Model 4 Working with Man, Woman, Child Practice Questions Answers Test with Solutions & More Shortcuts
time & work PRACTICE TEST [7 - EXERCISES]
Model 1 Basics on Time & Work
Model 2 Formula method ‘M1D1W1 = M2D2W2’
Model 3 Man leaves & joins
Model 4 Working with Man, Woman, Child
Model 5 Split & Fraction of work
Model 6 Efficiency of the worker
Model 7 Working with individual wages
Question : 6 [SSC CGL Prelim 1999]
5 men can do a piece of work in 6 days while 10 women can do it in 5 days. In how many days can 5 women and 3 men do it ?
a) 6 days
b) 4 days
c) 5 days
d) 8 days
Answer »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
Question : 7 [SSC CGL Tier-I 2015]
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 : 8 [SSC DEO 2009]
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 : 9 [SSC CGL Tier-I 2016]
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 : 10 [SSC CGL Prelim 1999]
If 6 men and 8 boys can do a piece of work in 10 days and 26 men and 48 boys can do the same in 2 days, then the time taken by 15 men and 20 boys to do the same type of work will be :
a) 6 days
b) 5 days
c) 4 days
d) 7 days
Answer »Answer: (c)
According to question,
(6M + 8B) × 10 = (26M + 48B) × 2
60M + 80B = 52M + 96B or, 1M = 2B
15M + 20B = (30 + 20)B
= 50 boys and 6M + 8B
= (12 + 8) boys = 20 boys
20 boys can finish the work in 10 days
50 boys can finish the work in 20 10 50 ´ days = 4 days
Using Rule 11If $A_1$ men and $B_1$ boys can do a certain work in $D_1$ days, Again, $A_2$ men and $B_2$ boys can do the same work in $D_2$ days, then, $A_3$ men and $B_3$ boys can do the same work inRequired 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
$A_1$ = 6, $B_1$ = 8, $D_1$ = 10
$A_2$ = 26, $B_2$ = 48, $D_2$ = 2
$A_3$ = 15, $B_3$ = 20
Required time = ${D_1D_2(A_1B_2 - B_1A_2)}/{D_1(A_1B_3 - A_3B_1) - D_2(A_2B_3 - A_3 B_2)}$ day
= ${10 × 2(6 × 48 - 8 × 26)}/{10(6 × 20 - 15 × 8) - 2(26 × 20 - 15 × 48)}$ days
= ${20(288 - 208)}/{10(120 - 120) - 2(520 - 720)}$
= ${20 × 80}/400$ = 4 days
IMPORTANT quantitative aptitude EXERCISES
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Top 199+ Basic Time and Work Aptitude MCQs with Answers »
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Top 200+ Time and Work Problem Solving By Formula Method »
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New 189+ Time & Work MCQs Practice Test for BANK Exams »
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Top 199+ Time & Work MCQs: Men Women Child for BANK Exam »
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Top 199+ Time and Work Questions Test on Fraction of Work »
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Top 199+ Time and Work Efficiency Based Aptitude MCQ Quiz »
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New 299+ Time and Work Aptitude Test On Individual Wages »
Model 4 Working with Man, Woman, Child Shortcuts »
Click to Read...Model 4 Working with Man, Woman, Child Online Quiz
Click to Start..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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