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

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