6 TIME AND WORK
Ifa person can complete a work in 'n' days then he can do 1/n
part of the work in one day.
The amount of work done be a person in I day is called his efficiency.
Example:
A can do a work in 10 days. Then the efficiency ofA is given by A=1/10.
Note: Number of days required to do a work =
work to be done /work per day.
Example 1:
IfA 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.
Example 2:
IfAand 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.
Note: Ifall the people do not work for all the time then the principle below can be used
mA +nB +oC= 1. (1 is the total work)
Here, m=no of days A worked
n=no of days B worked
O=no of days C worked
A,B,C=efficiencies
, Example 3:
IfAcan do a work in 12 days, B can do it in 18 days and C in 24 days. All the three startedthewoork. A
after two days and C left three days before the completion of the work. How many days are r e
days are required
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
2(1/12) + x(1/18) + (x-31/24) =
1
12 +4x+3(x-3) = 72
x 69/7 days.
Note: The ratio ofdividing wages= ratio ofefficiencies = ratio of parts of work done
Example 4:
A can do a work in 10 days and B can do it in30 days and Cin 60 days. Ifthe total wages for the work
Rs. 1800 what is the share of A?
Soln: Ratio of wages= 1/10:1/30: 1/60=6:2:1 (Multiplyingeachterm by LCM 60)
So total 9 equal parts in Rs. 1800=> each part= Rs. 200=> share ofA= 6 parts = Rs. 1200.
Note: When pipes are used filling the tank they are treated similar to the men working but some out.
pipes emptying the tank are present whose work will be considered negative.
Example 5:
A pipe can fill a tank in 5 hrs but because ofa leak a the bottom it takes l hr extra. In what time can the ea
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-Leakcan emptythe tank in 30hrs.
Example 6:
h e sam
Apipe Acan fill the tank in 10 hrs, B can fill it in 20 hrs and C can empty in 40 hrs. All are opened at
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.
Ifa person can complete a work in 'n' days then he can do 1/n
part of the work in one day.
The amount of work done be a person in I day is called his efficiency.
Example:
A can do a work in 10 days. Then the efficiency ofA is given by A=1/10.
Note: Number of days required to do a work =
work to be done /work per day.
Example 1:
IfA 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.
Example 2:
IfAand 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.
Note: Ifall the people do not work for all the time then the principle below can be used
mA +nB +oC= 1. (1 is the total work)
Here, m=no of days A worked
n=no of days B worked
O=no of days C worked
A,B,C=efficiencies
, Example 3:
IfAcan do a work in 12 days, B can do it in 18 days and C in 24 days. All the three startedthewoork. A
after two days and C left three days before the completion of the work. How many days are r e
days are required
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
2(1/12) + x(1/18) + (x-31/24) =
1
12 +4x+3(x-3) = 72
x 69/7 days.
Note: The ratio ofdividing wages= ratio ofefficiencies = ratio of parts of work done
Example 4:
A can do a work in 10 days and B can do it in30 days and Cin 60 days. Ifthe total wages for the work
Rs. 1800 what is the share of A?
Soln: Ratio of wages= 1/10:1/30: 1/60=6:2:1 (Multiplyingeachterm by LCM 60)
So total 9 equal parts in Rs. 1800=> each part= Rs. 200=> share ofA= 6 parts = Rs. 1200.
Note: When pipes are used filling the tank they are treated similar to the men working but some out.
pipes emptying the tank are present whose work will be considered negative.
Example 5:
A pipe can fill a tank in 5 hrs but because ofa leak a the bottom it takes l hr extra. In what time can the ea
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-Leakcan emptythe tank in 30hrs.
Example 6:
h e sam
Apipe Acan fill the tank in 10 hrs, B can fill it in 20 hrs and C can empty in 40 hrs. All are opened at
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.
