**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

M_{1}×η_{1}×D_{1}×H_{1} = W_{1} M_{2}×η_{2}×D_{2}×H_{2} = W_{2}

__M _{1}×η_{1}×D_{1}×H_{1}__ = constant

__M__

_{2}×η_{2}×D_{2}×H_{2}_{ }= constant

W

_{1}………(1) W

_{2}………(2)

From eqn (1) and (2)

**12 persons can complete a work in 5 days, then how many persons are required to complete the same work in 3 days?**

W

W

where R= rupees or salary and C = consumption.

__M___{1}× η_{1}× D_{1}× H_{1 }_{ }=__M___{2}× η_{2}× D_{2}× H_{2}W

_{1}W_{2 }__M___{1}× η_{1}× D_{1}× H_{1 }_{ }=__M___{2}× η_{2}× D_{2}× H_{2}W

_{1}/R_{1}/C_{1 }W_{2}/R_{2}/C_{2}where R= rupees or salary and C = consumption.

_{ }Important: When persons are the same their efficiency will be equal. If persons are different their efficiency will be different._{ }Example 1:**Solution:**Here M

_{1}= 12 and D

_{1}= 5, D

_{2 }= 3 days we have to find M

_{2}

M

_{1}× D

_{1}= M

_{2 }× D

_{2 }

_{ }12 × 5 = M

_{2}× 3

60 = M

_{2}× 3

M

_{2}= 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 M

_{1}= 18, D

_{1}= 6, H

_{1 }= 8 W

_{1}= 12 and M

_{2 }= 12,H

_{2 }= 6, W

_{2}=24 are given we have to find D

_{2 }

__M__

_{1}× η_{1}× D_{1}× H_{1 }_{ }=

__M__

_{2}× η_{2}× D_{2}× H_{2}W

_{1 }W

_{2 }

__18 × 6 × 8__=

__12 × D__

12 24

_{2 }× 6D

_{2}= 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}_{ }

__M__

_{1}×η_{1}×D_{1}×H_{1 }_{ }=

__M__

_{2}×η_{2}×D_{2}×H_{2}R

_{1}R

_{2 }Here M = no. of stoves

__6 × 12 × 4__=

__M__

600 900

_{2}× 6 × 8M

_{2}= 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:**

__M__

_{1}× D_{1}× H_{1 }_{ }=

__M__

_{2}× D_{2}× H_{2}W

_{1 }W

_{2 }

__Q × Q × Q__=

__P × D__

Q

_{2 }× P**P**

**D**

_{ }_{2}=

__Q__

^{2}_{ }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

__M__

_{1}×D_{1}×H_{1 }_{ }=

__M__

_{2}×D_{2}×H_{2}W

_{1 }W

_{2}

__5ω×6×9__=

__Xm×4×6__

1

**4**

_{ }__5×2×6×9__=

__X×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 language of questions change, 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?

(SBI Rural Business officers)

**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**

Share of P = __ η _{P} __ × total Rs.

__η__

_{(P+Q+R) }=

__3__× 6000

6

= 3000 Rs.

Share of Q =

__2__× 6000

__6__

= 2000 Rs.

Share of R =

__1__× 6000

6

= 1000 Rs.

**(2) Work Method**

Share of P = __work done by P __ × total Rs

work done by P+Q+R

= __3 × 4__ × 6000

24

= 3000 Rs.

Share of Q = __2 × 4__ × 6000

24

= 2000 Rs

Share of R = __1 × 4__ × 6000

24

= 1000 Rs

**Time & Work (Shortcut Approach) Part 1: Click here**

**Time & Work (Shortcut Approach) Part 2: Click here**

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