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<body bgcolor="aqua"><center><h2>Time and Work</h2></center>
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<strong>Important Facts:</strong>
1.If A can do a piece of work in n days, then A's 1 day work=1/n
2.If A's 1 day's work=1/n, then A can finish the work in n days.
Ex: If A can do a piece of work in 4 days,then A's 1 day's work=1/4.
If A's 1 day’s work=1/5, then A can finish the work in 5 days
3.If A is thrice as good workman as B,then: Ratio of work done by
A and B =3:1. Ratio of time taken by A and B to finish a work=1:3
4.Definition of Variation: The change in two different variables
follow some definite rule. It said that the two variables vary
directly or inversely.Its notation is X/Y=k, where k is called
constant. This variation is called direct variation. XY=k. This
variation is called inverse variation.
5.Some Pairs of Variables:
i)Number of workers and their wages. If the number of workers
increases, their total wages increase. If the number of days
reduced, there will be less work. If the number of days is
increased, there will be more work. Therefore, here we have
direct proportion or direct variation.
ii)Number workers and days required to do a certain work is an
example of inverse variation. If more men are employed, they
will require fewer days and if there are less number of workers,
more days are required.
iii)There is an inverse proportion between the daily hours of a
work and the days required. If the number of hours is increased,
less number of days are required and if the number of hours is
reduced, more days are required.
6.Some Important Tips:
More Men -Less Days and Conversely More Day-Less Men.
More Men -More Work and Conversely More Work-More Men.
More Days-More Work and Conversely More Work-More Days.
Number of days required to complete the given work=Total work/One
day’s work.
Since the total work is assumed to be one(unit), the number of days
required to complete the given work would be the reciprocal of one
day’s work.
Sometimes, the problems on time and work can be solved using the
proportional rule ((man*days*hours)/work) in another situation.
7.If men is fixed,work is proportional to time. If work is fixed,
then time is inversely proportional to men therefore,
(M1*T1/W1)=(M2*T2/W2)
<b>Problems</b>
1)If 9 men working 6 hours a day can do a work in 88 days. Then 6 men
working 8 hours a day can do it in how many days?
Sol: From the above formula i.e (m1*t1/w1)=(m2*t2/w2)
so (9*6*88/1)=(6*8*d/1)
on solving, d=99 days.
2)If 34 men completed 2/5th of a work in 8 days working 9 hours a day.
How many more man should be engaged to finish the rest of the work in
6 days working 9 hours a day?
Sol: From the above formula i.e (m1*t1/w1)=(m2*t2/w2)
so, (34*8*9/(2/5))=(x*6*9/(3/5))
so x=136 men
number of men to be added to finish the work=136-34=102 men
3)If 5 women or 8 girls can do a work in 84 days. In how many days can
10 women and 5 girls can do the same work?
Sol: Given that 5 women is equal to 8 girls to complete a work
so, 10 women=16 girls.
Therefore 10women +5girls=16girls+5girls=21girls.
8 girls can do a work in 84 days
then 21 girls ---------------?
answer= (8*84/21)=32days.
Therefore 10 women and 5 girls can a work in 32days
4)Worker A takes 8 hours to do a job. Worker B takes 10hours to do the
same job. How long it take both A & B, working together but independently,
to do the same job?
Sol: A's one hour work=1/8.
B's one hour work=1/10
(A+B)'s one hour work=1/8+1/10 =9/40
Both A & B can finish the work in 40/9 days
5)A can finish a work in 18 days and B can do the same work in half the
time taken by A. Then, working together, what part of the same work they
can finish in a day?
Sol: Given that B alone can complete the same work in days=half the time
taken by A=9days
A's one day work=1/18
B's one day work=1/9
(A+B)'s one day work=1/18+1/9=1/6
6)A is twice as good a workman as B and together they finish a piece of
work in 18 days.In how many days will A alone finish the work.
Sol: if A takes x days to do a work then
B takes 2x days to do the same work
=>1/x+1/2x=1/18
=>3/2x=1/18
=>x=27 days.
Hence, A alone can finish the work in 27 days.
7)A can do a certain work in 12 days. B is 60% more efficient than A. How
many days does B alone take to do the same job?
Sol: Ratio of time taken by A&B=160:100 =8:5
Suppose B alone takes x days to do the job.
Then, 8:5::12:x
=> 8x=5*12
=> x=15/2 days.
8)A can do a piece of work n 7 days of 9 hours each and B alone can do it
in 6 days of 7 hours each. How long will they take to do it working together
8 2/5 hours a day?
Sol: A can complete the work in (7*9)=63 days
B can complete the work in (6*7)=42 days
=> A's one hour's work=1/63 and
B's one hour work=1/42
(A+B)'s one hour work=1/63+1/42=5/126
Therefore, Both can finish the work in 126/5 hours.
Number of days of 8 2/5 hours each=(126*5/(5*42))=3days
9)A takes twice as much time as B or thrice as much time to finish a piece
of work. Working together they can finish the work in 2 days. B can do the
work alone in ?
Sol: Suppose A,B and C take x,x/2 and x/3 hours respectively finish the
work then 1/x+2/x+3/x=1/2
=> 6/x=1/2
=>x=12
So, B takes 6 hours to finish the work.
10)X can do ¼ of a work in 10 days, Y can do 40% of work in 40 days and Z
can do 1/3 of work in 13 days. Who will complete the work first?
Sol: Whole work will be done by X in 10*4=40 days.
Whole work will be done by Y in (40*100/40)=100 days.
Whole work will be done by Z in (13*3)=39 days
Therefore,Z will complete the work first.
<font size=3> <a href="timeandwork.html"><b>Top</b></a></font>
<b>Complex Problems</b>
1)A and B undertake to do a piece of workfor Rs 600.A alone can do it in
6 days while B alone can do it in 8 days. With the help of C, they can finish
it in 3 days, Find the share of each?
Sol: C's one day's work=(1/3)-(1/6+1/8)=1/24
Therefore, A:B:C= Ratio of their one day’s work=1/6:1/8:1/24=4:3:1
A's share=Rs (600*4/8)=300
B's share= Rs (600*3/8)=225
C's share=Rs[600-(300+225)]=Rs 75
2)A can do a piece of work in 80 days. He works at it for 10 days & then B alone
finishes the remaining work in 42 days. In how much time will A and B, working
together, finish the work?
Sol: Work done by A in 10 days=10/80=1/8
Remaining work=(1-(1/8))=7/8
Now, work will be done by B in 42 days.
Whole work will be done by B in (42*8/7)=48 days
Therefore, A's one day's work=1/80
B’s one day's work=1/48
(A+B)'s one day's work=1/80+1/48=8/240=1/30
Hence, both will finish the work in 30 days.
3)P,Q and R are three typists who working simultaneously can type 216 pages
in 4 hours In one hour , R can type as many pages more than Q as Q can type more
than P. During a period of five hours, R can type as many pages as P can
during seven hours. How many pages does each of them type per hour?
Sol:Let the number of pages typed in one hour by P, Q and R be x,y and z
respectively
Then x+y+z=216/4=54 ---------------1
z-y=y-x => 2y=x+z -----------2
5z=7x => x=5x/7 ---------------3
Solving 1,2 and 3 we get x=15,y=18, and z=21
4)Ronald and Elan are working on an assignment. Ronald takes 6 hours to
type 32 pages on a computer, while Elan takes 5 hours to type 40 pages.
How much time will they take, working together on two different computers
to type an assignment of 110 pages?
Sol: Number of pages typed by Ronald in one hour=32/6=16/3
Number of pages typed by Elan in one hour=40/5=8
Number of pages typed by both in one hour=((16/3)+8)=40/3
Time taken by both to type 110 pages=110*3/40=8 hours.
5)Two workers A and B are engaged to do a work. A working alone takes 8 hours
more to complete the job than if both working together. If B worked alone,
he would need 4 1/2 hours more to compete the job than they both working
together. What time would they take to do the work together.
Sol: (1/(x+8))+(1/(x+(9/2)))=1/x
=>(1/(x+8))+(2/(2x+9))=1/x
=> x(4x+25)=(x+8)(2x+9)
=> 2x2 =72
=> x2 = 36
=> x=6
Therefore, A and B together can do the work in 6 days.
6)A and B can do a work in12 days, B and C in 15 days, C and A in 20 days.
If A,B and C work together, they will complete the work in how many days?
Sol: (A+B)'s one day's work=1/12;
(B+C)'s one day's work=1/15;
(A+C)'s one day's work=1/20;
Adding we get 2(A+B+C)'s one day's work=1/12+1/15+1/20=12/60=1/5
(A+B+C)'s one day work=1/10
So, A,B,and C together can complete the work in 10 days.
7)A and B can do a work in 8 days, B and C can do the same wor in 12 days.
A,B and C together can finish it in 6 days. A and C together will do it in
how many days?
Sol: (A+B+C)'s one day's work=1/6;
(A+B)'s one day's work=1/8;
(B+C)'s one day's work=1/12;
(A+C)'s one day's work=2(A+B+C)'s one day's work-((A+B)'s one day
work+(B+C)'s one day work)
= (2/6)-(1/8+1/12)
=(1/3)- (5/24)
=3/24
=1/8
So, A and C together will do the work in 8 days.
8)A can do a certain work in the same time in which B and C together can do it.
If A and B together could do it in 10 days and C alone in 50 days, then B alone
could do it in how many days?
Sol: (A+B)'s one day's work=1/10;
C's one day's work=1/50
(A+B+C)'s one day's work=(1/10+1/50)=6/50=3/25
Also, A's one day's work=(B+C)’s one day's work
From i and ii ,we get :2*(A's one day's work)=3/25
=> A's one day's work=3/50
B's one day’s work=(1/10-3/50)
=2/50
=1/25
B alone could complete the work in 25 days.
9) A is thrice as good a workman as B and therefore is able to finish a job
in 60 days less than B. Working together, they can do it in:
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