# Some Important Point Related To Time & Work Aptitude Questions

This time we are going to solve work and time related shortcut tricks. we are also aware that this chapter is important for competitive exams from Time and work sections objective question are asked in various exam like IBPS/SBI, Bank PO, SSC, UPSC, etc. Time and Work sections are time consuming chapter.now, we are going to solve work and time illustrations with related important points.

Time and work aptitude questions are asked in every competitive exam. Placement papers for TCS, Infosys, Wipro, CTS, HCL, IBM or Bank exam or MBA exams like CAT, XAT, MAT, or other exams like GRE, GMAT tests always contain one or more aptitude questions from this section of quantitative aptitude. Problems on time and work which appear in CAT exams are quite advanced and complicated – But they can be solved easily if you know the basic formulas, shortcuts and tricks.

If know about these important points and short methods you can solve these question in seconds.Certain general important points are given underneath. the competitors will learn them truly valuable while taking care of issues identified with Time and Work.:

# Important Terms for Time and Work Formulas-

Before we plunge into time and work formulas, let’s quickly go through the different terms that make up time and work formulas-

Work (W) – The total amount effort or labour required to complete a given task is called work. It is represented by the letter ‘W’ and is always considered as a whole or one in every problem.

Person/ People (P) – The number of individuals involved in completing the given work is called people/person. It is represented by the letter ‘P’.

Time (T) – The exact count taken by the clock to complete a certain task is called time.  It is represented by the letter ‘T’. Time can be calculated in two ways- the number or days ‘D’ or the number of hours ‘H’.

Rate of Work (R) – The amount of work done in a unit of time is called the rate of work. It can be calculated on a daily or hourly basis, depending on the conditions given in the questions. It is represented by the letter ‘R’.

## Important Point Related To Time & Work-

1. While solving problems related to time and work we have to compare the work of several working agents (men,women and children). to solve this purpose, we need find out the amount of work done by each agent in a common unit time. this unit time is generally taken as an hour, or minute of the problem.

2. The total work is assumed to be 1.

3. If A can do a piece of work in n days, then A’s 1 day’s work = 1/n.

4. If one can do a piece of work in 12 days, then it means that he can do 1/12 of work in 1 day.

5. Conversely, if one can do 1/12 work in 1 day, it implies that he can complete the work in 12 days.

6. One day’s work = 1/No. of days Required to finish the work

7. No. of days required to finish the work = 1/one day’s work

8. If A does 1/15 of work in 1 day and B does 1/10 of the same work in 1 day then A + B together will do 1/15 + 1/10 i.e. 1/6 of that work in one day.

9. If A is thrice as good as a work man as B then :

A’s Work : B’s Work = 3 : 1

And ratio of the times taken by A and B to complete the work = 1 : 3 i.e. A will take 1/3 of the time taken by B to complete that work.

10. The time required to complete a certain peice of work is increased or decreased in the same ratio in which the number of men engaged to do a peice of work be changed in the ratio 4 : 5, the time required to complete that peice of work will change in the ratio 5 : 4.

11. Time and number of man each are always in dirrectly proportional to work.

All the above points can be summarised as given below:

if ‘M1’ persons can do ‘W1’ works in ‘D1’ days and ‘M2’ persons can do ‘w2’ works in ‘D2’ days then we have a very general formula in the relationship of:

M1D1W2 = M2D2W1

it is all in one formula and basic formula to solve these type questions.

we can also include that:

(a) more men less days and conversely more days less men.

(b) more men more work and conversely more work more men.

(c) more days more work and conversely more work more days.

12. If ‘A’ can do a peice of work in X days and ‘B’ can do it in Y days then ‘A’ and ‘B’ working together can do the same work in:

XY/X+Y days.

13. If A, B and c can do a work in X, Y and Z days respectively then all of them working together can finish the work in:

XYZ/XY+YZ+XZ days.

## Shortcuts for frequently asked time and work problems

• AA and BB can do a piece of work in a′a′ days and b′b′ days respectively, then working together:
1. They will complete the work in aba+baba+b days
2. In one day, they will finish (a+bab)th(a+bab)th part of work.
• If AA can do a piece of work in aa days, BB can do in bb days and CC can do in cc days then,
A, B and C together can finish the same work inabcab+bc+ca days A,

B and C together can finish the same work inabcab+bc+ca days

• If AA can do a work in xx days and AA and BB together can do the same work in yy days then, Number of days required to complete the work if B works
alone=xyxydaysNumber of days required to complete  the work if B works alone=xyx-ydays
• If AA and BB together can do a piece of work in xx days, BB and CC together can do it in yy days and CC and AA together can do it in zz days, then number of days required to do the same work:
1. If A, B, and C working together = 2xyzxy+yz+zx2xyzxy+yz+zx
2. If A working alone = 2xyzxy+yzzx2xyzxy+yz-zx
3. If B working alone = 2xyzxy+yz+zx2xyz-xy+yz+zx
4. If C working alone = 2xyzxyyz+zx2xyzxy-yz+zx
• If AA and BB can together complete a job in xx days.
If AA alone does the work and takes aa days more than AA and BB working together.
If BB alone does the work and takes bb days more than AA and BB working together.

Then,x=ab‾‾‾√ daysx=ab days
• If m1m1 men or b1b1 boys can complete a work in DD days, then m2m2 men and b2b2 boys can complete the same work in Dm1b1m2b1+m1b2Dm1b1m2b1+m1b2 days.
• If mm men or ww women or bb boys can do work in DD days, then 1 man, 1 woman and 1 boy together can together do the same work in Dmwbmw+wb+bmDmwbmw+wb+bm days
• If the number of men to do a job is changed in the ratio a:ba:b, then the time required to do the work will be changed in the inverse ratio. ie; b:ab:a
• If people work for same number of days, ratio in which the total money earned has to be shared is the ratio of work done per day by each one of them.
AABBCC can do a piece of work in xxyyzz days respectively. The ratio in which the amount earned should be shared is 1x:1y:1z=yz:zx:xy1x:1y:1z=yz:zx:xy
• If people work for different number of days, ratio in which the total money earned has to be shared is the ratio of work done by each one of them.

## Special cases of time and work problems

• Given a number of people work together/alone for different time periods to complete a work, for eg: AA and BB work together for few days, then CC joins them, after few days BB leaves the job. To solve such problems, following procedure can be adopted.
1. Let the entire job be completed in DD days.
2. Let sum of parts of the work completed by each person = 1.
3. Find out part of work done by each person with respect to DD. This can be easily found out if you calculate how many days each person worked with respect to DD.
4. Substitute values found out in Step 3 in Step 2 and solve the equation to get unknowns.
• A certain no of men can do the work in DD days. If there were mm more men, the work can be done in dddays less. How many men were there initially?
Let the initial number of men be MM
Number of man days to complete work = MDMD
If there are M+mM+m men, days taken = DdD-d
So, man days = (M+m)(Dd)(M+m)(D-d)
ie; MD=(M+m)(Dd)MD=(M+m)(D-d)
M(D(Dd))=m(Dd)M(D–(D-d))=m(D-d)

M=m(Dd)dM=m(D-d)d
• A certain no of men can do the work in DD days. If there were mm less men, the work can be done in dddays more. How many men were initially?
Let the initial number of men be MM
Number of man days to complete work = MDMD
If there are MmM-m men, days taken = D+dD+d
So, man days = (Mm)(D+d)(M-m)(D+d)
ie; MD=(Mm)(D+d)MD=(M-m)(D+d)
M(D+dD)=m(D+d)M(D+d–D)=m(D+d)

M=m(D+d)dM=m(D+d)d
• Given AA takes aa days to do work. BB takes bb days to do the same work. Now AA and BB started the work together and nn days before the completion of work AA leaves the job. Find the total number of days taken to complete work?
Let DD be the total number of days to complete work.
AA and BB work together for DnD-n days.
So, (Dn)(1a+1b)+n(1b)=1(D-n)(1a+1b)+n(1b)=1
D(1a+1b)nanb+nb=1D(1a+1b)–na-nb+nb=1

D=b(n+a)a+bD=b(n+a)a+b days.

## Illustrations Related To Time & Work-

Illustration 1: A can do a peice of work in 4 days 5 hours each and B can do it in 5 days 6 hours each. How long will they take to do it, working together 1/6 hours a day?

A can complete the work in 4 × 5 = 20 hours.

B can complete the work in 5 × 6 = 30 hours.

A’s 1 hours work = 1/20 and B’s 1 hour’s work = 1/30

(A + B)’s 1 hour’s work = (1/20 + 1/30) => 1/12

Both will finish the work in 12 hours

Number of days of 1/6 hrs. each = (12 × 1/6) => 2 days.

Illustration 2: 3 men and 5 boys can do a peice of work in 20 days while 5 men and 3 boys can do same work in 15 days. In how many days can 4 men and 2 boy do the work?

let 1 men’s 1 day’s work = X and 1 boy’s 1 day’s work = Y

Then, 3X + 5Y = 1/20

And 5X + 3Y = 1/15

Solving, We get :

X = 13/64

Y = 1/320

(4 men + 2 boy)’s 1 day’s work = (4 × 13/64 + 2 × 1/320) = 131/160

So, 4 men And 2 boy together can finish the work in 160/131 = 29 days

## Dr. R. S. Aggarwal Objective Type Questions, Try Yourself-

Question 1: A, B and c can complete a piece of work in 24, 6 and 12 days respectively. working together, they will complete the same work in:

(a) 1/24 days (b) 7/24 days (c) 3 3/7 days (d) 4 days

Question 2: A man can do a piece of work in 5 days, but with the help of his son, he can do it in 3 days. In what time can the son do it alone

(a) 6 1/2 days (b) 7 days (c) 7 1/2 days (d) 8 days