  # How to solve Time & Work – Part III

By BYJU'S Exam Prep

Updated on: September 25th, 2023 We have already posted articles related to the basic concepts of Time and Work. This post is in continuation with it. In this part 3, we will learn how to solve the problems based on Men, Women and Children; Work and Wages.

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Man, Hours Efficiency Day Work Rupees Consumption etc. terms are used in these types of questions. We know that
No. of persons (M) ∝ Work (W)
efficiency (η) ∝ Work (W)
no. of days (D) ∝ Work (W)
no. of hours (H) ∝ Work (W)

So, No. of persons × efficiency × no. of days × no. of hours per day = work
M1×η1×D1×H1 = W1         M2×η2×D2×H2 = W2
M1×η1×D1×H1 = constant  M2×η2×D2×H2 = constant
W1   ………(1)                      W2        ………(2)
From eqn (1) and (2)
M1 × η1 × D1 × H1   =   M2 × η2 × D2 × H2
W1                             W2

M1 × η1 × D1 × H1   =   M2 × η2 × D2 × H2
W1/R1/C1                              W2/R2/C2
where R= rupees or salary and C = consumption.
Important : When persons are same their efficiency will be equal. If persons are different their efficiency will be different.
Example 1:
12 persons can complete a work in 5 days, then how many persons are required to complete the same work in 3 days?
Solution: Here M1 = 12 and D1 = 5, D2 = 3 days we have to find M2
M1 × D1 = M2 × D2

12 × 5  = M2× 3
60     = M2 × 3
M2 = 20
Example 2: 18 persons can make 12 chairs in 6 days working 8 hours per day. In many days 12 persons can make 24 chairs working 6 hours per day?
Solution: Here M1 = 18, D1 = 6, H1 = 8 W1 = 12 and M2 = 12,H2 = 6, W2=24 are given we have to find D2

M1 × η1 × D1 × H1   =   M2 × η2 × D2 × H2
W1                                          W2
18 × 6 × 8 = 12 × D2 × 6
12               24
D2 = 24 days
Example 3: The expenditure of fuel is Rs.600 burning 6 stove for 12 days for 4 hours per day. How many stoves are required to burn 6 days for 8 hours making expenditure of Rs.900?
Solution:  Here R = Rupess
M1×η1×D1×H1   =   M2×η2×D2×H2
R1                        R2
Here M = no. of stoves
6 × 12 × 4 = M2 × 6 × 8
600             900
M2   =  9  Hence, 9 stoves are required.
Example 4: If Q persons can do Q units of work in Q days working Q hours per day then in how many days P persons can do P units of work, working P hours per day.
Solution: M1 × D1 × H1   =   M2 × D2 × H2
W1                           W2
Q × Q × Q      =   P × D2 × P
Q                       P
D2    =  Q2   days
P
Example 5:  If 5 women can do a work in 6 days working 9 hours per day. How many men are required to complete four times of work in 4 days working 6 hours per day. If 3 women can do the work in 6 hours that work can be done by 4 men in 3 hours.
Solution: In this type, we can see that efficiencies of men and women are different. So first we will calculate efficiency, check the last line of question.
work done by 3 women in 6 hours = work done by 4 men in 3 hours
3 ω × 6 = 4 m × 3
3ω = 2m or if ω = 2 then m = 3
ω = 2
m    3
where ω = eff. of women and m = effi. of men
using M1×D1×H1   =   M2×D2×H2
W1                             W2

5ω×6×9  =   Xm×4×6
1                     4

5×2×6×9X×3×4×6    (ω =2 and m = 3)
1                    4
X  = 30 men
Example 6: 4 women and 12 children together take 4 days to complete a piece of work. How many days will 4 children alone take to complete the piece of work if 2 women alone can complete the piece of work in 16 days?
Solution:   Let time take by 4 children to complete the work is X days.
work done by 4 women and 12 children in 4 days =work done by 2 women in 16 days = work done by 4 children in X days
(4ω+12C) × 4 = (2ω) × 16 = (4C) × X
By using
(4ω+12C) × 4 = (2ω) × 16
16ω+48C = 32ω
48C = 16ω
3C = ω
hence ω = 3 and C = 1
By using
(2ω) × 16 = (4C) × X   or  (4ω+12C) × 4 = (4C) × X
2×3× 16 = 4× 1 × X  or (4×3+12×1)× 4 = (4×1)× X
X = 24 days               or  (24) × 4 = 4 X
X = 24 days.

Important: There are so many shortcuts for this type of question, if the language of questions changes, students will be confused so we suggest you to go by this method. All approaches discussed above are the advanced part of Time and Work article 1 and 2.
Example 7: 4 men can complete a piece of work in 2 days. 4 women can complete the same piece of work in 4 days whereas 5 children can complete the same piece of work in 4 days. If 2 men, 4 women and 10 children work together, in how many days can the work be completed?
Solution: Let time taken by (2m+4ω+10C) is X days.
4m × 2 = 4ω × 4 = 5C × 4 = (2m+4ω+10C) × X ……………..(1)
Firstly, we will calculate efficiency of men, women and children.
8m = 16ω = 20C
2m = 4ω = 5C
m : ω : C   (to know the basics of ratio, check time and work article 2)
20 : 10 : 8
10 : 5 : 4
Now, putting in eqn (1)
4×10×2 = 4×5×4 = 5×4×4 = (2×10+4×5+10×4)× X
80 = (80) X
X = 1 day.

Example 8: 8 men and 4 women together can complete a piece of work in 6 days. Work done by a man in one day is double the work done by a woman in one day. If 8 men and 4 women started working and after 2 days, 4 men left and 4 new women joined. In how many more days will the work be completed? (IBPS PO)
Solution: Given, work of man in 1 day = 2×work of woman in 1 day
m × 1× 1 = 2 × ω  × 1
m = 2ω  So, m = 2 and ω = 1
Now, let total time take to complete the work is p days.
(8m+4ω)×6 = (8m+4ω)×2+(4m+8w)×(p-2)
(16+4)× 6 = (16+4)× 2 +(8+8)× (p-2)
120 = 40 + 16(p-2)
80 = 16 (p-2)
5 = p-2
p = 7
hence total time is 7 days but in question time taken by changed persons after 2 days is asked so answer will be 5 days.
Exam approach: As (8m+4ω) has worked for 2 days after that 4m left and 4ω joined, remaining work of 4 days of (8m+4ω) will be done by (4m+8ω) in X days.
so, (8m+4ω) × 4 = (4m+8ω)× X
(16+4) × 4 = (8+8)X
80  = 16X
X = 5 days
Example 9: If 5 women or 3 men or 12 children can complete a work in 6 days. In how many days 2 men, 3 women and 5 children can complete the same work.?
Solution: In this question, we can see that here ‘OR’ is used instead of ‘AND’.
So, we can write directly 5ω = 3m = 12C
ω   : m : C
12×3: 5×12: 5×3
12 : 20 : 5
5ω × 6 = 3m × 6 = 12C × 6 = (2m+3ω+5C)× X days
5× 12× 6 = 3×20×6 = 12×5×6 = (2×20+3×12+5×5)× X
360 = (40+36+25)X
X = 360/101 days.
Example 10:  A contractor wants to complete a project in 120 days and he employed 80 men. After 90 days ½ work is completed, then how many more persons he must hire to complete work on time?
Solution: Here given work has to be completed in 120 days and 80 men were employed earlier.
According to contractor,
In 120 days = 1 total work completed
In 90 days   = ¾ of work has to be completed
but we can see that it didn’t happen. In 90 days only ½ work is completed and remaining ½ work has to be completed in remaining 30 days.
Let more person hire are P.
90 × 80   = 30 × (P+80)
½                  ½
240 = P+80
P = 160 persons
Example 11:  P and Q undertake to do a piece of work in 6000Rs. P can do this in 8 days alone and B alone can do it in 12 days. With the help of R, they complete the work in 4 days. Find the part of P, Q and R individually?
Solution: Effici                  Days           total work
24/8 = 3          P………………8
24/12=2           Q…………….12             24
24/4 =6           P+Q+R……4      (LCM of 8,12and 4)
6 – (3+2)= 1       R
there are two methods :
(1) efficiency method
The share of P =   ηP      ×  total Rs.
η(P+Q+R)
=  3 × 6000
6
= 3000 Rs.
The share of Q = 2 × 6000
6
= 2000 Rs.
The share of R = 1  × 6000
6
=  1000 Rs.

(2) Work Method
The share of P = work done by  P            × total Rs
work done by P+Q+R
=  3 × 4 × 6000
24
=   3000 Rs.
The share of Q = 2 × 4 × 6000
24
= 2000 Rs
Share of R = 1 × 4 × 6000
24
=   1000 Rs

You can go through the previous parts of Time and Work series from the link below –

## How to solve Time & Work (Shortcut Approach) – Part II

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