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

Answer: (a)

1 man ≡ 2 women ≡ 3 boys

1 man + 1 woman + 1 boy

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

= $({6 + 3 + 2}/2)$ boys= $11/2$ boys

$M_1D_1 = M_2D_2$

3 × 88 = $11/2 × D_2$

$D_2 = {3 × 2 × 88}/11$ = 48 days

Using Rule 13,

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

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

Required time = $a/{A_1/A + B_1/B + C_1/C}$

= $88/{1/1 + 1/2 + 1/3} = 88/{{6 + 3 + 2}/6}$ = 48 days

Answer: (a)

3 men’s work = 5 women’s work

1 man’s work = $5/3$ women’s work

6 men’s work = $5/3$ × 6

=10 women’s work

6 men + 5 women = 15 women

5 women can do work in 12 days.

Hence, 15 women can do it in ${5 × 12}/15$ = 4 days

Using Rule 12,

Here, A = 3, B = 5, a = 12, $A_1$ = 6 and $B_1$ = 5

Required time = ${a(A × B)}/{A_1B + B_1A}$

= ${12(3 × 5)}/{6 × 5 + 5 × 3}$

= ${12 × 15}/45$ = 4 days

Answer: (d)

10 men ≡ 20 women

1 man = 2 women = 5 children

1 woman = 2 children

5 men + 5 women + 5 children

= 20 + 10 + 5 = 35 children

$M_1D_1 = M_2D_2$

40 × 7 = 35 × $D_2$

$D_2 = {40 ×7}/35$ = 8 months

5 men, 5 women and 5 children can do half of the work in 8 months

Required time = 4 months.

Using Rule 13,

Here, A = 10, B= 20, C = 40, a = 7

$A_1 = 5, B_1 = 5, C_1$ = 5

Time taken to do same work = $a/{A_1/A + B_1/B + C_1/C}$

= $7/{5/10 + 5/20 + 5/40}$

= $7/{1/2 + 1/4 + 1/8}$

= $7/{{4 + 2 + 1}/8}$ = 8 months

Half of the work they do in 4 months.

Answer: (c)

6m + 8w ≡ 10 days

2 (3m + 4w) ≡ 10 days

3m + 4w ≡ 20 days

[Since the workforce has become half of the original force, so number of days must be double].

Using Rule 14,

Let us assume efficiency of 6 men = efficiency of 8 men.

A = 6, a = 20, B = 8, b = 20

$A_1 = 3, B_1$ = 4

Required time = $1/{A_1/{A × a} + B_1/{B × b}}$

= $1/{3/{6 × 20} + 4/{8 × 20}}$

= $1/{1/40 + 1/40} = 40/2$ = 20 days

Answer: (b)

Using Rule 11,

According to the question,

1 man ≡ 2 women ≡ 4 boys

1 man + 1 woman + 1 boy

= (4 + 2 +1) boys = 7 boys

$M_1D_1 = M_2D_2$

7 × 7 = 1 × $D_2$

∴ $D_2$ = 49 days

Answer: (a)

Let 1 man’s 1 day’s work = x and

1 woman’s 1 day’s work = y

Then, 4x + 6y = $1/8$ and

$3x + 7y = 1/10$

From both equations,

we get y = $1/400$

10 women’s 1 day’s work = $10/400 = 1/40$

10 women will finish the work in 40 days.

Using Rule 11,

$A_1 = 4, B_1 = 6, D_1$ = 8

$A_2 = 3, B_2 = 7, D_2$ = 10

$A_3 = 0, B_3$ = 10

Required time = ${D_1D_2(A_1B_2 - A_2B_1)}/{D_1(A_1B_3 - A_3B_1) - D_2(A_2B_3 - A_3 B_2)}$

= ${8 × 10(4 × 7 - 3 × 6)}/{8(4 × 10 - 0 × 6) - 10(3 × 10 - 0 × 7)}$

= ${80 × 10}/20$ = 40 days