Time and Work - Shortcuts and Tricks

This post is a part of "Shortcut Techniques" series where I decided to share shortcut techniques related quantitative techniques that can be used in IBPS exam or any other competitive exam.

Time and work is an important topic in IBPS exams, including clerks, PO and specialist officers.

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 the equation we arrived in above question.

You need to understand one simple concept - If A can do a job in 10 days then in one day A can do 1/10th of the job.

Shortcut

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.

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%

Now let's 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

A's efficiency = 1/30 = 3.33%

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?

Solution -
Method 1
⇒ Efficiency of filling pipe = 20 minutes = 1/3 hour = 300%

⇒ 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 a base so the answer is 30 minutes.

Update - 09-09-2013 ( As Shobhna and Aswin are facing problem in solving this question, I am solving this question with the second method which is also very easy, hope this will make the solution lot easier.)

Method 2
⇒ 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

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.

You can find more tricks like this in quantitative aptitude section.

Comment below in case of any query, I promise to reply within 24 hours.

Update - Question requested by Chitra Salin

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 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 1 woman(y) = 2% per day

⇒ Efficiency of 10 women per day = 20%

So 10 women can complete the job in 100/20 = 5 days

Update Question requested by Praisy

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
⇒ 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

Taking equation 2 and 3 together
⇒ 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.