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

Answer: (b)

8 men ≡ 12 boys ⇒ 4 men ≡ 6 boys

20 men ≡ 30 boys

20 men + 6 boys = 36 boys

$M_1D_1 = M_2D_2$

12 × 16 = 36 × $D_2$

$D_2 = {12 × 16}/36 = 16/3 = 5{1}/3$ days

Using Rule 12,

Here, A = 8, B = 12, a = 16

$A_1 = 20, B_1$ = 6,

Required number of days = $a/{A_1/A + B_1/B} = 16/{20/8 + 6/12}$

= $16/{5/2 + 1/2} = {16 × 2}/6 = 5{1}/3$ days

Answer: (a)

According to the question,

20 men + 30 boys = 24 men + 16 boys

4 men = 14 boys ⇒ 2 men = 7 boys

2 men + 1 boy = 8 boys

2 men + 3 boys = 10 boys

By $M_1D_1 = M_2D_2$

10 × 10 = 8 × $D_2$

$D_2 = {10 × 10}/8 = 25/2 =12{1}/2$ days

Using Rule 11,

Here, $A_1 = 2, B_1 = 3, D_1 = 10$

$A_2 = 3, B_2 = 2, D_2$ = 8

$A_3 = 2, B_3$ = 1

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

= ${10 × 8(2 × 2 - 3 × 3)}/{10(2 × 1 - 2 × 3) - 8(3 × 1 - 2 × 2)}$

= ${80 × -5}/{-40 + 8} = 12{1}/2$ days

Answer: (c)

20 women complete 1 work in 16 days.

16 men complete same work in 15 days

16 × 15 men ≡ 20 × 16 women

3 men ≡ 4 women

∴ Required ratio = 4 : 3

Answer: (c)

3 men ≡ 7 women

7 men ≡ ${7 × 7}/3 = 49/3$ women

7 men + 5 women = $(49/3 + 5)$ women

= $({49 + 15}/3)$ women = $64/3$ women

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

${7 × 32}/1 = {64 × D_2}/{3 × 2}$

$D_2 = {7 × 32 × 3 × 2}/64$ = 21 days

Using Rule 12,

Here, A = 3, B = 7, a = 32

$A_1 = 7, B_1$ = 5

Required time = $a/{A_1/A + B_1/B}$

= $32/{7/3 + 5/7} = 32/64 × 21 = 21/2$

They do the twice work in

$21/2$ × 2 = 21 days

Answer: (c)

Using Rule 1,

4 men ≡ 6 women

1 men ≡ $6/4 = 3/2$ women

10 men + 3 women

= $10 × 3/2 + 3 = 18$ women

${M_1D_1T_1}/W_1 = {M_2D_2T_2}/W_2$

${6 × 12 × 7}/1 = {18 × D_2 × 8}/W_2$

$D_2 = {6 × 12 × 7 × 2}/{18 × 8}$ = 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

$A_1$ = 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