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

Question : 11 [SSC CHSL 2013]

If 4 men or 6 women can do a piece of work in 12 days working 7 hours a day; how many days will it take to complete a work twice as large with 10 men and 3 women working together 8 hours a day?

a) 8 days

b) 6 days

c) 7 days

d) 10 days

Answer: (c)

Using Rule 1,

4 men ≡ 6 women

1 men ≡ $6/4 = 3/2$ women

10 men + 3 women

= $10 × 3/2 + 3 = 18$ women

${M_1D_1T_1}/W_1 = {M_2D_2T_2}/W_2$

${6 × 12 × 7}/1 = {18 × D_2 × 8}/W_2$

$D_2 = {6 × 12 × 7 × 2}/{18 × 8}$ = 7 days

Question : 12 [SSC CHSL 2014]

3 men or 7 women can do a piece of work in 32 days. The number of days required by 7 men and 5 women to do a piece of work twice as large is

a) 27

b) 19

c) 21

d) 36

Answer: (c)

3 men ≡ 7 women

7 men ≡ ${7 × 7}/3 = 49/3$ women

7 men + 5 women = $(49/3 + 5)$ women

= $({49 + 15}/3)$ women = $64/3$ women

${M_1D_1}/W_1 = {M_2D_2}/W_2$

${7 × 32}/1 = {64 × D_2}/{3 × 2}$

$D_2 = {7 × 32 × 3 × 2}/64$ = 21 days

Using Rule 12,

Here, A = 3, B = 7, a = 32

$A_1 = 7, B_1$ = 5

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

= $32/{7/3 + 5/7} = 32/64 × 21 = 21/2$

They do the twice work in

$21/2$ × 2 = 21 days

Question : 13 [SSC CGL Tier-I 2012]

3 men and 4 boys can complete a piece of work in 12 days. 4 men and 3 boys can do the same work in 10 days. Then 2 men and 3 boys can finish the work in

a) 8 days

b) 17$1/2$ days

c) 5$5/11$ days

d) 22 days

Answer: (b)

12(3 men + 4 boys) ≡ 10 (4 men + 3 boys)

36 men + 48 boys = 40 men + 30 boys

4 men = 18 boys ⇒ 2 men = 9 boys

4 men + 3 boys

= 21 boys, who do the work in 10 days and

2 men + 3 boys = 12 boys

$M_1D_1 = M_2D_2$

21 × 10 = 12 × $D_2$

$D_2 = {21 × 10}/12 = 35/2 = 17{1}/2$ days

Using Rule 11,

Here, $A_1 = 3, B_1 = 4, D_1$ = 12

$A_2 = 4, B_2 = 3, D_2$ = 10

$A_3 = 2, B_3$ = 3

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

= ${12 × 10(3 × 3 - 4 × 4)}/{12(3 × 3 - 2 × 4) - 10(4 × 3 - 2 × 3)}$

= ${120 × - 7}/{12(9 - 8) - 10 × 6}$

= $ {-840}/{-48} = 17{1}/2$ days

Question : 14 [SSC CGL Tier-I 2013]

If 8 men or 12 boys can do a piece of work in 16 days, the number of days required to complete the work by 20 men and 6 boys is

a) 8$1/3$ days

b) 5$1/3$ days

c) 6$1/3$ days

d) 7$1/3$ days

Answer: (b)

8 men ≡ 12 boys ⇒ 4 men ≡ 6 boys

20 men ≡ 30 boys

20 men + 6 boys = 36 boys

$M_1D_1 = M_2D_2$

12 × 16 = 36 × $D_2$

$D_2 = {12 × 16}/36 = 16/3 = 5{1}/3$ days

Using Rule 12,

Here, A = 8, B = 12, a = 16

$A_1 = 20, B_1$ = 6,

Required number of days = $a/{A_1/A + B_1/B} = 16/{20/8 + 6/12}$

= $16/{5/2 + 1/2} = {16 × 2}/6 = 5{1}/3$ days

Question : 15 [SSC CGL Tier-I 2013]

2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days. In how many days can 2 men and 1 boy do the work ?

a) 12$1/2$ days

b) 8 days

c) 7 days

d) 2 days

Answer: (a)

According to the question,

20 men + 30 boys = 24 men + 16 boys

4 men = 14 boys ⇒ 2 men = 7 boys

2 men + 1 boy = 8 boys

2 men + 3 boys = 10 boys

By $M_1D_1 = M_2D_2$

10 × 10 = 8 × $D_2$

$D_2 = {10 × 10}/8 = 25/2 =12{1}/2$ 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

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

= ${80 × -5}/{-40 + 8} = 12{1}/2$ days

IMPORTANT quantitative aptitude EXERCISES

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