Model 3 Man leaves & joins Practice Questions Answers Test with Solutions & More Shortcuts

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: (b)

A’s one day’s work = $1/6$

B’s one day’s work = $1/12$

(A + B)’s one day’s work = $1/6 + 1/12 = {2 + 1}/12 = 1/4$

(A + B)’s three day’s work = $3/4$

Remaining work = $1 - 3/4 = 1/4$

Total required number of days

= $1/4 × 12/1 + 3$ = 3 + 3 = 6 days

Using Rule 20,

Here, m = 6, n= 12, and p = 3

Time taken by B = $\text"mn - p(m+n)"/ \text"m"$

= ${6 × 12 - (6 + 12) × 3}/6 = {72 - 54}/6$ = 3 days

Total number of days taken to finish the works = 6 days

Answer: (b)

Time taken by A

= ${8 × 12}/128 = {8 × 12}/4 = 24$ days

Work done of by B = $4/12 = 1/3$

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

Since, A can complete a work in 24 days

A can complete $2/3$ part of work in 24 × $2/3$ = 16 days

Answer: (c)

(A+B)’s 2 days’ work = $2(1/12 + 1/18) = 10/36$

Remaining work = $1 - 10/36 = 26/36$

Time taken by B to complete $26/36$ part of work

= $26/36$ × 18 = 13 days

Using Rule 20,

Here, m = 12, n = 18, p = 2

Time taken by B = $\text"mn - p(m+n)"/ \text"m"$

= ${12 × 18 - 2(12 + 18)}/12$

= ${216 - 60}/12$ = 13 days

Answer: (b)

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

= $1/9 + 1/12 = {4 - 3}/36 = 7/36$

Part of work done in first 10 days = $35/36$

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

Now it is the turn of A.

Time taken by A

= $1/36 × 9 = 1/4$ day

Total time = $10 + 1/4 = 10{1}/4$ days