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The following question based on time & work topic of quantitative aptitude
(a) 6 days
(b) 4 days
(c) 5 days
(d) 8 days
The correct answers to the above question in:
Answer: (c)
5 × 6 men = 10 × 5 women
3 men = 5 women
5 women + 3 men = 6 men
5 men complete the work in 6 days
6 men will complete the work in ${5 × 6}/6$ = 5 days
Using Rule 14
If 'A' men can do a certain work in 'a' days and 'B' women can do the same work in 'b' days, then the total time is taken when $A_1$ men and $B_1$ women work together isTime taken = $1/{A_1/{A . a} + B_1/{B . b}}$If A men do a certain work in 'a' days, B women do the same work in 'b' days and C boys do the same work in 'c' days then the total time taken when $A_1$ men, $B_1$ women and $C_1$ boys can work together is,
Total time taken = $1/{(A_1/{A . a} + B_1/{B . b} + C_1/{C . c})}$
Here, A = 5, a = 6, B = 10, b = 5, $A_1$ = 3, $B_1$ = 5
Time taken = $1/{A_1/{A × a} + B_1/{B × b}}$
= $1/{3/{5 × 6} + 5/{10 × 5}}$
= $1/{1/10 + 1/10}$ = 5 days
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Read more working with man woman child Based Quantitative Aptitude Questions and Answers
Question : 1
If 1 man or 2 women or 3 boys can do a piece of work in 44 days, then the same piece of work will be done by 1 man, 1 woman and 1 boy in
a) 26 days
b) 21 days
c) 24 days
d) 33 days
Answer »Answer: (c)
1 man ≡ 2 women ≡ 3 boys
1 man + 1 woman + 1 boy
≡ 3 boys + $3/2$ boys + 1 boy
≡ $(3 + 3/2 + 1)$ boys ≡ $11/2$ boys
By $M_1D_1 = M_2D_2$,
3 × 44 = $11/2 × D_2$
$D_2 = {2 × 3 × 44}/11$ = 24 days
Using Rule 13,
Here, A = 1, B= 2, C = 3, a = 44
$A_1 = 1, B_1 = 1, C_1$ = 1
Required time = $a/{A_1/A + B_1/B + C_1/C}$ days
= $44/{1/1 + 1/2 + 1/3} = {44 × 6}/11$ = 24 days
Question : 2
A man, a woman and a boy can complete a work in 20 days, 30 days and 60 days respectively. How many boys must assist 2 men and 8 women so as to complete the work in 2 days ?
a) 4
b) 8
c) 12
d) 6
Answer »Answer: (b)
Part of work done by 2 men and 2 women in 2 days.
= $2(2/20 + 8/30)$
= $2(1/10 + 8/30) = 2({3 + 8}/30)$
= $22/30 = 11/15$
= Remaining work =$1 - 11/15 = 4/15$
Work done by 1 boy in 2 days
= $2/60 = 1/30$
Number of boys required to assist = $4/15 × 30 = 8$
Using Rule 14,
Here, A = 1, B = 1, C = 1
a = 20, b = 30, c = 60
$A_1 = 2, B_1$ = 8
Required time = $1/{A_1/{A × a} + B_1/{B × b} + C_1/{C × c}$
2 = ${1/{2/{1 × 20} + 8/{1 × 30} + x/{1 × 60}$
2 = $10/{2/2 + 8/3 + x/6}$
2 = $10/{{6 + 16 + x}/6}$
22 + x = 30 ⇒ x = 8
∴ Number of boys = 8
Question : 3
18 men or 36 boys working 6 hours a day can plough a field in 24 days. In how many days will 24 men and 24 boys working 9 hours a day plough the same field ?
a) 6
b) 9
c) 10
d) 8
Answer »Answer: (d)
18 men ≡ 36 boys ⇒ 1 man ≡ 2 boys
24 men + 24 boys ≡ (24 + 12) men ≡ 36 men
$M_1D_1T_1 = M_2D_2T_2$
$18 × 24 × 6 = 36 × D_2$ × 9
$D_2 = {18 × 24 × 6}/{36 × 9}$ = 8 days
Question : 4
If 40 men or 60 women or 80 children can do a piece of work in 6 months, then 10 men, 10 women and 10 children together do half of the work in
a) 5$7/13$ months
b) 5$6/13$ months
c) 6 months
d) 11$1/13$ months
Answer »Answer: (a)
40 men ≡ 60 women ≡ 80 children
10 men ≡ $80/40 × 10$ = 20 children
10 women ≡ $80/60 × 10 = 40/3$ children
10 men + 10 women + 10 children
= $(20 + 40/3 +10)$ children
= $({60 + 40 + 30}/3)$ children = $130/3$ children
${M_1D_1}/W_1 = {M_2D_2}/W_2$
$D_2 = {80 × 6 × 13}/130 = 144/13$ months
Half of the work can do
= $144/13 × 1/2 = 72/13 = 5{7}/13$ months
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 = 40, B= 60, C = 80, a = 6
$A_1 = 10, B_1 = 10, C_1$ = 10
Time taken = $a/{A_1/A + B_1/B + C_1/C}$
= $6/{10/40 + 10/60 + 10/80}$
= $6/{1/4 + 1/6 + 1/8}$
= $6/{{6 + 4 + 3}/24} = 144/13$
Half of the work they do in = $1/2 × 144/13$ months
= $72/13 = 5{7}/13$ months
Question : 5
A man, a woman and a boy together finish a piece of work in 6 days. If a man and a woman can do the work in 10 and 24 days respectively. The days taken by a boy to finish the work is
a) 40
b) 30
c) 35
d) 45
Answer »Answer: (a)
Time taken by boy = x days
$1/10 + 1/24 + 1/x = 1/6$
$1/x = 1/6 - 1/10 - 1/24$
= ${20 - 12 - 5}/120 = 3/120 = 1/40$
x = 40 days
Using Rule 18,
Here , x = 6, y = 10, z = 24
Number of days = ${xyz}/{yz –x(y + z)}$ days
= ${6 × 10 × 24}/{10 × 24 - 6(10 + 24)}$
= $1440/{240 - 204} = 1440/36$ = 40 days
Question : 6
A man, a woman and a boy can complete a job in 3, 4 and 12 days respectively. How many boys must assist 1 man and 1 woman to complete the job in $1/4$ of a day?
a) 19
b) 1
c) 4
d) 41
Answer »Answer: (d)
1 man’s 1 day’s work = $1/3$
1 woman’s 1 day’s work = $1/4$
1 boy’s 1 day’s work = $1/12$
(1 man + 1 woman)’s $1/4$ day’s work = $1/4(1/3 + 1/4) = 7/48$
Remaining work = $1 - 7/48 = 41/48$
Now, 1 boy’s $1/4$ day’s work = $1/4 × 1/12 = 1/48$
$41/48$ work will be done by $41/48$ × 48 = 41 boys.
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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