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

Question : 1 [SSC CGL Prelim 2002]

If 5 men or 8 women can do a piece of work in 12 days, how many days will be taken by 2 men and 4 women to do the same work?

a) 13$1/3$ days

b) 15 days

c) 13$1/2$ days

d) 10 days

Answer: (a)

According to the question 5 men = 8 women

2 men = $8/5 × 2 = 16/5$ women

Total women = $16/5 + 4 = 36/5$ women

No. of days to do the same work

= ${8 × 12}/{36/5} = {8 × 12 × 5}/36$

= $40/3 = 13{1}/3$ days

Using Rule 12,

Here, A = 5, B = 8, a = 12, $A_1$ = 2 and $B_1$ = 4

Time taken = ${a(A × B)}/{A_1B + B_1A}$

= ${12(5 × 8)}/{2 × 8 + 4 × 5}$

= ${12 × 40}/36 = 33{1}/3$ days

Question : 2 [SSC CGL Prelim 2003]

If 3 men or 4 women can plough a field in 43 days, how long will 7 men and 5 women take to plough it ?

a) 9 days

b) 10 days

c) 11 days

d) 12 days

Answer: (d)

3 men = 4 women

1 man = $4/3$ women

7 men = ${7 × 4}/3 = 28/3$ women

7 men + 5 women = $28/3 + 5$

= ${28 + 15}/3 = 43/3$ Women

Now, $M_1D_1 = M_2D_2$

4 × 43 = $43/3 × D_2$ ,

where $D_2$ = number of days

$D_2 = {4 × 3 × 43}/43$ = 12 days.

Using Rule 12,

Here, A = 3, B = 4, a = 43, $A_1$ = 7 and $B_1$ = 5

Time taken = ${a(A × B)}/{A_1B + B_1A}$

= ${43(3 × 4)}/{7 × 4 + 5 × 3}$

= ${43 × 12}/43$ = 12 days

Question : 3 [SSC CGL Prelim 2000]

A man, a woman and a boy can complete a job in 3, 4 and 12 days respectively. How many boys must assist 1 man and 1 woman to complete the job in $1/4$ of a day?

a) 19

b) 1

c) 4

d) 41

Answer: (d)

1 man’s 1 day’s work = $1/3$

1 woman’s 1 day’s work = $1/4$

1 boy’s 1 day’s work = $1/12$

(1 man + 1 woman)’s $1/4$ day’s work = $1/4(1/3 + 1/4) = 7/48$

Remaining work = $1 - 7/48 = 41/48$

Now, 1 boy’s $1/4$ day’s work = $1/4 × 1/12 = 1/48$

$41/48$ work will be done by $41/48$ × 48 = 41 boys.

Question : 4 [SSC CGL Tier-I 2014]

A man, a woman and a boy together finish a piece of work in 6 days. If a man and a woman can do the work in 10 and 24 days respectively. The days taken by a boy to finish the work is

a) 40

b) 30

c) 35

d) 45

Answer: (a)

Time taken by boy = x days

$1/10 + 1/24 + 1/x = 1/6$

$1/x = 1/6 - 1/10 - 1/24$

= ${20 - 12 - 5}/120 = 3/120 = 1/40$

x = 40 days

Using Rule 18,

Here , x = 6, y = 10, z = 24

Number of days = ${xyz}/{yz –x(y + z)}$ days

= ${6 × 10 × 24}/{10 × 24 - 6(10 + 24)}$

= $1440/{240 - 204} = 1440/36$ = 40 days

Question : 5 [SSC CGL Tier-I 2014]

If 40 men or 60 women or 80 children can do a piece of work in 6 months, then 10 men, 10 women and 10 children together do half of the work in

a) 5$7/13$ months

b) 5$6/13$ months

c) 6 months

d) 11$1/13$ months

Answer: (a)

40 men ≡ 60 women ≡ 80 children

10 men ≡ $80/40 × 10$ = 20 children

10 women ≡ $80/60 × 10 = 40/3$ children

10 men + 10 women + 10 children

= $(20 + 40/3 +10)$ children

= $({60 + 40 + 30}/3)$ children = $130/3$ children

${M_1D_1}/W_1 = {M_2D_2}/W_2$

$D_2 = {80 × 6 × 13}/130 = 144/13$ months

Half of the work can do

= $144/13 × 1/2 = 72/13 = 5{7}/13$ months

Using Rule 13
If A men or B boys or C women can do a certain work in 'a' days, then $A_1$ men, $B_1$ boys and $C_1$ women can do the same work in
Time taken = $a/{A_1/A + B_1/B + C_1/C}$

Here, A = 40, B= 60, C = 80, a = 6

$A_1 = 10, B_1 = 10, C_1$ = 10

Time taken = $a/{A_1/A + B_1/B + C_1/C}$

= $6/{10/40 + 10/60 + 10/80}$

= $6/{1/4 + 1/6 + 1/8}$

= $6/{{6 + 4 + 3}/24} = 144/13$

Half of the work they do in = $1/2 × 144/13$ months

= $72/13 = 5{7}/13$ months

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