Work Problems
1. If a person can do a piece of work in ‘m’ days, he can do 1/m of the work in 1 day.
2. If the number of persons engaged to do a piece of work be increased (or decreased) in a certain ratio the time required to do the same work will be decreased (or increased) in the same ratio.
3. If A is twice as good a workman as B, then A will take half the time taken by B to do a certain piece of work.
4. Time and work are always in direct proportion.
5. If two taps or pipes P and Q take ‘m’ and ‘n’ hours respectively to fill a cistern or tank, then the two pipes together fill (1/m + 1/n) part of the tank in 1 hour and the entire tank is filled in 1/(1/m + 1/n) = mn/(m+n) hours.
Example 1. If 12 man can do a piece of work in 36 days. In how many days 18 men can do the same work?
Solution: 12 men can do a work in 36 days. 18 men can do the work in 12/18 x 36 = 24 days.
Note:If the number of men is increased, the number of days to finish the work will decrease.
Example 2. A and B can finish a work in 12 days. B and C can finish the same work in 18 days. C and A can finish in 24 days. How many days will take for A, B and C combined together to finish the same amount of work?
Solution: A and B can finish the work in 12 days.
(A+B) Can finish in 1 day 1/12 of the work.
Similarly (B+C) can finish in 1/18 of the work.
(C+A) can finish in 1/24 of the work.
2(A+B+C) can finish in 1 day (1/12 + 1/18 + 1/1/24) of the work.
= (6 + 4 + 3)/72 = 13/72 of the work.
(A + B + C) can finish in 1 day = 13/144 of the work
Therefore (A + B + C) can together finish the work in 144/13 = 11 1/3 days.
Example 3: A and B can do a piece of work in 12 days. B and C can do it in 20 days. If A is twice as good a workman as C, then in what time will B alone do it?
Solution: (A+B) can do a work in 12 days.
(A+B) in one day can do 1/12 of the work --- (1)
Similarly (B+C) in one day can do 1/20 of the work --- (2)
Since A=2C, in (1) put A=2C
Therfore (2C+B) in one day can do 1/12 of the work.
i.e., [(B+C)+C] in one day can do 1/12 of the work.
C alone can do in one day 1/12 - 1/20 = (5-3)/60 = 2/60 = 1/30 of work --- (3)
From (2), using (3), B alone can do in one day 1/20 - 1/30 = (3-2)/60 = 1/60 of the work.
Therefore B alone can do the entire work in 60 days.
Pipes and Cisterns
Inlet: A pipe connected with a tank or a cistern or a reservoir, that fills it, is known as an inlet.
Outlet: A pipe connected with a tank or a cistern or a reservoir, emptying it, is known as an outlet.
Formulae:
(i) If a pipe can fill a tank in x hours, then:
(a) Part filled in 1 hour = 1/x
(b) If a pipe can empty a full tank in y hours, then: Part emptied in hour = 1/y
(ii) If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where y > x), then on opening both the pipes, the net part filled in 1 hour = [ 1/x – 1/y].
Example 1: Two pipes A and B can fill a tank in 36 hours and 45 hours respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank?
Solution: Part filled by A In 1 hour = 1/36
Part filled by B in 1 hour = 1/45
Part filled by (A + B) In 1 hour = [1/36 + 1/45] = 9/180 = 1/20
Hence, both the pipes together will fill the tank in 20 hours.
Example 2: A pipe can fill a tank in 16 hours. Due to a leak in the bottom,it is filled in 24 hours. If the tank is full, how much time will the leak take to empty it ?
Solution: Work done by the leak in 1 hour = [ 1/16 – 1/24 ]=1/48
:. Leak will empty the full cistern in '48 hours.
Example 3: A cistern is filled by pipe A in 10 hours and the full cistern can be leaked out by an exhaust pipe B in 12 hours. If both the pipes are opened, in what time the cistern is full?
Solution: Work done by the inlet in 1 hour: = 1/10
Work done by the outlet in 1 hour = 1/12
Net part filled in 1 hour = [ 1/10 – 1/12 ] = 1/60
The cistern will be full in 60 hours.
Example 4: Two pipes can fill a cistern in 14 hours and 16 hours respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom, 32 minutes extra are taken for the cistern to be filled up. When the cistern is full in what time will the leak empty it ?
Solution: Work done by .the two pipes in 1 hour = [ 1/14 + 1/16 ] = 15/112
Time taken by these pipes to fill the tank = 112/15 hours
= (7 hrs. 28 min.)
Due to leakage, time taken = (7 hrs 28 min.) + 32 min. = 8 hrs
Work done by ( two pipes + leak) in 1 hour = 1/8
Work done by the leak m 1 hour = [ 15/112 – 1/8 ] = 1/112
Leak will empty the full cistern in 112 hours.
Example 5: Pipes A and B can fill a tank in 20 hours and 30 hours respectively and pipe C can empty the full tank in 40 hours. If all the pipes are opened together, how much time will be needed to make the tank full ?
Solution: Net part filled m 1 hour = [1/20 + 1/30 – 1 /40] = 7/120
The tank will be full in 120/7 = 17 ½
