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

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

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

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

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

Here, D = 55, a = 6, d = 11

No of people initially = $\text"a(D - d)"/\text"d"$

= ${6(55 - 11)}/11$ = 24

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: (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)

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.