Time And Work - Aptitude MCQ Questions and Solutions with Explanations

(Ramesh & Rahman)'s 1 day's work
$$ = \frac{1}{{20}} + \frac{1}{{25}} = \frac{{5 + 4}}{{100}} = \frac{9}{{100}}$$
∴ Their 10 day's work $$ = \frac{{90}}{{100}} = \frac{9}{{10}}$$
∴ Remaining work $$ = 1 - \frac{9}{{10}} = \frac{1}{{10}}$$
∵ Suresh does $$\frac{1}{{10}}$$ work in 3 days
∴ Time taken by Suresh in doing 1 work = 3 × 10 = 30 days

$$\eqalign{ & {\text{From }}{{\text{M}}_1}{{\text{D}}_1} = {{\text{M}}_2}{{\text{D}}_2} \cr & \Rightarrow \frac{{{{\text{M}}_1}{{\text{D}}_1}}}{{{{\text{M}}_2}{{\text{D}}_2}}} = \frac{5}{6} \cr & \Rightarrow \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = \frac{5}{6} \cr & \Rightarrow \frac{{\left( {x - 1} \right)}}{{\left( {x + 2} \right)}} = \frac{5}{6} \cr & \Rightarrow 6x - 6 = 5x + 10 \cr & \Rightarrow x = 16 \cr} $$

Let to work Total Work = 36
One day work of A = $$\frac{{36}}{{36}}$$ = 1 unit/day
One day work of B = $$\frac{{36}}{{18}}$$ = 2 unit/day
One day work of C = $$\frac{{36}}{{12}}$$ = 3 unit/day
In 3 days cycle total work done is A, A + B, A + C = 1 + (1 + 2) + (1 + 3) = 8 unit/Cycle
∴ 32 units of the work completed in 4 cycle and reminder 4 units of works in next two days.
1 cycle = 3 days
∴ 4 cycle = 12 days
And remaining 4 unit work done in next two days. In days 13, A will work 1 unit and in day 14, A and B will work 3 units.
Total number the day required to complete the work in the given condition is 14 days

$$\frac{{250}}{{100}}A \to 15{\text{ days}}$$
A → 60 days. (Whole work)
Let A do 60 work in 60 days.
A do 15 day work, remaining work completed by A in 45 days. But B do in 30 days.
∴ 45A = 30B
$$\frac{{\text{A}}}{{\text{B}}} = \frac{2}{3}\,\,\,\left( {{\text{efficiency}}} \right)$$
A + B = 5
Total work = A × 60 = 2 × 60 = 120
50% of 120 = 60
A + B = $$\frac{{60}}{5}$$ = 12

450 men × 20 = x men × 30
x = 300 men

⇒ A can do $$\frac{1}{3}$$ of a work in 30 day
⇒ A completed work in 90 days
⇒ B can do $$\frac{2}{5}$$ of the same work in 24 days
⇒ B can completed work in 60 days
⇒ Let the total work = LCM (90, 60) = 180 unit
⇒ Efficiency of A = 3 unit/day
⇒ Efficiency of B = 2 unit/day
⇒ They both worked for 20 days, work done in 20 days = 20 × 5 = 100 unit
⇒ Remaining work = 180 - 100 = 80 unit
⇒ Remaining work done by C in 8 days
⇒ Efficiency of C = 10 unit/day
⇒ Efficiency of (A + B + C) = 15 unit/day
⇒ Work completed = $$\frac{{180}}{{15}}$$ = 12 days
∴ If all worked together, the work complete in 12 days.

2 days work of X = 2 × 2 = 4
After Y joined X and Y work 6 days = 3 × 6 = 18
After 8 days from begging - X, Y and Z work together $$ = \frac{{92 - 22}}{7} = \frac{{70}}{7} = 10{\text{ days}}$$
Total days work of X = 10 + 8 = 18 days

\[\begin{array}{*{20}{c}} {}&{\left( {A + C} \right)}&:&B \\ {{\text{Time}}}&1&{}&3 \\ {{\text{Efficiency}}}&3&{}&1 \end{array}\]
$$B\,{\text{alone}} = \frac{{10 \times 4}}{1} = 40\,{\text{days}}$$

A = 20 days
B = 30 days

(A + B) × 9 + C × 6 = 60
(3 + 2) × 9 + C × 6 = 60
C = $$\frac{5}{2}$$ unit
C = $$\frac{{60 \times 2}}{5}$$  = 24 days
(B + C) × x = 60 × $$\frac{3}{4}$$
$$\left( {2 + \frac{5}{2}} \right){\text{x}} = 45$$
9x = 90
x = 10 days

Two days work of A and B = 3 × 2 = 6
Remaining work = 10 - 6 = 4
Now as per question
$$\eqalign{ & \frac{4}{{1 + {\text{C}}}} = 3 \cr & \frac{4}{3} = 1 + {\text{C}} \cr & {\text{C}} = \frac{1}{3} \cr} $$
So that 40% of total work done by $${\text{C}} = \frac{{10 \times 40\% }}{{\frac{1}{3}}} = 12$$