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

#### TIME & WORK PRACTICE TEST [7 - EXERCISES]

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

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

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

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

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

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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