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The following question based on time & work topic of quantitative aptitude
(a) 20 days
(b) 24 days
(c) 36 days
(d) 72 days
The correct answers to the above question in:
Answer: (d)
Suppose a man can complete the work in x days and that boys in y days.
According to question
$24/x + 24/y$ = 1 … (i) × 13
$26/x + 20/y = 1$ … (ii) × 12
$312/x + 312/y = 13$
$312/x + 240/y = 12$
Solving these two equations we get,
$72/y$ = 1 ⇒ y = 72 days
Boys alone can complete the work in 72 days
Using Rule 9If A and B together can finish a certain work in 'a' days. They worked together for 'b' days and then 'B'(or A) left the work. A (or B) finished the rest work in 'd' days, thenTotal time taken by A (or B) alone to complete the work= $\text"ad"/ \text"a - b"$ or $\text"bd"/ \text"a - b"$ days
Here, a = 24, b = 24 - 6 = 18, d = 26 days
Total time taken by B alone to complete the work
= $\text"bd"/ \text"a - b"- 6$
(Since, man has work d or 6 days)
= ${18 × 26}/{24 - 18} - 6$ = 78 - 6 = 72 days
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Read more man leaves and joins Based Quantitative Aptitude Questions and Answers
Question : 1
A can complete a piece of work in 18 days, B in 20 days and C in 30 days. B and C together start the work and are forced to leave after 2 days. The time taken by A alone to complete the remaining work is
a) 12 days
b) 15 days
c) 16 days
d) 10 days
Answer »Answer: (b)
(B + C)’s 2 days’ work = $2(1/20 + 1/30)$
= $2({3 + 2}/60) = 1/6$ part
Remaining work = $1 - 1/6 = 5/6$ part
Time taken by A to complete this part of work
= $5/6$ × 18 = 15 days
Question : 2
A certain number of persons can complete a piece of work in 55 days. If there were 6 persons more, the work could be finished in 11 days less. How many persons were originally there ?
a) 24
b) 30
c) 22
d) 17
Answer »Answer: (a)
Originally, let there be x men
Now, more men, less days
(x + 6) : x : : 55 : 44
So, ${x + 6}/x = 55/44 = 5/4$
or 5x = 4x + 24 ⇒ x = 24
Using Rule 23Some people finish a certain work in 'D' days. If there were 'a' less people, then the work would be completed in 'd'days more, what was the number of people initially?Required number = $\text"a(D - d)"/\text"d"$ people
Here, D = 55, a = 6, d = 11
No of people initially = $\text"a(D - d)"/\text"d"$
= ${6(55 - 11)}/11$ = 24
Question : 3
A and B can do a job in 6 and 12 days respectively. They began the work together but A leaves after 3 days. Then the total number of days needed for the completion of the work is :
a) 5 days
b) 6 days
c) 9 days
d) 4 days
Answer »Answer: (b)
A’s one day’s work = $1/6$
B’s one day’s work = $1/12$
(A + B)’s one day’s work = $1/6 + 1/12 = {2 + 1}/12 = 1/4$
(A + B)’s three day’s work = $3/4$
Remaining work = $1 - 3/4 = 1/4$
Total required number of days
= $1/4 × 12/1 + 3$ = 3 + 3 = 6 days
Using Rule 20,
Here, m = 6, n= 12, and p = 3
Time taken by B = $\text"mn - p(m+n)"/ \text"m"$
= ${6 × 12 - (6 + 12) × 3}/6 = {72 - 54}/6$ = 3 days
Total number of days taken to finish the works = 6 days
Question : 4
A and B can do a piece of work in 28 and 35 days respectively. They began to work together but A leaves after sometime and B completed remaining work in 17 days. After how many days did A leave ?
a) 9 days
b) 8 days
c) 7$5/9$ days
d) 14$2/5$ days
Answer »Answer: (b)
Let A worked for x days.
According to question
$x/28 + (x + 17)/35 = 1$
${5x + 4(x + 17)}/140 = 1$
5x + 4x + 68 = 140
9x = 140 - 68 = 72
x = 8
∴ A worked for 8 days
Question : 5
40 men can complete a work in 40 days. They started the work together. But at the end of each 10th day, 5 men left the job. The work would have been completed in
a) 53$1/3$ days
b) 52 days
c) 50 days
d) 56 $2/3$ days
Answer »Answer: (d)
For the first 10 days 40 men worked.
Now, 40 men can complete the work in 40 days
1 man will complete the same work in 1600 days
1 man’s 1 day’s work = $1/1600$
Part of work done in first 10 days = $1/4$
For the next 10 days 35 men worked.
Part of the work done
= ${1 × 35 × 10}/1600 = 7/32$
For the next 10 days, 30 men worked
Part of the work done
= ${30 × 10}/1600 = 3/16$
For the next 10 days, 25 men worked. Part of the workdone
= ${25 × 10}/1600 = 5/32$
Similarly, part of the work done by 20 men in next 10 days
= ${20 × 10}/1600 = 1/8$
Work done in 50 days
= $1/4 + 7/32 + 3/16 + 5/32 + 1/8$
= ${8 + 7 + 6 + 5 + 4}/32 = 30/32 = 15/16$
Remaining work = $1 - 15/16 = 1/16$
Now 15 men remain to work
15 men’s 1 day’s work= $15/1600$
Time taken to complete $1/16$ part of work
= $1600/15 × 1/16 = 20/3 = 6{2}/3$ days
Total time = $50 + 6{2}/3 = 56{2}/3$ days
Question : 6
8 men can do a work in 12 days. After 6 days of work, 4 more men were engaged to finish the work. In how many days would the remaining work be completed?
a) 3
b) 4
c) 5
d) 2
Answer »Answer: (b)
Using Rule 1If $M_1$ men can finish $W_1$ work in $D_1$ days and $M_2$ men can finish $W_2$ work in $D_2$ days then, Relation is${M_1D_1}/{W_1} = {M_2D_2}/{W_2}$ andIf $M_1$ men finish $W_1$ work in $D_1$ days, working $T_1$ time each day and $M_2$ men finish $W_2$ work in $D_2$ days, working $T_2$ time each day, then${M_1D_1T_1}/{W_1} = {M_2D_2T_2}/{W_2}$
Work done by 8 men in 6 days
= $6/12 = 1/2$
Remaining work
= $1 - 1/2 = 1/2$
4 more men are engaged.
Total number of men = 8 + 4 = 12
By work and time formula
$W_1/{M_1D_1} = W_2/{M_2D_2}$, we have
$1/{8 × 12} = {1/2}/{12 × D_2}$
$D_2 = 1/2 × {8 × 12}/12$ = 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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