Practice MCQ Questions and Answer on Time and Work

Solution: (M + B)'s 1 day's work =$$\frac{1}{{24}}$$ (M + B)'s 20 day's work + M's 6 day's work = 1 $$\eqalign{ & \Rightarrow {\text{M's 6 day's work}} \cr & = \left( {1 - \frac{1}{{24}} \times 20} \right) \cr & = \frac{4}{{24}} = \frac{1}{6} \cr & \Rightarrow {\text{M's 1 day's work}} \cr & = \frac{1}{6} \times \frac{1}{6} \cr & = \frac{1}{{36}} \cr & \therefore {\text{B's 1 day's work}} \cr & = \frac{1}{{24}} - \frac{1}{{36}} \cr & = \frac{1}{{72}} \cr} $$ Hence, the boy alone can do the work in 72 days.

Solution: $$\eqalign{ & {\text{C's 3 day's work}} \cr & = \left( {\frac{1}{{15}} \times 3} \right) \cr & = \frac{1}{5} \cr & \left( {{\text{B}} + {\text{C}}} \right){\text{'s 2 day's work}} \cr & = \left[ {\left( {\frac{1}{{12}} + \frac{1}{{15}}} \right) \times 2} \right] \cr & = \left( {\frac{3}{{20}} \times 2} \right) \cr & = \frac{3}{{10}} \cr & \therefore {\text{Remaining work}} \cr & = \left[ {1 - \left( {\frac{1}{5} + \frac{3}{{10}}} \right)} \right] \cr & = \left( {1 - \frac{1}{2}} \right) \cr & = \frac{1}{2} \cr & \left( {{\text{A}} + {\text{B}} + {\text{C}}} \right){\text{'s 1 day's work}} \cr & = \left( {\frac{1}{{10}} + \frac{1}{{12}} + \frac{1}{{15}}} \right) \cr & = \frac{{15}}{{60}} \cr & = \frac{1}{4} \cr} $$ $$\frac{1}{4}$$ work is done by A, B ans C in 1 day. ∴ $$\frac{1}{2}$$ work is done by A, B and C in $$\eqalign{ & = \left( {4 \times \frac{1}{2}} \right) \cr & = {\text{2 days}} \cr & {\text{Total number of days}} \cr & = \left( {3 + 2 + 2} \right) \cr & = 7 \cr} $$

Solution: Let time taken by A alone in doing work be x days ∴ Time taken by B alone = 3x days $$\eqalign{ & {\text{A's 1 day's work}} = \frac{1}{x} \cr & {\text{B's 1 days work}} = \frac{1}{{3x}} \cr & \because {\text{A and B together finish}} \cr & = \frac{2}{5}{\text{work in 9 days}}{\text{.}} \cr} $$ ∴ Time taken by A and B in doing whole work $$\eqalign{ & = \frac{{9 \times 5}}{2} \cr & = \frac{{45}}{2}{\text{ days}} \cr} $$ According to given information we get $$\eqalign{ & \therefore \frac{1}{x} + \frac{1}{{3x}} = \frac{2}{{45}} \cr & \Rightarrow \frac{{3 + 1}}{{3x}} = \frac{2}{{45}} \cr & \Rightarrow \frac{4}{{3x}} = \frac{2}{{45}} \cr & {\text{By cross - multiply we get }} \cr & \Rightarrow 2 \times 3x = 4 \times 45 \cr & \Rightarrow x = \frac{{4 \times 45}}{{2 \times 3}} \cr & \Rightarrow x = 30{\text{ days}} \cr & {\text{Time taken by A}} \cr & = x{\text{ days}} \cr & = {\text{30 days}} \cr & \therefore {\text{Time taken by B}} \cr & = 3x{\text{ days}} \cr & = 3 \times 30 \cr & = 90{\text{ days}} \cr} $$

Solution: $$\eqalign{ & \Rightarrow {{\text{R}}_{\text{x}}} \cr & = \frac{{80}}{{20}} \cr & {\text{ = 4 pages/hour}} \cr & \Rightarrow {{\text{R}}_{{\text{x + y}}}} \cr & = \frac{{135}}{{27}} \cr & {\text{ = 5 pages/hour}} \cr & \Rightarrow {{\text{R}}_{\text{y}}} \cr & = {{\text{R}}_{{\text{x + y}}}} - {{\text{R}}_{\text{x}}} \cr & = 5 - 4 \cr & = 1{\text{ pages/hour}} \cr & \therefore {\text{y can copy 20 pages in}} \cr & = \frac{{{\text{20p}}}}{{{\text{1p/h}}}} \cr & = {\text{ }}20{\text{ hours}} \cr} $$

Solution: Time taken by A =x hours. Therefore taken by A, B and C together = (x - 6) Time taken by B = (x - 5) Time taken by C = 2(x - 6) Now, rate of work of A + Rate of work of B + Rate of work of C = Rate of work of ABC. $$ \Rightarrow \frac{1}{x} + \frac{1}{{x - 5}} + \frac{1}{{2\left( {x - 6} \right)}} = \frac{1}{{x - 6}}$$ On solving above equation, we get x = 3, $$\frac{{40}}{6}$$ When x = 3, the expression (x - 6) becomes negative, thus it's not possible. $$ \Rightarrow x = \frac{{40}}{6}$$ Time taken by A & B together = $$\frac{1}{{\frac{3}{{20}} + \frac{3}{5}}}$$ = $$\frac{4}{3}$$ hours

Solution: A : B = 1 : 2 B : C = 1 : 2 (Multiply by 2) B : C = 2 : 4 A : B : C = 1 : 2 : 4

Solution: (5M + 8W) × 34 = (4M + 18W) × 28 85M + 136W = 56M + 252W 29M = 116W $$\frac{{\text{M}}}{{\text{W}}} = \frac{4}{1}$$ (5M + 8W) × 34 = (3M + 5W) × x (5 × 4 + 8 × 1) × 34 = (3 × 4 + 5 × 1) × x 28 × 34 = 17 × x x = 56 days

Solution: According to the question, $$\eqalign{ & {\text{2m}} + {\text{3w}} = {\text{4m}} \cr & {\text{3w}} = {\text{4m}} - {\text{2m }} \cr & {\text{3w}} = {\text{2m}} \cr & {\text{3m}} + {\text{3w}} = {\text{3m}} + {\text{2m}} \cr & {\text{3m}} + {\text{3w}} = {\text{5m}} \cr} $$ 4 men can do work in 20 days 1 men can do work in 20 × 4 days 5 men can do work in $$\frac{{{\text{20}} \times {\text{4}}}}{5}$$  = 16 days $$\eqalign{ & {\bf{Alternate:}} \cr & \left( {2{\text{m}} + {\text{3w}}} \right) \times {\text{20}} = {\text{4m}} \times {\text{20}} \cr & \frac{{\text{m}}}{{\text{w}}} = \frac{3}{2} \cr & {\text{Total work }} \cr & {\text{ = }}\left( {{\text{2}} \times {\text{3}} + {\text{3}} \times {\text{2}}} \right) \times 20 \cr & = 240\,{\text{units}} \cr & {\text{5 men efficiency}} \cr & = 5 \times 3 \cr & = 15 \cr & {\text{Required number of days}} \cr & = \frac{{240}}{{15}} \cr & = 16{\text{ days}} \cr} $$

Solution: Vimal's 1 day work = $$\frac{1}{{20}}$$ Since, Vimal and Kamal can together complete in 12 days i.e. (Vimal + Kamal)'s 1 day work = $$\frac{1}{{12}}$$ Then, Kamal's 1 day work, $$ = \frac{1}{{12}} - \frac{1}{{20}} \Rightarrow \frac{2}{{60}} \Rightarrow \frac{1}{{30}}$$ If Kamal Works only for half a day daily, then his 1 day work becomes $$\frac{1}{2} \times \frac{1}{{30}}$$  = $$\frac{1}{{60}}$$ Therefore, 1 day work of both Vimal and Kamal, $$ = \frac{1}{{20}} + \frac{1}{{60}} \Rightarrow \frac{4}{{60}} \Rightarrow \frac{1}{{15}}$$ Hence, the work will be completed in 15 days.

Solution: 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 $$\eqalign{ & = \frac{{{\text{Total work}}}}{{{\text{Efficiency}}}} \cr & = \frac{{12}}{2} \cr & = {\text{6 days }} \cr} $$