Time and work Tricks for Competitive Exams Part – 5th

In the previous post of Time and work we discussed about some questions of basic concepts of this chapter. Now we will see some of the some more various frequently asked important question of this chapter in competition like SSC, Bank (IBPS), and Railways. This chapter is one of the easiest chapters for students to solve with less calculations.

Type 2^nd  :- Questions based One day work Concept:-

In these type of questions you just have to read the questions carefully to analyse the given data in the question and then calculate the required answer with the help of one day work concept or formula given in this post, which ever you feel more comfortable to solve the question.

Q1. A can do a work in 10 days, B can do the same work in 15 days and worker C can do the same work in 30 days. How long would they take to complete the same work, working together.

Solution:- First of all you should read this question carefully and analyse it carefully in such manner:-

Method 1^st :- The work in this question is same means it is constant and always be considered as 1 (one) work only. Now the procedure for this type of question is :-

worker “A” can do a piece of work in “10 days”

Therefore the one day work of worker “A” = (1/10) work

worker “B” can do a piece of work in “15 days”

Therefore the one day work of worker “B” = (1/15) work

worker “C” can do a piece of work in “30 days”

Therefore the one day work of worker “C” = (1/30) work

Now The one day work done “A”, “B” and “C”, working together:- One day work of (A + B + C) = [one day work of worker “A” + one day work of worker “B” + one day work of worker “C” ]

One day work of (A+B) = (1/10) + (1/15) + (1/30)

On solving (taking LCM and adding them) = (6/30) work

On simplifying this fraction:- One day work of (A+B) = (1/5) work

Therefore the number of days to complete the (1) work = [Total work/one day work]

Number of days = [1/(1/5)] = 5 days

Hence the required Number of days = 5 Days.

Method 2^nd :- This method is direct formula based, let see and memorise this formula:-

If “A” can do a work in D₁ days, “B” can do the same work in D₂ days and “B” can do the same work in D₃

Then “A”, “B” and “C” can do the same work, working together in :-

D= [D₁×D₂×D₃/( D₁D₂+D₂D₃+ D₃D₁ )] days

Let us analyse our given question for data:- D₁ = 10 days, D₂ = 15 days, D₃ = 30 days & D =?

Formula :- D = [D₁×D₂×D₃/( D₁D₂+D₂D₃+ D₃D₁ )] days

                    D = [10×15×30/(10×15 + 15×30 + 30×10)] days

                    D = [10×15×30/(150 + 450 + 300)] days

                    D = [10×15×30/(900)] days.

On solving   D = 5 days.               

Hence the required Number of days = 5 Days.

Q2. A  & B can do a work together in 12 days, B & C can do the same work in 15 days while C & A can do the same work in 20 days. How long would they take to complete the same work, working separately?

Method 1^st :- The work in this question is same means it is constant and always be considered as 1 (one) work only. Now the procedure for this type of question is :-

worker “A” & “B” can do a piece of work in “12 days”

Therefore the one day work of worker “A” & “B” together = (1/12) work

worker “B” & “C” can do the same piece of work in “15 days”

Therefore the one day work of worker “B” & “C” together = (1/15) work

worker “C” & “A” can do the same piece of work in “20 days”

Therefore the one day work of worker “C” & “A” together = (1/20) work

Mathematically we can say that

    A + B = 1/12 .................................. eq.no.1

    B + C = 1/15 .................................. eq.no.2

    C + A = 1/20 ................................. eq.no.3

Adding the above three equations we get:-

                                       2{A + B + C} = (1/12) + (1/15) + (1/20)

On taking LCM we get      2{A + B + C} = 12/60

On solving we get              2{A + B + C} = 1/5

                                                {A + B + C} = 1/10.  ........................eq.no.4

Eq.no.4 implies that the work done by worker A,B & C in one day, working together = 1/10

Now one day work of worker “A” = one day work (A+B+C) – one day work (B+C)

                                                              = (1/10) – (1/15)

         one day work of worker “A” = (1/30) (taking LCM & solving)

Therefore worker “A” can complete the work in 30 days.

Now one day work of worker “B” = one day work (A+B+C) – one day work (C+A)

                                                              = (1/10) – (1/20)

         one day work of worker “B” = (1/20) (taking LCM & solving)

Therefore worker “B” can complete the work in 20 days.

Now one day work of worker “C” = one day work (A+B+C) – one day work (A+B)

                                                              = (1/10) – (1/12)

         one day work of worker “C” = (1/60) (taking LCM & solving)

Therefore worker “C” can complete the work in 60 days.

Hence Required answers are :-

“A” can complete the work in 30 days.

“B” can complete the work in 20 days.

“C” can complete the work in 60 days.

(Note you may say that this is so long but dear this is the only way to explain the whole concept of the question to you now its upto you that how fast and short handedly you can solve these type of questions)                                                                        

The Various Questions related to this and other concept which are frequently asked in the examinations will be presented to your for self study on this blog soon, so keep visiting and sending your suggestions and requirements too to us through the forum.

More types of questions will be followed soon on this blog........................