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Monday, April 21, 2014
MACP ON HIERARCHICAL BASIS INSTEAD OF NEXT GRADE PAY BASIS – HON’BLE DELHI HIGH COURT POSTPONES THE CASE TO MAY 16TH 2014
MACP ON HIERARCHICAL BASIS INSTEAD OF NEXT GRADE PAY BASIS – HON’BLE DELHI HIGH COURT POSTPONES THE CASE TO MAY 16TH 2014
It is informed by All India Association of Administrative Staff (Nongazetted) that the next hearing on appeal filed by the Government against the Principal CAT Order for granting MACP on hierarchical Basis instead of Next Grade Pay Basis has been postponed to 16th May 2014.
Earlier in O.A. No.864/2014 dated 12.03.2014 CAT Principal Bench New Delhi had decided that MACP has to be granted in the nex promotional scale.
It is also opined by General Secretary, aiamshq (All India Association of Administrative Staff – Non gazetted ) to the extent that If the Hon’ble supreme court accept the decision of Principal Bench CAT Delhi in its merits for granting MACP in promotion hierarchy it will be implemented to all Central Government Employees.
BJP wants to raise Income Tax slab to Rs 5 lakh
BJP leader Arun Jaitley, the man tipped to be India’s next finance Minister says he wants raise in the Income Tax ceiling from Rs 2 lakh to Rs 5 lakh, which he claimed will benefit thirty million people.
Jaitley claimed that the Vajpayee government kept the rate of interest at 7 to 8 per cent which Congressled UPA raised to 1314 per cent.
“Due to this, Trade and Industry are running under loss and closing down, and production has become costly,” he said, adding, that is why country like China and Thailand are moving ahead of India.
Jaitley said,”The savings in the pockets of ordinary person by reducing ceiling on Income Tax will lead to increased purchase which in turn will lead to increased VAT and Excise Duty thereby increasing the Revenue“.
Speaking on Amritsar’s development, Jaitley said Economy of the constituency is dependent on tourists and small businesses.
He emphasised that with the investment of Rs 300400 crore, 23 elevated roads, better roads with construction of 1012 new bridges together with provision for few concessions and grants to increase tourism by providing special tax concessions to hotels the economy of Amritsar can be strengthened.
Along with this special attraction of Golden Temple, Durgiyana Temple, Wagha Border, improvement in specialised foods of Amritsar will attract more tourists and all round efforts will be made to increase their numbers from 1 lakh per day to 23 lakh per day.
Jaitley said, “I will take care of each and every problem of Guru Ki Nagri and ensure that release of funds from Delhi is not a problem for the development of Amritsar”.
Inputs with PTI
Friday, April 18, 2014
Excess pay given due to wrong pay fixation shall not be recovered from the employee
High Court today said that if excess pay was granted to an employee due to erroneous pay fixation done by the department and not due to any misrepresentation by him, that amount shall not be recovered from employee from the retiral benefits, that too after retirement.
He directed the authorities not to make any recovery on the ground of any wrong fixation done during the service, as it was not the petitioner’s mistake.
The judge also directed that the amount be refunded if any already recovered from the petitioner within six weeks.
Razack submitted he was appointed GradeII constable on June 1, 1971. After two promotions, he was listed in ‘C’ list of head constables fit for promotion to SubInspector’s post on Sep 22, 1985. As there was no vacancy, he was not promoted at that time.
Later, he was promoted temporarily as SI on September 15, 1987 and also paid salary by fixing his pay to the post. He was regularised in the post with effect from August 17 1992, promoted as Inspector on October 15,2003 and retired from service on January 31, 2009.
When the pension proposals were sent, the Accountant General opined that the benefit of fixation was given from September 15, 1987 and should be given only from the date of regularisation of service (August 7, 1992).
The Police Commissioner then passed an order to recover Rs 36,312 from the DCRG of Razack towards alleged excess payment due to the fixation.
Citing a judgment of a division bench, which held such recovery was bad, the judge said he was also of the view that the judgement applies to the facts of this particular case.
“Any wrong fixation that was said to have been made in 1987 shall not be sought to be recovered, after retirement in 2009, more particularly, when it is not the case of the authorities that the wrong fixation was done at the instance of the petitioner by way of misrepresentation.”
“Even, if there was any error, the petitioner cannot be made to suffer after retirement, by way of recovery,” the judge said.
PTI
Thursday, April 17, 2014
POSTAL ASSISTANT EXAM MATERIAL QUANTITATIVE APTITUDETIME & WORK
Disclaimer: All the Information provided in this post are prepared & compiled by A. Praveen Kumar, SPM, Papannapet SO502303, Telangana State for in good faith of Postal Assistant Exam Aspirants. Author of blog does not accepts any responsibility in relation to the accuracy, completeness, usefulness or otherwise, of 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 M_{1 }persons can do W_{1} work in D_{1} days and M_{2} persons can do W_{2} work in D_{2} days, then
9. If M_{1 }persons can do W_{1} work in D_{1} days for h_{1} hours and M_{2} persons can do W_{2} work in D_{2} days for h_{2} 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
N_{1}H_{1}D_{1}E_{1}W_{2}_{ }= N_{2}H_{2}D_{2}E_{2}W_{1}
Where:
N_{1} and N_{2} = number of person
H_{1} and H_{2} = Hours worked by per person per day (assumed constant)
D_{1} and D_{2} = days
E_{1} and E_{2} = Efficiency
W_{1} and W_{2}= 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 –
N_{1}H_{1}D_{1}E_{1}W_{2}_{ }= N_{2}H_{2}D_{2}E_{2}W_{1}
16*12*8*1*3 = N_{2}*8*24*0.5*1
N_{2} = (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 –
N_{1}H_{1}D_{1}W_{2}_{ }= N_{2}H_{2}D_{2}W_{1}
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)/(xy)
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(Dd)/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: 1/10^{th} 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
= 1/15^{th} 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, 1/4^{th} 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
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 threefourth 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 dÃ¡ys 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 dÃ¡ys.
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 dÃ¡ys.
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

= (1/31/5)

=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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