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The following question based on time & work topic of quantitative aptitude
(a) 2 days
(b) 4 days
(c) 16 days
(d) 8 days
The correct answers to the above question in:
Answer: (b)
Work done by A, B and C in 1 day
= $1/10 + 1/12 + 1/15 = {6 + 5 + 4}/60$
= $15/60 = 1/4$
Required time = 4 days
Using Rule 3,
Time Taken = ${xyz}/{xy + yz + zx}$
= ${10 × 12 × 15}/{10 × 12 + 12 × 15 + 15 × 10}$
= $1800/{120 + 180 + 150}$
= $1800/450$ = 4 days
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Read more basics on time and work Based Quantitative Aptitude Questions and Answers
Question : 1
A and B can do a work in 12 days, B and C in 15 days and C and A in 20 days. If A, B and C work together, they will complete the work in :
a) 15$2/3$ days
b) 10 days
c) 5 days
d) 7$5/6$ days
Answer »Answer: (b)
According to question,
A and B can do a work in 12 days
(A + B)'s one day's work = $1/12$
Similarly, (B + C)'s one day's work = $1/15$
and (C + A)'s one day's work = $1/20$
On adding all three,
2 (A + B + C)'s one days's work
= $1/12 + 1/15 + 1/20 = {10 + 8 + 6}/120 = 1/5$
(A + B + C)'s one days's work = $1/10$
A, B and C together can finish the whole work in 10 days.
Using Rule 5If A and B can do a work in 'x' days, B and C can do the same work in 'y' days, C and A can do the same work in 'z' days. Then total time taken, when A, B and C work together =$2/{(1/x + 1/y + 1/z)}$ OR ${2xyz}/{xy + yz + zx}$ days
Time taken = ${2 × 12 × 15 × 20}/{12 × 15 + 15 × 20 + 20 × 12}$
= ${24 × 300}/{180 + 300 + 240} = 7200/720$ = 10 days.
Question : 2
A work can be completed by P and Q in 12 days, Q and R in 15 days, R and P in 20 days. In how many days P alone can finish the work?
a) 60 days
b) 30 days
c) 10 days
d) 20 days
Answer »Answer: (b)
(P + Q)’s 1 day’s work = $1/12$ ...(i)
(Q + R)’s 1 day’s work = $1/15$ ...(ii)
(R + P)’s 1 day’s work = $1/20$ ...(iii)
Adding all three equations,
2 (P + Q + R)’s 1 day’s work = $1/12 + 1/15 + 1/20$
${5 + 4 + 3}/60 = 12/60 = 1/5$
(P + Q + R)’s 1 day’s work = $1/10$ ...(iv)
P’s 1 day’s work = $1/10 - 1/15 = {3 - 2}/30 = 1/30$
P alone will complete the work in 30 days.
Question : 3
A particular job can be completed by a team of 10 men in 12 days. The same job can be completed by a team of 10 women in 6 days. How many days are needed to complete the job if the two teams work together?
a) 18 days
b) 9 days
c) 4 days
d) 6 days
Answer »Answer: (c)
According to question,
10 men’s one day’s work = $1/12$
1 man one day’s work =$1/{12 × 10} = 1/120$
Similarly, 1 woman one day’s work = $1/{6 × 10} = 1/60$
(1 man + 1 woman)’s one day’s work = $1/120 + 1/60$
= ${1 + 2}/120 = 3/120 = 1/40$
(10 men + 10 women)’s one day’s work = $10/40 = 1/4$
Therefore, both the teams can finish the whole work in 4 days.
Question : 4
A and B can do a piece of work in 8 days, B and C can do it in 24 days, while C and A can do it in 8$4/7$ days. In how many days can C do it alone?
a) 10 days
b) 30 days
c) 60 days
d) 40 days
Answer »Answer: (c)
(A + B)’s 1 day’s work = $1/8$
(B + C)’s 1 day’s work = $1/24$
(C + A)’s 1 day’s work = $7/60$
On adding all three,
2 (A + B + C)’s 1 day’s work = $1/8 + 1/24 + 7/60$
= ${15 + 5 + 14}/120 = 34/120$
(A + B + C)’s 1 day’s work = $17/120$
C’s 1 day’s work
= $17/120 - 1/8 = {17 - 15}/120 = 1/60$
C alone will complete the work in 60 days.
Using Rule 19,
C alone can do in= ${2xyz}/{xy - yz + zx}$
= ${2 × 8 × 24 × 60/7}/{8 × 24 - 24 × 60/7 + 60/7 × 8$
= ${23040/7}/{192 - 1440/7 + 480/7}$
= ${23040/7}/{{1344 - 1440 + 480}/7}$
= $23040/7 × 7/384$ = 60 days
Question : 5
A and B together can do a piece of work in 5 days and A alone can do it in 8 days. B alone can do the same piece of work in
a) 16$4/5$ days
b) 13$1/3$ days
c) 11$1/3$ days
d) 12$3/5$ days
Answer »Answer: (b)
(A + B)’s 1 day’s work = $1/5$
A’s 1 day’s work = $1/8$
B’s 1 day’s work = $1/5 - 1/8$
= ${8 - 5}/40 = 3/40$
B alone will complete the work in $40/3 = 13{1}/3$ days.
Using Rule 4,
Time taken by B = ${5 × 8}/{8 - 5}$
= $40/3 = 13{1}/3$ days
Question : 6
If A and B together can finish a piece of work in 20 days, B and C in 10 days and C and A in 12 days, then A, B, C jointly can finish the same work in
a) $7/60$ days
b) 8$4/7$ days
c) 4$2/7$ days
d) 30 days
Answer »Answer: (b)
(A + B)’s 1 day’s work = $1/20$
(B + C)’s 1 day’s work = $1/10$
(C + A)’s 1 day’s work = $1/12$
On adding all three,
2 (A + B + C)’s 1 day’s work = $1/20 + 1/10 + 1/12$
= ${3 + 6 + 5}/60 = 14/60 = 7/30$
(A + B + C)’s 1 day’s work = $7/60$
Hence, the work will be completed in $60/7 = 8{4}/7$ days.
Using Rule 5,
Time taken = ${2xyz}/{xy + yz + zx}$
= ${2 × 20 × 10 × 12}/{20 × 10 + 10 × 12 + 12 × 20}$
= $4800/{200 + 120 + 240}$
= $4800/560 = 60/7 = 8{4}/7$ 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’
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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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