Sunday , September 24 2017
Recent Post
Home / Crack IBPS PO Bag / Time and Work Tricks & Tips

# Time and Work Tricks & Tips

Time and Work is a very easy topic and at least 2 Questions are asked from this topic in Any Exam whether it’s Clerk or PO or SSC.What actually the Time and Work Topic is all about? In simple words It’s how much time a person will take to Complete a Given Work.

Here we are providing some cases of time and work and efficient method to solve it. Hope this would be helpful to all aspirants.

These Tricks are shared by one of the our ardent BA reader Insomniac.We wish you good luck for your future.

## Note -: In conventional Method work is always treated as 1

Example: So if I say that a person can complete a work in 15 days that means he will do 1/15 work in one day, It’s simple maths.
Now another person does the same work in 30  days. So he will do 1/30 work in 1 day.
Now if i ask in how many days they will complete the work together. What we gonna do is Add their 1 day of work
like 1/15 + 1/30 = (2+1)/30 = 3/30 = 1/10
Now this 1/10 we got is actually their 1 day work, So if they do 1/10 work in one day then it’s simple they will complete the whole work in 10 days.
Now that was the conventional method and I believe that you all know how to solve Questions through Conventional method.
So now lets move on to the Faster method i.e efficiency method.
In efficiency method the Work is not treated in numerical value, Like in Conventional method the work is 1 but here the work is treated as percentage.
So by common sense the work is always treated as 100%
So when i say a person completes a work in 15 days it means he will do 100/15 % work in 1 day i.e 6.66% work in 1 day
If another person does the work in 30 days that means he will do 3.33% work in 1 day.
And together they will do 6.66 + 3.33 = 9.99 or 10% work in one day So in how many days they will do the complete work, that will be 100/10 = 10 days.
Now it may sound difficult That we have to convert Each value in % but don’t worry you don’t have to convert each value, You just have to learn all the values till 1/30 and then it will be a cakewalk.

Now we will take Some standard Cases Of time and work and you all are free to ask any problem if you have in any case.

### Question- A completes the work in 10 days and B completes the work in 15 days In how many days they will complete the work.

Conventional Method
Work done by A in 1 day = 1/10
Work Done by B in 1 day = 1/15
Work done By A & B together in 1 day = 1/10 + 1/15 = (3+2)/30 = 5/30 = 1/6

As A & B Completes 1/6 work in one day So they will complete the whole work in 6 Days.

Efficiency method.
Efficiency of A =100/10 = 10%
Efficiency of B = 100/15 = 6.66%
Efficiency of A & B Together = 10+ 6.66 = 16.66%
So the time taken by A & B together to Complete the work will be 100/16.66 = 6 Days.

### Case -2 A can do a work in X days and B can do it Y days, In how many days the work is completed if they work alternatively Started by A.

Now in these type of question the person are not actually working together, what happens in this type of question is that A works for 1 day and then on 2nd day B work and on 3rd again A work and on Fourth again B works and so on till the work is completed.
For example A can do a work in 10 days B can do it 15 days and how many days they will finish it if The work is started by A
So again work done by A in one day = 1/10
”     ”          ”        ”      ”  B      ”      ”      = 1/15
Now the work done by Togther will be = 1/10 + 1/15 = 1/6 [ Note now this 1/6 work is not done by them in 1 day but in 2 days Actually, See A worked for 1 day and did 1/10 work on the second day B worked and finished the 1/15 work So in total 2 days they did 1/6 work]
So in 2 days they did 1/6 work so in how many days they will complete the whole work, Simple 12 days.
Efficienecy Method
A’s Efficiency = 10%
B’s Efficiency = 6.66%
A + B Efficiency = 16.66%
Work done by A and B in 2 days [ remember 2 days because they are not working together but working alternatively] = 16.66%
So time taken by them to complete 100% work = 100/(16.66 = 6 [ but always remember multiply this by 2, Beacause 16.66% work is done by them in 2 days and not in 1 day.
So The answer will be 6*2 = 12 days.

### Question- A can do a work in 10 days and B can do it in 15 days, A works for 2 days and then leaves, In how many days will the work be completed?

Now here we can se that A leaves after 2 days that means A and B only worked for 2 days and the remaining work is done by B alone.
So first we have to calculated the work done by A and B together in 2 days.
So work done be A in 1 day = 1/10
“”   ”         ”        ”  B ”  ”   ”     = 1/15
Work done by A & B together is 1 day = 1/10 + 1/15 = 1/6
Work done by A & B together in 2 days = (1/6) * 2 = 1/3
So remaining work = 1 – 1/3 = 2/3
Now this 2/3 work has to be done by B alone.
So it takes 15 days for B to do Complete work, How much time it will be taken by B to do 2/3 work ? So it will be 15*(2/3) = 10 days
So the work will be completed in 2 + 10 days = 12 days
Efficiency method
A’s efficiency = 10%
B’s Efficiency = 6.66%
Total a+b = 16.66%
work done by A and B together in 2 days = 16.66*2 = 33.33%
Work remaining = 66.66%
time taken by B to complete 66.66% work = 66.66/6.666 = 10 days
Total time = 10+2 = 12 days

### A can do a piece of Work in 10 days and B can do it in 15 days, In how many days will the work be completed if B leaves 2 days before the completion on work.

Now in this question B leaves before 2 days
Together they can do the work in what = 1/10 + 1/15 = 1/6
That means 6 Days.
That means Together they could have completed the work in 6 days but B works only till 4th day and The remaining work will be done by A alone
So they worked together for 4 days in all So work done by them in 4 days = (1/6)*4 = 4/6 = 2/3
remaining work = 1/3
Now this 1/3 work will be done by A alone
Now A can do the complete work in 10 days, So the time taken by him to do 1/3 work = 10 * (1/3) = 10/3 days or 3.33 days
So total time taken = 4+ 3.33 days = 7.33 days
Efficiency method
A’s efficiency = 10%
B’s efficiency = 6.66%
Total = 16.66%
Work will be completed in 6 days
Work done in 4 days = 66.66%
remaining = 33.33%
A will complete the remaining in = 33.33/10 = 3.33
Total = 4+3.33 = 7.33

### Question: A can do a work in 10 days B can do it in 15 days, In how many days The work will get completed if B leaves 2 days before the Actual Completion of Work.what is the difference between this Actual completion of work and Completion of Work?

See in last example the work was supposed to get completed in 6 days, So we just Solved the question taking into consideration that B leaves 2 days before the completion of work i.e B worked for 4 days and the rest 2 days work was don by A alone and Completes that work in 3.33 days Total 7.33 days.
So if i ask In this question If B left 2 days before the actual completion then it means B should have left on 5.33 days Got it ?
Now back to the question.
Now just imagine that the work gets completed in x days.
So A would work for x days[ A works for the whole time ]
And B would work for x-2 days[ because B left 2 days before the actual completion of work]
So now According to Question
x/10 + (x-2)/15 = 1 [ Beacuse work is always one ]
(3x+2x-4)/30 = 1
5x -4 = 30
5x = 34
x = 6.8 days.
So the work will be completed in 6.8 Days.
It can also be asked That after how many days B left, So the answer would be Simple 6.8 – 2 = 4.8 days
Efficiency Method
A’s Efficiency = 10%
B’s Efficiency = 6.66%
Let the no. of days be x
so Accordring to question
10x + 6.66(x-2) = 100[ Work is always 100% in efficiency method ]
10x + 6.66x – 13.33 = 100
16.66x = 113.33
x = 113.33/16.66 = 6.8

### SHORT CUT TRICKS FOR WORK AND TIME

In Time And Work Type-1 I am going to discuss some important Time and Work Problem Shortcut Tricks which are frequently asked in Bank PO, SSC CGL, SSC CHSL, Railways etc Exam.

Trick-1: If A can do a certain work in x days and B can complete same work in y days if they work together then they will complete the work in

For Example:

(Q1)A can do a certain work in 10 days and B can complete same work in 15 days if they work together then in how many days works gets done.

Solution:

(A+B)’s 1 day work=3+2=5 unit

Time taken by (A+B)=30/5

= 6 days Ans.

Rough:[L.C.M of 10 and 15=30 now 30/10=3 unit and 30/15=2 unit]

BY TRICK:

Trick-2: If A, B, and C can do a certain work in x days, y days, and z days respectively. if they work together then they will complete the work in

For Example:

(Q1) Ram, Mohan, and Sita can do a work in 6, 12 and 24 days respectively. In what time will they finish the work while working together?

Solution:

(Ram+Mohan+Sita) 1 day work=4+2+1

=7 unit

BY TRICK:

Trick-3: A and B together can do a work in x days, B and C together in y days, while C and A together in z days. If they all work together, the work will be completed in.

For Example:

(Q1)A and B together can do a work in 8 days, B and C together in 6 days, while C and A together in 10 days. If they all work together, the work will be completed in.

Solution:

(A+B)+(B+C)+(C+A) 1 day work=15+20+12=47 units

i.e 2(A+B+C)1 day work=47 units

∴ (A+B+C)1 day work=47/2 units

Time Taken by(A+B+C)=120/(47/2)

BY TRICK:

Common Type Of Time and Work Problem Shortcut Tricks That Are Frequently Asked In Various Competitive Exam

Trick-4: A and B together can do a piece of work in x days. If A alone can do it in y days, in how many days B can alone complete the same piece of work

For Example:

(Q1)A and B together can do a piece of work in 4 days. If A alone can do it in 6 days, in how many days B can alone complete the same piece of work?

Solution:

Time taken by B=12/1

= 12 Days Ans.

By Trick:

(Q2)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?

Solution:

Time taken by son alone=15/2

=7.5 Days Ans.

By Trick:

(Q3)A can do a piece of work in 16 days and B can do the same piece of work in 12 days. With the help of C, they did in 4 days only. Then, C alone can do the job in?

Solution:

Time taken by C=48/5 days Ans.

Common Type Of Time and Work Problem Shortcut Tricks That Are Frequently Asked In Various Competitive Exam

Trick-5: If A and B together can complete a piece of work in x days; B and C together in y days; and C and A together in z days. How many days will each take separately to complete the same work?

(Q1)A and B together can complete a piece of work in 24 days; B and C together in 30 days; and C and A together in 40 days. How many days will each take separately to complete the same work?

Solution:

(A+B)+(B+C)+(C+A) 1 day work=5+4+3=12 units

i.e 2(A+B+C)1 day work=12 units

∴ (A+B+C)1 day work=12/2 units

=6 units.

Efficiency of A=(A+B+C)-(B+C)

=6 Units – 4 Units

=2 Units

Efficiency of B=(A+B+C)-(C+A)

=6 Units – 3 Units

=3 Units

Efficiency of C=(A+B+C)-(A+B)

=6 Units – 5 Units

=1 Units

Time taken by A=120/2

=60 Days

Time taken by B=120/3

=40 Days

Time taken by A=120/1

=120 Days

BY TRICK:

(Q2)A and B can do a given piece of work in 8 days, B and C can do the same work in 12 days and A, B, C complete it in 6 days. Number of days required to finish the work by A and C is.

Solution:

Efficiency of A=(A+B+C)-(B+C)

=4 Units – 2 Units

=2 Units

Efficiency of C=(A+B+C)-(A+B)

=4 Units – 3 Units

=1 Units

(A+C)’s 1 day work=2+1=3 unit

Time taken by( A+C)=24/3

=8 days Ans.

The common and frequently asked a question of Time and work problem with solution and Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-2.

For Example:

(Q1)Ram can complete a piece of work in 15 days. He worked for 5 days and left. Mohan alone completed the remaining work in 20 days. How long would Mohan take to complete the whole work?

Solution:

Let Ram per day done 1 unit work

Total work done by Ram in 15 days=15 unit

Total work done by Ram in 5 days=5 unit

Remaining work=15-5=10 unit

work done by Mohan in 20 days=10 unit

work done by Mohan in 1 day=10/20=0.5

Time taken by Mohan=15/0.5

=30 days Ans.

(Q2)Ram alone can complete a piece of work in 40 days. He worked for 8 days and left. Sita alone completed the remaining work in 16 days. How long would Ram and Sita together take to complete the work?

Solution:

Let Ram per day done 1 unit work

Total work done by Ram in 40 days=40 unit

Total work done by Ram in 8 days=8 unit

Remaining work=40-8=32 unit

work done by Sita in 16 days=32 unit

work done by Sita in 1 day=32/16

=2 unit per day

Total work done by Ram and Sita in 1 day=(1+2)=3unit per day

Time taken by Ram+Sita=40/3

=13 ¹/3 Ans.

The common and frequently asked a question of Time and work problem with solution and Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-2.1.

(Q1)A can finish a work in 18 days and B can do the same work in 15 days. B worked for 10 days and left the job. In how many days, A alone can finish the remaining work?

Solution:

Let A= A unit work per day

B= B unit work per day

Now,

[Concept- above equation से हमलोगों ने find out कर लिया कि A एक दिन में 5 unit कम करता है and B एक दिन में 6 unit कम करता है| अब A का value eqn.-1 में रखकर या B का value eqn.-2 में रखकर total work निकाल सकते है ]

Put the value of A in eqn-1, you will get total work

so, total work=5×18=90 unit

Total work done by B in 10 days=10×6=60 unit

Remaining work=90-60

=30 unit

Time taken by A=30/5

=6 days Ans.

Method-2

Total work done by B in 10 days=10×6=60 unit

Remaining work=90-60=30 unit

=30 unit

Time taken by A=30/5

=6 days Ans.

The common and frequently asked a question of Time and work problem with solution and Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-2.2.

(Q1)A can do a certain work in 10 days and B can do same work in 12 days. They started work together but after 5 days A left and B finish the remaining work. In how many days B can complete the remaining work?

Solution:

Let A= A unit work per day

B= B unit work per day

Now,

[Concept- above equation से हमलोगों ने find out कर लिया कि A एक दिन में 6 unit कम करता है and B एक दिन में 5 unit कम करता है| अब A का value eqn.-1 में रखकर या B का value eqn.-2 में रखकर total work निकाल सकते है ]

Put the value of A in eqn-1, you will get total work

so, total work=6×10=60 unit

Total work done by A and B in 1 day=(6+5)=11 unit

[Concept: 5 days बाद A चला गया it means A के जाने के पहले दोनों(A and B) 5 days तक एक साथ कम किया होगा तो दोनों मिलकर 5 दिन में कितना कम किया ये हमलोग निकाल लेंगे or फिर उसके बाद total work में से A and B का 5 days का work minus कर देंगे तो इससे Remaining work निकल जाएगा]

work done by A and B in 5 days=11×5=55 unit

Remaining work=60-55

=5 unit

Time taken by B=5/5

=1 day Ans.

Method-2

Total work done by A and B in 1 day=(6+5)=11 unit

work done by A and B in 5 days=11×5=55 unit

Remaining work=60-55

=5 unit

Time taken by B=5/5

=1 day Ans.

(Q2)A and B can complete a work in 15 days and 10 days respectively. They started doing the work together but after 2 days B had to leave and A alone completed the remaining work. The whole work was completed in:

Solution:

Let A= A unit work per day

B= B unit work per day

Now,

[Concept- above equation से हमलोगों ने find out कर लिया कि A एक दिन में 2 unit कम करता है and B एक दिन में 3 unit कम करता है| अब A का value eqn.-1 में रखकर या B का value eqn.-2 में रखकर total work निकाल सकते है ]

Put the value of A in eqn-1, you will get total work

so, total work=2×15=30 unit

Total work done by A and B in 1 day=(2+3)=5 unit

[Concept:2 days बाद B चला गया it means B के जाने के पहले दोनों(A and B) 2 days तक एक साथ कम किया होगा तो दोनों मिलकर 2 दिन में कितना कम किया ये हमलोग निकाल लेंगे or फिर उसके बाद total work में से A and B का 2 days का work minus कर देंगे तो इससे Remaining work निकल जाएगा]

Total work done by A and B in 2 days=5×2=10 unit

Remaining work=30-10

=20 unit

Time taken by A=20/2

=10 days

Hence, total time taken=(2+10)=12days Ans.

Method-2

Total work done by A and B in 1 day=(2+3)=5 unit

Total work done by A and B in 2 days=5×2=10 unit

Remaining work=30-10

=20 unit

Time taken by A=20/2

=10 days

Hence, total time taken=(2+10)=12days Ans.

(Q3)A and B can do a piece of work in 45 and 40 days respectively. They began the work together but A left after some days and B finished the remaining work in 23 days. A left after.

Solution:

Let A= A unit work per day

B= B unit work per day

Now,

[Concept- above equation से हमलोगों ने find out कर लिया कि A एक दिन में 8 unit कम करता है and B एक दिन में 9 unit कम करता है| अब A का value eqn.-1 में रखकर या B का value eqn.-2 में रखकर total work निकाल सकते है ]

Put the value of A in eqn-1, you will get total work

so, total work=8×45=360 unit

[Concept:A कुछ days के बाद चला गया it means A के जाने के बाद सिर्फ B बचा|अब B remaining work को 23 days में complete कर देता है| B एक दिन में 9 unit काम करता है तो 23 दिन में =23×9=207 unit. A के जाने पहले दोनों(A and B) एक साथ कम किया होगा पर कितना दिन तक एक साथ work किया है ये हमलोगों को मालूम नहीं है पर हमलोग ये तो मालूम कर सकते है कि दोनों (A+B)ने मिलकर कितना unit work किया है| so total work done by (A+B)= 360-207=153 unit.]

Total work done by B in 23days=23×9=207 unit

Total Work done by (A+B)=360-207

=153 unit

Total work done by A+B in 1 day=(8+9)=17 unit

Time taken by A+B=153/17

=9 days

hence, A left after 9 days Ans.

Method-2

Total work done by B in 23days=23×9=207 unit

Total Work done by (A+B)=360-207

=153 unit

Total work done by A+B in 1 day=(8+9)=17 unit

Time taken by(A+B)=153/17

=9 days

hence, A left after 9 days Ans.

(Q4)A can do a piece of work in 20 days while B can finish it in 25 days. B started working and A joined him after 10 days. The whole work is completed in.

Solution:

Let A= A unit work per day

B= B unit work per day

Now,

[Concept- above equation से हमलोगों ने find out कर लिया कि A एक दिन में 5 unit कम करता है and B एक दिन में 4 unit कम करता है| अब A का value eqn.-1 में रखकर या B का value eqn.-2 में रखकर total work निकाल सकते है ]

Put the value of A in eqn-1, you will get total work

so, total work=5×20=100 unit

[Concept- Now 10 दिन बाद A ने join किया तो B अकेला 10 days तक work किया है so total work done by B in 10 days =10×4=40 unit. Remaining work=100-40=60 unit, 60 unit work अब A and B दोनों मिलकर करेगा|]

work done by B in 10 days=10×4=40 unit

Remaining work=100-40=60 unit

Total work done by A+B in 1 day=(5+4)=9 unit

Time taken by(A+B)=60/9

=20/3=6 ²/3 days

Total time=10+6 ²/3

=16 ²/3 days Ans.

Method-2

work done by B in 10 days=10×4=40 unit

Remaining work=100-40=60 unit

Total work done by A+B in 1 day=(5+4)=9 unit

Time taken by(A+B)=60/9

=20/3=6 ²/3 days

Total time=10+6 ²/3

=16 ²/3 days Ans.

(Q5) A and B can together finish a work in 30 days. They worked at it for 20 days and then B left. The remaining work was done by A alone in 20 more days. A alone can finish the work in.

Solution:

Let work done by (A+B) in per day=1 unit

Total work done by (A+B) in 30 days=30 unit

Total work done by (A+B) in 20 days=20 unit

Now,

Remaining work=30-20=10 unit

work done by A in 20 days=10 unit

∴work done by A in 1 day=10/20 unit

=0.5 unit

Time taken by A=30/0.5

=60 days Ans.

(Q6) A can do a piece of work in 24 days. When he had worked for 4 days. B joined him. If complete work was finished in 16 days, B can alone finish that work in.

Solution:

Let work done by A in per day=1 unit

Total work done by A in 24 days=24 unit

[Concept:जब A 4 days का काम कर चुका था तो B join किया| so work done by A in 4 days=4 unit, then remaining work=24-4=20 unit| 20 unit दोनों(A+B) ने मिलकर किया है| अब question में कह रहा है total work i.e 24 unit work 16 दिन में complete हो रहा है जिसमे 4 days सिर्फ A ने work किया है or remaining 12 days A or B दोनों ने मिलकर किया है |]

work done by A in 4 days=4 unit

remaining work=24-4=20 unit

work done by A+B in 12 days=20 unit

work done by A+B in 1 days=20/12

Now,

A+B=20/12

1+B=20/12

∴ B=2/3

Time taken by B=24/(2/3)

=36 days Ans.

The common and frequently asked a question of Time and work problem with solution and Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-2.3.

(Q1)A and B alone can complete work in 9 days and 18 days respectively. They worked together; however 3 days before the completion of the work A left. In how many days was the work completed?

Solution:

Let A= A unit work per day

B= B unit work per day

Now,

[Concept- above equation से हमलोगों ने find out कर लिया कि A एक दिन में 2 unit कम करता है and B एक दिन में 1 unit कम करता है| अब A का value eqn.-1 में रखकर या B का value eqn.-2 में रखकर total work निकाल सकते है ]

Put the value of A in eqn-1, you will get total work

so, total work=2×9=18 unit

[Concept: A work complete होने के 3 days पहले चला जाता है तो A के चले जाने के बाद सिर्फ B ने 3 days तक work किया है so total work done B in 3 days=3×1=3 unit. total work=2×9=18 unit so remaining work=18-3=15 unit| 15 unit work A and B ने मिलकर किया है]

work done by B in 3 days=3×1=3 unit

Remaining work=18-3=15 unit

Total work done by A+B in 1 day=(2+1)=3 unit

Time taken by(A+B)=15/3

=5 days

Total time=3+5

=8 days Ans.

Method-2

work done by B in 3 days=3×1=3 unit

Remaining work=18-3=15 unit

Total work done by A+B in 1 day=(2+1)=3 unit

Time taken by(A+B)=15/3

=5 days

Total time=3+5

=8 days Ans.

The common and frequently asked a question of Time And Work Problem with solution and Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-3.

For Example:

(Q1)A, B, and C can do a piece of work in 10, 12, and 15 days respectively.They began the work together but B leaves after two days, how long would it take A and C to finish the Remaining work?

Solution:

Note: [L.C.M of 10,12,15=60 and 60/10=6, 60/12=5, 60/15=4]

Total work done by A, B, and C in 1 day=(6+5+4)unit=15 unit

[Concept: B 2 days बाद work छोड़ देता है, it means B के  work छोडने से पहले A, B, and C तीनों 2 days तक एक साथ work किया होगा| so total work done by A, B, and C in 2 days=15×2=30 unit. Remaining work=60-30=30 unit. जब B work छोड़ दिया तो सिर्फ अब A and C बचा है तो total work done by A and C in 1 day=6+4=10 unit ]

Total work done by A, B, and C in 2 day=15×2=30 unit

Remaining work=60-30=30 unit

When B left the job then

Total work done by A and C in 1 day=6+4=10 unit

Time taken by A and C=30/10

=3 days Ans.

(Q2)A, B, and C can do a piece of work in 20, 25, and 30 days respectively.They began the work together but C leaves after 3 days. In how many days the whole work was completed?

Solution:

Note: [L.C.M of 20,25,30=300 and 300/20=15, 300/25=12, 300/30=10]

Work done by A, B, and C in 1 day=(15+12+10)unit=37 unit

[Concept: C 3 days बाद work छोड़ देता है, it means C के  work छोडने से पहले A, B, and C तीनों 3 days तक एक साथ work किया होगा| so total work done by A, B, and C in 3 days=37×3=111 unit. Remaining work=300-111=189 unit. जब C work छोड़ दिया तो सिर्फ अब A and B बचा है तो total work done by A and B in 1             day=15+12=27 unit.]

Total work done by A, B, and C in 3 day=37×3=111 unit

Remaining work=300-111=189 unit

When C left the work then

Work done by A+B in 1 day=15+12=27 unit

Time taken by (A+B)=189/27

=7 days

The whole work was completed in=3+7

=10 days Ans.

(Q3)A, B, and C together can do a piece of work in 40 days. After working with A, B and C for 16 days, A leaves and then B and C complete the remaining work in 40 days more. A alone could do the work in.

Solution:

Let work done by (A+B+C) in per day=1 unit

Total work done by (A+B+C) in 40 day=40 unit

Total work done by (A+B+C) in 16 day=16 unit

Now,

Remaining work=40-16=24 unit

[Concept: 24 unit work अब सिर्फ B and C करेगा जिसे करने में B and C को 40 days लग रहा है क्योंकि A ने work छोड़ दिया है |]

work done by B and C in 40 days =24 unit

work done by B and C in 1 day =24/40

=0.6 unit

Efficiency(eff.) of A=Eff. of(A+B+C)-Eff. of (B+C)

= 1-0.6=0.4 unit per day

Time taken by A=40/0.4=100 days Ans.

(Q4)A, B, and C can do a job in 6 days, 12 days and 15 days respectively. After 1/8 of the work is completed, C leaves the job. Rest of the work is done by A and B together. Time taken to finish the work is.

Solution:

[Concept: 60×(1/8)=7.5 unit. जब 7.50 unit work complete हो गया तब C ने job छोड़ दिया now remaining work=60-7.50=52.50 unit. 52.50 unit work सिर्फ A and B मिलकर कर रहा है|]

Now,

60×(1/8)=7.5 unit

Remaining work=60-7.50=52.50 unit

Work done by A +B in 1 days=10+5=15 unit

Time taken by (A+B)=52.50/15

=7/2=3 ½ days Ans.

The common and frequently asked a question of Time And Work Problem with solution and Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-3.1.

(Q1)A, B, and C can do a piece of work in 15, 20, and 12 days respectively. They began the work together but A left 5 days before completion of the work. In how many days was the work done?

Solution:

Note: [L.C.M of 15,20,12=60 and 60/15=4, 60/20=3, 60/12=5]

[Concept: A work complete होने के 5 days पहले चला जाता है तो work complete होने में जो 5 days बचा था तो उसमे सिर्फ B and C ने ही work किया so, work done by B and C in 1 days=3+5=8 unit ∴work done by B and C in 5 days=8×5=40 unit. remaining work 60-40=20 unit. 20 unit work A, B, and C ने मिलकर किया होगा ]

Work done by B and C in 1 day=(3+5)unit=8 unit

Total work done by B and C in 5 day=8×5=40 unit

Remaining Work=60-40=20 unit

Work done by A, B, and C in 1 day=(4+3+5)unit=12 unit

Now,

20 unit work is done by A, B and C so,

Time taken by (A+B+C)=20/12

=5/3 days

Total time=5+(5/3)

=20/3 days Ans.

(Q2)A, B, and C can do a piece of work in 24, 30, and 40 days respectively. They began the work together but C left 4 days before completion of the work. In how many days was the work done?

Solution:

[Concept: C work complete होने के 4 days पहले चला जाता है तो work complete होने में जो 4 days बचा था तो उसमे सिर्फ A and B ने ही work किया so, work done by A and B in 1 days=5+4=9 unit ∴work done by A and B in 4 days=9×4=36 unit. remaining work 120-36=84 unit. 84 unit work A, B, and C ने मिलकर किया होगा ]

Work done by A and B in 1 day=(5+4)unit=9 unit

Total work done by A and B in 4 day=9×4=36 unit

Remaining Work=120-36=84 unit

Work done by A, B, and C in 1 day=(5+4+3)unit=12 unit

Now,

84 unit work is done by A, B and C so,

Time taken by (A+B+C)=84/12

=7 days

Total time=4+7=11 days Ans.

The common and frequently asked a question of Time And Work Problem with solution and Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-3.2.

(Q1)A, B, and C can complete a work separately in 24, 36, and 48 days respectively. They started work together but C left after 4 days of start and A left 3 days before the completion of the work. In how many days will the work be completed.

Solution:

[Concept: C work start होने के 4 days बाद work छोड़ कर चला जाता है| it means C के जाने से पहले A, B, and C ने 4 days तक एकसाथ work किया है so, work done (A+B+C) in 1 day=(6+4+3)=13 unit ∴work done (A+B+C) in 4 day=13×4=52 unit Remaining work=144-52=92 unit.]

Work done (A+B+C) in 1 day=(6+4+3)=13 unit

Work done (A+B+C) in 4 day=13×4=52 unit

Remaining work=144-52=92 unit.

[concept: C तो पहले ही जा चुका है अब सिर्फ A and B बचा है. now question में कह रहा है A work complete होने से 3 days पहले चला जाता है तो last 3 days सिर्फ B ने ही work किया होगा so work done by B  in 3 days=3×4=12 unit. remaining work=92-12=80 unit. 80 unit work A and B ने मिलकर किया है |]

Now,

Work done by B  in 3 days=3×4=12 unit

Remaining work=92-12=80 unit

Work done by (A+B) in 1 day=(6+4)=10 unit

Time taken by (A+B)=80/10

=8 days

Total time=(4+3+8)=15 days Ans.

(Q2)A, B, and C can do a work separately in 16, 32, and 48 days respectively. They started the work together but B left off 8 days and C 6 days before the completion of the work. In what time is the work finished?

Solution:

[Concept B work complete होने के 8 days before चला जाता है तो work करने के लिए अब सिर्फ A and C बचा| so work done by (A+C) in 1 day=6+2=8 unit ∴ work done by (A+C) in 2 day=8×2=16 unit. remaining work=96-16=80 unit. यहाँ पर A and C का 2 days का ही work निकाला गया है क्योंकि question में कह रहा है कि work complete होने से 6 days before फिर C चला जाता है that means B के जाने के 2 days बाद ही C चला जाता है| अब last का 6 days work करने के लिए सिर्फ A ही बच जाता है| so work done by A in 6 day=6×6=36 unit. remaining work=80-36=44 unit.ये 44 unit work A, B and C ने मिलकर किया है |]

Work done by (A+C) in 1 day=6+2=8 unit

∴ Work done by (A+C) in 2 day=8×2=16 unit.

Remaining work=96-16=80 unit.

Work done by A in 6 day=6×6=36 unit

Remaining work=80-36=44 unit.

Time taken by (A+B+C)=44/11

=4 days

Total days=2+6+4=12 days Ans.

Method-2

Let the whole work be completed in x days

According to the question,

(Q3)A, B and C can complete a work in 10, 12 and 15 days respectively. A left the work 5 days before the work was completed and B left 2 days after A had left. Number of days required to complete the whole work is:

Solution:

[Concept- A work complete होने के 5 days before चला जाता है तो work करने के लिए अब सिर्फ B and C बचा| so work done by (B+C) in 1 day=5+4=9 unit ∴ work done by (B+C) in 2 day=9×2=18 unit. remaining work=60-18=42 unit. यहाँ पर B and C का 2 days का ही work निकाला गया है क्योंकि question में कह रहा है कि B left 2 days after A had left means A के जाने के 2 days बाद ही B चला जाता है that means work complete होने में 3 days बचा ही था कि B work छोड़ चला जाता है | अब last का 3 days work करने के लिए सिर्फ C ही बच जाता है| so work done by C in 3 day=4×3=12 unit. remaining work=42-12=30 unit.ये 30 unit work A, B and C ने मिलकर किया है |]

Work done by (B+C) in 1 day=5+4=9 unit

∴ Work done by (B+C) in 2 day=9×2=18 unit.

Remaining work=60-18=42 unit

now,

Work done by C in 3 day=4×3=12 unit

Remaining work=42-12=30 unit

Time taken by (A+B+C)=30/15

=2 days

Total days=2+3+2=7 days Ans.

Method-2

Let the whole work be completed in x days

According to the question,

The common and frequently asked a question of Time And Work Problem with solution and Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-3.

For Example:

(Q1)A, B, and C can do a piece of work in 10, 12, and 15 days respectively.They began the work together but B leaves after two days, how long would it take A and C to finish the Remaining work?

Solution:

Note: [L.C.M of 10,12,15=60 and 60/10=6, 60/12=5, 60/15=4]

Total work done by A, B, and C in 1 day=(6+5+4)unit=15 unit

[Concept: B 2 days बाद work छोड़ देता है, it means B के  work छोडने से पहले A, B, and C तीनों 2 days तक एक साथ work किया होगा| so total work done by A, B, and C in 2 days=15×2=30 unit. Remaining work=60-30=30 unit. जब B work छोड़ दिया तो सिर्फ अब A and C बचा है तो total work done by A and C in 1 day=6+4=10 unit ]

Total work done by A, B, and C in 2 day=15×2=30 unit

Remaining work=60-30=30 unit

When B left the job then

Total work done by A and C in 1 day=6+4=10 unit

Time taken by A and C=30/10

=3 days Ans.

(Q2)A, B, and C can do a piece of work in 20, 25, and 30 days respectively.They began the work together but C leaves after 3 days. In how many days the whole work was completed?

Solution:

Note: [L.C.M of 20,25,30=300 and 300/20=15, 300/25=12, 300/30=10]

Work done by A, B, and C in 1 day=(15+12+10)unit=37 unit

[Concept: C 3 days बाद work छोड़ देता है, it means C के  work छोडने से पहले A, B, and C तीनों 3 days तक एक साथ work किया होगा| so total work done by A, B, and C in 3 days=37×3=111 unit. Remaining work=300-111=189 unit. जब C work छोड़ दिया तो सिर्फ अब A and B बचा है तो total work done by A and B in 1             day=15+12=27 unit.]

Total work done by A, B, and C in 3 day=37×3=111 unit

Remaining work=300-111=189 unit

When C left the work then

Work done by A+B in 1 day=15+12=27 unit

Time taken by (A+B)=189/27

=7 days

The whole work was completed in=3+7

=10 days Ans.

(Q3)A, B, and C together can do a piece of work in 40 days. After working with A, B and C for 16 days, A leaves and then B and C complete the remaining work in 40 days more. A alone could do the work in.

Solution:

Let work done by (A+B+C) in per day=1 unit

Total work done by (A+B+C) in 40 day=40 unit

Total work done by (A+B+C) in 16 day=16 unit

Now,

Remaining work=40-16=24 unit

[Concept: 24 unit work अब सिर्फ B and C करेगा जिसे करने में B and C को 40 days लग रहा है क्योंकि A ने work छोड़ दिया है |]

work done by B and C in 40 days =24 unit

work done by B and C in 1 day =24/40

=0.6 unit

Efficiency(eff.) of A=Eff. of(A+B+C)-Eff. of (B+C)

= 1-0.6=0.4 unit per day

Time taken by A=40/0.4=100 days Ans.

(Q4)A, B, and C can do a job in 6 days, 12 days and 15 days respectively. After 1/8 of the work is completed, C leaves the job. Rest of the work is done by A and B together. Time taken to finish the work is.

Solution:

[Concept: 60×(1/8)=7.5 unit. जब 7.50 unit work complete हो गया तब C ने job छोड़ दिया now remaining work=60-7.50=52.50 unit. 52.50 unit work सिर्फ A and B मिलकर कर रहा है|]

Now,

60×(1/8)=7.5 unit

Remaining work=60-7.50=52.50 unit

Work done by A +B in 1 days=10+5=15 unit

Time taken by (A+B)=52.50/15

=7/2=3 ½ days Ans.

The common and frequently asked a question of Time And Work Problem with solution and Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-3.1.

(Q1)A, B, and C can do a piece of work in 15, 20, and 12 days respectively. They began the work together but A left 5 days before completion of the work. In how many days was the work done?

Solution:

Note: [L.C.M of 15,20,12=60 and 60/15=4, 60/20=3, 60/12=5]

[Concept: A work complete होने के 5 days पहले चला जाता है तो work complete होने में जो 5 days बचा था तो उसमे सिर्फ B and C ने ही work किया so, work done by B and C in 1 days=3+5=8 unit ∴work done by B and C in 5 days=8×5=40 unit. remaining work 60-40=20 unit. 20 unit work A, B, and C ने मिलकर किया होगा ]

Work done by B and C in 1 day=(3+5)unit=8 unit

Total work done by B and C in 5 day=8×5=40 unit

Remaining Work=60-40=20 unit

Work done by A, B, and C in 1 day=(4+3+5)unit=12 unit

Now,

20 unit work is done by A, B and C so,

Time taken by (A+B+C)=20/12

=5/3 days

Total time=5+(5/3)

=20/3 days Ans.

(Q2)A, B, and C can do a piece of work in 24, 30, and 40 days respectively. They began the work together but C left 4 days before completion of the work. In how many days was the work done?

Solution:

[Concept: C work complete होने के 4 days पहले चला जाता है तो work complete होने में जो 4 days बचा था तो उसमे सिर्फ A and B ने ही work किया so, work done by A and B in 1 days=5+4=9 unit ∴work done by A and B in 4 days=9×4=36 unit. remaining work 120-36=84 unit. 84 unit work A, B, and C ने मिलकर किया होगा ]

Work done by A and B in 1 day=(5+4)unit=9 unit

Total work done by A and B in 4 day=9×4=36 unit

Remaining Work=120-36=84 unit

Work done by A, B, and C in 1 day=(5+4+3)unit=12 unit

Now,

84 unit work is done by A, B and C so,

Time taken by (A+B+C)=84/12

=7 days

Total time=4+7=11 days Ans.

The common and frequently asked a question of Time And Work Problem with solution and Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-3.2.

(Q1)A, B, and C can complete a work separately in 24, 36, and 48 days respectively. They started work together but C left after 4 days of start and A left 3 days before the completion of the work. In how many days will the work be completed.

Solution:

[Concept: C work start होने के 4 days बाद work छोड़ कर चला जाता है| it means C के जाने से पहले A, B, and C ने 4 days तक एकसाथ work किया है so, work done (A+B+C) in 1 day=(6+4+3)=13 unit ∴work done (A+B+C) in 4 day=13×4=52 unit Remaining work=144-52=92 unit.]

Work done (A+B+C) in 1 day=(6+4+3)=13 unit

Work done (A+B+C) in 4 day=13×4=52 unit

Remaining work=144-52=92 unit.

[concept: C तो पहले ही जा चुका है अब सिर्फ A and B बचा है. now question में कह रहा है A work complete होने से 3 days पहले चला जाता है तो last 3 days सिर्फ B ने ही work किया होगा so work done by B  in 3 days=3×4=12 unit. remaining work=92-12=80 unit. 80 unit work A and B ने मिलकर किया है |]

Now,

Work done by B  in 3 days=3×4=12 unit

Remaining work=92-12=80 unit

Work done by (A+B) in 1 day=(6+4)=10 unit

Time taken by (A+B)=80/10

=8 days

Total time=(4+3+8)=15 days Ans.

(Q2)A, B, and C can do a work separately in 16, 32, and 48 days respectively. They started the work together but B left off 8 days and C 6 days before the completion of the work. In what time is the work finished?

Solution:

[Concept B work complete होने के 8 days before चला जाता है तो work करने के लिए अब सिर्फ A and C बचा| so work done by (A+C) in 1 day=6+2=8 unit ∴ work done by (A+C) in 2 day=8×2=16 unit. remaining work=96-16=80 unit. यहाँ पर A and C का 2 days का ही work निकाला गया है क्योंकि question में कह रहा है कि work complete होने से 6 days before फिर C चला जाता है that means B के जाने के 2 days बाद ही C चला जाता है| अब last का 6 days work करने के लिए सिर्फ A ही बच जाता है| so work done by A in 6 day=6×6=36 unit. remaining work=80-36=44 unit.ये 44 unit work A, B and C ने मिलकर किया है |]

Work done by (A+C) in 1 day=6+2=8 unit

∴ Work done by (A+C) in 2 day=8×2=16 unit.

Remaining work=96-16=80 unit.

Work done by A in 6 day=6×6=36 unit

Remaining work=80-36=44 unit.

Time taken by (A+B+C)=44/11

=4 days

Total days=2+6+4=12 days Ans.

Method-2

Let the whole work be completed in x days

According to the question,

(Q3)A, B and C can complete a work in 10, 12 and 15 days respectively. A left the work 5 days before the work was completed and B left 2 days after A had left. Number of days required to complete the whole work is:

Solution:

[Concept- A work complete होने के 5 days before चला जाता है तो work करने के लिए अब सिर्फ B and C बचा| so work done by (B+C) in 1 day=5+4=9 unit ∴ work done by (B+C) in 2 day=9×2=18 unit. remaining work=60-18=42 unit. यहाँ पर B and C का 2 days का ही work निकाला गया है क्योंकि question में कह रहा है कि B left 2 days after A had left means A के जाने के 2 days बाद ही B चला जाता है that means work complete होने में 3 days बचा ही था कि B work छोड़ चला जाता है | अब last का 3 days work करने के लिए सिर्फ C ही बच जाता है| so work done by C in 3 day=4×3=12 unit. remaining work=42-12=30 unit.ये 30 unit work A, B and C ने मिलकर किया है |]

Work done by (B+C) in 1 day=5+4=9 unit

∴ Work done by (B+C) in 2 day=9×2=18 unit.

Remaining work=60-18=42 unit

now,

Work done by C in 3 day=4×3=12 unit

Remaining work=42-12=30 unit

Time taken by (A+B+C)=30/15

=2 days

Total days=2+3+2=7 days Ans.

Method-2

Let the whole work be completed in x days

According to the question,

The common and frequently asked a question of Time And Work Problem with solution and Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-4.

For Example:

(Q1)2 men and 3 boys can complete a work in 10 days, while 3 men and 2 boys can complete in 8 days. In how many days can 2 men and 1 boy complete the same work?

Solution:

Let 1 man=m unit work per day

1 women=w unit work per day

Now,

Note:[ 1 man=7 unit work per day 1 boy=2 unit work per day]

[Concept:eqn-1 or eqn-2 में m and b का value रखने पर जो value प्राप्त होगा वह total work होगा |]

Put the value of m and b in eqn-1

=(2×7+3×2)×10

=200 unit i.e total work

Now,

work done by 2 men and 1 boy in 1 day

=2m+1b=(20×3+6×2)

=16 unit

so, total time taken by 2 men and 1 boy=200/16

=12 ½ days Ans.

(Q2)10 man and 15 women together can complete a piece of work in 6 days. A man alone can complete the work in 100 days. In how many days can one woman alone complete the work?

Solution:

Let 1 man=m unit work per day

1 women=w unit work per day

Note:[1 man=9 unit work per day1 woman=4 unit work per day]

[Concept:eqn-1 or eqn-2 में m and w का value रखने पर जो value प्राप्त होगा वह total work होगा |]

Put the value of m in eqn-2

=100×9

=900 unit i.e total work

Now,

1 woman=4 unit work per day

so, the total time taken by 1 woman=900/4

=225 days Ans.

The common and frequently asked a question of Time And Work Problem with solution and Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-4.1.

(Q1)If 8 men or 12 boys can do a piece of work in 16 days, the number of days required to complete the work by 20 men and 6 boys is.

Solution:

Let 1 man=m unit work per day

1 boy=b unit work per day

Now,

Note:[ 1 man=3 unit work per day 1 boy=2 unit work per day]

[Concept:eqn-1 or eqn-2 में m and b का value रखने पर जो value प्राप्त होगा वह total work होगा |]

Put the value of m in eqn-1

=(8×3)×16

=384 i.e total work

Now,

work done by 20 men and 6 boy in 1 day

=20m+6b=(20×3+6×2)

=72 unit

so, the total time taken by 20 men and 6 boy=384/72

=5 ¹/3 days Ans.

(Q2)If 4 men or 6 women can do a piece of work in 12 days working 7 hours a day; how many days will it take to complete a work twice as large with 10 men and 3 women working together 8 hours a day?

Solution:

Let 1 man=m unit work per day

1 boy=b unit work per day

Now,

Note:[ 1 man=3 unit work per day 1 women=2 unit work per day]

[Concept:eqn-1 or eqn-2 में m and w का value रखने पर जो value प्राप्त होगा वह total work होगा |]

Put the value of m in eqn-1

=(4×3)×84

=1008 unit i.e total work

Now,

work done by 10 men and 3 women in 1 day

=10m+3w=(10×3+3×2)

=36 unit

so, the total time taken by 10 men and 3women=(2×1008)/(36×8)

=7 days Ans.

(Q3)If 1 men or 2 women or 3 boys can do a piece of work in 44 days, then the same piece of work will be done by 1 man, 1 woman and 1 boy in.

Solution:

Let 1 men=m unit work per day

1 woman=w unit work per day

1 boy=b unit work per day

put the value of m in eqn-1

=(1×6)×44

=264 unit i.e total work

work done by 1 man, 1 women and 1 boy in 1 day

=1m+1w+1b=(1×6+1×3+1×2)

=11 unit

so, the total time taken by 1 man, 1 women and 1 boy=264/11

=24 days Ans.

The common and frequently asked a question of Time And Work Problem with solution and Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-4.2.

(Q1)Some staff promised to do a job in 18 days, but 6 of them went on leave. So the remaining men took 20 days to complete the job. How many men were there originally?

Solution:

Let number of men originally=x

∴ M1D1 = M2D2

⇒x×18 = (x-6)×20

⇒-2x = -144

∴ x=60 men Ans.

(Q2)A certain number of men can do a piece of work in 40 days. If there were 45 men more the work could have been finished in 25 days. Find the original number of men employed in the work.

Solution:

Let number of men originally=x

∴ M1D1 = M2D2

⇒x×40 = (x+45)×25

⇒15x =1125

∴ x=75 men Ans.

(Q3)A certain number of men can do a work in 40 days. If there were 8 men more, it could be finished in 10 days less. How many men were there initially?

Solution:

Let number of men originally=x

∴ M1D1 = M2D2

⇒x×40 = (x+8)×30

⇒10x =240

∴ x=24 men Ans.

The common and frequently asked a question of Time And Work Problem with solution and Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-4.3.

(Q1)A company employed 200 workers to complete a certain work in 150 days. If only one-fourth of the work has been done in 50 days, then in order to complete the whole work in time, the number of additional workers to be employed was.

Solution:

Let total work=1

200 workers will do 1/4 work in 50 days

Remaining work=1-(1/4)=3/4

Remaining time=150-50=100 days

Now,

x=100 Ans.

(Q2)A contractor was engaged to construct a road in 16 days. After working for 12 days with 20 labours it was found that only 5/8th of the road had been constructed.To complete the work in stipulated time the number of extra labours required is.

Solution:

Let total work=1

20 workers will do 5/8 work in 12 days

Remaining work=1-(5/8)=3/8

Remaining time=16-12=4 days

Let number of extra labours=x

Now,

⇒ 20×12×3 = (20+x)×4×5

∴ x= 16 labours Ans.

(Q3)Mohan completes 2/3 of his work in 10 days. Time he will take to complete 3/5 of the same work is.

Solution:

The common and frequently asked a question of Time And Work Problem with solution and Shortcut Tricks for Bank PO, SSC CGL, SSC CHSL, Railway Exam Type-4.4.

(Q1)15 men take 20 days to complete a job working 8 hours a day. The number of hours a day should 20 men take to complete the job in 12 days.

Solution:

(Q2)If 90 men can do a certain job in 16 days, working 12 hours per day, then the part of that work which can be completed by 70 men in 24 days, working 8 hours per day is.

Solution:

(Q3)8 workers can build a wall 18m long, 2m broad and 12m high in 10 days, working 9 hours a day. Find how many workers will be able to build a wall 32m long, 3m broad and 9m high in 8 days working 6 hours a day.

Solution:

Mantra to Crack Time and Work Questions

1. Work from Days:

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

No. of days = total work / work done in 1 day

Days from Work: If A’s 1 day’s work =1/n then A can finish the work in n days.

2. Relationship between Men and Work.

More men  ——- can do ——->   More work

Less men  ——- can do ——->   Less work

3. Relationship between Work and Time

More work   ——– takes——>   More Time

Less work   ——– takes——>   Less Time

4. Relationship between Men and Time

More men   ——- can do in ——->   Less Time

Less men   ——- can do in ——->   More Time

5. If M1 persons can do W1 work in D1 days and M2 persons can do W2 work in D2 days, then

6. If M1 persons can do W1 work in D1 days for h1 hours and M2 persons can do W2 work in D2 days for h2 hours, then

7. If A can do a work in ‘x’ days and B can do the same work in ‘y’ days, then the number of days required to complete the work if A and B work together is

8. If A can do a work in ‘x’ days and A + B can do the same work in ‘y’ days, then the number of days required to complete the work if B works alone is

Sample Questions –

1. Shambhu is twice as good as workman as Bablu and together they finish a piece of work in 18 days. Find the total number of days in which Bablu can finish the work.

a) 27 days

b) 54 days

c) 56 days

d) 68 days

Explanation:- As per question, Shambhu does twice the work as done by Bablu. So A:B = 2:1. Also (Shambhu+Bablu) one day work is 1/18

To get days in which Bablu will finish the work, lets calculate work done by Bablu in 1 day =(118 ∗13 )=154. So Bablu will complete the work in 54 days

2. Ritu can complete a piece of work in 5 days, but with the help of her son she can do it in 3 days. Find the time taken by the son alone to complete the work.

a) 7.5 days

b) 13 days

c) 11 days

d) 9 days

Explanation:- In this type of question, where we have one person work and together work done. Then we can easily get the other person work just by subtracting them. As,
Son’s one day work = (1/3 −1/5 )=(5−3/15 )=2/15

So son will do whole work in 2/15 days = 7.5 days
which is =7.5 days

3. Two pipes can fill the cistern in 10 hr and 12 hr respectively, while the third pipe can empty it in 20 hr. Simultaneously, if all the pipes are opened then the cistern will be filled in

a) 7.5 hr

b) 8 hr

c) 5 hr

d) 10 hr

Explanation:- Work done by all the tanks working together in 1 hour.
⇒ 1/10+1/12-1/20=2/15. Hence, tank will be filled in 15/2= 7.5 hour.

4. Mr. Chawla is on tour and he has Rs 360 for his expenses. If he exceeds his tour by 4 days he must cut down daily expenses by Rs 3. The number of days of Mr. Chawla’s tour programme is

a) 28 Days

b) 24 Days

c) 22 Days

d) 20 Days

Explanation:- Let Mr. Chawla under takes a tour of x days.

Then, expenses for each day =360/x

360/x+4=360/x–3

x=20 and −24

Hence, x= 20 days.

#### 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 1.

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 2.

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 3.

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.

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

Update – Question requested by Chitra Salin

#### Question 4.

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

## About “LCM method for time and work problems”

Let us look at the steps involved in solving time and work problems using LCM method.

Further process from step 3 will be depending upon the situation given in the problem.

It has been explained clearly in the example problems given below.

Let us look at some example problems on “LCM method for time and work problems”.

## Examples on “LCM method for time and work problems”

Example 1 :

A can do a piece of work in 8 days. B can do the same in 14 days. In how many days can the work be completed if A and B work together?

Solution :

Let us find LCM for the given no. of days “8” and “14”.

L.C.M of (8, 14) = 56

Therefore, total work = 56 units

A can do =  56 / 8 =  7 units/day

B can do = 56 / 14 =  4 units/day

(A + B) can do = 11 units per day

No. of days taken by (A+B) to complete  the same work

= 56 / 11 days

Let us look at the next example on “LCM method for time and work problems”

Example 2 :

A and B together can do a piece of work in 12 days and A alone can complete  the work in 21 days. How long will B alone to complete  the same work?

Solution :

Let us find LCM for the given no. of days “12” and “21”.

L.C.M of (12, 21) =  84

Therefore, total work = 84 units.

A can do =  84 / 21 =  4 units/day

(A+B) can do = 84 / 12 =  7 units/day

B can do = (A+B) – A = 7 – 4 = 3 units/day

No. of days taken by B alone  to complete  the same work

= 84 / 3

= 28 days

Let us look at the next example on “LCM method for time and work problems”

Example 3 :

A and B together can do a piece of work in 110 days. B and C can do it in 99 days. C and A can do the same  work in 90 days. How  long would each take to complete  the work ?

Solution :

Let us find LCM for the given no. of days “110”, “99” and “90”.

L.C.M of (110, 99, 90) =  990

Therefore, total work = 990 units.

(A + B) = 990/110 = 9 units/day  ———>(1)

(B + C) = 990/99 = 10 units/day  ———>(2)

(A + C) = 990/90 = 11 units/day  ———>(3)

By adding (1), (2) & (3), we get,

2A + 2B + 2C = 30 units/day

2(A + B + C) = 30 units/day

A + B + C = 15 units/day ———>(4)

(4) – (1) ====> (A+B+C) – (A+B) = 15 – 9 = 6 units

C can do = 6 units/day ,  C will take = 990/6 = 165 days

(4) – (2)====> (A+B+C) – (B+C) = 15 – 10 = 5 units

A can do = 5 units/day,   A will take = 990/5 = 198 days

(4) – (3) ====> (A+B+C) – (A+C) = 15 – 11 = 4 units

B can do = 4 units/day,   B will take = 990/4 = 247.5 days

Let us look at the next example on “LCM method for time and work problems”

Example 4 :

A and B can do a work in 15 days. B and C can do it in 30 days. C and A can do the same  work in 18 days. They all work together for 9 days and then A left. In how many days can B and C finish remaining work?

Solution :

Let us find LCM for the given no. of days “15”, “30” and “18”.

L.C.M of (15, 30, 18) = 90 units

Therefore, total work = 90 units.

(A + B) = 90/15 = 6 units/day  ———>(1)

(B + C) = 90/30 = 3 units/day  ———>(2)

(A + C) = 90/18 = 5 units/day  ———>(3)

By adding (1), (2) & (3), we get,

2A + 2B + 2C = 14 units/day

2(A + B + C) = 14 units/day

A + B + C = 7 units/day ———>(4)

A, B and C all work together for 9 days.

No. of units completed in these 9 days = 7×9 = 63 units

Remaining work to be completed by B and C = 90 – 63 = 27 units

B and C will take = 27/3 = 9 days   [ Because (B+C) = 3 units/day ]

Hence B and C will take 9 days to complete the remaining work.

Let us look at the next example on “LCM method for time and work problems”

Example 5 :

A and B each working alone can do a work in 20 days and 15 days respectively. They started the work together, but B left after sometime and A finished the remaining work in 6 days. After how many days from the start, did B leave?

Solution :

Let us find LCM for the given no. of days “20” and “15”.

L.C.M of (20, 15) = 60 units

Therefore, total work = 60 units.

A can do = 60/20 = 3 units/day

B can do = 60/15 = 4 units/day

(A + B) can do = 7 units/day

The work done by A alone in 6 days = 6×3 = 18 units

Then the work done by (A+B) = 60 – 18  = 42 units

Initially, no. of days worked by A and B together

= 42/7 = 6 days

Let us look at the next example on “LCM method for time and work problems”

Example 6 :

A is 3 times as fast as B and is able to complete the work in 30 days less than B. Find the time in which they can complete  the work together.

Solution :

A & B working capability ratio = 3 : 1

A & B time taken ratio = 1 : 3

From the ratio, time taken by A = k and time taken by B = 3k

From “A takes 30 days less than B”, we have

3k – k = 30

2k = 30 ===> k = 15

Time (A) = 15 days, Time (B) = 3×15 = 45 days

LCM (15, 45) = 45

Total work = 45 units

A can do =  45 / 15 =  3 units/day

B can do = 45 / 45 =  1 unit/day

(A + B) can do = 4 units per day

No. of days taken by (A+B) to complete  the same work

= 45 / 4  = 11 1/4 days

Let us look at the next example on “LCM method for time and work problems”

Example 7 :

A and B working separately can do a piece of work in 10 and 8 days  respectively. They work on alternate days starting with A on the first day. In how many days will the work be completed ?

Solution :

Let us find LCM of the given no. of days “10” and “8”

LCM of (10, 8) = 40

Total work = 40 units

A can do = 40/10 = 4 units/day

B can do = 40/8 = 5 units/day

On the first two days,

A can do 4 units on the first day and B can do 5 units on the second day. (Because they are working on alternate days)

Total units completed in the 1st day and 2nd day = 9 units —-(1)

Total units completed in the 3rd day and 4th day = 9 units —-(2)

Total units completed in the 5th day and 6th day = 9 units —-(3)

Total units completed in the 7th day and 8th day = 9 units —-(4)

By adding (1),(2),(3) & (4), we get 36 units.

That is, in 8 days 36 units of the work completed.

Remaining work = 40 – 36 = 4 units

These units will be completed by A on the 9th day.

Hence the work will be completed in 9 days.

Let us look at the next example on “LCM method for time and work problems”

Example 8 :

Two pipes A and B can fill a tank in 16 minutes and 20 minutes respectively. If both the pipes are opened simultaneously, how long will it take to complete  fill the tank ?

Solution :

Let us find LCM of the given no. of minutes “16” and “20”

LCM of (16, 20) = 80

Total work = 80 units

A can fill = 80/16 = 5 units/min

B can fill = 80/20 = 4 units/min

(A+B) can fill = 9 units/min

No. of minutes taken by (A+B) to fill the tank

= 80/9 = 8 8/9 minutes

Let us look at the next example on “LCM method for time and work problems”

Example 9 :

Pipe A can fill a tank in 10 minutes. Pipe B can fill the same tank in 6 minutes. Pipe C can empty the tank in 12 minutes. If all of them work together, find the time taken to fill the empty tank.

Solution :

Let us find LCM of the given no. of minutes “10”, “6” and “12”

LCM of (10,6, 12) = 60

Total work = 60 units

A can fill = 60/10 = 6 units/min

B can fill = 60/6 = 10 units/min

(A+B) can fill = 16 units/min

C can empty = 60/12 = 5 units/min

If all of them work together,

(6 + 10 – 5)  =  11 units/min will be filled

If all of them work together, time taken to fill the empty tank

= 60/11 = 5 5/11 minutes

Let us look at the next example on “LCM method for time and work problems”

Example 10 :

A water tank is two-fifth full. Pipe A can fill a tank in 10 minutes and pipe B can empty it in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely ?

Solution :

Let us find LCM of the given no. of minutes “10” and “6”.

LCM of (10, 6) = 30

Total work = 30 units (to fill the empty tank)

So, the work completed already = (2/5)x30 = 12 units

Out of 30 units, now the tank is 12 units full.

A can fill = 30/10 = 3 units/min (filling)

B can empty  = 30/6 = 5 units/min (emptying)

If both the pipes are open, the tank will be emptied

2 units/minute

No of minutes taken to empty the tank (already two-fifth filled)

= 12/2 =  6 minutes

Formula required for Time and Work problems:

1. If ‘A’ can do the work in ‘n’ days , Then A’s one day work is 1/n

Tricks and Tips to solve Aptitude problems on Time and Work:

For Example, ‘A’ can do the work in 2 days, in one day he does half of the work (that is 1/2 work). It is simple, he can do the work in 2 days, that means first days he completed 50% of the work and second day finishing 50% of the work.

2. If A’s 1 day’s work is 1/n, then A can finish the work in ‘n’ days.

Tricks and Tips to solve Aptitude problems on Time and Work

For example, A can finish the work in 3 days that means in first day he finishing 1/3rd of work, second day 1/3rd of work and third day 1/3rd of work.

### Example Problems on Time and Work with solutions :

If A can finish a work in 2 days, B can finish the work in 3 days, then How many days together takes to finish the work?

General Method to solve Aptitude problems on Time and work:

A) as per first formulae A’s one day work is 1/2
B’s one day work is 1/3
Now A and B work together in one day is (1/2)+(1/3) = 5/6

In one day A and B together work in one days is 5/6, So number of days they take together to finish the work ids 6/5. (as per second formula)

Tricks and tips to solve Aptitude problems on Time and Work:

a)If A can finish a work in x days, B can finish the work in y days, then and A and B together takes time to finish the work is xy/(x+y)

Check the answer with general method, A in 2 days, B in 3 days then A+B in 2*3/(2+3) = 6/5 days

b)  If A can finish a work in x days, B can finish the work in y days, C and finish the work in z days then now together A&B&C finish the job in (x*y*z) /(xy+yz+zx)

### Example problems on Time and Work with solutions :

1. If A can finish a work in 3 days, B can finish the work in 6 days, then How many days together takes to finish the work?

Tips and Tricks to solve Time and work problems:

As per short cut method answer is 3*6/(3+6) = 18/9 = 2 days

One more trick to solve this type of Aptitude problems on time and work

Tricks and tips: Let Assume the job is 6 units (L.C.M of 3 and 6) (whatever it may be like making burgers or walking kilometers etc)

A–> 3 days – > 6/3 = 2 units per day
B–> 6 days – > 6/6 = 1 unit per day
————————————————-
A+B -> 3 units per day

The total job is 6 units, so  A and B together complete the work in 6/3 = 2 days.

2. If A can finish a work in 12 hours, B can finish the same work in 15 hours, then How many minutes together takes to finish the work?

Tricks to solve Time and work problems Aptitude :

Let Assume the job is 60 units (L.C.M of 12 and 15)

A– > 12 hours – > 60/12 = 5 units per day
B– >  15 hours – > 60/15 = 4 unit per day
————————————————-
A+B – > 9 units per day

The total job is 60 units, so  A and B together complete the work in 60/9 = 20/3 hours
= 20*60/3 minutes = 400 minutes

3. A can do the work in 20 days, B can do the same work in 30 days. A,B and C together completed work in 10 days. In how many days C alone can do the work?

Tricks and Tips to solve Time and work
Aptitude problems:

Total work = 60 units  (LCM of 20, 30,10)
A can do 3 units per day
B can do 2 units per day
A & B & C can do 6 units per day

A– > 3
B— > 2
c- > ?
———
A + B + C –> 6

So C can do 1 unit per day (6-3-2)

Total job is 60 units , So C alone the complete job in 60/1 = 60 days

4. A and B together can complete a work in 12 hours. B and C together can complete a same work in 15 hours. C and A together can complete a same work in 20 hours. In how many hours A & B & C together complete the work?
A) 5                        B) 6                        C) 10                      D) 12

Time and work problems tricks:

Total work = 60 units  (LCM of 12, 15,20)
A & B– > 60/12 = 5 units/day
B & C— > 60/15 = 4 units/day
C & A– > 60/20 = 3 units/day
———————————————————————–
2 (A + B + C) = 12 units/ day
= >(A + B + C) = 6 units/ day

Total job is 60 units. So A & B & C together complete the work in 60/6 = 10 days

If the same question asking like following

5. A and B together can complete a work in 12 hours. B and C together can complete a same work in 15 hours. C and A together can complete a same work in 20 hours. In How many hours B alone complete the task?

Tricks and Tips to solve Aptitude problems on Time and work with examples:

As the previous question answer (A + B + C) = 6 units/ day
B one day work is (A + B + C) one day work – C & A one day work
B = 6 – 3 = 3 units per day
Total job is 60 units. So B alone can complete the 60/3 = 20 days

6. P can do a work in 12 hours and Q can do the same work in 18 hours. P started the work and Q joinedthe work after 6 hours. How much time in all they take to complete the work?
a) 3 hrs 36 mins                 b) 3 hrs 48 mins                 c) 9 hrs 36 mins                 d) 9 hrs 48 mins

Tricks to solve problems on Time and work:

Total work = 36 units (LCM of 12 and 18)
P — > 36/12 = 3 units/ hour
Q — > 36/18 = 2 units/ hour
———————————————-
P + Q — > 5 units/ hour

P can alone do the work for first 6 hours so 6*3 = 18 units
Total work is 36 units. In 6 days 18 units are completed. So the remaining work is 36 – 18 = 18 units.
So remaining work completed by P and Q is 18/5 hour
Total time taken for completing the work is 6 + 18/5 = 9.6 hours = 9 hours 36 minutes

7. Suresh can do work in 24 days and Naresh can do the same work in 40 days. Suresh started the work and Naresh replaced him after 12 days. In how many days from the beginning 75% of the work will be completed?
A) 10                      B) 12                      c) 18                       D) 22

Shortcut tricks to solve Time and work problems:

Total work = 120 units (LCM of 24 and 40)

Suresh — > 120/24 = 5 units/ day
Naresh — > 120/40 = 3 units/ day
———————————————-
Suresh + Naresh — > 8 units/ day

In first 12 days suresh can complete 60 units of work (5 units per day)
Remaining work is 120 units – 60 units = 60 units
But the question is “In how many days from the beginning 75% of the work will be completed?”
So 120*(75/100) = 90 units
Now the remaining work is 90 units – 60 units = 30 units
So naresh complete the remaining work in 30/3 = 10 days

Don’t select the option which has 10. The question is “In how many days from the beginning 75% of the work will be completed?.

That meaning total number of days = days taken by Suresh + days taken by naresh
= 12 + 10 = 22 days

## Understanding Time and Work through Examples

Now we will go through some examples to understand this trick closely.

Q1. Anil can do a job in 10 days. Babu can do the same job in 5 days. In how many days they can complete the job if they work together?

1. 6 days
2. 3.33 days
3. 7.25 days
4. 7.5 days
5. None of these

Ans: 3.33 days

Solution:

Anil can do a job in 10 days,

So efficiency of Anil = 100/10 = 10%

Similarly, Babu’s efficiency = 100/5 = 20%

Combined efficiency of Anil + Babu per day becomes = 20 + 10 = 30%

Now, we have to find out the number of days taken by both Anil and Babu to do 100% work,

Since they can do 30% wok in 1 day,

So, they will 100% work in 100/30 = 3.33 days

Q2. Anil and Babu can do a job in 8 days. Babu and Charan can do the same job in 12 days. Anil, Babu and Charan can do the job in 6 days. In how many days Anil and Charan can complete the job?

1. 6 days
2. 8 days
3. 10 days
4. 12 days
5. None of these

Ans: 8 days

Solution:

Anil and Babu can do a job in 8 days

∴ Combined efficiency of Anil + Babu = 100/8 = 12.5% ……………………………………(1)

Similarly, combined efficiency of Babu + Charan = 100/12 = 8.33%……………………..(2)

Similarly, combined efficiency of Anil + Babu + Charan = 100/6 = 16.66%……………..(3)

So now, if we want to calculate the efficiency of Anil and Charan together, we need to first calculate the individual efficiency of Anil and Charan.

So we first deduct equation 1 from equation 3 to get efficiency of Charan. Then we deduct equation 2 from equation 3 we will get efficiency of Anil.

Let’s calculate the efficiency of Charan = (Anil + Babu + Charan) – (Anil + Babu) = 16.66 – 12.5 = 4.16%

In the same way, we calculate the efficiency of Anil = 16.66 – 8.33 = 8.33%

So, combined efficiency of Anil and Charan = 8.33%  + 4.16%  ≈ 12.5%

∴ They need = 100/12.5 = 8 days

So they can complete the job in 8 days.

NOTE: This method looks long but only because it has been solved with all steps. When you attempt time and work questions in the exam paper, try to solve it directly using this technique, noting down only the figures, while performing all calculations (especially fractions) mentally. You simply need to get into the habit of using this shortcut method for time and work, as it is easier to solve mentally. This will save you precious time during the exam.

Most of the aptitude questions on time and work can be solved if you know the basic correlation between time, work and man-hours which you have learnt in your high school class.

• Analogy between problems on time and work to time, distance and speed:
1. Speed is equivalent to rate at which work is done
2. Distance travelled is equivalent to work done.
3. Time to travel distance is equivalent to time to do work.
• Man – Work – Hour Formula:
1. More men can do more work.
2. More work means more time required to do work.
3. More men can do more work in less time.
4. MM men can do a piece of work in TT hours, then Total effort or work =MT man hoursTotal effort or work =MT man hours.
5. Rate of work * Time = Work DoneRate of work * Time = Work Done
6. If AA can do a piece of work in DD days, then AA‘s 1 day’s work = 1D1D.
Part of work done by AA for tt days = tDtD.
7. If AA‘s 1 day’s work = 1D1D, then AA can finish the work in DD days.
8. MDHW=ConstantMDHW=Constant
Where,

M = Number of men
D = Number of days
H = Number of hours per day
W = Amount of work
9. If M1M1 men can do W1W1 work in D1D1 days working H1H1 hours per day and M2M2 men can do W2W2work in D2D2 days working H2H2 hours per day, then
M1D1H1W1=M2D2H2W2M1D1H1W1=M2D2H2W2
10. If AA is xx times as good a workman as BB, then:
1. Ratio of work done by AA and BB = x:1x:1
2. Ratio of times taken by AA and BB to finish a work = 1:x1:x ie; AA will take (1x)th(1x)th of the time taken by BB to do the same work.

## 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 daysA, 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 CCand 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 dd days 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.

## Frequently asked questions in quantitative aptitude test on time and work

• Given A takes x days to do work. B takes y days to do the same work. If A and B work together, how many days will it take to complete the work.
• If A and B together can do a piece of work in x days, B and C together can do it in y days and C and A together can do it in z days, find how many days it takes for each of them to complete the work if they worked individually. How many days will it take to complete the work if they worked together?
• Give A is n times efficient than B. Also A takes n days less than B to complete the work. How many days will it take to complete the work if they worked together?
• Given A takes x days to do work. B takes y days to do the same work. Now A & B together begins a work. After few days one of them leaves. Also given the other takes n more days to complete the work.
1. Find total number of days to complete the work.
2. How many days did they work together?
• Given A takes x days to do work. B takes y days to do the same work. A started the work and B joined him after n days.
1. How long did it take to complete the work?
2. How many days did they work together? Or How long did B work?
• Case 5 with 3 people joining work one after the other.
• Given A takes x days to do work. B takes y days to do the same work. If A and B works on alternate days ie A alone works on first day, B alone works on next day and this cycle continues, in how many days will the work be finished
• Given A alone can complete a job in x days and also B is b% efficient than A. How many days will it take to complete work if B works alone.
• Problems where combinations of workers [men, women, girls and boys] take some days to do a work. These problems are solved using man days concept.
1. You have to calculate for another combination of them to complete the work.
2. How long will one set of people take to complete the entire work?
3. A certain combination starts the job and after few days leaves the work. Find the number of people from the category who are required to finish the remaining work.
• Problems related to wages from work. How much each person earns from the work done.

Ex. 1: A can do a work in 6 days and B can do the same work in 5 days. The contract for the work is Rs 220. How much shall B get if both of them work together?

Solution:

Method 1: A’s 1 day’s work = 1/6;

B’s 1 day’s work = 1/5

Ratio of their wages = 1/6 : 1/5

= 5:6

B’s share = 220 / 5+6 * 6

= Rs. 120

Shortcut trick for time and wages question Method 2: As wages distributed in inverse proportion of number of days, their share should be in the ratio 5:6

Government Jobs in India 2017-18 Presented by GovtJobsPortal.IN. Daily 10 New Govt Jobs. Stay Tuned in with Us for latest jobs, Results and Study Material for Government Exam.

B’s share = 220/11 * 6 = 120

Ex. 2: A man can do a work in 10 days. With the help of a boy he can do the same work in 6 days. If they get Rs 50 for that work, what is the share of that boy?

Solution:

The boy can do the work in 10*6 / 10-6 = 15 days.

Man’s share : boy’s share = 15 : 10 = 3 : 2

Man’s share = 50/5 * 3 = Rs 30

Ex. 3: A, B and C can do a work in 6, 8 and 12 days respectively. Doing that work together they get an amount of Rs. 1350. What is the share of B in that amount?

Solution: A’s one day’s work = 1/6

B’s one day’s work = 1/8

C’s one day’s work = 1/12

A’s share : B’s share : C’s share = 1/6 : 1/8 : 1/12

Multiplying each ratio by the LCM of their denominations, the rations become 4 : 3 : 2

B’s share = 1350/9 * 3 = 450

Shortcut trick: A’s share : B’s share : C’s share =

B’s time * C’s time : A’s time * C’s time : A’s time * B’s time

= 96 : 72 : 48 = 4 : 3 : 2

B’s share = 1350/9 * 3

= 450 Rs.

Ex. 4: two men undertake to do a piece of work for Rs 200. One alone could do it in 6 days. With the assistance of a boy they finish it in 3 days. How should the money be divided?

Solution: 1st man’s 3 days’ work = 3/6

2nd man’s 3 days’ work = 3/8

The boy’s 3 days work = 1 – (3/6 + 3/8) = 1/8

Their share will be in the ratio 3/6 : 3/8 : 1/8 = 4 : 3 : 1

1st man’s share = 200/8 * 4 = Rs. 100

2nd man’s share = 200/8 * 3 = Rs 75

The boy’s share = 200/8 * 1 = Rs. 25

Time and Work is yet another easy topic and almost all the questions are predictable. Please go through the following solved examples and I am sure that 90% questions would be similar to these examples. Most probably I will complete Time and Work in 3 parts.

Note: In the complete Time and Work series, Efficiency would mean “Work Done in 1 day”, and efficiency has been denoted by small letters, e.g. “a” means “Efficiency of A”.

1. 1)

Let the total work be 8 units (because 8 is the LCM of 4 and 8)
Efficiency of x (Work done by x in 1 hour) = 8/4 = 2 units
Efficiency of y (Work done by y in 1 hour) = 8/8 = 1 unit
Work done by (x + y) in 1 hour = 3 units
3 units work in completed in 1 hour. Hence 8 units work will be completed in 8/3 hours or 160 minutes.

1. 2)

Let total work be 60 units (LCM of 10 and 12)
Raj completes the work in 12 days. Hence efficiency or per day work of Raj = 60/12 = 5 units
Raj and Ram take 10 days to complete the work, hence their efficiency = 60/10 = 6 units
Now Efficiency of Ram = (Efficiency of Raj and Ram) – (Efficiency of Raj) = 6 – 5 = 1 unit
That means Raj completes 1 unit of work per day. So to perform 60 units of work, he will take 60 days.

1. 3)

Let total work = 120 units
Efficiency of A + B = 120/15 = 8 units
Efficiency of B + C = 120/12 = 10 units
Efficiency of C + A = 120/10 = 12 units
Adding all the above 3 equations –
2 * (A + B + C) = 30
Efficiency of (A + B + C) = 15 units
Efficiency of B + C = 120/12 = 10 units
Hence Efficiency of A = Efficiency of (A + B + C) – Efficiency of (B + C) = 15 – 10 = 5 units
Hence time taken by A to do 120 units of work = 120/5 = 24 days

1. 4)

Let the total work be 16 units.
Efficiency of first pipe = 16/4 = +4 units
Efficiency of second pipe = 16/16 = -1 units [negative sign because this pipe is emptying the tank] When both the pipes are opened together, their efficiency = (+4) + (-1) = +3 units [The positive sign indicates that when both the pipes are opened together, their net result will fill the tank] 3 units of work is done in 1 hour
16 units of work is done in 16/3 hours

Note : In questions where one pipe is emptying the cistern while another is filling it, you must put a positive or negative sign before the efficiency. But in questions where both the pipes are emptying the cistern or both the pipes are filling the cistern, you can take the efficiency of both the pipes as positive.

1. 5)

Let the total work be 15 units.
Efficiency of first pipe = 15/3 = +5 units
Efficiency of second pipe = 15/3.75 = +4 units
Efficiency of third pipe = 15/1 = -15 units
Efficiency of all the three pipes = 5 + 4 – 15 = -6 units
If all the pipes are opened, it will take 15/6 or 5/2 hours to empty the cistern, but the cistern is already half empty, hence only 5/4 hours are required to empty it.

1. 6)

Let the total work = 60 units
Efficiency of A = 60/20 = +3 units
Efficiency of B = 60/30 = -2 units
Now total work to be performed is 60 units. When 57 units work is complete, A will take 1 more minute to add 3 units and hence will make it a total of 60 units.
Hence time taken to fill the tank = Time taken to perform 57 units of work + 1 minute
Now A and B are opened alternatively. That means for the first minute only A is opened, for the second minute A is closed and B is opened, then for third minute again B is closed and A is opened and so on.
So for each 2 minutes cycle, work done = Efficiency of A + Efficiency of B = +3 + (-2) = 1 unit
1 unit work is done in 2 minutes, so 57 units work is done in 114 minutes
Time taken to fill the tank = 114 + 1 = 115 minutes
Explanation : We have to perform a total of 60 units of work. For the 1st minute – A adds 3 units of work, but in the 2nd minute, B adds (-2) units of work and hence makes total work for 2 minutes = (+3) + (-2) = 1 unit. So effectively in 2 minutes, we are just adding 1 unit of work. Hence in 4 minutes, 2 units of work will be performed and in 6 minutes 3 units of work will be performed. Same sequence will continue till 57 units. As soon as 57 units of work is done (in 114 minutes), it will be A’s turn to do the work. A will add 3 units of work(in 1 minute) and hence take the total work from 57 units to 60 units. B won’t be needed any more.

1. 7)

Let the total work be 240 units.
40 men complete the work in 6 months. Hence 10 men can complete the work in 6*4 = 24 months. Hence, Efficiency of 10 men = 240/24 = 10 units
60 women complete the work in 6 months. Hence 10 women can complete the work in 6*6 = 36 months. Hence, Efficiency of 10 women = 240/36 = 20/3 units
80 boys complete the work in 6 months. Hence 10 boys can complete the work in 6*8 = 48 months. Hence, Efficiency of 10 boys = 240/48 = 5 units
Efficiency of 10 men + Efficiency of 10 women + Efficiency of 10 boys = 10 + 20/3 + 5 = 65/3 units
So, 10 men, 10 women and 10 boys complete 65/3 units of work in 1 month. To complete 120 units(half of the work), they will take = 120 * 3/65 = 72/13 months

1. 8)

Let the total work be 112 units and the efficiency of 1 man and 1 woman be m and w respectively
2m + w = 112/14 = 8
4w + 2m = 112/8 = 14
Solve the equations and you will get w = 2 and m = 3
Hence the wage of woman = 2/3 * 180 = Rs. 120

1. 9)

A and C complete 19/23 of the work. Hence B does 4/23 of the work
Amount paid to B = 4/23 * 575 = Rs. 100

1. 10)

Let A and B complete the work in x days
Then A will complete the work in (x + 8) days and B will complete the work in (x + 4.5) days. Now,
1/(x + 8) + 1/(x + 4.5) = 1/x
Solve the equation and you will get x = 6 hours

1. 11)

The question is same the previous one.
Let A, B and C take ‘x’ days to do the job. Then,
A takes (x + 6) days, B takes (x + 1) days and C taken 2x days
1/(x + 6) + 1/(x + 1) + 1/2x = 1/x
1/(x + 6) + 1/(x + 1) = 1/x – 1/2x
1/(x + 6) + 1/(x + 1) = 1/2x               …    (1)
Solve it and you will get x = 2/3
From equation (1) you can see that A and B take 2x days to complete the work

In this article, I have solved questions where some workers quit in the middle, as well as some questions on efficiencies.

Note: In the complete Time and Work series, Efficiency would mean “Work Done in 1 day”, and efficiency has been denoted by small letters, e.g. “a” means “Efficiency of A”.

1. 1)

If Pratibha finishes the work in X days, then Sonia will take 3X days to finish the same work
Given 3X – X = 60
Or X = 30
Pratibha takes 30 days and Sonia takes 90 days

1. 2)

Let the total work be 24 units.
Efficiency of Sunil = 24/4 = 6 units (Since Sunil takes 4 days to complete the work)
Efficiency of Ramesh = 6 * 1.5 = 9 units (Since Ramesh is 1.5 times efficient as Sunil)
Efficiency of Dinesh = 24/6 = 4 units ((Since Sunil takes 6 days to complete the work))

Efficiency of (Sunil + Ramesh + Dinesh) = 6 + 9 + 4 = 19 units
Time required to finish the complete work = 24/19 days

1. 3)

Let the total work be 15 units. Efficiency of A = a and Efficiency of B = b
A and B complete the work in 5 days.
Hence efficiency of A and B = 15/5 = 3 units
So, a + b = 3 … (1)
New efficiency of A = 2a
New efficiency of B = b/3
With new efficiency the work was completed in 3 days.
So, 2a + b/3 = 15/3 = 5 … (2)
Solve (1) and (2), you will get a = 12/5 = 2.4 units
So A will complete 15 units work in 15/2.4 or 25/4 days

1. 4)

Let the total work be 24 units
Given, 3*Efficiency of A = Efficiency of B + Efficiency of C
3a = b + c
A, B and C compete the work in 24 days.
Hence, a + b + c = 24/24 = 1 or 4a = 1 [Put b + c = 3a] a = 1/4 = 0.25 unit
A completes 0.25 unit work in 1 day. So to complete 24 units of work, he will take 24/0.25 = 96 days

1. 5)

Let the total work be 7 units. Since they all complete the work in 7 days, so their total efficiency = 7/7 = 1 unit
Let efficiency of boy = x
Then efficiency of women = 2x
Efficiency of man = 4x
x + 2x + 4x = 1
7x = 1 or x = 1/7
The boy completes 1/7 work in 1 day, so to complete 7 units of work, he will take 49 days

1. 6)

A does 1/2 as much work as B in 3/4 of the time. Hence A will do (1/2 + 1/2) or complete work in (3/4 + 3/4) or 1.5 times more time than B.
A = 1.5B  (where A = no. of days taken by A to finish the work and B = no. of days taken by B to finish the work)
Also A*B/(A+B) = 18
Put A = 1.5B in the above equation and solve
B = 30 days

1. 7)

Let the total work = 60 units
Efficiency of A = 60/20 = 3 units
Efficiency of B = 60/30 = 2 units
Efficiency of (A + B) = 5 units
Work done by A and B in 7 days = 5*7 = 35 units
Work left = 60 – 35 = 25 units
C completes 25 units of work in 10 days. Hence he will complete 60 units of work in 10* 60/25 = 24 days

1. 8)

Let total work be 120 units.
Efficiency of A = 120/6 = 20 units
Efficiency of B = 120/12 = 10 units
Efficiency of C = 120/15 = 8 units
Work left = 7/8 * 120 = 105 units
Efficiency of A + B = 30 units
Hence time taken by A and B to complete 105 units of work = 105/30 = 3.5

1. 9)

Let the total work = 80 units
Efficiency of (A + B + C) = 80/40 = 2 units
Work done by (A + B + C) in 16 days = 16 * 2 = 32 units
Remaining work = 80 – 32 = 48 units
B and C complete the remaining work (48 units) in 40 days.
Efficiency of B + C = 48/40 = 1.2 units
Efficiency of A = Efficiency of (A + B + C) – Efficiency of (B + C) = 2 – 1.2 = 0.8 unit
Time taken by A to complete the whole work = 80/0.8 = 100 days

1. 10)

Let the total work = 360 units
Efficiency of A = 360/45 = 8 units
Efficiency of B = 360/40 = 9 units
Efficiency of A + B = 17 units
Let A left after x days, that means A and B worked together for x days. Total work done by A and B together = 17x
Then the remaining work is finished by B in 23 days. Hence work done by B alone = 23 * 9 = 207 units
So, 17x + 207 = 360
Or x = 9 days

1. 11)

This question appeared in SSC Tier-2 2015, and stumped many candidates. Although there is nothing tricky about it.

Let the total work be 60 units.

p + q = 60/6 = 10

q + r = 60*7/60 = 7

Given, Total work done = 3 days work of P + 6 days work of Q and R

60 = 3*p + 6*(7)

p = 6

Hence time taken by P to complete the work = 60/6 = 10 days

p + q = 10, hence q = 4

q + r = 7, hence r = 3

Hence time taken by R to complete the work = 60/3 = 20 days

Difference = 20 – 10 = 10 days

Q. 12)  4 Men and 6 Women working together can complete the work in 10 days. 3 men and 7 women working together will complete the same work in 8 days. In how many days 10 women will complete this work?

One day work for a man = 1/m
One day work for a woman = 1/w

In one day, 4 men and 6 women will do 1/10 of the work. Hence,
4/m + 6/w = 1/10    …  (i)

Similarly,
3/m + 7/w = 1/8      …  (ii)
Multiply equation (i) with 3 and equation (ii) with 4
12/m + 18/w = 3/10
12/m + 28/w = 1/2
Subtract the equations
10/w = 1/5
So 10 women will complete the work in 5 days

Here W  is the work. For e.g., if 5 men are cutting 10 trees in 2 days, working 4 hours per day. Then,
M = 5, D = 2, H = 4 and W = 10.

1. 1)

H1 = 6
D1 = 18
D2 = 12
H2 = ?
We know, H1 * D1 = H2 * D2
6 * 18 = H2 * 12
H2 = 9 hours

1. 2)

M1 = 15
D1 = 20
H1 = 8
M2 = 20
D2 = 12
H2 = ?
We know, M1 * D1 * H1 = M2 * D2 * H2
15*20*8 = 20*12*H2
H2 = 10 hours

1. 3)

In this question, Taps = Men
Number of taps required = 20*9/15 = 12

1. 4)

Let there be X number of men.
X men can finish a piece of work in 100 days. Hence total work = 100X
If there were (X – 10) men, it would have taken 110 days to finish the work. Total work in this case = 110(X – 10)
Total work remains the same. Hence,
100X = 110(X – 10)
X = 110

1. 5)

Efficiency of Subhash = 50/10 = 5 per hour
Efficiency of Subhash and Prakash = 300/40 = 7.5 per hour
Efficiency of Prakash = (Efficiency of Subhash and Prakash) – (Efficiency of Subhash) = 7.5 – 5 = 2.5
So Prakash can copy 2.5 pages per hour. To copy 30 pages, he would require 30/2.5 or 12 hours.

1. 6)

40 men can finish a work in 60 days. Hence, total work = 40*60 = 2400

Let the 10 men left after X days.

For X days, all the 40 men worked. Total work performed = 40X

Now when 10 men quit, only 30 men were left to do the work and they took (70 – X) more days to finish it.

Total work done by 30 men = 30*(70 – X)

Now, 40X + 30*(70 – X) = 2400

X = 30 days

1. 7)

Let the total work = 360 units

Efficiency of A = 360/45 = 8 units

Efficiency of B = 360/40 = 9 units

Efficiency of A + B = 17 units

Let A left after X days.

For X days, both A and B worked. Hence work performed = 17X

B worked for 23 days. Hence work performed by B = 23*9 = 207 units

Now, 17X + 207 = 360

X = 9 days

1. 8)

B and C together do 8/23 of the work, hence A does (1 – 8/23) or 15/23 of the work.

A should be paid = 15*5290/23 = Rs. 3450

Note : In this question, they have asked the wages of A. Had they asked the wages of B, firstly you would have calculated the work performed by B with the formula-

Work done by B = (Portion of work done by A and B) + (Portion of work done by B and C) – 1

Work done by B = 19/23 + 8/23 – 1 = 4/23

Wages of B = 4*5290/23 = Rs. 920

1. 9)

This is a very famous question. A company employed 200 workers to complete a certain work in 150 days. Here the total work is not 200*150 because 200 workers and 150 days was only a plan. In reality, only 1/4th of the work has been done in 50 days. So if they go with the same pace, 200 workers will take 200 days to complete the work.
So total work = 200*200 units
200 workers have worked for 50 days. Hence they have finished 200*50 units of work.
Remaining work = 200*200 – 200*50 = 200*150
Let the number of additional workers required = X.
Now (200+X) workers will work for 100 days to finish the work as per the schedule.
Work they need to perform = (200 + X)*100
Now, (200 + X)*100 = 200*150
X = 100

Q. 10) A contractor undertook to finish a certain work in 124 days and employed 120 men.After 64 days,he found that he had already done 2/3 of work. How many men can be discharged now so that the work may finish in time?
A) 56             B) 44               C) 50               D) 60

120 workers finish 2/3 of the work in 64 days. So to complete the whole work, workers will take 64*3/2 or 96 days.
Total work to be performed = 120*96
Now the workers have already finished 2/3 of the work and only 1/3 work has to be performed.
Remaining work = 120*96/3 = 120*32
Let the contractor discharges X men. Remaining workers = 120-X. These workers will continue the work for (124-64) or 60 days. Hence,
120*32 = (120-X)*60
X = 56

Method 2

M1 = 120, D1 = 64, W1 = 2/3
M2 = 120 – x, D2 = 60, W2 = 1/3
(M1*D1)/W1 = (M2*D2)/W2
120*64*3/2 = (120 – x)*60*3
x = 56

1. 11)

Let the second pipe fills the pool in X hours. Then first pipe takes (X+5) hours and the third pipe takes (X-4) hours to fill the pool. Now, 1st and 2nd pipe together take the same time to the fill the pool as the 3rd pipe alone. Hence,

1/(X+5) + 1/(X) = 1/(X – 4)

Solve this quadratic equation and you will get X = 10 hours

That means second pipe takes 10 hours to fill the pool while the third pipe takes 6 hours. Together they will take 10*6/(10 + 6) hours to fill the pool.

I have covered almost all the possible questions from Time and Work that are asked by SSC. If you have any doubt in this topic, please drop a comment.

# Practice Problems On Time and Work

• A & B can separately finish the work in 30 days and 50 days respectively. They worked together and A left the work, so B complete the remaining work in 10 days. Find after how many days A left the work?
A) 10days
B) 12days
C) 15days
D) 20days
E) 18days

Option C
Solution:

A = 30………..5
B = 50……….3    (LCM = 150)
A+ B = 8
B’s work in 10days = 3*10 =30
Means they together did ……120
120/8= 15days
• A is 20% more efficient than B and 50% more efficient than C. if they together can do a work in 24 days then find in how many days B alone can do the work?
A) 60days
B) 72days
C) 90days
D) 180days
E) 100days

Option B
Solution:

.                  A ………B   |   A………C
Efficiency    6……….5    |   3………2
Days            5……….6   |   2………3
Days A : B : C
.        10 12 15
A = 10……….6
B = 12………..5 ….        (LCM = 60)
C = 15………..4
A+B+ C = 15
60/15=4
4=24
1= 6
B=12= 12*6 = 72days
• A can make 10000 papers in an hour B can make 8000 papers in an hour. Find in how many days they both can make 5,90,000 papers, if A do work for 7 hours and B do work for 6 hours?
A) 4days
B) 3days
C) 5days
D) 6days
E) 7days

Option C
Solution:
A’s 1hr work = 10,000
7hr work = 70,000
B’s 1hr work = 8000
6hr work = 48,000
Total work of A& B of 1day = 70,000+48,000= 1,18,000
So 590000/118000 = 5days
• 7 men and 5 women can do a work in 6 days. Also 6 men and 7 women can do same work in 6 days. Find in how many days will 2 men & 2 women can finish the work?
A) 19days
B) 15days
C) 10days
D) 14days
E) 22days

Option A
Solution:

7M + 5W =6
42M+30W = 1……………….(1)
6M +7W =6
36M + 42W =1……………..(2)
FROM (1) & (2)
. 42M + 30W =1
-36M (-)+ 42W =(-)1
6M -12W =0
6M =12W
M =2W…….PUT IT IN EQUATION(1)
S0 .. 14W +5W= 6
19W =6
1W = 6*19
6W = 6*19/6 =19DAYS
• A & B can do a work in 18 days. They started work together and A left after 7 days and B did the remaining work in 33 days. Find in how many days A can alone do the work?
A) 18
B) 54
C) 27
D) 36
E) 32

Option C
Solution:

let total work = 18
Efficiency of A &B’s work of 1day =1
In 7days they complete =7 work
Remaining = 18-7 = 11
B do 11 work = 33 days
1 work = 3days
18work = 54days
So ……. A+B = 18…….3                (LCM = 54)
.                 B = 54……..1
So A = 3-1 = 2
A = 54/2 =27days
• A tap can fill a tank in 16 hrs but due to a leak it takes 6 hrs more. If leakage withdraw 9ltr water in an hour than find the quantity of tank?
A) 520ltr
B) 528ltr
C) 536ltr
D) 544ltr
E) 576ltr

Option B
Solution:

A = 16 ………….11                (LCM = 176)
A+B=22 …………8
So B = 11-8 = 3
176/3 * 9= 528ltr.
• A & B can do a work in 35 days and 45 days respectively. They worked for 10 days and after then they complete the work with the help of C in 15 days. If they all get Rs 770. Then find the share of C?
A) 330
B) 270
C) 190
D) 170
E) 250

Option D
Solution:

A = 35…………..11          (LCM = 385)
B = 45……………9
A+B=20
20*10 =200
Remaining 385-200=185
This work is completed in 5days. So 185/5=37
A&B 5days work = 20*5=100
C’s 5days work =85
A………….B………………C
11*15     9*15              85
165         135               85
33            27                17
C’s share = 17/77  * 770 = 170
• 1 man or 2 women or 3 children can do a work in 55 days. Find in how many days 1 man and 1 woman and 1 child can do the work?
A) 30days
B) 24days
C) 25days
D) 28days
E) 32days

Option A
Solution:

1M = 55   1
2W =55    1            55
2C =55     1
We need 1day work of … 1M +1W+1C= 1+1/2+1/3 =11/6
So …….. 55/(11/6) = 55/11 * 6=30 DAYS
• A can a work in 50 days and B is 50% efficient than A. find in how many days A and B together can complete the work?
A) 30
B) 40
C) 50
D) 33(1/3)
E) 16(2/3)

Option D
Solution:

B is 50% efficient than A
.                   A………..B
Efficiency      2……….. 1
Days             1…………2
1 == 50  . So 2 == 100
A= 50……….2               (LCM = 100)
B =100………1
A+ B = 3
100/3 =33(1/3) days.
• A and B can do a work in 60days. B and C can do same work in 40 days and C and A can do same work in 50 days. Find in how many days will A,Band C work together will finish the work?
A) 32(16/37)
B) 42(14/37)
C) 33(7/37)
D) 43(9/37)
E) 39(7/37)

Option A
Solution:

A+B = 60…………10
B+C = 40………….15…………….(LCM =.600)
C+A = 50………….12
2(A+B+C)            37
A+B+C = 37/2
= 600/37 * 2 = 32(16/37)

• A is thrice as good a workman as B and therefore able to finish a job in 48 days less than B working together ,they can do it in how many days together ?
A) 13 days
B) 15 days
C) 18 days
D) 12 days
E) 116 days

Option  C
Solution:
.                   A:B
Efficiency=3:1
Time=1:3
Multiplying by 24 on both the sides ,
A’s=24 days
B’s=72 days
Therefore,2 units =48 days
1 units =24 days
Total work=No. of days *Efficiency
.                                       72*1
.                                        =72
One day work of A and B is 3+1=4 units
A and B will complete the work in =72/4=18days.
• Three men –A ,B and C working together can do a job 6 hours less time than A did alone ,1 hour less time than B alone and half the time needed by C .In how many days will A finish the work alone ?
A) 20/3 days
B) 23/4 days
C) 22/5 days
D) 33/6 days
E) 27/8 days

Option A
Solution:
A+B+C          A           B            Cx hr.            x+6       x+1       2xTaking LCM = 2x (x+1)(x+6)
Taking efficiency of A and B ;
2x(x+1)(x+6)/(2x2 +2x+2x+12x)        = 2x/1
3x+7x-6=0
X=-3(ignore)
X=2/3
A will finish its work in (x+6)=20/3 days
• A work is started by a man on the first day. Each subsequent day a new person joined the work and it is known that the total work will completed on the 11th day. If from the starting day 6 men working on that work and no new men added later, in how many days the work got completed?
A) 15 days
B) 12 days
C) 14  days
D) 11 days
E)  None of these.

Option D
Solution:
1day work of a man is 1 unit. If a new person joined the work on second day, 2 units of work get completed. Similarly 3 units on 3rd day, 4 units on 4th day so on…
Then for all the eleven days the total work = 1 + 2 + 3+ ………….11 = 66 units (Use formula
N(N+1)/2)
Now 6 men /day work = 6 units/day.
They can complete 66 units of work in =
66/6
=11 days
• Two men can complete a piece of work in 3 days while 3 women can complete the same work in 4 days and 4 children can complete the same work in 6 days. Then find in how many days 1 men ,1 women and 2 children can complete the same work ?
A) 4 days
B) 3 days
C) 5 days
D) 2 days
E) None of these.

Option B
Solution:
2M *3 = 3W*4 = 4C*6
.      1M =  2W =   4C
LCM=4
1 Man’s efficiency= 4units/day
2 Women’s efficiency=2units/day
4 Children’s efficiency= 1 units/day
Total work =(2*4)*3=24 days
(1man+1woman+2children)=4+2+2=8
(1man+1woman+2children) complete the work =24/8=3 days
• 30 men are supposed to do a work in 38 days. After 25 days, 5 more men were employed on work for which the work is completed in 1 day before . If 5 more men were not worked then how many days took in delay?
A) 1 day
B) 2 days
C) 3 days
D) 4 days
E) None of these.

Option A
Solution:
30Men * 25 days = 750
35 Men * 12 Days = 420
Total =750+420=1170
Now,1170/30 =39 days
1 day delay
• A group of men decided to do a job in 4 days but 20 men dropped out everyday ,the job was completed at the end of the 7th day .Find the men who are in the work initially ?
A) 155
B) 135
C) 120
D) 140
E)  160

Option D
Solution:
Total work = M * 4 = 4M
M + (M+20) +…….
7/2 [2M +6(-20)] =4M
M=140
• A printer A can print one thousand books in 15 hours ,printer B can print the same number of books in 10 hours and printer C can print the same number of books in 12 hours . If all the printers are started to print the books at 8 A.M, After sometime printer A is closed at 9 A.M and printer B and printer C remains working. Find at what time the printing will be completed ?
A) 4(3/11)hours
B) 3(1/11)hours
C) 5(1/11)hours
D) 3(5/11)hours
E) None of these.

Option C
Solution:
Let printing completed in be T hours
Then A ‘s 1 hour work ,B’s T hours work and C;s T hours work =Total work
1/15 +T/10+T/12 =1
T= 5(1/11)
Hence ,the printing of books will be completed at 5(1/11)hours
• Ramesh and Ram can do a piece of work in 24 and 30 days respectively. They both started and worked for 6 days. Ram then leaves the work and another their friend Rohit joins the work and completed the remaining work with Ramesh in 11 days . Find how many days are taken by Rohit alone to finish the work?
A) 110 days
B) 132 days
C) 150 days
D) 120 days
E) None of these.

Option D
Solution:
(1/24 + 1/30) *6 +(1/24 + 1/Rohit ) *11 = 1
Therefore ,Rohit takes 120 days to finish the work.
• A woman has her three daughters. First and second can take 24 and 30 days resp. to complete a work .In how many days third one takes to complete the work. If woman can complete the whole work alone in 3(3/11) days .The efficiency of woman is double than her three daughters.
A) 22 days
B) 12 days
C) 13 days
D) 21 days
E) 19 days

Option B
Solution:
LCM = 72 (if we are taking woman and her two daughters )
Here it is given
Woman Three daughters
Time taken = 1 : 2
Efficiency= 2 : 1Three daughters, let P +Q+R=11
3+2+R=11
R=6 days
Her third daughter complete the work in =72/6= 12 Days
• A contractor takes a road construction project to finish it in 40 days and for that he engaged 200 men. After 30 days he employed 100 more men in this project, then the work finished on time. Find if the 100 more men would not worked then how many more days required to finish the work ?
A) 8 days
B) 10 days
C) 12 days
D) 7 days
E) None of these.

Option E
Solution:
100 * 10 days = 1000
Now 1000/200 = 5 days (Initial total no. of men engaged in the project)
Hence ,5 more days required to finish the work if 100 more men would not joined .

• A cistern can be filled by two pipes separately in 6 and 9 mins respectively. Both pipes are opened together for a certain time but being clogged, only 5/6 of full quantity water flows through the first and only 3/4 through the second pipe. The obstructions, however, being suddenly removed, the cistern is filled in 2 mins from that moment. How long was it before the full flow began?
A) 3 min
B) 2 min
C) 1 min
D) 2.5 min
E) 1.5 min

Option  B
Solution:
total units …….36
first pipe ………36/6 = 6 units
second………….36/9 = 4units
now, ( 5/6 * 6 + 3/4* 4 ) T + 2 ( 6+4 ) = 36 >> T =2
• Ram and mohan together can complete typing a book of 1575  pages in 25  days working 15 hrs per day. Ram is 20% more efficient than Mohan. A page contains an average of 275 words, then how many words can ram type in an hour?
A) 525
B) 600
C) 625
D)630
E) 645

Option D
Solution:
Ram : mohan = 6:5
R+M = 11
R+M = 1575 * 275 / 15* 25 = 1155 words in 1 hour
ram will type = 1155 * 6/11 = 630 words in 1 hour
• Subhash can copy 70 pages in 16 hours ; Subhash and Prakash together can copy 275  pages in 40 hours. In how much time can Prakash copy 30 pages ?
A) 15 hr.
B) 12 hr.
C) 14  hr.
D) 18 hr.
E)  None of these.

Option B
Solution:
Subhash can copy 70 pages in 16 hours so In 40 hours he can copy 70*2.5 = 175 pages. Hence prakash can copy 100 pages in 40 hours . Thus , he can copy 30 pages in 30% of the time i.e 12 hours.
• A and B together can do a piece of work in 12 days which B and C together can do in 16 days. After A has been working at it for 5 days and B for 7 days, C finishes it in 13 days. In how many days could each do the work by himself ?
A) 24, 12, 36
B) 24, 16, 12
C) 16, 48, 24
D) 24, 36, 12
E) None of these.

Option C
Solution:
Total work  = 48 ,… A+B = 4 ….B+C = 3
now, 5A + 7B + 13C = 48
split it
5A +5B + 2B + 2C + 11C = 48
so 5*4 + 2*3 + 11c = 48
so 11c = 22 …… c = 2
so c alone = 48/2 = 24
A — 48/3 = 16
B— 48/1 =48
• 24 men take 12 days to complete a piece of work . They worked for a period of 4 days . After that , they were joined by 8 more men . How many more days will be taken by them to complete the remaining work?
A) 8 days
B) 9 days
C) 7 days
D) 6 days
E) None of these.

Option D
Solution:
24*12 – 24*4 =( 24+8 )x
By solving we get x = 6 days
• In two days A, B and C together can finish 1/4 of a work and in another 2 days B and C together can finish 1/5 part of the work. Then A alone can complete the whole work in?
A) 10 days
B) 20 days
C) 40 days
D) 35 days
E)  None of these.

Option C
Solution:
work ……. 20
a+b+c = 5 in 2 days
b+c = 4 in 2 days
a = 1 in 2 days >> 40 days total
or
2(a+b+c) =1/4 =>a+b+c =1/8
2(b+c) =1/5 =>b+c =1/10 >  a =1/8-1/10 =1/40 =40 days
• A team of 100 men is supposed to do a work in 60 days. After 35 days, only 5/12 of the work was completed, so to complete the work before 40 more men were employed. If 40 men were not employed, how many extra days were required to complete the work by earlier number of men?
A) 11 days
B) 15 days
C) 12 days
D) 14 days
E) None of these.

Option D
Solution:
100 men completed 5/12 work in 35 days
So 100 men can complete the remaining 7/12 work in 49 days:
Use M1*D1*W2 = M2*D2*W1
100*35*(7/12) = 100*D2* (5/12)
D2 = 49 days
But after 35 days, 40 more men were employed, so 140 men now and they completed 7/12 work in
By M1*D1*W2 = M2*D2*W1
100*35*(7/12) = 140*D2*(5/12)
D2 = 35 days
So extra days = 49-35 = 14 days
• Two typist of varying skills can do a job in 6 minutes if they work together. If the first typist typed alone for 4 minutes and then second typed for 6 minutes , they would be left with  1/5 of the whole work. How many minutes would it take the slower typist to complete the work alone ?
A) 10 min
B) 12 min
C) 15 min
D) 20 min
E) None of these.

Option C
Solution:
The first typist types for 4 minutes and second one for 6 minutes , the work left would be the work the first typist can do in 2 minutes. Thus the time taken by the first typist to do the work would be 10 minutes and his rate of work would be 10% per minute . Since both can do whole work in 6 minutes their combined efficiency = 100/6 = 16.66% > second typist = 6.66%
so he would take = 100/6.66 = 15 minutes.
• Tap A  can fill a tank with water in 10 hrs. Tap B fills the same tank with milk in 12.5 hrs. A man who wanted to fill the tank with the mixture opens tap A first , which already contains 8% milk of its own capacity. After two hours he opened tap B till the tank gets filled completely. In what proportion should he mix this solution with the other one containing water and milk in the ratio 2 : 3, so that the new solution will contain half milk and half water?
A) 2:3
B) 1:1
C) 1:2
D) 2:1
E) 1:3

Option B
Solution:
Total work  – 50
a== 5
b== 4
it already contain 8% milk = 4 lit
a—- in 2 hrs = 5*2 = 10 lit water
so, total fill = 14 lit. remain. = 36 lit
done by a+b in 36/9 = 4 hrs
so,water added by a — 4*5 = 20 litre
milk added by b —- 4*4 = 16 litre
so. Total water ==== 10 + 20 = 30 litre
Total milk ===== 4 + 16 === 20 litre
W:M = 3:2
now 3/5______________2/5
____________1/2_______1:1
• A, B, C complete a work in 15, 20 and 30 days . They work together for sometime after which C left. A total of 18000 rs is paid for the work and B gets 6000 rs more than C.For how many days did A work ?
A) 8 days
B) 10 days
C) 12 days
D) 7 days
E) None of these.

Option A
Solution:
Total work  = 60
A — 4 ,, B– 3 ,,,, C — 2
let A &B work for x days
Then work done by A = 4x
B = 3x
C =60- (4x+3x)
So ratio of their share
4x : 3x : 60-7x
Difference between b and c = 3x – (60-7x) = 10x – 60
( 10x – 60)/60 * 18000 = 6000
so,x =8

• B takes twice time as A to complete a work and C takes thrice time as B to complete a work. If Rs6000 is given to them to complete a work together then B gets how much amount?
A) Rs1800
B) Rs3600
C) Rs600
D) Rs3000
E) Rs1200

Option A
Solution:

.               B……A              B…….A
.               2…….1               1………3
Days                     A…..B…….C
.                             1…….2……6
Efficiency             6…….3…..1
.                              3/10*6000 =1800
• A & B can do a piece of work in 80days. B & C can do same work in 50days and C & A can do same work in 60days. Find in how many days they all together can complete that work?
A) 40 (40/59)
B) 60 (40/59)
C) 36 (40/59)
D) 25 (40/59)
E) 26 (40/59)

Option A
Solution:

LCM = 2400
A + B = 80………….2400/80 = 30
B + C = 50…………………………..48
C + A = 60…………………………..40
2(A + B + C) = 118
A+ B + C = 59
So 2400/59  days
• A & B separately can do a piece of work in 9days and 12days respectively. If they work for a day alternatively, A starts the work, in how many days will the work will get completed?
A) 12(1/4)
B) 10(1/4)
C) 8(1/6)
D) 10(5/6)
E) 9(1/6)

Option B
Solution:

A=9        4
B=12      3          [LCM=36] 2 days alternate            (4+3) = 7 days
2*5                                    7*5
10days                             35days
now A’s turn so-  10(1/4) days
• 4men and 6boys earn Rs1600 in 5days, 3men and 7boys earn Rs1740 in 6days, in what time will 7men and 6boys earn Rs3760?
A) 4days
B) 6days
C) 8days
D) 10days
E) 5days

Option C
Solution:

4M + 6B= 1600/5=320……………(1)
3M + 7B= 1740/6=290……………(2)
FROM EQUATION (1) AND (2) WE GET
(4*7)B – (3*6)B = 290*4 – 320*3
B = Rs20……put in (1)
M = Rs50
now required number of days
3760/(7*50+6*20) = 3760/470 = 8days
• A tap take 42hrs extra to fill a tank due to a leakage equivalent to half of its inflow. The inlet pipe alone can fill the tank in how many hour?
A) 42hrs
B) 21hrs
C) 36hrs
D) 28hrs
E) 30hrs

Option A
Solution:

.               Without leak……………………With leak
Efficiency             2………………………….1
Time                      1…………………………..2
.                                          +1 == 42hours
So 42hours
• A tank can be filled with two pipes in 30minutes and 45minutes. When the tank was empty the two pipes A and B were opened. After some time, the first pipe A was closed and the tank was filled in 24minutes. After how much time from the start was the first pipe A closed?
A) 16 days
B) 15 days
C) 14 days
D) 12 days
E) 10 days

Option C
Solution:

A = 30, B = 45
LCM = 90
So, A = 90/30 = 3
. B = 90/45 = 2
B worked for all 24 days means did 24*2 = 48 units work
Remaining work = 90-48 = 42
This work is done by A, so 42/3 = 14 days
• A boy and a girl together fill a cistern with water. The boy pours 3ltr of water in every 2minutes and the girl pours 2ltr of water in every 3minutes. How much time will it take to fill 91ltr of water in the cistern?
A) 36min
B) 42min
C) 48min
D) 44min
E) 45min

Option B
Solution:

boy 2min = 3ltr
girl 3min = 2ltr
make time same
6min………..13ltr
*7…………….*7
42min………91ltr
• A and B together can complete a work in 30days and B alone can do it in 60days. Find in how many days A alone can do the work?
A) 40
B) 60
C) 120
D) 90
E) 110

Option B
Solution:

A+B…30……2
B……60……1……………(LCM=60)
A  ……………1
A = 60/1 = 60
• A and B can do a work in 18days. They started together but after 8days A left the work. If B does the remaining work in 20days, then in how many days A will do the work alone?
A) 18
B) 24
C) 36
D) 48
E) 50

Option C
Solution:

. A+B = 18
Let total work  =18
1day work of A & B=1
8days work of A&B =8
remaining =18 -8 =10
B does 10 work in 20days
So     18 work in 36days
A+B= 18
B      =36
LCM = 36
A+B= 18……..36/18 =2
B      =36………36/36 = 1
So A= 1
So days A   = 36/1   = 36
• A can write 75pages in 25hrs. A and B together can write 135pages in 27hrs.  In what time can B write 42pages?
A) 17
B) 19
C) 23
D) 21
E) 20

Option D
Solution:

A can write 75/25 =3pages in 1hr
A+B can  135/27 = 5pages in 1hr
B can write 5-3 = 2page in 1hr
42/2 =21hrs