Time And Work - Aptitude MCQ Questions and Solutions with Explanations

L.C.M. of Total Work =60 unit
One day work of A = $$\frac{{60}}{{12}}$$ = 5 unit/day
One day work of B = $$\frac{{60}}{{20}}$$ = 3 unit/day
$$\eqalign{ & {\text{5}}\left( {{\text{A}} + {\text{B}}} \right) + {\text{3C}} = {\text{60 units}} \cr & {\text{5}} \times {\text{8}} + {\text{3C}} = 60 \cr & 3{\text{C}} = 20 \cr & {\text{C}} = \frac{{20}}{3}{\text{ units/day}} \cr & {\text{Time taken by C}} \cr & = \frac{{60}}{{\frac{{20}}{3}}} \cr & = \frac{{60 \times 3}}{{20}} \cr & = 9{\text{ days}} \cr} $$

$$\eqalign{ & {\text{Total work}} = 100 \cr & {\text{Remaining work}} \cr & = 100 - 37 \cr & = 63 \cr & {\text{5A}} = {\text{4B}} \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{4}{5}{\text{ efficiency}} \cr & {\text{Total efficiency of A}} + {\text{B}} = 9 \cr & {\text{Work done by in 7 days}} \cr & = 9 \times 7 \cr & = 63 \cr & \therefore {\text{Time taken by B}} \cr & = \frac{{100}}{5} \cr & = 20{\text{ days}} \cr} $$

$$\eqalign{ & 20\% = \frac{1}{5}\left[ {{\text{efficiency}} \propto \frac{1}{{{\text{days}}}}} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{A }}:{\text{ B}} \cr & {\text{Efficiency}}\,\,\,\,\,\,4{\text{ }}:{\text{ }}5 \cr & {\text{Days}}\,\,\,\,\,\,\,\,\,\,\mathop {\mathop 5\limits_{{\text{ }} \downarrow \times 2} }\limits_{10\,{\text{days}}} :\mathop {\mathop 4\limits_{{\text{ }} \downarrow \times 2} }\limits_{{\bf{8}}\,{\bf{days}}} \cr} $$

Let a man complete 1 piece of work in a day.
Then total work = 50 units
Then by statement 1 st day = one man × 1 work/day = 1
Then by statement 2nd day = two men × 1 work/day = 2
Then by statement 3rd day = three men × 1 work/day = 3
Let the whole work will be completed in N days.
Then total work 1 + 2 + 3 + ..... + N = 50
$$\eqalign{ & \frac{{{\text{N}}\left( {{\text{N}} + {\text{1}}} \right)}}{2} = 50 \cr & {\text{N}}\left( {{\text{N + 1}}} \right){\text{ = 100}} \cr & {\text{Then, N}} = 10{\text{ days}}\left( {{\text{approx}}} \right) \cr} $$

L.C.M. of Total Work =60
One day work of A = $$\frac{{60}}{{20}}$$ = 3 unit/day
One day work of B = $$\frac{{60}}{{15}}$$ = 4 unit/day
One day work of A + B + C = $$\frac{{60}}{{5}}$$ = 12 unit/day
$$\eqalign{ & {\text{C's efficiency}} \cr & = 12 - 3 - 4 \cr & = 5 \cr & {\text{C will complete total work in}} \cr & = \frac{{60}}{5} \cr & {\text{ = 12 days}} \cr} $$

$$\eqalign{ & 25\% = \frac{1}{4}\left[ {{\text{efficiency}} \propto \frac{1}{{{\text{days}}}}} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Sashi }}:{\text{ Tanya}} \cr & {\text{Efficiency}}\,\,\,\,\,\,\,4{\text{ }}:{\text{ }}5 \cr & {\text{Days}}\,\,\,\,\,\,\,\,\,\,\,\mathop {\mathop 5\limits_{{\text{ }} \downarrow \times 4} }\limits_{20} :\mathop {\mathop 4\limits_{{\text{ }} \downarrow \times 4} }\limits_{{\bf{16}}} \cr} $$

Let the required no of days be x.
18 Men = 36 Boys
1 Man = 2 Boys
∴ 24 Men = 48 Boys
According to the question,
M₁D₁H₁ = M₂D₂H₂
36 × 24 × 6 = (48 + 24) × x × 9
36 × 24 × 6 = 72 × 9 × x
x = $$\frac{{36 \times 6 \times 24}}{{72 \times 9}}$$
x = 8 days

$$\eqalign{ & \frac{1}{3}{\text{work in 5 days}} \cr & {\text{then a complete work in}} \cr & = 5 \times 3 = 15{\text{ days}} \cr & {\text{B}} \to \frac{2}{5}{\text{work in 10 days}} \cr & {\text{then B completes work in}} \cr & = 10 \times \frac{5}{2} = {\text{25 days}} \cr} $$
L.C.M. of Total Work =75
One day work of A = $$\frac{{75}}{{15}}$$ = 5 unit/day
One day work of B = $$\frac{{75}}{{25}}$$ = 3 unit/day
$$\eqalign{ & \left( {{\text{A}} + {\text{B}}} \right){\text{ can do work}} \cr & = \frac{{75}}{8} = 9\frac{3}{8}{\text{ days}} \cr} $$

L.C.M. of Total Work = 45
One day work of A = $$\frac{{45}}{{5}}$$ = 9 unit/day
One day work of B = $$\frac{{45}}{{15}}$$ = 3 unit/day
One day work of A + B + C = $$\frac{{45}}{{3}}$$ = 15 unit/day
$$\eqalign{ & {\text{Efficiency of C}} \cr & = 15 - \left( {9 + 3} \right) \cr & = 3 \cr & {\text{C's amount}} \cr & = \frac{{250}}{{15}} \times 3 \cr & = {\text{Rs}}{\text{. 50}} \cr} $$

$$\eqalign{ & {\text{B}}:{\text{A}} \cr & \,1:2 \to {\text{Efficiency ratio}} \cr & {\text{Total work}} = 16 \times \left( {1 + 2} \right) = 48 \cr} $$
Number of days taken by A to complete the work
$$\eqalign{ & = \frac{{48}}{2} \cr & = 24{\text{ days}} \cr} $$