HomeQuantitative TechniquesQuestions for Practice on Time and Work for CLAT 2020

Questions for Practice on Time and Work for CLAT 2020

Questions for Practice on Time and Work for CLAT 2020

Questions for Practice on Time and Work for CLAT 2020

Question 1-3:​​ Roman can craft a Wood work in 24 days which Sonam can do in 18 days. ​​ Answer the following questions:

  • In how many days they can together complete the work?

  • Who is more Efficient and by how much percent?

  • If they on work alternate days starting with Roman, when will be the work complete?​​ 

Question 4-6:​​ Mili starts working on​​ a project and works on it for 20​​ days and complete 40% of the project. After some days, Nita joins the project and together they complete the project in another 12​​ days.​​ 

  • How long would Mili​​ take to complete the job alone?

  • If they started to work together from the beginning, how long they would have taken to complete the job?

  • What is the ratio of time that Nita would take to that of the time Mili would take if they worked alone?

Question 7-8:​​ Pipe A can fill a tank in 2.5 hours. There is a leak​​ at the bottom of the tank and hence it takes twice longer to fill the tank.​​ 

  • In how many hours the full tank will get empty, if pipe A is closed?

  • If another pipe B takes 2 hours​​ to completely fill​​ the​​ tank, in how many hours they can together fill the tank​​ without any leakage?

Question 9-10:​​ If three taps are opened together, a tank gets filled in 10​​ hours. Filling alone, one tap takes 5 hours, another takes 10 hours.​​ 

  • At what rate 3rd​​ tap can fill the tank?

  • ​​ If only 1 and 3rd​​ tap were open, how long it will take to fill the tank? ​​ 

Answer and Solutions:

  • Ans: 10 +2/7days

Sol:​​ LCM of 18 and 24 is 72, we can assume that there is a total of 72 units of work has to be done in total. So in 1 day, Roman can do 72/24 =3 units of work in one day while Sonam can complete 72/18 = 4 units of work in one day.

Hence in one day, together they can do 3+4 = 7 units of work.

Therefore, 72 units of work can be done in = 72/7 = 10 +2/7days

  • Ans:​​ Sonam by 33.33%​​ 

Sol:​​ Roman’s one day work = 3 units

Sonam’s one day work = 4 units

Efficiency in terms of per day work = 4-3/3 *100 = 33.33 %

Sonam is 33.33% more efficient than Roman.

  • Ans:​​ 20 + 2/3 days

Sol:​​ If Roman started working, on first day he will complete 3 units of work, and next day Sonam will complete 4 units of work. Hence in 2 days they will complete 7 units of work.​​ 

Hence, in 20 days (i.e. 10 complete cycles) time 7*10=70 units of work. Remaining work is 2 units done by Roman on 11th​​ day. But roman can do 3 units of work in one day, 2 units left will take 2/3 time. Hence,​​ in total they will take​​ 20+​​ 2/3 days to complete the work.

  • Ans:​​ 50 days

Sol:​​ If Mili​​ can do 40% of work in 20 days, she can complete the next 40% of work in another 20 days, and 20% work in 10 days. Hence for completing the whole work alone she would take 20+20+10 days = 50 days.​​ 

  • Ans:​​ 20 days

Sol:​​ In 12 days the remaining​​ 60% of the work is done. Hence for​​ completing 100%​​ of work​​ they would need 12*100/60 = 20 days

  • Ans:​​ 2:3

Sol:​​ Mili alone can do the work alone in 50 days. And together they would take 20 days; Let Nita would take b days.

1/50 + 1/b = 1/20 => b = 100/3 days

Ratio of time they would take alone =

Nita alone time / Mili alone time= 100/50*3 = 2/3

  • Ans:​​ 10 hours

Sol:​​ When pipe a works along with leakage, the total time taken is 2*2.5=5 hours

So, 1/2.5 – 1/x =1/5​​ 

X= 2.5 hours

LCM of 2.5 and 5 is 5. So, work required to be done is 5 units, in 1 hour, A can do 5/2.5 = 2 units of work.

Work done per hour = 1 unit. Leakage per hour = 5/2.5 = 2​​ 

So, negative work done =2.5-2 = 0.5 unit in 1 hour.

Hence to complete the task of 5 unit = 5/0.5 = 10 hours.

  • Ans:1.111 hours

Sol:​​ A takes 2.5 hours alone, while another pipe takes 2 hours;

1/A=1/2.5 + ½ = (4+5)/10= 9/10

Hence, together they take = 10/9= 1.111 hours

  • ​​ Ans:​​ 5 hours negative work​​ 

Sol:​​ If tap​​ A take 5 hours and pipe B takes 10 hours to completely fill the tank.

Together all three​​ took 9 hours to fill, hence 1/10​​ = 1/a+1/b +1/c​​ 

1/c = 1/10​​ -1/10 - 1/5​​ => x= -5,​​ 

This shows that tap 3 can empty a tank in 5 hours. As negative sign is showing that third tap is doing leakage work. ​​ 

  • Ans:​​ zero

Sol:​​ If only 1st​​ and 3rd​​ tap are open then, 1/5 – 1/5 = 0. This suggests that, the tank will never be filled as both​​ the taps are cutting each other​​ work.

Read the theory on Simple and Compound Interest here.

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Aditya Anand
Aditya is 93.1% sure that he knows Japanese. We think he speaks Japanese in Bhojpuri accent.

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