Problems on Time and Work

1) Problem: 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?

Solution: 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) Problem: 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?

Solution: 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) Problem: 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?

Solution: Given that 5 women is equal to 8 girls to complete a work. So, 10 women = 16 girls. Therefore 10 women + 5 girls = 16 girls + 5 girls = 21 girls. 8 girls can do a work in 84 days then 21 girls can do a work in (8*84/21) = 32 days. Therefore 10 women and 5 girls can a work in 32 days

4) Problem: Worker A takes 8 hours to do a job. Worker B takes 10 hours to do the same job. How long it take both A & B, working together but independently, to do the same job?

Solution: 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) Problem: 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?

Solution: Given that B alone can complete the same work in days = half the time taken by A = 9 days
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) Problem: 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.

Solution: 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) Problem: 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?

Solution: 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) Problem: 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?

Solution: 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)) = 3 days

9) Problem: 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 ?

Solution: 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) Problem: 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?

Solution: 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.

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