How To Solve Time And Work Problems – Tips and Tricks

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Time and Work Problems are one of the most asked questions in the competitive examinations and aptitude tests by various organisations.  As you know its very important to solve aptitude problems quickly to get good marks in the examinations. These time and work problems could be very annoying if you do not know how to solve it. But the best part of these problems are that once you know the method to solve them , its the easiest question you can get in the examination and you could be assure of securing full marks in time and work problems. Now let’s start with the solving methodology of time and work problem :

Let’s take examples :

Example 1 :

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 ?

Now there are 2 methods of solving it :

Method 1 :  ( Time consuming )

• If A can do a piece of work in n days, then A’s 1 day work = 1/n
• If A’s 1 day work = 1/n, then A can finish the work in n days

Lets solve the given problem using this formula:

Since A take 8 hours to do a job, A’s 1 hour work = 1/8 , Similarly B’s 1 hour work = 1/10

So if they work together , A and B ‘s 1 hour work =  A’s 1 hour work + B’s 1 hour work = 1/8 + 1/10 = 9/40

Therefore, A and B together can complete the whole job in 1/(9/40) = 40/9 hours

Method 2 : ( Efficient and Easy )

• If A can do a piece of work in n days, then A’s 1 day work = (100/n) percentage of total work
• If A’s 1 day work = N percentage, then A can finish the work in (100/N) days

Let’s solve the given problem using this formula:

Since A takes 8 hours to do a job, A’s 1 hour work = (100/8)% = 12.5%

Similarly B’s 1 hour work = (100/10)% = 10%

So if they work together, A and B ‘s 1 hour work = 12.5% + 10% = 22.5%

Therefore, A and B together can complete the whole job in (100/22.5 ) hours = 40 1000 / 225 9 = 40/9 hours

So Method 2 is very efficient in solving Time and Work problems especially during competitive exams where the shortcut methods are required to solve the questions quickly.

Example 2 :

A is twice as good a workman as B and together they can finish a piece of work in 18 days. In how many days will A alone finish the work ?

Method 1 :  ( Time consuming )

• If A can do a piece of work in n days, then A’s 1 day work = 1/n
• If A’s 1 day work = 1/n, then A can finish the work in n days
• If A is X times as good a workman as B, then
Ratio of Work Done by A and B = X:1
Ratio of Time Taken By A and B to Finish the work = 1:X

Let’s solve the problem using this formula :

(A’s 1 day work ) : (B’s 1 day work) = 2 : 1

Given, (A+B)’s 1 day work = 1/18

Divide 1/18 in to 2:1,

therefore A’s 1 day work = ((1/18) * 2/(2+1)) = (1/18) * (2/3) = 1/27

Thus, A can finish the work in 27 days.

Method 2 : ( Efficient and Easy )

• If A can do a piece of work in n days, then A’s 1 day work = (100/n) percentage of total work
• If A’s 1 day work = N percentage, then A can finish the work in (100/N) days
• If A is X times as good a workman as B, then
Ratio of Work Done by A and B = X:1
Ratio of Time Taken By A and B to Finish the work = 1:X

Let’s solve the given problem using this method:

(A’s 1 day work ) : (B’s 1 day work) = 2 : 1

Given, (A+B)’s 1 day work = 1/18 = 5.55%

Divide 5.55% into 2:1

Therefore, A’s 1 day work = (5.55 *(2/3)) % = 3.7 %

Thus, A can complete the work in (100/3.7) = 27 days

Example 3 : Pipes and Cisterns

A tank can be filled by pipe A in 30 minutes. There is a leakage which can empty it in 60 minutes. In how many minutes tank can be filled?

Formula for Pipes and Cisterns :

• If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where y > x), then on opening both the pipes, then the net part filled in 1 hour = ((1/x) – (1/y))
• If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where x > y), then on opening both the pipes, then the net part emptied in 1 hour = ((1/y) – (1/x))

Solving the above problems using pipes and cisterns formula and time and work methodologies :

Using Method 1 :

Net Part filled in 1 minute = (1/30)

Net Part emptied in 1 minute = (1/60)

Therefore, Net Part filled in 1 minute = 1/30 – 1/60 = 1/60

Therefore, Tank can be filled in 1/(1/60) hours = 1 hour.

Using Method 2 :

Assume Efficiency of pipe filled 1 hour as 100%

Efficiency of filling pipe = 30/60 hour = 1/2 hour = twice efficient of the supposed value 1 hour = 200%

Efficiency of Emptying of Tank = 60/60 hour = 1 hour = 100%

Therefore, Efficiency of Net Part Filled is (200-100)% = 100%

Since 100% is the efficiency of pipe filled in 1 hour, Therefore, Tank can be filled up in 1 hour.

Example 4 :

If 2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days. In how many days 2 men and 1 boy do the work ?

Using Method 1 :

Let 1 Man’s 1 day work = x and 1 Boy’s 1 day work = y

Then, 2x + 3y = 1/10 and 3x + 2y = 1/8

Solving above equation will give, x = 7/200 and y = 1/100

Therefore, (2 Men + 1 Boy )’s 1 day work = 2*(7/200)+1*(1/100) = 2/25

Thus, 2 men and 1 boy can do the work in 1/(2/25) days = 12.5 days

Using Method 2 :

Let 1 Man’s 1 day work = x and 1 Boy’s 1 day work = y

Then, Equations in terms of Efficiency

2x + 3y = 10% , 3x+2y = 12.5%

Solving it, will give x = 3.5%, y = 1%

Therefore, (2 Men + 1 Boy)’s 1 day work = 2*3.5+1*1 = 8%

Thus, 2 men and 1 boy can do the work in (100/8 ) days = 12.5 days

Example 5 :

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.

Using Method 1 :

C’s 1 day work = 1/3 – (1/6+1/8) = 1/24

Therefore, Ratio of Their 1 day work = A:B:C = 1/6 : 1/8 : 1/24  ( take LCM and put it in the numerator) = 4:3: 1

Thus, A’s share = Rs (600 * 4/(4+3+1)) = 600 * 4/8 = Rs 300

Similary B’s Share = Rs 225, C’s Share = Rs 75

Using Method 2 :

Efficiency of C’s 1 day work = 33.3 – (16.6 + 12.5)  = 4.2 %

Ratio of their 1 day work efficiency = 16.6 : 12.5 : 4.2 => 4 : 3 : 1 ( always solve it using round values)

And Answer would be done same as done in the Method 1. In this type of questions Method 1 is preferred.

The good part of using Method 1 is that you can get the exact values for your answers whereas for Method 2 you need to use round up values to solve the problems.  Below is given table which you should remember to fasten up your calculation while solving the problem using Method 2.

No of Days Required To Finish The Work                                       Work Done in 1 day  Efficiency of Work 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%
12 1/12 8.33%
13 1/13 7.7%
14 1/14 7.1%
15 1/15 6.66%

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