Model 4  Working with Man, Woman, Child Practice Questions Answers Test with Solutions & More Shortcuts

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: (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: (d)

16 men = 20 women ⇒ 4 men = 5 women.

Now, according to question,

16 men complete the work in 25 days.

1 man one day’s work = $1/{25 × 16}$

4 men one day’s work = $4/{25 × 16} = 1/100$

Similarly,

1 woman one day’s work =$1/{25 × 20}$

5 women one day’s work = $5/{25 × 20} = 1/100$

28 men = $28/4$ × 5 = 35 women

[28 men + 15 women]

50 women one day’s work men= $50/{25 × 20} = 1/10$

Therefore, 28 men and 15 women can complete the whole work in 10 days.

Using Rule 12,

A = 16, B = 20, a = 25, $A_1$ = 28, $B_1$ = 15

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

= ${25(16 × 20)}/{28 × 20 + 15 × 16}$

= ${25 × 320}/{560 + 240}$

= ${25 × 320}/800$ = 10 days

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