Quantitative Aptitude Notes on Time & Work (Part II)

We have already posted the article related to the basic concepts of Time and Work Part I. This post is in continuation with it. Here we will help you understand the use of Work Constant approach.

While attempting questions, you must remember some important points – 

1. Read questions carefully.

2. Words like together, alone, before complete, after complete etc. are important.

3. Data provided in the question should be read carefully.

How to approach and express data while taking the exam?

If there are two persons A and B. If A begins and both work alternate days.It means

1st day	2nd day	3rd day	4th day	5thday	6thday	7thday	8thday	9thday	10thday	…..
A	B	A	B	A	B	A	B	A	B	…..

 Before going forward, we will discuss some basics of Ratios to make calculation easy for this topic and these basics will also help you in other topics like Partnership, time and distance, average, allegation etc.

TIME AND WORK EXAMPLES:

Example 5: A and B can do a work in 8 days and 12 days respectively.

                     A                B              

Time (T)         8 days      12 days        

Efficiency (η)     12             8              

        η             3               2

total work  = T_A× η_A  = T_B× η_B
total work =  8 × 3 =  12 × 2 
total work = 24 units

(a). If both A and B work alternatively and  A begins, then in how many days work will be completed?

According to question A begins and both and b work alternately.

Days	1st day	2nd day	3rd day	4th day	5th day	6th day	7th day	8th day	9th day	10th /2 day	Total Work
Person	A	B	A	B	A	B	A	B	A	B
Work Unit	3	2	3	2	3	2	3	2	3	1	24

Hence work will be completed in 9 (1/2) days.

But this type of approach is not helpful in exams. We will go by basic but in a smarter way.

If we make a pair of A and B, we can say that both together will work 5 units but in 2 days.

2 days = 5 unit of work

2 days × 4 = 5 units of work × 4 

8 Days   = 20 units of work, i.e. we can say that up to 8^th day 20 units of work will be done

But on 9th day it’s a chance of A and he will do his 3 units of work. So up to 9^th day 23 units of work will be done. Now 1 unit of work will remain. On 10^th day it’s a chance of B.

As above B do 2 unit of work in a day. So, he will do 1 unit of work in (1/2) days.

So total time taken to complete the work is 9 (1/2) days.

(b). If A and B both work for 4 days then A leaves. In how many days total work will be completed.

(Eff_A+B× t_A+B )+( Eff._B × t_B) = total work
(5 × 4)+(2 × t_B) = 24
20 + 2 × t_B  = 24
2 × t_B = 4
t_B = 2 days , hence total time will be ( 4+2 )days i.e. 6 days.

Important: Same question can be framed in many ways.

Way 1 : B works for 2 days and after that A joins B, then in how many days the work will be completed.

Solution: B works only for 2 days and then A joins B, i.e. both will work together after 2 days.
(t_B × η_B) + (t_A+B× η_A+B) = total work
(2 × 2) + (t_A+B × 5) = 24
(t_A+B × 5) = 20
t_A+B  = 4 days   hence total time is (2+4) days i.e. 6 days

Way 2 : A and B both work together and A takes leaves for 2 days, then in how many days the work will be completed.

Solution:  When both A and B are working together and A takes leaves for two days it means B has to work alone for 2 days .

Let total time to complete the work is t days.

So,  η­_A+B × (t-2) + η_B × 2 days = 24
5 × (t-2) + 2 × 2 = 24
t-2 = 4
t = 6 days Hence total time taken to complete the work is 6 days.

Don’t confuse between t_A+B and T_A+B  (as mentioned in article1). Both are different.

Example 6: X can do a work in 6 days, Y can do it in 8 days and Z can do it in 12 days.

(a). If X starts the work and X,Y,Z works in alternate days ,then in how many days the work will be completed?

Solution: Here X will start work on 1^st  day, then Y will work on 2^nd day and z will work on 3^rd day.        

                  X :   Y  :  Z                         
Time           6 :   8  : 12                

Efficiency    12 × 8: 6 × 12 : 8 × 6              

η               96 :  72 : 48                           
η                 4 :  3  :  2                             

total work = η_X× T_X= η_Y×T_Y =η_Z×T_Z        

total work = 6 × 4 = 8 × 3 = 12 × 2 = 24 units

On 1^st day, work done by X = 4 unit

On 2^nd day, work done by Y = 3 unit

On 3^rd day, work done by Z = 2 unit

Total work in 3 days done by X,Y and Z = 9 unit

Now again X will come then Y,then Z and so on till work is completed.

                  In 3 days   =    9 units

                        × 2           × 2

                 In 6 days    =   18 units

X will work on 7^th day   = 4 units

                                  =  22 units

Now we need (24-22) units = 2 units work more but Y can do 3 unit of work in one day. So 2 unit will be done in (2/3) day.

               +(2/3) day          + 2 unit

Total days=7 (2/3) days       24 units

Hence total work will be done in 7 (2/3) days.

(b). If all started together and after completion of (3/4)^th work, Y left and remaining work is done by X and Z together. Then in how many days work will be completed?

Solution: Total work is 24 units then (3/4)^th work is 18 units.

One day work of (X+Y+Z) = 9 units so in 2 days 18 units of work will be done by (X+Y+Z) together.

After this Y left, X+Z worked together and 6 units of work remained.
One day work of X+Z = (4+2) units = 6 units. So in 3 days total work will be completed.

• Time & Work (Shortcut Approach) Part 1: Click here

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