Model 4  Working with Man, Woman, Child Section-Wise Topic Notes With Detailed Explanation And Example Questions

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

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

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

$A_1$ = 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

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

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

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

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

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