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

Answer: (d)

3 men = 4 women

1 man = $4/3$ women

7 men = ${7 × 4}/3 = 28/3$ women

7 men + 5 women = $28/3 + 5$

= ${28 + 15}/3 = 43/3$ Women

Now, $M_1D_1 = M_2D_2$

4 × 43 = $43/3 × D_2$ ,

where $D_2$ = number of days

$D_2 = {4 × 3 × 43}/43$ = 12 days.

Using Rule 12,

Here, A = 3, B = 4, a = 43, $A_1$ = 7 and $B_1$ = 5

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

= ${43(3 × 4)}/{7 × 4 + 5 × 3}$

= ${43 × 12}/43$ = 12 days

Answer: (d)

1 man’s 1 day’s work = $1/3$

1 woman’s 1 day’s work = $1/4$

1 boy’s 1 day’s work = $1/12$

(1 man + 1 woman)’s $1/4$ day’s work = $1/4(1/3 + 1/4) = 7/48$

Remaining work = $1 - 7/48 = 41/48$

Now, 1 boy’s $1/4$ day’s work = $1/4 × 1/12 = 1/48$

$41/48$ work will be done by $41/48$ × 48 = 41 boys.

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