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

Answer: (b)

12 (3 men + 4 boys) = 10 (4 men + 3 boys)

36 men + 48 boys = 40 men + 30 boys

4 men = 18 boys or 2 men = 9 boys

4 men + 3 boys

= 21 boys who do the work in 10 days

and, 2 men + 3 boys = 12 boys

$M_1D_1 = M_2D_2$

21 × 10 = 12 × $D_2$

$D_2 = {21 × 10}/12 = 35/2 = 17{1}/2$ days

Using Rule 11,

Here, $A_1 = 3, B_1 = 4, D_1 = 12$

$A_2 = 4, B_2 = 3, D_2$ = 10

$A_3 = 2, B_3$ = 3

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

= ${12 × 10(3 × 3 - 4 × 4)}/{12(3 × 3 - 2 × 4) - 10(4 × 3 - 2 × 3)}$

= ${-120 × 7}/{12 - 60} = {-840}/{-48} = 70/4$

= $35/2 = 17{1}/2$ days

Answer: (a)

5 men can do 1 work in 14 days.

3 men will do $3/5$ work in 14 days.

Remaining work = $1 - 3/5 = 2/5$

5 women do $2/5$ work in 14 days.

Time taken by 5 women in doing 1 work

= ${14 × 5}/2 = 35$ days

(5 men + 5 women)’s 1 day’s work

= $1/14 + 1/35 = {5 + 2}/70 = 7/70 = 1/10$

∴Required time = 10 days.

Answer: (a)

Work done by 1 woman in 1 day

= $1/8 - 1/10 = {5 - 4}/40 = 1/40$

One woman will complete the work in 40 days.

Here, x = 10 and y = 8

Woman can do work in = $({xy}/{x - y})$ days

= $({10 × 8}/{10 - 8})$ = 40 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: (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: (b)

12(3 men + 4 boys) ≡ 10 (4 men + 3 boys)

36 men + 48 boys = 40 men + 30 boys

4 men = 18 boys ⇒ 2 men = 9 boys

4 men + 3 boys

= 21 boys, who do the work in 10 days and

2 men + 3 boys = 12 boys

$M_1D_1 = M_2D_2$

21 × 10 = 12 × $D_2$

$D_2 = {21 × 10}/12 = 35/2 = 17{1}/2$ days

Using Rule 11,

Here, $A_1 = 3, B_1 = 4, D_1$ = 12

$A_2 = 4, B_2 = 3, D_2$ = 10

$A_3 = 2, B_3$ = 3

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

= ${12 × 10(3 × 3 - 4 × 4)}/{12(3 × 3 - 2 × 4) - 10(4 × 3 - 2 × 3)}$

= ${120 × - 7}/{12(9 - 8) - 10 × 6}$

= $ {-840}/{-48} = 17{1}/2$ days