Time and Work - Examples

Q
If A can do a work in 10 days, B can do it in 20 days and C in 30 days in how many days will the three together do it?

Soln:
The efficiencies are A = 1/10, B = 1/20 and C = 1/30
So work done per day by the three = 1/10 + 1/20 + 1/30 = 11/60 => No of days = 60/11 = 5.45 days.


Q
If A and B can do a work in 10 days , B and C can do it in 20 days and C and A can do it in 40 days in what time all the three can do it?

Soln:

A+B = 1/10
B+C = 1/20
C+A = 1/40
Adding all the three we get 2(A+B+C) = 7/40 => A+B+C = 7/80 => No of days = 80/7 days.

If A can do a work in 12 days, B can do it in 18 days and C in 24 days. All the three started the work. A left after two days and C left three days before the completion of the work. How many days are required to complete the work?

Soln:

Let the total no of days be x.

A worked only for 2 days, B worked for x days and C worked for x-3 days.

So, mA + nB + oC = 1
ð      2(1/12) + x(1/18) + (x-3)(1/24) = 1
ð      12 + 4x + 3(x-3) = 72
ð      x = 69 / 7 days.

Note:

The ratio of dividing wages = ratio of efficiencies = ratio of parts of work done

Q:

A can do a work in 10 days and B can do it in 30 days and C in 60 days. If the total wages for the work is Rs. 1800 what is the share of A?

Soln:

Ratio of wages = 1/10 : 1/30 : 1/60 = 6 : 2 : 1  (Multiplying each term by LCM 60)

So total 9 equal parts in Rs. 1800 => each part = Rs. 200 => share of A = 6 parts = Rs. 1200.
Applying the same logics to pipes and cistern

Q:

A pipe can fill a tank in 5 hrs but because of a leak a the bottom it takes 1 hr extra. In what time can the leak alone empty the tank?

Soln:

Let the filling pipe be A.
A = 1 / 5.

But with the leak L,  A – L = 1 / 6   ( A-L because leak is outlet)

So, 1/L = 1 / 5 – 1/ 6 = 1/30 => Leak can empty the tank in 30 hrs.

Q:

A pipe A can fill the tank in 10 hrs, B can fill it in 20 hrs and C can empty in 40 hrs. All are opened at the same time. After how many hours shall the pipe B be closed such that the tank can be filled in 10 hrs?

Soln:

Let the pipe B be closed after x hrs.

Then A worked for 10 hrs, B worked for x hrs and C worked for 10 hrs.

mA + nB – oC = 1    (since C is outlet)

10(1/10) + x(1/20) – 10(1/40) = 1

x = 5 hrs.