If A does a work in a days, then in one day A does \( \frac{1}{a} \) of the work. If B does a work in b days, then in one day B does \( \frac{1}{b} \) of the work. Then, in one day, if A and B work together, then their combined work is \( \frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab} \)
For example, if A can do a work in 10 days and B can do the same work in 12 days, then the work will be completed in how many days. One day’s work = 1/10 + 1/12 = (12 + 10)/120. Then the number of days required to complete the work is 120/22.
Instead of taking the value of the total work as 1 unit of work, we can also look at the total work as 100 per cent work. In such a case, the following rule applies: If A does a work in 'a' days, then in one day A does \( \frac{100}{a} \)% of the work. If B does a work in 'b' days, then in one day B does \( \frac{100}{b} \)% of the work. Then, in one day, if A and B work together, then their combined work is \( \frac{100}{a} + \frac{100}{b} \).
Q. If A can do a work in 10 days and B can do the same work in 12 days, then the work will be completed in how many days
If A can do a work in 10 days (so means 10% work) and B can do the same work in 12 days (so 8.33% work so 18.33% work in a day in 5 days 91.66% work so leaves 8.33% work to be done so which can be done in 8.33/18.33 of a day = 5/11 of a day (since both the numerator and the denominator are divisible by 1.66), then the work will be completed in 5 5/11 days.
The Concept of Negative Work
Suppose, that A and B are working to build a wall while C is working to break the wall. In such a case, the wall is being built by A and B while it is being broken by C. Here, if we consider the work as the building of the wall, we can say that C is doing negative work
A can build a wall in 10 days and B can build it in 5 days, while C can completely destroy the wall in 20 days. If they start working at the same time, in how many days will the work be completed
The net combined work per day here is: A’s work + B’s work – C’s work = 10% + 20% – 5% = 25% work in one day
Hence, the work will get completed (100% work) in 4 days
WORK EQUIVALENCE METHOD : Work rate × Time = Work done (or work to be done)
A contractor estimates that he will finish the road construction project in 100 days by employing 50 men. However, at the end of the 50th day, when as per his estimation half the work should have been completed, he finds that only 40% of his work is done. (a) How many more days will be required to complete the work? (b) How many more men should he employ in order to complete the work in time?
The contactor has completed 40% of the work in 50 days. If the number of men working on the project remains constant, the rate of work also remains constant. Hence, to complete 100% work, he will have to complete the remaining 60% of the work. For this he would require 75 more days
In order to complete the work on time, it is obvious that he will have to increase the number of men working on the project
50 men working for 50 days then 50 × 50 = 2500 man-days
2500 man-days has resulted in 40% work completion. Hence, the total work to be done in terms of the number of man-days is got by using unitary method: Work left = 60% = 2500 × 1.5 = 3750 man-days
This has to be completed in 50 days. Hence, the number of men required per day is 3750/50 = 75 men. Since, 50 men are already working on the project, the contractor needs to hire 25 more men
Work as volume of work
In certain cases, the unit of work can also be considered to be in terms of the volume of work. For example, building of a wall of a certain length, breadth and height.
In such cases, the following formula applies:
\( \frac{L_1 * B_1 * H_1}{L_2 * B_2 * H_2} = \frac{m_1 * t_1 * d_1}{m_2 * t_2 * d_2} \)
where L, B and H are respectively the length, breadth and height of the wall to be built, while m, t and d are respectively the number of men, the amount of time per day and the number of days. Further, the suffix 1 is for the first work situation, while the suffix 2 is for the second work situation
Q. 20 men working 8 hours a day can completely build a wall of length 200 meters, breadth 10 metres and height 20 metres in 10 days. How many days will 25 men working 12 hours a day require to build a wall of length 400 meters, breadth 10 metres and height of 15 metres.
\( \frac{L_1 * B_1 * H_1}{L_2 * B_2 * H_2} = \frac{m_1 * t_1 * d_1}{m_2 * t_2 * d_2} \)
Then we get (200 × 10 × 20)/(400 × 10 × 15) = (20 × 8 × 10)/(25 × 12 × d_{2})
d_{2} = 8 days
Equating Men, Women and Children This is directly derived from the concept of efficiencies
8 men can do a work in 12 days while 20 women can do it in 10 days. In how many days can 12 men and 15 women complete the same work.
Total work to be done = 8 × 12 = 96 man-days or total work to be done = 20 × 10 = 200 woman-days
Since, the work is the same, we can equate 96 man-days = 200 woman-days. Hence, 1 man-day = 2.08333 woman-days
Now, if 12 men and 15 women are working on the work we get 12 men are equal to 12 × 2.08333 = 25 women
Hence, the work done per day is equivalent to 25 + 15 women working per day
That is, 40 women working per day
Hence, 40 × no. of days = 200 woman days. Number of days = 5 days
Ans .
200
He will complete the work in 20 days. Hence, he will complete ten times the work in 200 days.
Ans .
4
6 men for 12 days means 72 mandays. This would be equal to 4 men for 18 days
Ans .
9 4/7
A’s one day work will be 5%, while B will do 6.66 % of the work in one day. Hence, their total work will be 11.66% in a day. In 8 days they will complete Æ 11.66 × 8 = 93.33% This will leave 6.66% of the work. This will correspond to 4/7 of the ninth day since in 6.66/11.66 both the numerator and the denominator are divisible by 1.66.
Ans .
6 18/47
A’s work = 5% per day B’s work = 6.66% per day C’s work = 4% per day. Total no. of days = 100/15.66 = 300/47 = 6(18/47)
Ans .
60
N + A = 10% N = 8.33% Hence A = 1.66% = 60 days.
Ans .
30
The ratio of the wages will be the inverse of the ratio of the number of days required by each to do he work. Hence, the correct answer will be 3:2 = 30
Ans .
22 2/7
24 man days + 18 women days = 20 man days + 28 woman days = 4 man days = 10 woman days. = 1 man day = 2.5 woman days Total work = 24 man days + 18 woman days = 60 woman days + 18 woman days = 78 woman days. Hence, 1 man + 1 woman = 3.5 women can do it in 78/3.5 = 156/7 = 22(2/7) days.
Ans .
cant say
The data is insufficient, since we only know that the work gets completed in 200 boy days and 300 women days.
Ans .
20
A = 10%, B = 5% and Combined work is 20%. Hence, C’s work is 5% and will require 20 day
Ans .
8
In 5 days, A would do 25% of the work. Since, B finishes the remaining 75% work in 10 days, we can conclude that B’s work in a day = 7.5% Thus, (A + B) = 12.5% per day. Together they would take 100/12.5 = 8 days
Ans .
40
A = 20%, B = 10% and A + B + C = 50%. Hence, C = 20%. Thus, in two days, C contributes 40% of the total work and should be paid 40% of the total amount
Ans .
15
Total man days required = 600 man days. If 5 workers leave the job after ‘n’ days, the total work would be done in 35 days. We have to find the value of ‘n’ to satisfy: 20 × n + (35 – n) × 15 = 600. Solving for n, we get 20n – 15n + 35 × 15 = 600 5n = 75 n = 15.
Ans .
10.5
Let the time taken by Arun be ‘t’ days. Then, time taken by Vinay = 2t days. 1/t + 1/2t = 1/7 = t = 10.5
Ans .
12
Subhash can copy 200 pages in 40 hours (reaction to the first sentence). Hence, Prakash can copy100 pages in 40 hours. Thus, he can copy 30 pages in 30% of the time: i.e. 12 hours
Ans .
12
30X = 20 (X + 6) = 10X = 120 = X = 12
Ans .
11
Sashi = 4%, Rishi = 5%. In five days, they do a total of 45% work. Rishi will finish the remaining 55% work in 11 more days.
Ans .
7
Raju = 10%, Vicky = 8.33% and Tinku = 6.66%. Hence, total work for a day if all three work = 25%. In 2 days they will complete, 50% work. On the third day onwards Raju doesn’t work. The rate of work will become 15%. Also, since Vicky leaves 3 days before the actual completion of the work, Tinku works alone for the last 3 days (and must have done the last 6.66 × 3 = 20% work alone). This would mean that Vicky leaves after 80% work is done. Thus, Vicky and Tinku must be doing 30% work together over two days. Hence, total time required = 2 days (all three) + 2 days (Vicky and Tinku) + 3 days (Tinku alone)
Ans .
144/17
Sambhu requires 16 days to do the work while Kalu requires 18 days to do the work. (1/16 + 1/18) × n = 1 so n = 288/34 = 144/17
Ans .
6
Let Anjay take 3t days, Vijay take 2t days and Manoj take 6t days in order to complete the work. Then we get: 1/3t + 1/2t + 1/6t = 1 so t = 1. Thus, Manoj would take 6t = 6 days to complete the work
Ans .
68
After 100 days and 4500 mandays, only 1/6 th of the work has been completed. You can use the product change algorithm of PCG to solve this question. 100 × 45 = 16.66% of the work. After this you have 200 days (i.e. 100% increase in the time available) while the product 200 × no. of men should correspond to five times times the original product. This will be got by increasing the no. of men by 150% (300/200).
Ans .
4.8
Since the ratio of money given to Apurva and Amit is 2:3, their work done would also be in the same ratio. Thus, their time ratio would be 3:2 (inverse of 2:3). So, if Apurva takes 12 days, Amit would take 8 days and the total number of days required (t) would be given by the equation: (1/12 + 1/8)t = 1 so t = 24/5 = 4.8 days
Ans .
42
Raju being twice as good a workman as Vijay, you can solve the following equation to get the required answer: 1/R + 1/2R = 1/14. Solving will give you that Vijay takes 42 days
Ans .
15
40n = 30 (n + 5) so n = 15
Ans .
2 : 1
12 × 5 man days + 16 × 5 Boy days = 13 × 4 man days + 24 × 4 Boy days so 8 man days = 16 Boy days 1 man day = 2 Boy days. Required ratio of man’s work to boy’s work = 2 : 1
Ans .
5
A’s rate of working is 10 per cent per day while B’s rate of working is 5 per cent per day. In 5 days they will complete 75 per cent work. Thus the last 25 per cent would be done by B alone. Working at the rate of 5 per cent per day, B would do the work in 5 days.
Ans .
20
Work equivalence method: 30 × 5 × 16 = 20 × 6 × n Gives the value of n as 20 days
Ans .
345
A + V + S = 1 A + V = 19/23 V + S = 8/23 Æ A + 2V + S = 27/23 (2)–(1) gives us: V = 4/23.
Ans .
22.5
Interpret the starting statement as: Anmol takes 30 days and Vinay takes 90 days. Hence, the answer will be got by: (1/30 + 1/90) * n = 1 Alternatively, you can also solve using percentages as: 3.33 + 1.11 = 4.44% is the daily work. Hence, the no. of days required is 100/4.44 = 22.5 days
Ans .
55
Total work = 15 × 210 = 3150 mandays. After 100 days, work done = 15 × 100 = 1500 mandays. Work left = 3150 – 1500 = 1650 mandays. This work has to be done with 30 men working each day. The number of days (more) required = 1650/30 = 55 days
Ans .
20
Ajay’s daily work = 4.1666%, Vijay’s daily work = 3.33% and the daily work of all the three together is 8.33%. Hence, Pradeep’s daily work will be 0.8333%. Hence, he will end up doing 10% of the total work in 12 days. This will mean that he will be paid Rs. 20.
Ans .
6
After 27 days, food left = 4 × 200 = 800 soldier days worth of food. Since, now there are only 80 soldiers, this food would last for 800/80 = 10 days. Number of extra days for which the food lasts = 10 – 4 = 6 days.
Ans .
40
Total work of Anju, Manju and Sanju = 16.66% Anju’s work = 10% Manju’s work = 4.166% Sanju’s work = 2.5% So Sanju can reap the field in 40 days.
Ans .
84
Ajay + Vijay = 1/28 and Ajay + Vijay + Manoj = 1/21. Hence, Manoj = 1/21 – 1/28 = 1/84. Hence, Manoj will take 84 days to do the work.
Ans .
20
A + M = 8.33, M + B = 6.66 and A = 2B a so A’s 1 days work = 3.33%, M’s = 5% and B’s = 1.66%. Thus, Mohan would require 100/5 = 20 days to complete the work if he works alone
Ans .
15
A + V = 16.66% and A = 10% so V = 6.66%. Consequently Vijay would require 100/6.66 = 15 days to do it alone
Ans .
30
The rate of filling will be 20% and the net rate of filling (including the leak) is 16.66%. Hence, the leak accounts for 3.33% per hour. i.e. it will take 30 hours to empty the tank
Ans .
10 min, 15 min
A + B = 16.66%. From here solve this one using the options. Option (c) fits the situation as it gives us A’s work = 10%, B’s work = 6.66% as also that B takes 5 minutes more than A (as stipulated in the problem).
Ans .
4
A + B = 5.55 + 11.11 = 16.66. In two days, 33.33% of the work will be done. C adds 16.66% of work to that of A and B. Hence, the rate of working will go to 33.33%. At this rate it would take 2 more days to complete the work. Hence, in total it will take 4 days to complete the entire work.
Ans .
32
24 × 8 × 10 = N × 10 × 6 so N = 32
Ans .
32
n × 20 = (n – 12) × 32 so n = 32.
Ans .
9
12 × 18 = 12 × 6 + 16 × t so t = 9
Ans .
15
(A + B)’s work = C’s work. Also if A takes ‘a’ days B would take ‘a – 5’ days and C would take ‘a – 9’ days. Solving through options, option ‘c’ fits. A (15 days) so A’s work = 6.66% B (10 days) so B’s work = 10% C (6 days) so C’s work = 16.66%
Ans .
15
The cistern fills in 6 hours normally, means that the rate of filling is 16.66% per hour. With the leak in the bottom, the rate of filling becomes 10% per hour (as it takes 10 hours to fill with the leak). This means that the leak drains out water at the rate of 6.66% per hour. This in turn means that the leak would take 100/6.66 = 15 hours to drain out the entire cistern.
Ans .
Waste pipe emptying the tank is 5 h
Since the net work of the three taps is 10% and the first and second do 20% + 10% = 30%. Hence, the third pipe must be a waste pipe emptying at the rate of 20% per hour. Hence, the waste pipe will take a total of 5 hours to empty the tank
Ans .
15
A’s work = 10% B’s negative work = 6.66% (A + B)’s work = 3.33% To fill a half empty tank, they would take 50/3.33 = 15 hours.
Ans .
16.5
The work rate would be 10% on the first day, 5% on the second day and 2.5% on the third day. For every block of 3 days there would be 17.5% work done. In 15 days, the work completed would be 17.5 × 5 = 87.5%. On the sixteenth day, work done = 10% Æ 2.5% work would be left after 16 days. On the 17th day the rate of work would be 5% and hence it would take half of the 17th day to complete the work. Thus, it would take 16.5 days to finish the work in this fashion
Ans .
108
(A + B) = 2C. Also,(A + C) = 3B 36(A + B + C) = 1 Solving for C, we get: 36 (2C + C) = 1 so 108 C = 1 C = 1/108 Hence, C takes 108 days
Ans .
72
A + B + C = 19%. In the first two hours they will do 38 % of the work. Further, for the next two hours work will be done at the rate of 15% per hour. Hence, after 4 hours 68% of the work will be completed, when tap B is also closed. The last 32% of the work will be done by A alone. Hence, A does 40% (first 4 days) + 32% = 72% of the work.
Ans .
20
Without the leak: Rate of work = 20% + 5% = 25%. Thus, it would have taken 4 hours to complete the work. Due to the leak the filling gets delayed by 1 hour. Thus, the tank gets filled in 5 hours. This means that the effective rate of filling would be 20% per hour. This means that the rate at which the leak empties the tank is 5% per hour and hence it would have taken 20 hours to empty a filled tank
Ans .
22
In 6 days A would do 25% of the work and in 8 days B would do 25% of the work himself. So, C has to complete 50% of the work by himself. In all C would require 30 days to do 50% of the work. So, he would require 22 more days
Q. Worker A takes 8 hours to do a job. Worker B takes 10 hours to do the same Job.How long should it take both A and B, working together but independently, to do the same job?
A. A does 1/8 work in an hour and B does 1/10 so both do (1/8 + 1/10) in one hour. 18/80 work in an hour so they need 80/18 to do the work completely.
Q. A and B together can complete a piece of work in 4 days. If A alone can complete the same work in 12 days, in how many days can B alone complete that work?
A. A can do 1/12 in a day, B can do 1/x. (1/12 + 1/x) = 1/4 from given data. Solving it we get value of x.
Q. 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 hours a day?
A. A can do work in 63 hrs so 1/63 per hour. B can do work in 42 hrs so 1/42 per hour. Both can do work in (1/63 + 1/42) = (2+3/126) = 5/126 work in 1 hr so they need 126/5 hrs or 25.2 hrs so if 8 hrs / day are taken then 4 days.
Q. A and B can do a piece of work in 18 days; B and C can do it in 24 days A and C can do it in 36 days. In how many days will A, Band C finish it together?
A. A and B can do 1/18 work per day; B and C can do 1/24 work per day; A and C can do 1/36 work per day.
Adding all three we get 2(A+B+C) = (4+3+2)/72 = 1/8 work per day or 8 days to do work together.
Q. A can do a certain job in 12 days. B
is 60% more efficient than A. How many days does B alone
take to do the same job?
A. B is 60% more efficient than A so if B takes 100 hrs then A needs 160 hrs so time taken ratio of A : B = 160 : 100 = 8 : 5.
B's time = A's time * 5 / 8 = 12 * 5 / 8 = 7 1/2 days.
Q. A can do a piece of work in 80 days. He works at it for 10 days B alone finishes the remaining work in 42 days. In how much time will A and B working together, finish the work?
A. A does 1 / 80 work in 1 day and so 10/80 work in 10 days. 70/80 work is left.
B does 70/80 work in 42 days so in 1 day he does 70 /42 * 80 = 1 / 48 work.
A and B together do (1/80 + 1/48) in one day. Reciprocal of result will give total days needed.
Q. A and B undertake to do a piece of work for Rs. 600. A alone can do it in 6 days while B alone can do it in 8 days. With the help of C, they finish it in 3 days. find the share of each.
A. A, B, C do (1/6 + 1/8 + 1/x) = 1/3 so 1/x = 1/3 - 1/6 - 1/8
= (8 - 4 - 3) / 24
= 1/24 work per day so C can finish job in 24 days.
So ratio of 1 days work is = 1/6 : 1/8 : 1/24 = 4:3:1 . This can be used to get their share of the money.
Q. A and B working separately can do a
piece of work in 9 and 12 days respectively, If they work
for a day alternately, A beginning, in how many days, the
work will be completed?
A. A work in 1 day is 1/9 and B's work is 1/12 so since they work alternately in 2 days work by them is (1/9+1/12) = (4+3)/36 = 7/36.
Work done in 10 days is = 7/36 * 5 = 35/36.
On last day A's turn and 1/36 work remains which A can do in 1/4 day. Total days needed are 10 1/4 days.
If a pipe can fill a tank in 'x' hours then in 1 hour it can fill 1/x. If a pipe can empty a tank in y hours then in 1 hour it can empty 1/y.
If one pipe fills a tank in 'x' hrs and second pipe empties it in 'y' hrs then in 1 hour net part filled is 1/x - 1/y (x>y).
If one pipe fills a tank in 'x' hrs and second pipe empties it in 'y' hrs then in 1 hour net part filled is 1/y - 1/x (y>x).
Q. Two pipes can fill
a tank in 10 hours and 12
hours respectively while a third,
pipe empties the full tank
in 20 hours. If all the
three pipes operate simultaneously, in how much
time will the tank be filled?
A. Total work by all pipes together = (1/10 + 1/12 - 1/20) = (6+5-3)/60 = 2/15
So it takes 7.5 hrs to fill tank.
Q. Two pipes can fill a cistern in 14 hours and 16 hours respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom it took 32 minutes more to fill the cistern.When the cistern is full, in what time will the leak empty it?
A.Pipe A and B can fill tank in (1/14 + 1/16) = (8+7)/112 = 15/112 work in 1 hr. So total time needed is 112/15 i.e. 7 hrs 28 mins. But when leak is there it took 32 mins more so 8 hrs.
15/122 - 1/x = 1/8 so we can get value of x.
Q. Two pipes A and B can fill a tank in 36 min. and 45 min. respectively. A water pipe C can empty the tank in 30 min. First A and B are opened. after 7 min, C is also opened. In how much time, the tank is full?
A. A and B can together fill in 1 min = (1/36 + 1/45) = (5+4)/180 = 1/20 so in 7 mins 7/20 is filled. Remaining 13/20 will be filled by all three.
(1/36 + 1/45 - 1/30) = (5+4- 6)/180 = 3/180 = 1/60. So they need to fill 13/20 or 39/60. Which they can do in 39 mins as they do 1/60 in 1 min.
Q. Two pipes A,B can fill a tank in 24 min. and 32 min. respectively. If both the pipes are opened simultaneously, after how much time B should be closed so that the tank is full in 18 min.?
A. Suppose B is closed in 'x' min.
Part of tank filled in 'x' min by both + part of tank filled in (18-x) min by A = 1.
x * ( 1/24 + 1/32) + (18-x) * (1/24) = 1 Solving this
we can get 'x'.
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