Model 6  Efficiency of the worker Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (b)

In second case, the efficiency of a man is twice to that in the first case.

$M_1D_1 = M_2D_2$

10 × 20 = 2 × 20 × $D_2$

$D_2 = {10 × 20}/{2 × 20}$ = 5 days.

Answer: (d)

Time taken by B

= ${21 × 100}/140$ = 15 days

Here, x = 21, R = 40%

Time taken By B = $x × 100/{100 + R}$ days

= $21 × 100/140$ days = 15 days

Answer: (b)

Using Rule 2,

Let time taken by P = x days

Then, time taken by Q = 3x days

3x - x = 48 ⇒ x = 24

(P + Q)’s 1 day’s work

= $1/24 + 1/72 = {3 + 1}/72 = 1/18$

Required time = 18 days

Answer: (b)

Using Rule 2,

Let time taken by B in completing the work = x days

Time taken by A = (x - 10) days

⇒ $1/x + 1/{x - 10} = 1/12$

⇒ ${x - 10 + x}/{x(x - 10)} = 1/12$

⇒ $24x - 120 = x^2 - 10x$

⇒ $x^2 - 34x + 120$ = 0

⇒ $x^2 - 30x - 4x + 120$ = 0

⇒ $x (x - 30) - 4 (x - 30)$ = 0

⇒ $(x - 4) (x - 30)$ = 0

⇒ $x = 30$ because $x ≠ 4$