
As 1 days work = (1/60) - (1/120) = 1/120
A alone can fini the work in 120 days.
(B+C)s 1 days work = 1/120,
(A+C)s 1 days work= 1/90.
Adding 2(A+B+C)s 1 days work [(1/72) + (1/120) + (1/90)] = 12/360 = 1/30
(A+B+c)s 1 day’s work = 1/60.
As 1 days work = (1/60) - (1/120) = 1/120
A alone can fini the work in 120 days.
20. A does half as much work as B in three-fourth of the time. If
together they take 18 days to complete the work, how much time shall B take to
do
Sol: Suppose B takes x dáys to do the
work.
A takes (2*(3/4)*x) = 3x/2 days to do it.
(A+B)s 1 days work= 1/18
1/x + 2/3x = 1/18 or x = 30.
A takes (2*(3/4)*x) = 3x/2 days to do it.
(A+B)s 1 days work= 1/18
1/x + 2/3x = 1/18 or x = 30.
21.45 men
can complete a work in 16 days. Six days after they started working, 30 more
men joined them. How many days will they now take to complete the remaining
work?.
Sol: (45 * 16) men can complete the work in
1 day.
1 mans 1 days work = 1/720
45 mens 6 days work; = (1/16) * 6 = 3 / 8
Remaining work = (1 - (3/8)) = 5/8
75 mens 1 days work = 75/720 = 5/48
Now 5/48 work is done by them in 1 day.
5/8 work is done by them in (48/5) * (5x8) = 6 day
1 mans 1 days work = 1/720
45 mens 6 days work; = (1/16) * 6 = 3 / 8
Remaining work = (1 - (3/8)) = 5/8
75 mens 1 days work = 75/720 = 5/48
Now 5/48 work is done by them in 1 day.
5/8 work is done by them in (48/5) * (5x8) = 6 day
22. A
and B can do a piece of work in 45 days and 40 days respectively. They began to
do the work together but A leaves after some days and then B completed the
remaining work in 23 days. The number of days after which A left the work
was
Sol: (A+B)s 1 days work = (1/45) + (1/40) = 17/360.
Work done by B in 23 days =(1/40) * 23 =.23/40
remaining work = (1 — 23/40) = 17/40 .
Now, work was done by (A + B) in 1 day
17 /40 work was done by (A + B) in (1 * (360/17) * (17/40))= 9 days.
A left after 9 dáys.
Work done by B in 23 days =(1/40) * 23 =.23/40
remaining work = (1 — 23/40) = 17/40 .
Now, work was done by (A + B) in 1 day
17 /40 work was done by (A + B) in (1 * (360/17) * (17/40))= 9 days.
A left after 9 dáys.
23.4 men and 6 women finish a job in 8 days, while 3 men and 7
women finish it in 10 days. In how many days will 10 women working together
finish it?
Sol: Let 1 mans 1 days work = x and 1
womans 1 days work = y.
Then, 4x + 6y = 1/8 and 3x+7y = 1/10
solving, we get y = 1/400
10 womens 1 day work = 10/400 = 1/40
10 women will finish the work in 40 days.
Then, 4x + 6y = 1/8 and 3x+7y = 1/10
solving, we get y = 1/400
10 womens 1 day work = 10/400 = 1/40
10 women will finish the work in 40 days.
24.12
children take 16 days to complete a work which can be completed by 8 adults in
12 days. 16 adults started working and after 3 days 10 adults left and 4
children joined them. How many days will it take them to complete the remaining
work?
Sol: Let 1 child’s 1 day’s work = x and 1
adults 1 days work = y.
Then, 12x = 1/16 or x = 1/192 and 8y= 1/12 or Y = 1/196. Work done in 3 days = (16 * (1/96)*3)= 1/2
Work left = (1 - (1/2)) = 1/2
(6 adults + 4 children)s 1 days work = (6/96) + (4/192) = 1/12
1/12 work is done by them in 1 day.
1/2 work is done by them in (12 * (1/2)) = 6 days.
Then, 12x = 1/16 or x = 1/192 and 8y= 1/12 or Y = 1/196. Work done in 3 days = (16 * (1/96)*3)= 1/2
Work left = (1 - (1/2)) = 1/2
(6 adults + 4 children)s 1 days work = (6/96) + (4/192) = 1/12
1/12 work is done by them in 1 day.
1/2 work is done by them in (12 * (1/2)) = 6 days.
25. Twelve
men can complete a work in 8 days. Three days after they started the work, 3
more men joined. In how many days will all of them together complete the
remaining work?
Sol: 1 mans 1 days = 1/96
12 men’s 3 days work = (1/8) * 3 = 3/8.
Remaining work 1 — (3/8) = 5/8 15 mens 1 days work = 15/96
Now, 15/96 work is done by them in 1 day
5/8 work is done by them in (96/15) * (5/8) = 4 days
12 men’s 3 days work = (1/8) * 3 = 3/8.
Remaining work 1 — (3/8) = 5/8 15 mens 1 days work = 15/96
Now, 15/96 work is done by them in 1 day
5/8 work is done by them in (96/15) * (5/8) = 4 days
26. A can do of the work in 5 days and B can do of the work in
10 days. In how many days both A and B together can do the work?
Sol:
Whole work will be done by A in (5 x 3) = 15 days.
Whole work will be done by B m (10 * (5/2)) = 25 days.
As 1 day’s work = 1/15 and Bs 1 days work = 1/25
(A+B)s 1 daYs work=((1/15) + (1/25) ) = 8/75
So, both together can finish it in 75/8 = 9 and 3/8 days days
Whole work will be done by B m (10 * (5/2)) = 25 days.
As 1 day’s work = 1/15 and Bs 1 days work = 1/25
(A+B)s 1 daYs work=((1/15) + (1/25) ) = 8/75
So, both together can finish it in 75/8 = 9 and 3/8 days days
27.A and B can do a Piece of work in 5 days ;B and c can do it in 7
day; A and C can do it in 4 days. Who among these will take the least time if
put to do it alone?
Sol:(A+B)s 1 days work = 1/4
and (A+C)’s 1 day’s work = 1/4
2(A+B+C)s 1 days work = (1/5 + 1/7 + 1/4) = 83/140
(A+B+C)s 1 days work = 83/280
Cs 1 days work = (83/280 - 1/5) = 27/280
As 1 days work = (83/280 - 1/7) = 43/280
Bs 1 days work = (83/280 - 1/4) = 13/280
Thus time taken by A, B, C is 280/43 days, 280/13 days, 280/27 days respectively.
Clearly, the time taken by A is least.
and (A+C)’s 1 day’s work = 1/4
2(A+B+C)s 1 days work = (1/5 + 1/7 + 1/4) = 83/140
(A+B+C)s 1 days work = 83/280
Cs 1 days work = (83/280 - 1/5) = 27/280
As 1 days work = (83/280 - 1/7) = 43/280
Bs 1 days work = (83/280 - 1/4) = 13/280
Thus time taken by A, B, C is 280/43 days, 280/13 days, 280/27 days respectively.
Clearly, the time taken by A is least.
28. A does half as much work as B and C does half as much work
as A and B together. If C alone can finish the work in 40 days, then together
all will finish the work in:
Sol:
C alone can finish the work in 40 days.
(A + B)can do it in 20 days
(A + B)s 1 days wok = 1/20.
As 1 days work : Bs 1 days Work = 1/2 : 1 = 1:2.
A’s 1 day’s work = (1/20) * (1/3) = (1/60). [Divide 1/20 in the raio 1:2] Bs 1 days work = (1/20) * (2/3) = 1/30
(A+B+c)S 1 day’s work = (1/60) + (1/30) + (1/40) = 9/120 = 3/40
All the three together will finish it in 40/3 = 13 and 1/3 days.
(A + B)can do it in 20 days
(A + B)s 1 days wok = 1/20.
As 1 days work : Bs 1 days Work = 1/2 : 1 = 1:2.
A’s 1 day’s work = (1/20) * (1/3) = (1/60). [Divide 1/20 in the raio 1:2] Bs 1 days work = (1/20) * (2/3) = 1/30
(A+B+c)S 1 day’s work = (1/60) + (1/30) + (1/40) = 9/120 = 3/40
All the three together will finish it in 40/3 = 13 and 1/3 days.
29. A, B and
C can do a piece of work in 11 days, 20 days and 55 days respectively, working
alone. How soon can the work be done if A is assisted by B and C on alternate
days?
Sol:
(A+B)s 1 days Work = (1/11) + (1/20) = 31/220
(A+C)S 1 days work = (1/11) + (1/55) = 6/55
Work done in 2 days = (31/220) + (6/55) = 55/220 = 1/4
Now, 1/4 work is done in 2 days.
Whole work will be done in (4 x 2) = 8 days.
(A+C)S 1 days work = (1/11) + (1/55) = 6/55
Work done in 2 days = (31/220) + (6/55) = 55/220 = 1/4
Now, 1/4 work is done in 2 days.
Whole work will be done in (4 x 2) = 8 days.
29. A can do
a piece of work in 80 days. He works at it for 10 days and then B alone
finishes the remaining work in 42 days. The two together could complete the
work in
Sol: Work done by A in 10 days = (1/80) *
10 = 1/8
Remaining work=(1 — (1/8)) = 7/8
Now, 7/8 work is done by B in 42 days
Whole work will be done by B in 42 * (8/7) = 48 days.
As 1 days work = 1/80 and Bs 1 days work = 1/48
(A + B)’s 1 days work = [(1/80) + (1/48)] = 8/240 = 1/30
so, both will finish the work in 30 days.
Remaining work=(1 — (1/8)) = 7/8
Now, 7/8 work is done by B in 42 days
Whole work will be done by B in 42 * (8/7) = 48 days.
As 1 days work = 1/80 and Bs 1 days work = 1/48
(A + B)’s 1 days work = [(1/80) + (1/48)] = 8/240 = 1/30
so, both will finish the work in 30 days.
30.A and B
can separately do a piece of work in 20 and 15 days respectively. They worked
together for 6 days, after which B was replaced by C. If the work was finished
in next 4 days, then the number of days in which C alone could do the work will
be
Sol: (A + B)s 6 days work = 6 [(1/20) +
(1/15)] = 7/20
(A+C)s 4days work = (3/10)*(A+C)s 1days work = 1/20 Cs 1 days work = (3/40) - (1/20) = 1/40
Hence, C alone can finish the work in 40 days.
(A+C)s 4days work = (3/10)*(A+C)s 1days work = 1/20 Cs 1 days work = (3/40) - (1/20) = 1/40
Hence, C alone can finish the work in 40 days.
31.A can do
a piece of work in 7 days of 9 hours each and B can do it in 6 days of 7 hours
each. How long will they take to do it, working together 8*(2/5) hours a day?
Sol:A can complete the work in (7 * 9) hrs
= 63 hrs.
B can complete the work in (6 * 7) hrs = 42 hrs.
As 1 hours work = 1/63 and Bs 1 hours work = 1/42
(A+B)s 1 hours work = [(1/63) + (1/42)] = 5/126
Both will finish the work in (126/5) hrs
Number of days of8*(2/5) hrs each = (126/5) * (5/42) = 3 days.
B can complete the work in (6 * 7) hrs = 42 hrs.
As 1 hours work = 1/63 and Bs 1 hours work = 1/42
(A+B)s 1 hours work = [(1/63) + (1/42)] = 5/126
Both will finish the work in (126/5) hrs
Number of days of8*(2/5) hrs each = (126/5) * (5/42) = 3 days.
32. A
father can do a job as fast as his two sons working together. If one son does
the job in 3 hours and the other in 6 hours, how many hours does it take the
father to do the job’?
Sol: Fathers 1 hours work= (1/3) + (1/6) =
1/2
Time taken by father to complete the work = 2 hours.
Time taken by father to complete the work = 2 hours.
33. A can do a piece of work in 15 days and B alone can do
it in 10 days. B works at it for 5 days and then leaves. A alone can finish the
remaining work in : .
Sol: Bs 5 days work = ((1/10) * 5) = 1/2
Remaining work = (1-(1/2)) = 1/2
A can do 1/2 work in 7and half days.
Remaining work = (1-(1/2)) = 1/2
A can do 1/2 work in 7and half days.
34. 12 men can complete a work in 18 days. Six days after they
started working, 4 men joined them. How many days will all of them take to
finish the remaining work?
Sol: (12 * 18) men can complete the work in
1 day.
1 mans 1 days work = 1/216
12 men’s 6 days work = (1/18)*6 = 1/3
remaining work = 1 - (1/3) = 2/3
16 mens 1 days work = 16/216 = 2/27
2/27 work is done by them in 1 day.
2/3 work is done by them in (27/2)*(2/3) = 9 days.
1 mans 1 days work = 1/216
12 men’s 6 days work = (1/18)*6 = 1/3
remaining work = 1 - (1/3) = 2/3
16 mens 1 days work = 16/216 = 2/27
2/27 work is done by them in 1 day.
2/3 work is done by them in (27/2)*(2/3) = 9 days.
35. 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:.
Sol: Work done by A and B in 20 days =
(1/30 * 20) = 2/3
Remaining work= ( 1 -2/3) = 1/3 Now,1/3 work is done by A in 20days
Whole work will be done by A in (20x3) = 60 days.
Remaining work= ( 1 -2/3) = 1/3 Now,1/3 work is done by A in 20days
Whole work will be done by A in (20x3) = 60 days.
36. A can do a piece of work in 30 days while B alone can do it
in 40 days. In how many days can A and B working together do it?
Sol:As l days work = 1/30 and Bs l days
work= 1/40
(A + B)s 1 days work = (1/30 + 1/40) = 7/120
Both together will finish the work in 120/7 = 17.14 days.
(A + B)s 1 days work = (1/30 + 1/40) = 7/120
Both together will finish the work in 120/7 = 17.14 days.
37. A piece of work can be done by 6 men and 5 women in 6 days
or 3 men and 4 women in 10 days. It can be done by 9 men and 15 women in
Sol: Let 1 mans 1 days work =x and 1 womans
1 days work = y. Then, 6x+5y = 1/6 , 3x+4y = 1/10
On solving, we get x = 1/54 and y = 1/90
(9 men + 15 women)s 1 days work = (9/54) + (15/90) = 1/3
They will finish the work in 3 days.
On solving, we get x = 1/54 and y = 1/90
(9 men + 15 women)s 1 days work = (9/54) + (15/90) = 1/3
They will finish the work in 3 days.
38. A can do
a certain job in 25 days which B alone can do in 20 days. A started the work
and was joined by B after 10 days. The number of days taken in completing the work
was
Sol: Work done by A in l0 days = (1/25) *10
= 2/5.
Remaining work = 1 - (2/5) = 3/5
(A+B)s 1 days work = (1/25) + (1/20) = 9/100
9/100 work is done by them in 1 day.
hence 3/5 work will be done by them in (100/9) * (3/5) = 20/3days.
Total time taken = (10 + 20/3) = 16 * (2/3) days.
Remaining work = 1 - (2/5) = 3/5
(A+B)s 1 days work = (1/25) + (1/20) = 9/100
9/100 work is done by them in 1 day.
hence 3/5 work will be done by them in (100/9) * (3/5) = 20/3days.
Total time taken = (10 + 20/3) = 16 * (2/3) days.
39.Ram can
do a piece of work in 8 days which Shyam can finish in 12 days. If they work at
it on alternate days with Ram beginning, in how many days, the work will be
finished?.
Sol: (Ram + Shyam)s 2 days work = (1/8) +
(1/12) = 5/24
Their 8 days work = (5/24) * 4 = 5/6
Their 8 days work = (5/6) + (1/8) = 23/24
Remaining work = (1 - (23/24))
Now it is Shyam’S turn.
1/12 work is done by him in 1 day.
1/24 work is done by him in (12 * (1/24)) = 1/2 day.
Total time taken = 9 and half days.
Their 8 days work = (5/24) * 4 = 5/6
Their 8 days work = (5/6) + (1/8) = 23/24
Remaining work = (1 - (23/24))
Now it is Shyam’S turn.
1/12 work is done by him in 1 day.
1/24 work is done by him in (12 * (1/24)) = 1/2 day.
Total time taken = 9 and half days.
40. A, B and
C are employed to do a piece of work for Rs. 529. A and C are supposed to
finish 19/23 of the work together. How much shall be paid to B ?
Sol: Work done by B = 1 - (19/23) =
4/23
(A + C) : B = (19/23) : (4/23) = 19:4
Bs share = RS.529*(4/23) = 92
(A + C) : B = (19/23) : (4/23) = 19:4
Bs share = RS.529*(4/23) = 92
41. A can do a certain job in 12 days. B is 60% more efficient
than A. The number of days, it takes B to do the same piece of work, is.
Ratio of times taken by A and B = 160:100 =
8:5
8 : 5 :: 12 : x or 8x = 5*12 or x = 7 and half days.
8 : 5 :: 12 : x or 8x = 5*12 or x = 7 and half days.
42. A and B
together can complete a piece of work in 35 days while A alone can complete the
same work in 60 days. B alone will be able to complete the same work in:
|
Sol:(A+B)s 1 days work = 1/35 and As 1
days work = 1/60
Bs 1 days work =(1/35 - 1/60) = 5/420 = 1/84 |
43.A man can do a piece of work in 5 days, but with the help of
his son, he can do it in 3 days. In what time can the son do it alone?
|
Son's 1 day's work
|
= (1/3-1/5)
|
|
=2/15.
|
|
|
The son alone can do the work in
|
= 15/2
|
|
=7x1/2 days.
|
44.Sakshi can do a piece of work in 20 days. Tanya is 25% more
efficient than Sakshi. The number of days taken by tanya to do the same piece
of work is
|
Ratio of times taken by Sakshi and
Tanya=125:100
|
||
|
= 5 : 4.
|
||
|
Suppose Tanya takes x days to do the
work.
|
||
|
5 : 4 ::20 : x
|
||
|
x<=> (4×20/5)
|
||
|
<=> 16 days.
|
45.Kim can do a work in 3 days while David can do the same work
in 2 days. Both of them finish the work together and get Rs.150. What is the
share of Kim?
|
Kim's wages : David's wages
|
= Kim's 1 day's work : David's 1 day's
work
|
|
= 1/3 : 1/2
|
|
|
= 2 : 3
|
|
|
Kim's share
|
= Rs.(2/5×150)
|
|
=Rs.60.
|
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