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

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

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.

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

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

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

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

Answer: (a)

1 man = 2 women ≡ 3 boys 1 man + 1 woman + 1 boy

= $(3 + 3/2 + 1)$ boys = $11/2$ boys

$M_1D_1 = M_2D_2$

$3 × 88 = 11/2 × D_2$

$D_2 = {2 × 3 × 88}/11$ = 48 days

Here, A = 1, B= 2, C = 3, a = 88

$A_1 = 1, B_1 = 1, C_1$ = 1

Time taken = $a/{A_1/A + B_1/B + C_1/C}$

= $88/{1/1 + 1/2 + 1/3}$

= ${88 × 6}/{6 + 3 + 2}$ = 48 days

Answer: (c)

2 × 10 men + 3 × 10 women = 3 × 8 men + 2 × 8 women

20 men + 30 women = 24 men + 16 women

4 men = 14 women or 2 men = 7 women

2 men + 3 women = 10 women

2 men + 1 woman = 8 women

$M_1D_1 = M_2D_2$

10 × 10 = 8 × $D_2$

$D_2 = 25/2 = 12{1}/2$ 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.

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

= ${80(4 - 9)}/{10(2 - 6) - 8(3 - 4)}$

= ${-400}/{-40 + 8} = {-400}/{-32} = 25/2 = 12{1}/2$ days