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
(a) 12 days
(b) 9 days
(c) 13 days
(d) 15 days
The correct answers to the above question in:
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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Read more working with man woman child Based Quantitative Aptitude Questions and Answers
Question : 1
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 »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 : 2
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 »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 13If 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 inTime 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 : 3
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
Answer »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
Question : 4
If 10 men or 20 boys can make 260 mats in 20 days, then how many mats will be made by 8 men and 4 boys in 20 days?
a) 280
b) 260
c) 240
d) 520
Answer »Answer: (b)
10 men ≡ 20 boys ⇒ 1 man ≡ 2 boys
8 men + 4 boys = (16 + 4) boys = 20 boys
Hence, 8 men and 4 boys will make 260 mats in 20 days.
Question : 5
If 10 men or 20 women or 40 children can do a piece of work in 7 months, then 5 men, 5 women and 5 children together can do half of the work in :
a) 5 months
b) 6 months
c) 4 months
d) 8 months
Answer »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.
Question : 6
3 men or 5 women can do a work in 12 days. How long will 6 men and 5 women take to finish the work?
a) 4 days
b) 20 days
c) 10 days
d) 15 days
Answer »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
time & work Shortcuts and Techniques with Examples
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Model 1 Basics on Time & Work
Defination & Shortcuts … -
Model 2 Formula method ‘M1D1W1 = M2D2W2’
Defination & Shortcuts … -
Model 3 Man leaves & joins
Defination & Shortcuts … -
Model 4 Working with Man, Woman, Child
Defination & Shortcuts … -
Model 5 Split & Fraction of work
Defination & Shortcuts … -
Model 6 Efficiency of the worker
Defination & Shortcuts … -
Model 7 Working with individual wages
Defination & Shortcuts …
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