Time And Work - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{From, }} \cr & \frac{{{{\text{m}}_1} \times {{\text{d}}_1} \times {{\text{t}}_1}}}{{{{\text{w}}_1}}} = \frac{{{{\text{m}}_2} \times {{\text{d}}_2} \times {{\text{t}}_2}}}{{{{\text{w}}_2}}} \cr & {\text{Let extra workers be x}} \cr & \Rightarrow \frac{{20 \times 12}}{{\frac{5}{8}}} = \frac{{\left( {20 + x} \right) \times 4}}{{\frac{3}{8}}} \cr & \Rightarrow 4 \times 12 = \frac{{\left( {20 + x} \right) \times 4}}{3} \cr & \Rightarrow 36 = 20 + x \cr & \Rightarrow x = 16 \cr & \Rightarrow {\text{Extra workers }} = {\text{16}} \cr} $$

$$\eqalign{ & {\text{B's}}\,{\text{10}}\,{\text{day's}}\,{\text{work}} \cr & = \left( {\frac{1}{{15}} \times 10} \right) = \frac{2}{3} \cr & {\text{Remaining}}\,{\text{work}} \cr & = \left( {1 - \frac{2}{3}} \right) = \frac{1}{3} \cr & {\text{Now}},\frac{1}{{18}}{\text{work}}\,{\text{is}}\,{\text{done}}\,{\text{by}}\,{\text{A}}\,{\text{in}}\,{\text{1}}\,{\text{day}} \cr & \therefore \frac{1}{3}\,{\text{work}}\,{\text{is}}\,{\text{done}}\,{\text{by}}\,{\text{A}}\,{\text{in}} \cr & \left( {18 \times \frac{1}{3}} \right) = 6\,{\text{days}} \cr} $$

Total work = 12
Efficiency 1 units/day = 12 days time (A)
After 3 days A finishes 3 units.
∴ Work left =12 - 3 = 9 units
9 units of work, 3 units/day = 3 days (A + B)
(A + B)'s one day work = 3 units
A's one day work = 1 unit
B's one day work = 3 - 1 = 2 units
∴ B completes whole work in
$$\eqalign{ & = \frac{{{\text{Total work}}}}{{{\text{Efficiency}}}} \cr & = \frac{{12}}{2} \cr & = {\text{6 days }} \cr} $$

According to the question,
Let the remaining work is complete in D days
$${\text{4}}{{\text{5}}_{{\text{men}}}} \times {\text{1}}{{\text{6}}_{{\text{days}}}}$$    = $$\left( {{\text{4}}{{\text{5}}_{{\text{men}}}} \times {{\text{4}}_{{\text{days}}}}} \right)$$    + $${\left( {{\text{45}} + {\text{36}}} \right)_{{\text{men}}}}$$   $$ \times {\text{D}}$$
$$\eqalign{ & \Rightarrow \frac{{540}}{{81}} = {\text{D}} \cr & \Rightarrow {\text{D}} = 6\frac{2}{3}{\text{ days}} \cr} $$

Suppose A takes x days to finish the work alone and B take y days to finish the work alone.
$$\eqalign{ & {\text{Then,}}\frac{2}{x} + \frac{9}{y} = 1.....(i) \cr & {\text{And,}}\frac{3}{x} + \frac{6}{y} = 1 \cr & \Leftrightarrow \frac{1}{x} + \frac{2}{y} = \frac{1}{3} \cr & \Leftrightarrow \frac{2}{x} + \frac{4}{y} = \frac{2}{3}.....({\text{ii)}} \cr & {\text{Subtracting (ii) from (i), }} \cr & {\text{We get}}:\frac{5}{y} = \frac{1}{3}{\text{or }}y = 15 \cr & {\text{Putting }}y = 15{\text{ in (i), }} \cr & {\text{We get}}:\frac{2}{x} = \frac{2}{5}{\text{or }}x = 5 \cr} $$
Hence, A alone takes 5 days while B alone takes 15 days to finish the work.

Suppose, Asha takes x days to complete the task alone
while Sudha takes y days to complete it alone
Since Sudha is more efficient than Asha, we have x > y
$$\eqalign{ & {\text{Asha's 1 day's work}} = \frac{1}{x} \cr & {\text{Sudha's 1 day's work}} = \frac{1}{y} \cr & \left( {{\text{Asha}} + {\text{Sudha}}} \right){\text{'s 1 day's work}} \cr & = \frac{1}{x} + \frac{1}{y} \cr & = \frac{{x + y}}{{xy}} \cr} $$
If Asha and Sudha each does half of the work alone, time taken
$$\eqalign{ & = \left( {\frac{x}{2} + \frac{y}{2}} \right){\text{ days }} \cr & = \left( {\frac{{x + y}}{2}} \right){\text{ days}}{\text{.}} \cr & \therefore \frac{{x + y}}{2} = 45 \cr & \Rightarrow x + y = 90 \cr & {\text{From (i) and (ii), }} \cr & {\text{We have}}:\frac{{xy}}{{20}} = 90{\text{ or }}xy = 1800 \cr & {\text{Now, }}xy = 1800{\text{ and }}x + y = 90 \cr & \Rightarrow x = 60,{\text{ }}y = 30{\text{ }}\left[ {\because x > y} \right]{\text{ }} \cr} $$
Hence, Asha alone will take 60 days to complete the task.

$$\eqalign{ & {\text{B's 3 day's work}} \cr & = \left( {\frac{1}{{21}} \times 3} \right) \cr & = \frac{1}{7} \cr & {\text{Remaining work}} \cr & = \left( {1 - \frac{1}{7}} \right) \cr & = \frac{6}{7} \cr & \left( {{\text{A}} + {\text{B}}} \right){\text{'s 1 day's work}} \cr & = \left( {\frac{1}{{14}} + \frac{1}{{21}}} \right) \cr & = \frac{5}{{42}} \cr} $$
Now, $$\frac{5}{{42}}$$ work is done by A and B in 1 day
$$\eqalign{ & \therefore \frac{6}{7}{\text{ work is done by A and B in}} \cr & = \left( {\frac{{42}}{5} \times \frac{6}{7}} \right) \cr & {\text{ = }}\frac{{36}}{5}{\text{ days}} \cr & {\text{Hence, total time taken}} \cr & = \left( {3 + \frac{{36}}{5}} \right){\text{days}} \cr & = 10\frac{1}{5}{\text{days}}{\text{.}} \cr} $$

L.C.M. of total days = 24
One day work of A + B = $$\frac{{24}}{8}$$ = 3 units/day
One day work of B = $$\frac{{24}}{12}$$ = 2 units/day
One day work of C = $$\frac{{24}}{12}$$ = 2 units/day
A and B work for 4 days they completed
$$\eqalign{ & = 3 \times 4 \cr & = 12{\text{ units}} \cr & \therefore {\text{Work left}} \cr & = 24 - 12 \cr & = {\text{12 units}} \cr & {\text{B's 2 days work}} \cr & = 2 \times 2 \cr & = {\text{4 units}} \cr & \therefore {\text{Work left}} \cr & = 12 - 4 \cr & = {\text{8 units}} \cr & {\text{Now, }} \cr & {\text{C's complete the work in}} \cr & = \frac{8}{2} \cr & = 4{\text{ days}} \cr} $$

	(T.W.) 60
2 units/day	↓
	30 days (A + B)

$$\eqalign{ & \left( {{\text{A }} + {\text{ B}}} \right){\text{ worked for 20 days}} \cr & {\text{So, they completed}} \cr & 20 \times 2 = 40{\text{ units}} \cr & {\text{Work left}} = 60 - 40 = {\text{20 units}} \cr} $$

	(T.W.) 2 0
1 units/day	↓
	20 days (A)

A's efficiency 1 units/day A will complete the whole work in
$$\eqalign{ & = \frac{{{\text{Total work}}}}{{{\text{Efficiency}}}} \cr & = \frac{{60}}{1} \cr & = 60{\text{ days}} \cr & \cr & {\bf{Alternate:}} \cr & \left( {{\text{A}} + {\text{B}}} \right) \times {\text{10}} = {\text{A}} \times {\text{20}} \cr & \frac{{{\text{A + B}}}}{{\text{A}}} = \frac{2}{1} \cr & {\text{Total work}} \cr & = 2 \times 30 \cr & = {\text{60 units}} \cr & {\text{Time taken by A alone}} \cr & = \frac{{60}}{1} \cr & = 60{\text{ days}} \cr} $$

$$\eqalign{ & 16{\text{ women}} \times {\text{12}} = {\text{12 men}} \times {\text{8}} \cr & {\text{2 women}} = {\text{1 man}} \cr & {\text{Total work}} \cr & = {\text{12 men}} \times {\text{8 days}} \cr & = {\text{96 units}} \cr & {\text{16 men do work in 3 days}} \cr & = 16 \times 3 \cr & = 48{\text{ units}} \cr & {\text{Work left }}96 - 48 = 48{\text{ units}} \cr} $$
16 men - 10 men left = 6 men + 4 women join
$$\eqalign{ & {\text{6men}} + {\text{2 men}} = {\text{8 men}} \cr & {\text{8 men will do 48 units in}} \cr & = \frac{{48}}{8} \cr & = 6{\text{ days}} \cr} $$