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

Answer: (d)

A’s 1 day’s work = $1/12$

B’s 1 day’s work = $1/18$

Part of work done by A and B in first two days

= $1/12 + 1/18 = {3 - 2}/36 = 5/36$

Part of work done by A and B in 14 days = $35/36$

[14 days to be taken randomly]

Remaining work = $1 - 35/36 = 1/36$

Now A will work for 15th day.

A will do the $1/36$ work in $1/36 × 12 = 1/3$ day.

Total Work will be done in $14{1}/3$ days.

Answer: (a)

(A+B)’s 1 day’s work = $1/30$

(A + B)’s 20 day’s work = $20/30 = 2/3$

Remaining work = $1 - 2/3 = 1/3$

Now, $1/3$ part of work is done by A in 20 days.

Whole work will be done by A alone in 20 × 3 = 60 days.

Answer: (b)

Let the work be completed in x days.

According to the question,

${x - 5}/10 + {x - 3}/12 + x/15$ = 1

${6x - 30 + 5x - 15 + 4x}/60$ = 1

15x - 45 = 60

15x = 105 ⇒ x = $105/15$ = 7

Hence, the work will be completed in 7 days.

Answer: (a)

Let the work be completed in x days.

According to the question,

A worked for (x –3) days, while B worked for x days.

${x - 3}/9 +x/18$ = 1

${2x - 6 + x}/18$ = 1

3x–6 = 18

3x = 18 + 6 = 24 ⇒ x = $24/3$ = 8 days

Using Rule 8,

Here, x = 9, y = 18, m = 3

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

=${(9 + 3) × 18}/{9 + 18}$

= ${12 × 18}/27$ = 8 days

Answer: (d)

Work done by B in 9 days = $9/12 = 3/4$ part

Remaining work

= $1 - 3/4 = 1/4$ which is done by A

Time taken by A = $1/4 × 20$ = 5 days

Answer: (d)

(A + B)’s day’s work = $1/30$

(B + C)’s 1 day’s work = $1/24$

(C + A)’s 1 day’s work = $1/20$

2 (A + B + C)’s 1 day’s work = $1/30 + 1/24 + 1/20$

= ${4 + 5 + 6}/120 = = 15/120 = 1/8$

(A + B + C)’s 1 day’s work = $1/16$

(A + B + C)’s 10 days’ work = $10/16 = 5/8$

Remaining work = $1- 5/8 = 3/8$

This part of work is done by A alone.

Now A’s 1 day’s work = $1/16 - 1/24$

= ${3 - 2}/48 = 1/48$

The required no. of days

=$3/8 × 48$ = 18 days