Important Information
In this post, we shall discuss about the Time and Work. Lets start with some basic formulas.
Work = Strength X Time
=> Strength X Time = Work
=> (Strength X Time ) / Work = 1 🙂
So, For every Case
(Strength X Time) / Work = 1 [ No matter the values]
So, (SxT)/W will be same for all the cases.
I mean st/w = ST/W
Keep this point in your mind.
Now lets see some more important formulas
If Days are Given :
If A can do some work in n days, then he can do 1/n work in One day
If Work is Given :If A can do 1/n work in One day, he can finish it in n days
If A is TWICE as good a work man as B, then
The ratio of work done by A and B = 2:1
The ratio of time taken by A and B to finish the work = 1:2 (Please Don’t be confused)
Time and Work Trick
One simple technique is using days in denominator while solving questions. For example, A can do a job in 3 days and B can do the same job in 6 days. In how much time they can do the job together.
Solution – 1/3 + 1/6 = 1/2, hence 2 days is the answer.
Examiner can set the question in opposite way and can ask you how much time A or B alone will take to complete the job. It is quite easy to calculate said question by putting values in equation we arrived in above question.
You need to understand one simple concept – If A can do a job in 10 day then in one day A can do 1/10th of job.
Shortcut
Number of days
required to complete the work |
Work that can be done per day |
Efficiency in Percent |
| n |
1/n |
100/n |
| 1 |
1/1 |
100% |
| 2 |
1/2 |
50% |
| 3 |
1/3 |
33.33% |
| 4 |
1/4 |
25% |
| 5 |
1/5 |
20% |
| 6 |
1/6 |
16.66% |
| 7 |
1/7 |
14.28% |
| 8 |
1/8 |
12.5% |
| 9 |
1/9 |
11.11% |
| 10 |
1/10 |
10% |
| 11 |
1/11 |
9.09% |
A’s efficiency = 1/30 = 3.33%
Solution –
⇒ Efficiency of filling pipe = 20 minutes = 1/3 hour = 300%
⇒ Efficiency of filling pipe = 100/20 = 5%
⇒ Efficiency of leakage pipe = 100/60 = 1.66%
⇒ Net filling efficiency = 3.33%
So tank can be filled in = 100/3.33% = 30 minutes
⇒ Efficiency of 1 woman(y) = 2% per day
⇒ Efficiency of B and C = 1/30 per day = 3.33% per day______________2
⇒ Efficiency of C and A = 1/30 per day = 3.33% per day______________3
⇒ B + C = 3.33% and C + A = 3.33%
⇒ C and 3.33% will be removed. Hence A = B
⇒ Efficiency of A = B = 5%/2 = 2.5% = 1/40
⇒ Efficiency of C = 3.33% – 2.5% = 0.833% = 1/120
⇒ A can do the job in 40 days and C can do the job in 120 days he they work alone.
⇒ Ratio of number of days in which A and C can complete the job 1:3.
Time and Work – Problems 2
6. A, B and C can do a job in 20 days, 30 days and 60 days respectively. If they work together, in how many days will the work be finished?
7. Two taps A and B can fill a tank in 10 hours and 15 hours respectively. a third tap C can empty the full tank in 12 hours. How many hours will be required if all of them are opened simultaneously to fill in an empty tank completely?
8. A and B can do a job in 12 days. B and C in 15 days and C and A in 20 days. In how many days can they finish it if they work TOGETHER?
9. A and B can do a job in 12 days. B and C can do the same job in 15 days. C and A in 20 days. In how many days can A alone finish the whole task???
But according to our Problem, 2A = (1/12)-(1/15)+(1/20)
=>A = 1/30 (this is A’s one day work but we need A’s total work)
=> A = 30 Days
10. A and B can do a piece of work in 20 days. A alone can do it in 30 days. In how many days can B alone do it?
Best trick that I use in exams myself is by finding the efficiency of workers in percent. If A can do a job in 2 days then he can do 50% in a day.
Solve Questions with this Trick
Question – A take 5 days to complete a job and B takes 10 days to complete the same job. In how much time they will complete the job together ?
Solution –
A’s efficiency = 20%, B’s efficiency = 10%. If they work together they can do 30% of the job in a day. To complete the job they need 3.33 days.
Question – A is twice as efficient as B and can complete a job 30 days before B. In how much they can complete the job together ?
Solution –
Let efficiency percentage as x
A’s efficiency = 2x and B’s efficiency = x
A is twice efficient and can complete the job 30 days before B. So,
A can complete the job in 30 days and B can complete the job in 60 days
B’s efficiency = 1/60 = 1.66%
Both can do 5% ( 3.33% + 1.66% ) of the job in 1 day.
So the can complete the whole job in 20 days (100/5)
Question – A tank can be filled in 20 minutes. There is a leakage which can empty it in 60 minutes. In how many minutes tank can be filled?
Method 1
⇒ Efficiency of leakage = 60 minutes = 100%
We need to deduct efficiency of leakage so final efficiency is 200%. We are taking 100% = 1 Hour as base so answer is 30 minutes.
( As Priya and Teja are facing problem in solving this question, I am solving this question with second method which is also very easy, hope this will make the solution lot easier.)
Method 2
You can change the base to minutes or even seconds.
You can solve every time and work question with this trick. In above examples I wrote even simple calculations. While in exams you can do these calculations mentally and save lots of time.
Question – 4 men and 6 women working together can complete the work within 10 days. 3 men and 7 women working together will complete the same work within 8 days. In how many days 10 women will complete this work ?
Solution –
Let number of men =x, number of women = y
⇒ Efficiency of 4 men and 6 women = 100/10 = 10%
⇒ so, 4x+6y = 10
Above equation means 4 men and 6 women can do 10% of a the job in one day.
⇒ Efficiency of 3 men and 7 women = 100/8 = 12.5%
so, 3x+7y = 12.5
By solving both equations we get, x = -0.5 and y = 2
⇒ Efficiency of 10 women per day = 20%
So 10 women can complete the job in 100/20 = 5 days
Question – A and B together can complete a task in 20 days. B and C together can complete the same task in 30 days. A and C together can complete the same task in 30 days. What is the respective ratio of the number of days taken by A when completing the same task alone to the number of days taken by C when completing the same task alone?
Solution –
⇒ Efficiency of A and B = 1/20 per day = 5% per day ________________1
Taking equation 2 and 3 together
Time and Work – Problems 1
1. 36 men can do a piece of work in 25 days. In how many days can 30 men do it?
Here , the work is same.. So,
(S X T) / W = (s X t) / w
= > (S X T) / W = (s X t) / w [ Cancel W for both sides]
Now, substitute the values…
= > 36 X 25 = 30 X t
= > (36 X 25) / 30 = t => t = 900/3 = 30
2. 32 men can do a piece of work in 15 days working for 6 hours a day. In how many days will 40 men can finish it if they work for 8 hours a day?
32 X 15 X 6 = 40 X d X 8 = 9
3. If 16 men can build a wall of 52 m long in 25 days working for 8 hours a day, in how many days can 64 men build a similar wall of 260m long working for 10hrs a day?
As we know that (Strength X Time) / Work = 1 (Read details Here)
We can write =>
(16 * 200) / 52 = (64 * X * 10) / 260
=> X = 25
4. A man engaged 10 laborers to make 320 toys in 5 days. After 3 days he found that only 120 toys were made. How many additional men should he engage to finish the work in time?
here, equate the complete work with the remaining work
(10 X 3) / 120 = (10+X) x 2 / 200
= > 10+X = 25 => X = 15
5. A can do a task in 20 days and B can do it in 30 days. In how many days can they finish it if they work together?
A’s one day’s work = 1/20
B’s one day’s work = 1/30
=> Their one day’s work = (1/20)+(1/30) = (3+2)/60 = 5/60 = 1/12
This is their one day’s work TOGETHER.
So, obviously the number of days will be = 12
Short Cut : calculate Product/Sum = (20 X 30) / 50 = 12
(1/20)+(1/30)+(1/60) = (3+2+1)/60 = 6/60 = 1/10
So, the number of days is = 10
Here, first two are Inlets and the last one is Outlet,
So, (1/10)+(1/15)-(1/12) = (6+4-5)/60 = 5/60 = 1/12
So here, A+B’s One day’s work = > 1/ (A+B) = 1/12
B+C’s one day’s work => 1/ (B+C) = 1/15
C+A’s one day’s work = > 1/(C+A) = 1/20
Just Add them => 1/( 2A+2B+2C) = 12/60 = 1/5
=>1/(A+B+C) = 1/10 [this is their one day’s work TOGETHER]
So, they can finish it in 10 days 🙂
Here we need A, so take a pair which is NOT HAVING A and subtract it from the others,
so, A+B-(B+C)+C+A = A+B-B-C+C+A = 2A
Per day work of A and B = 1/20
So, B’s one day work = 1/20 – 1/30 = (3-2)/60 = 1/60
Short Cut : Product/diff = 600/10 = 60 🙂
Leave a Comment