Model 3 Man leaves & joins 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

Questions : 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

The correct answers to the above question in:

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

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Read more man leaves and joins Based Quantitative Aptitude Questions and Answers

Question : 1

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: (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 : 2

A man and a boy can complete a work together in 24 days. If for the last six days man alone does the work then it is completed in 26 days. How long the boy will take to complete the work alone ?

a) 20 days

b) 24 days

c) 36 days

d) 72 days

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 9
If 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, then
Total 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

Question : 3

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: (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 : 4

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: (b)

Using Rule 1
If $M_1$ 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
${M_1D_1}/{W_1} = {M_2D_2}/{W_2}$ and
If $M_1$ 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.

Question : 5

X alone can complete a piece of work in 40 days. He worked for 8 days and left. Y alone completed the remaining work in 16 days. How long would X and Y together take to complete the work ?

a) 14 days

b) 15 days

c) 16$2/3$ days

d) 13$1/3$ days

Answer: (d)

Part of the work done by X in 8 days.

=$8/40 = 1/5$

[Since, work done in 1 day = $1/40$]

Remaining work = $1 - 1/5 = 4/5$

This part of work is done by Y in 16 days.

Time taken by Y in doing 1 work

= ${16 × 5}/4$ = 20 days

Work done by X and Y in 1 day

= $1/40 + 1/20 = {1+2}/40 = 3/40$

Hence, both together will complete the work in $40/3$

i.e.$13{1}/3$ days.

Question : 6

A and B can do a work in 18 and 24 days respectively. They worked together for 8 days and then A left. The remaining work was finished by B in :

a) 5$1/3$ days

b) 8 days

c) 10 days

d) 5 days

Answer: (a)

Since, A can finish the work in 18 days.

A’s one day’s' work = $1/18$

Similarly, B’s one day’s work = $1/24$

(A + B)’s 8 days’ work = $(1/18 + 1/24) × 8 = 7/72 × 8 = 7/9$

Remaining work = $1 - 7/9 = 2/9$

Time taken to finish the remaining work by B is $2/9 × 24 = 16/3 = 5{1}/3$ days

Using Rule 20
A can do a certain work in 'm' days and B can do the same work in 'n' days. They worked together for 'P' days and after this A left the work, then in how many days did B alone do the rest of work ?
Required time = ${mn - P(m + n)}/m$ days
when after 'P' days B left the work, then in how many days did A alone do the rest of work?
Required time = ${mn - P(m + n)}/n$ days

Here, m = 18, n= 24 and p = 8

Required Time = ${18 × 24 - 8(18 + 24)}/18$

= ${432 - 336}/18 = 96/18$

= $16/3 = 5{1}/3$ days

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