Time And Work - Aptitude MCQ Questions and Solutions with Explanations

Janardan completes $$\frac{2}{3}$$ of work in 10 days
Janardan completes 1 of work in
$$\eqalign{ & = \frac{{10 \times 3}}{2} \cr & = 15{\text{ days}} \cr} $$
Janardan completes $$\frac{3}{5}$$ of work in
$$\eqalign{ & = 15 \times \frac{3}{5} \cr & = 9{\text{ days}} \cr} $$

(A + B)'s 1 day work
$$\eqalign{ & = \frac{1}{{12}} + \frac{1}{{15}} \cr & = \frac{{5 + 4}}{{60}} \cr & = \frac{9}{{60}} = \frac{3}{{20}} \cr} $$
(A + B + C)'s 1 day work = $$\frac{1}{5}$$
∴ C's 1 day work
$$\eqalign{ & = \frac{1}{5} - \frac{3}{{20}} \cr & = \frac{{4 - 3}}{{20}} \cr & = \frac{1}{{20}} \cr} $$
Ratio of their work
$$\eqalign{ & = \frac{1}{{12}}:\frac{1}{{15}}:\frac{1}{{20}} \cr & = 5:4:3 \cr} $$
$$\eqalign{ & {\text{A's}}\,{\text{share}} = \frac{5}{{12}} \times 960 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{. }}400 \cr} $$

Length of cloth produced by A and B in 10 hours
$$ = 300000{\text{ metres}}$$
Length of cloth produced by B in 10 hours
$$\eqalign{ & = \left( {\frac{{300000}}{{15}} \times 10} \right) \cr & = 200000{\text{ metres}} \cr} $$
∴ Length of cloth produced by A in 10 hours
$$\eqalign{ & = \left( {300000 - 200000} \right) \cr & = 100000{\text{ metres}} \cr} $$

(X + Y)'s 1 day's work
$$\eqalign{ & = \left( {\frac{1}{{16}} + \frac{1}{{16}}} \right) \cr & = \frac{2}{{16}} \cr & = \frac{1}{8} \cr} $$
∴ Z's 1 day's work
(X + Y + Z)'s 1 day's work - (X + Y)'s 1 day's work
$$\eqalign{ & = \frac{1}{6} - \frac{1}{8} \cr & = \frac{1}{{24}} \cr} $$
∴ Z alone can finish the work in 24 days.

$$\eqalign{ & \frac{3}{{10}}\left( {{\text{B}} + {\text{C }}} \right) = 2{\text{ days}} \cr & \left( {{\text{B}} + {\text{C }}} \right) = 2 \times \frac{{10}}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{20}}{3}{\text{ days}} \cr & \frac{1}{2}\left( {{\text{A}} + {\text{B}} + {\text{C }}} \right) = 2{\text{ days}} \cr & {\text{A}} + {\text{B}} + {\text{C}} = {\text{4 days}} \cr} $$
L.C.M of total work = 20
One day work of A + B + C = $$\frac{{20}}{4}$$ = 5 unit/day
One day work of B + C = $$\frac{{20}}{{\frac{{20}}{3}}}$$ = 3 unit/day
A = 5 - 3 = 2
A alone will complete the work
$$\eqalign{ & = \frac{{20}}{2}{\text{days}} \cr & = {\text{10 days}} \cr} $$

Since we know efficiency and time are inversely proportional to each other.

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{P}}:{\text{Q}} \cr & {\text{Efficiency }}3:4 \cr & {\text{Time }}\,\,\,\,\,\,\,\,\,\,{\text{ }}4:3 \cr} $$

$$\eqalign{ & {\text{5M}} \times {\text{6 days}} = {\text{10W}} \times {\text{5 days}} \cr & {\text{3M}} = {\text{5W}} \cr & \frac{{\text{M}}}{{\text{W}}} = \frac{5}{3} \cr & 1{\text{M}}\left( {{\text{work}}} \right) = 5{\text{ units/day}} \cr & {\text{1W}}\left( {{\text{work}}} \right) = 3{\text{ units/day}} \cr & {\text{Hence, }} \cr & {\text{Total work}} \cr & = \left( {{\text{5M}} \times {\text{6}}} \right) \cr & = {\text{5}} \times {\text{5}} \times {\text{6}} \cr & = {\text{150 units}} \cr & {\text{Required time for }}\left( {{\text{5W}} + {\text{3M}}} \right) \cr & = \frac{{{\text{Total work}}}}{{{\text{Work done/day}}}} \cr & = \frac{{150}}{{\left( {5 \times 3 + 3 \times 5} \right)}} \cr & = \frac{{150}}{{30}} \cr & = 5{\text{ days}} \cr} $$

$$\eqalign{ & \left( {{\text{A}} + {\text{B}} + {\text{C}}} \right){\text{'s 1 day's work}} = \frac{1}{4} \cr & {\text{A's 1 day's work}} = \frac{1}{{16}} \cr & {\text{B's 1 day's work}} = \frac{1}{{12}} \cr & \therefore {\text{C's 1 day's work}} \cr & = \frac{1}{4} - \left( {\frac{1}{{16}} + \frac{1}{{12}}} \right) \cr & = \left( {\frac{1}{4} - \frac{7}{{48}}} \right) \cr & = \frac{5}{{48}} \cr & {\text{So, C alone can do the work in }} \cr & = \frac{{48}}{5} \cr & = 9\frac{3}{5}{\text{ days}} \cr} $$

Whole work will be done by A in
$$\eqalign{ & = \left( {5 \times 3} \right) \cr & = 15{\text{ days}} \cr} $$
Whole work will be done by B in
$$\eqalign{ & = \left( {10 \times \frac{5}{2}} \right) \cr & = 25{\text{ days}} \cr} $$
$$\eqalign{ & {\text{A's 1 day's work}} = \frac{1}{{15}} \cr & {\text{B's 1 day's work}} = \frac{1}{{25}} \cr & \left( {{\text{A}} + {\text{B}}} \right){\text{'s 1 day's work}} \cr & {\text{ = }}\left( {\frac{1}{{15}} + \frac{1}{{25}}} \right) \cr & = \frac{{16}}{{150}} \cr & = \frac{8}{{75}} \cr} $$
∴ A and B together can complete the work in
$$\eqalign{ & = \frac{{75}}{8} \cr & = 9\frac{3}{8}{\text{days}}{\text{.}} \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & {\text{3m}} \times {\text{16}} = {\text{6w}} \times {\text{16}} \cr} $$
$$\frac{{\text{m}}}{{\text{w}}} = $$ $$\frac{{2 \to {\text{Efficiency of men}}}}{{1 \to {\text{Efficiency of women}}}}$$
$$\eqalign{ & {\text{Total work}} = 3 \times 2 \times 16 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 96{\text{ units}} \cr} $$
$${\text{One day work of}}$$    $$\left( {{\text{12m}} + {\text{8w}}} \right)$$
$$\eqalign{ & = 12 \times 2 + 8 \times 1 \cr & = 32{\text{ units}} \cr} $$
$${\text{Total time taken by}}$$    $$\left( {{\text{12m}} + {\text{8w}}} \right)$$
$$\eqalign{ & = \frac{{96}}{{32}} \cr & = 3{\text{ days}} \cr} $$