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

Answer: (b)

Part of work done by B in 10 days

= 10 × $1/15 = 2/3$

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

Time taken by A = $1/3 × 18$ = 6 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.

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)

Let A and B worked together for x days

According to the question,

Part of work done by A for (x + 10) days

+ part of work done by B for x days = 1

${x + 10}/20 + x/30$ = 1

${3x + 30 + 2x}/60 = 1$

5x + 30 = 60

5x = 30 ⇒ x = $30/5$ = 6 days

Here, m =20, n= 30, p = x and time taken by A alone = 10

10 = ${mn - p(m + n)}/n$

10 = ${30 × 20 - x(30 + 20)}/30$

300 = 600 - x 50

50x = 300 x = 6

B worked for 6 days

Answer: (c)

B’s 1 day's work

= $1/12 - 1/20 = {5 - 3}/60 = 1/30$

B's $1/2$ day’s work = $1/60$

(A + B)'s 1 day's work

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

[Since, B works for half day daily]

Hence, the work will be completed in 15 days.

Answer: (b)

Part of the work done by A and B in 4 days

= $4(1/12 + 1/15) = 4({5 + 4}/60)$

= $4 × 9/60 = 3/5$

Remaining work = $1 - 3/5 = 2/5$

Time taken by B to complete the remaining work

= $2/5 × 15$ = 6 days

Using Rule 20,

Here, m = 12, n = 15, p = 4

B alone do the works in= ${mn - p(m + n)}/m$

= ${12 × 15 - 4(12 + 15)}/12$

= ${180 - 108}/12 = 72/12$ = 6 days