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

MOST IMPORTANT quantitative aptitude - 7 EXERCISES

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The following question based on time & work topic of quantitative aptitude

Questions : 6 men or 12 women can do a piece of work in 20 days. In how many days can 8 men and 16 women do twice as big as this work ?

(a) 15 days

(b) 2 days

(c) 5 days

(d) 10 days

The correct answers to the above question in:

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

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Read more working with man woman child Based Quantitative Aptitude Questions and Answers

Question : 1

If 3 men or 6 women can do a piece of work in 16 days, in how many days can 12 men and 8 women do the same piece of work?

a) 3 days

b) 4 days

c) 5 days

d) 2 days

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

Using Rule 12
If A men or B boys can do a certain work in 'a' days, then $A_1$ men and $B_1$ boys can do the same work in
Time taken = $a/{A_1/A + B_1/B} = {a(A . B)}/{A_1B + B_1A}$ 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

Question : 2

12 men and 16 boys can do a piece of work in 5 days; 13 men and 24 boys can do it in 4 days, then the ratio of the daily work done by a man to that of a boy is

a) 1 : 3

b) 2 : 1

c) 3 : 1

d) 5 : 4

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

Question : 3

A man, a woman and a boy can together complete a piece of work in 3 days. If a man alone can do it in 6 days and a boy alone in 18 days, how long will a woman alone take to complete the work?

a) 24 days

b) 9 days

c) 21 days

d) 27 days

Answer: (b)

Work done by 1 woman in 1 day

= $1/3 - 1/6 - 1/18 = {6 - 3 - 1}/18 = 1/9$

Woman will do the work in 9 days.

Using Rule 18
A, B and C can do a certain work together within 'x' days. While, any two of them can do the same work separately in 'y' and 'z' days, then in how many days can 3rd do the same work?
Required time = ${xyz}/{yz - x(y + z)}$days

Here, x = 3, y = 6 and z = 18

Required time = ${xyz}/{yz - x(y + z)}$ days

= ${3 × 6 × 18}/{6 × 18 - 3(6 + 18)}$

= $324/{108 - 3 24}$

= $324/{108 - 3 × 24} = 324/36$ = 9

Question : 4

If 1 man or 2 women or 3 boys can complete a piece of work in 88 days, then 1 man, 1 woman and 1 boy together will complete it in

a) 48 days

b) 36 days

c) 42 days

d) 54 days

Answer: (a)

1 man = 2 women ≡ 3 boys 1 man + 1 woman + 1 boy

= $(3 + 3/2 + 1)$ boys = $11/2$ boys

$M_1D_1 = M_2D_2$

$3 × 88 = 11/2 × D_2$

$D_2 = {2 × 3 × 88}/11$ = 48 days

Using Rule 13
If A men or B boys or C women can do a certain work in 'a' days, then $A_1$ men, $B_1$ boys and $C_1$ women can do the same work in
Time taken = $a/{A_1/A + B_1/B + C_1/C}$

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

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

Time taken = $a/{A_1/A + B_1/B + C_1/C}$

= $88/{1/1 + 1/2 + 1/3}$

= ${88 × 6}/{6 + 3 + 2}$ = 48 days

Question : 5

2 men and 3 women can do a piece of work in 10 days while 3 men and 2 women can do the same work in 8 days. Then, 2 men and 1 woman can do the same work in

a) 13 days

b) 12 days

c) 12$1/2$ days.

d) 13$1/2$ days

Answer: (c)

2 × 10 men + 3 × 10 women = 3 × 8 men + 2 × 8 women

20 men + 30 women = 24 men + 16 women

4 men = 14 women or 2 men = 7 women

2 men + 3 women = 10 women

2 men + 1 woman = 8 women

$M_1D_1 = M_2D_2$

10 × 10 = 8 × $D_2$

$D_2 = 25/2 = 12{1}/2$ 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.

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

= ${80(4 - 9)}/{10(2 - 6) - 8(3 - 4)}$

= ${-400}/{-40 + 8} = {-400}/{-32} = 25/2 = 12{1}/2$ days

Question : 6

8 children and 12 men complete a certain piece of work in 9 days. Each child takes twice the time taken by a man to finish the work. In how many days will 12 men finish the same work ?

a) 12 days

b) 9 days

c) 13 days

d) 15 days

Answer: (a)

Using Rule 1,

2 children ≡ 1 man

8 children + 12 men ≡ 16 men

$M_1D_1 = M_2D_2$

16 × 9 = 12 × $D_2$

$D_2 = {16 × 9}/12$ = 12 days.

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