Model 3 Man leaves & joins Section-Wise Topic Notes With Detailed Explanation And Example Questions

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.

Answer: (b)

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.

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

Answer: (c)

Work done by (A + C) in 2 days

= $2(1/10 + 1/20) = 2({2 + 1}/20) = 6/20 = 3/10$

Remaining work = $1 - 3/10 = 7/10$

(B + C)’s 1 day’s work

= $1/15 + 1/20 = {4 + 3}/60 = 7/60$

Time taken by (B + C) to finish $7/10$ part of the work

= $60/7 × 7/10$ = 6 days

Total time = 2 + 6 = 8 days

Answer: (c)

Let the work be finished in x days.

According to the question,

A worked for x days while B worked for (x - 3) days

$x/18 + {x - 3}/12 = 1$

${2x + 3x - 9}/36 = 1$

5x - 9 = 36

5x = 45 ⇒ x = $45/5$ = 9

Hence, the work was completed in 9 days.

Here, x = 18, y = 12, m = 3

Total time taken = $({y + m}/{x + y})x$

= $({12 + 3}/{18 + 12})$ × 18 = 9 days

Answer: (d)

A’s 1 day’s work = $1/20$

A’s 4 days’ work = $4/20 = 1/5$

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

This part is completed by A and B together.

Now, (A + B)’s 1 day’s work = $1/20 + 1/12$

= ${3 + 5}/60 = 8/60 = 2/15$

Now, $2/15$ work is done by (A + B) in 1 day.

$4/5$ work is done in

= $15/2 × 4/5$ = 6 days.

Hence, the work lasted for 4 + 6 = 10 days.