Model 4  Working with Man, Woman, Child Practice Questions Answers Test with Solutions & More Shortcuts

Question : 6 [SSC CGL Prelim 1999]

5 men can do a piece of work in 6 days while 10 women can do it in 5 days. In how many days can 5 women and 3 men do it ?

a) 6 days

b) 4 days

c) 5 days

d) 8 days

Answer: (c)

5 × 6 men = 10 × 5 women

3 men = 5 women

5 women + 3 men = 6 men

5 men complete the work in 6 days

6 men will complete the work in ${5 × 6}/6$ = 5 days

Using Rule 14
If 'A' men can do a certain work in 'a' days and 'B' women can do the same work in 'b' days, then the total time is taken when $A_1$ men and $B_1$ women work together isTime taken = $1/{A_1/{A . a} + B_1/{B . b}}$If A men do a certain work in 'a' days, B women do the same work in 'b' days and C boys do the same work in 'c' days then the total time taken when $A_1$ men, $B_1$ women and $C_1$ boys can work together is,
Total time taken = $1/{(A_1/{A . a} + B_1/{B . b} + C_1/{C . c})}$

Here, A = 5, a = 6, B = 10, b = 5, $A_1$ = 3, $B_1$ = 5

Time taken = $1/{A_1/{A × a} + B_1/{B × b}}$

= $1/{3/{5 × 6} + 5/{10 × 5}}$

= $1/{1/10 + 1/10}$ = 5 days

Question : 7 [SSC CGL Tier-I 2015]

If 1 man or 2 women or 3 boys can do a piece of work in 44 days, then the same piece of work will be done by 1 man, 1 woman and 1 boy in

a) 26 days

b) 21 days

c) 24 days

d) 33 days

Answer: (c)

1 man ≡ 2 women ≡ 3 boys

1 man + 1 woman + 1 boy

≡ 3 boys + $3/2$ boys + 1 boy

≡ $(3 + 3/2 + 1)$ boys ≡ $11/2$ boys

By $M_1D_1 = M_2D_2$,

3 × 44 = $11/2 × D_2$

$D_2 = {2 × 3 × 44}/11$ = 24 days

Using Rule 13,

Here, A = 1, B= 2, C = 3, a = 44

$A_1 = 1, B_1 = 1, C_1$ = 1

Required time = $a/{A_1/A + B_1/B + C_1/C}$ days

= $44/{1/1 + 1/2 + 1/3} = {44 × 6}/11$ = 24 days

Question : 8 [SSC DEO 2009]

A man, a woman and a boy can complete a work in 20 days, 30 days and 60 days respectively. How many boys must assist 2 men and 8 women so as to complete the work in 2 days ?

a) 4

b) 8

c) 12

d) 6

Answer: (b)

Part of work done by 2 men and 2 women in 2 days.

= $2(2/20 + 8/30)$

= $2(1/10 + 8/30) = 2({3 + 8}/30)$

= $22/30 = 11/15$

= Remaining work =$1 - 11/15 = 4/15$

Work done by 1 boy in 2 days

= $2/60 = 1/30$

Number of boys required to assist = $4/15 × 30 = 8$

Using Rule 14,

Here, A = 1, B = 1, C = 1

a = 20, b = 30, c = 60

$A_1 = 2, B_1$ = 8

Required time = $1/{A_1/{A × a} + B_1/{B × b} + C_1/{C × c}$

2 = ${1/{2/{1 × 20} + 8/{1 × 30} + x/{1 × 60}$

2 = $10/{2/2 + 8/3 + x/6}$

2 = $10/{{6 + 16 + x}/6}$

22 + x = 30 ⇒ x = 8

∴ Number of boys = 8

Question : 9 [SSC CGL Tier-I 2016]

18 men or 36 boys working 6 hours a day can plough a field in 24 days. In how many days will 24 men and 24 boys working 9 hours a day plough the same field ?

a) 6

b) 9

c) 10

d) 8

Answer: (d)

18 men ≡ 36 boys ⇒ 1 man ≡ 2 boys

24 men + 24 boys ≡ (24 + 12) men ≡ 36 men

$M_1D_1T_1 = M_2D_2T_2$

$18 × 24 × 6 = 36 × D_2$ × 9

$D_2 = {18 × 24 × 6}/{36 × 9}$ = 8 days

Question : 10 [SSC CGL Prelim 1999]

If 6 men and 8 boys can do a piece of work in 10 days and 26 men and 48 boys can do the same in 2 days, then the time taken by 15 men and 20 boys to do the same type of work will be :

a) 6 days

b) 5 days

c) 4 days

d) 7 days

Answer: (c)

According to question,

(6M + 8B) × 10 = (26M + 48B) × 2

60M + 80B = 52M + 96B or, 1M = 2B

15M + 20B = (30 + 20)B

= 50 boys and 6M + 8B

= (12 + 8) boys = 20 boys

20 boys can finish the work in 10 days

50 boys can finish the work in 20 10 50 ´ days = 4 days

Using Rule 11
If $A_1$ men and $B_1$ boys can do a certain work in $D_1$ days, Again, $A_2$ men and $B_2$ boys can do the same work in $D_2$ days, then, $A_3$ men and $B_3$ boys can do the same work in
Required time = ${D_1D_2(A_1B_2 - A_2B_1)}/{D_1(A_1B_3 - A_3B_1) - D_2(A_2B_3 - A_3 B_2)}$ days

$A_1$ = 6, $B_1$ = 8, $D_1$ = 10

$A_2$ = 26, $B_2$ = 48, $D_2$ = 2

$A_3$ = 15, $B_3$ = 20

Required time = ${D_1D_2(A_1B_2 - B_1A_2)}/{D_1(A_1B_3 - A_3B_1) - D_2(A_2B_3 - A_3 B_2)}$ day

= ${10 × 2(6 × 48 - 8 × 26)}/{10(6 × 20 - 15 × 8) - 2(26 × 20 - 15 × 48)}$ days

= ${20(288 - 208)}/{10(120 - 120) - 2(520 - 720)}$

= ${20 × 80}/400$ = 4 days

IMPORTANT quantitative aptitude EXERCISES

Model 4  Working with Man, Woman, Child Shortcuts »

Click to Read...

Model 4  Working with Man, Woman, Child Online Quiz

Click to Start..

Recently Added Subject & Categories For All Competitive Exams

100+ Quadratic Equation Questions Answers PDF for Bank

Quadratic Equation multiple choice questions with detailed answers for IBPS RRB SO. more than 250 Attitude practice test exercises for all competitive exams

03-Jul-2024 by Careericons

Continue Reading »

IBPS Aptitude Linear Equations MCQ Questions Answers PDF

Linear equations multiple choice questions with detailed answers for IBPS RRB SO. more than 250 Attitude practice test exercises for all competitive exams

03-Jul-2024 by Careericons

Continue Reading »

New 100+ Compound Interest MCQ with Answers PDF for IBPS

Compound Interest verbal ability questions and answers solutions with PDF for IBPS RRB PO. Aptitude Objective MCQ Practice Exercises all competitive exams

02-Jul-2024 by Careericons

Continue Reading »

100+ Mixture and Alligation MCQ Questions PDF for IBPS

Most importantly Mixture and Alligation multiple choice questions and answers with PDF for IBPS RRB PO. Aptitude MCQ Practice Exercises all Bank Exams

02-Jul-2024 by Careericons

Continue Reading »