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

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: (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: (a)

3m = 6w ⇒ 1m = 2w

12m + 8w = (12 × 2w) + 8w = 32w

6 women can do the work in 16 days.

32 women can do the work in ${16 × 6}/32 = 3$ days

Here, A = 3, B = 6, a = 16, $A_1$ = 12, $B_1$ = 8

Time taken = ${a(A × B)}/{A_1B + B_1A}$

= ${16(3 × 6)}/{12 × 6 + 8 × 3}$

= ${16 × 18}/96$ = 3 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