Type of Problems
Type 1: Calculating Ratio
If 'A' is 'x' times as good a workman as 'B', then:
 The ratio of work done by A & B in equal time = x: 1

The ratio of time taken by A & B to complete the work = 1: x.
This means that 'A' takes (1/x^{th}) time as that of 'B' to finish the same amount of work.
So, A is twice good a workman as B means that
a) A does twice as much work as done by B in equal time i.e. A: B = 2:1
b) A finishes his work in half the time as B.
Example Problems
P1. Dev completed the school project in 20 days. How many days will Arun take to complete the same work if he is 25% more efficient than Dev?
 10 days
 12 days
 16 days
 15 days
 5 days
Answer: Let the days taken by Arun to complete the work be x
The ratio of time taken by Arun and Dev = 125:100 = 5:4
5:4 :: 20:x
⇒ x = {(4×20) / 5}
⇒ x = 16
So, Ans = 16 Days
Type 2: Calculating Combined Work
a) If 'A' and 'B' can finish the work in 'x' & 'y' days respectively, then
A's one day work = 1/x
B's one day work = 1/y
(A + B)'s one day work = 1/x + 1/y = (x+y)/xy
So, together A and B can finish work in (xy)/(x+y) days.
b) If 'A', 'B' & 'C' can complete the work in x, y & z days respectively, then
(A+B+C)’s 1 day work = 1/x + 1/y + 1/z = (xy+yz+xz)/xyz
So, together A and B can finish work in (xyz)/(xy+yz+xz) days.
c) If A can do a work in 'x' days and if the same amount of work is done by A & B together in 'y' days, then:
A's one day work = 1/x
(A+B)'s one day work = 1/y
B’s one day work = (1/y)(1/x) = (xy)/xy
So, B alone will take (xy)/(xy) days.
d) If A & B together perform some part of work in 'x' days, B & C together perform it in 'y' days and C & A together perform it in 'z' days, then:
(A+B)'s one day work = 1/x
(B+C)'s one day work = 1/y
(C+A)'s one day work = 1/z
=> (1/x + 1/y + 1/z) = 2(A+B+C)’s one day's work.
So, (A+B+C)’s one day's work = (1/x + 1/y + 1/z) / 2.
=> (A+B+C) will together complete work in 2/ (1/x + 1/y + 1/z) days.
If A works alone, then deduct A's work from the total work of B & C to find the time taken by A alone.
For A working alone, time required = A's work  (A+B+C)'s combined work
= 2/ (1/x  1/y + 1/z)
= (2xyz)/ (xy+yzzx)
Similarly,
For B working alone, time required = (2xyz)/ (xy+yz+zx)
For C working alone, time required = (2xyz)/ (xyyz+zx)
Example Problems
P1. Sonal and Preeti started working on a project and they can complete the project in 30 days. Sonal worked for 16 days and Preeti completed the remaining work in 44 days. How many days would Preeti have taken to complete the entire project all by herself?
 20 days
 25 days
 55 days
 46 days
 60 days
Answer:
Let the work done by Sonal in 1 day be x
Let the work done by Preeti in 1 day be y
Then, x+y = 1/30 ——— (1)
⇒ 16x + 44y = 1 ——— (2)
Solving equations (1) and (2),
x = 1/60
y = 1/60
Thus, Preeti can complete the entire work in 60 days
P2. Time taken by A to finish a piece of work is twice the time taken by B and thrice the time taken by C. If all three of them work together, it takes them 2 days to complete the entire work. How much work was done by B alone?
 2 days
 6 days
 3 days
 5 days
 Cannot be determined
Answer:
Time taken by A = x days
Time taken by B = x/2 days
Time Taken by C = x/3 days
⇒ {(1/x) + (2/x) + (3/x) = 1/2
⇒ 6/x = 1/2
⇒ x = 12
Time taken by B = x/2 = 12/2 = 6 days
P3. To complete a piece of work, Samir takes 6 days and Tanvir takes 8 days alone respectively. Samir and Tanvir took Rs.2400 to do this work. When Amir joined them, the work was done in 3 days. What amount was paid to Amir?
 Rs. 300
 Rs. 400
 Rs. 800
 Rs. 500
 Rs. 100
Answer: Total work done by Samir and Tanvir = {(1/6) + (1/8)} = 7/24
Work done by Amir in 1 day = (1/3) – (7/24) = 1/24
Amount distributed between each of them = (1/6) : (1/8) : (1/24) = 4:3:1
Amount paid to Amir = (1/24) × 3 × 2400 = Rs.300 = (1) option
Type 3: Calculating ManWorkHour Related Problems
Note: (M*D*H)/W = Constant
where,
M: Number of Men
D: Number of Days
H: Number of Hours
W: Amount of Work done
If men are fixed, work is proportional to time.
If work is fixed, time is inversely proportional to men.
Thus,
(M1*T1)/W1 = (M2*T2)/W2
Example Problems
P1. 10 men working 6 hours a day can complete work in 18 days. How many hours a day should 15 men work for 12 days so that they can complete double the work?
Answer:
Original work = 10 × 18 × 6 mandays.
New work = 10 × 18 × 6×2
Let x hours per day 15 men take.
According to work equivalence; 10 × 18 × 6×2 = 15 × 12 × 𝑥
Therefore x = 12 hr/day
P2. A contractor undertakes to complete a job in 100 days and employs 200 men to complete the work. After 50 days he finds that only 40% of the work is completed. To complete the work in time, how many men should he hire?
Answer:
Work to be done in 50 days = 200×50 =10000 mandays
10000 mandays are only 40% of the work.
Remaining work = 100  40 = 60%
40% work = 10000 mandays
60% work = (10000/40)×60 = 15000 mandays
You have only 50 more days left.
Let n be the number of men required to complete the work.
Therefore; 50×n = 15000 and n = 300 men.
Hence; 300  200 = 100 men need to hire.
In this blog, we learned about some important formulas/insights about time and work. Then we learned about the different types of problems. We covered examples/sample problems for each type to understand these concepts better.
You can learn more about permutations and combinations from here and practice similar problems on the Coding Ninjas Studio.