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

Question : 31 [SSC CGL Prelim 2004]

6 men or 12 women can do a piece of work in 20 days. In how many days can 8 men and 16 women do twice as big as this work ?

a) 15 days

b) 2 days

c) 5 days

d) 10 days

Answer: (a)

6 men = 12 women ⇒ 1 man = 2 women

Now, 8 men + 16 women

= (8 × 2 + 16) women = 32 women

12 women can do a work in 20 days.

1 woman can do the work in 20 × 12 days.

32 women can do the twice work in = ${20 × 12 × 2}/32$ = 15 days.

Using Rule 12,

Here, A = 6, B= 12, a = 20, $A_1$ = 8, $B_1$ = 16

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

= ${20(6 × 12)}/{8 × 12 + 16 × 6}$

= ${20 × 72}/192 = 15/2$

They will do the twice as big work in $2 × 15/2$ days = 15 days

Question : 32 [SSC CGL Prelim 2000]

If 3 men or 6 women can do a piece of work in 16 days, in how many days can 12 men and 8 women do the same piece of work?

a) 3 days

b) 4 days

c) 5 days

d) 2 days

Answer: (a)

3m = 6w ⇒ 1m = 2w

12m + 8w = (12 × 2w) + 8w = 32w

6 women can do the work in 16 days.

32 women can do the work in ${16 × 6}/32 = 3$ days

Using Rule 12
If A men or B boys can do a certain work in 'a' days, then $A_1$ men and $B_1$ boys can do the same work in
Time taken = $a/{A_1/A + B_1/B} = {a(A . B)}/{A_1B + B_1A}$ days

Here, A = 3, B = 6, a = 16, $A_1$ = 12, $B_1$ = 8

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

= ${16(3 × 6)}/{12 × 6 + 8 × 3}$

= ${16 × 18}/96$ = 3 days

Question : 33 [SSC CGL Tier-I 2016]

12 men and 16 boys can do a piece of work in 5 days; 13 men and 24 boys can do it in 4 days, then the ratio of the daily work done by a man to that of a boy is

a) 1 : 3

b) 2 : 1

c) 3 : 1

d) 5 : 4

Answer: (b)

Work done by 12 men + 16 boys in 5 days

≡ Work done 13 men + 24 boys in 4 days

(60 men + 80 boys)’s 1 day’s work ≡ (52 men + 96 boys)’s 1 day’s work

(60 - 52) men ≡ (96 - 80) boys

8 men ≡ 16 boys ⇒ 1 man ≡ 2 boys

∴ Required ratio = 2 : 1

Question : 34 [SSC CGL Prelim 2005]

A man, a woman and a boy can together complete a piece of work in 3 days. If a man alone can do it in 6 days and a boy alone in 18 days, how long will a woman alone take to complete the work?

a) 24 days

b) 9 days

c) 21 days

d) 27 days

Answer: (b)

Work done by 1 woman in 1 day

= $1/3 - 1/6 - 1/18 = {6 - 3 - 1}/18 = 1/9$

Woman will do the work in 9 days.

Using Rule 18
A, B and C can do a certain work together within 'x' days. While, any two of them can do the same work separately in 'y' and 'z' days, then in how many days can 3rd do the same work?
Required time = ${xyz}/{yz - x(y + z)}$days

Here, x = 3, y = 6 and z = 18

Required time = ${xyz}/{yz - x(y + z)}$ days

= ${3 × 6 × 18}/{6 × 18 - 3(6 + 18)}$

= $324/{108 - 3 24}$

= $324/{108 - 3 × 24} = 324/36$ = 9

Question : 35 [SSC CHSL 2013]

Three men can complete a piece of work in 6 days. Two days after they started the work, 3 more men joined them. How many days will they take to complete the remaining work ?

a) 3 days

b) 1 days

c) 2 days

d) 4 days

Answer: (c)

Work done in two days = $1/6 × 2$

= $1/3$, remaining work = $2/3$

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

${3 × 2}/{1/3} = {6 × D_2}/{2/3}$

$D_2 = {3 × 2 × 2}/6$ = 2 days

Using Rule 10
If food is available for 'a' days for 'A' men at a certain place and after 'b' days. 'B' men join, then the remaining food will serve total men for
Required time = ${A(a - b)}/{(A + B)}$ days
If food is available for 'a' days for 'A' men at a certain place, and after 'b' days 'B' men leave then the remaining food will serve remaining men for
Required time = ${A(a - b)}/{(A - B)}$ days

Here, A = 3, a = 6, b = 2, B = 3

Required time = ${A(a - b)}/{(A + B)}$

= ${3(6 - 2)}/{3 + 3}$

= ${3 × 4}/6$ = 2 days

IMPORTANT quantitative aptitude EXERCISES

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