Time and Work

Type of Problems

Type 1: Calculating Ratio

If 'A' is 'x' times as good a workman as 'B', then:

1. The ratio of work done by A & B in equal time = x: 1
2. 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? 

1. 10 days
2. 12 days
3. 16 days
4. 15 days
5. 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) = (x-y)/xy

So, B alone will take (xy)/(x-y) 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+yz-zx)

 

Similarly,

For B working alone, time required = (2xyz)/ (-xy+yz+zx)

For C working alone, time required = (2xyz)/ (xy-yz+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? 

1. 20 days
2. 25 days
3. 55 days
4. 46 days
5. 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?

1. 2 days
2. 6 days
3. 3 days
4. 5 days
5. 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?

1. Rs. 300
2. Rs. 400
3. Rs. 800
4. Rs. 500
5. 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 Man-Work-Hour 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 man-days. 

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 man-days 

10000 man-days are only 40% of the work. 

Remaining work = 100 - 40 = 60% 

40% work = 10000 man-days 

60% work = (10000/40)×60 = 15000 man-days

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.
 

Key Takeaways

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.