Practice MCQ Questions and Answer on Time and Work

Solution: 10 men = 20 boys →1 men = 2 boys 8 men = 2 × 8 boys = 16 boys Then, (16 boys + 4 boys) = 20 boys can make 260 mats in 20 days Now, It can be calculated by work equivalence method: 20 × 260 × 20 = x × 20 × 20 x = 260 mats

Solution: Total work = 80 × 60 + 60 × 20 = 4800 + 1200 = 6000 60 men × d = 6000 d = 100 days Extra days = 100 - 80 = 20 days

Solution: Pipe A can fill the empty tank in = 0.9 hours So work rate of the Pipe A = $$\frac{{100}}{{0.9}}$$ % per hour Pipe B can empty the tank in = 0.7 hours Negative Work rate of B = $$\frac{{100}}{{0.7}}$$ % per hour. (B is removing water, so, taken as negative work) Tank fill per hour = $$\frac{{100}}{{0.7}} - \frac{{100}}{{0.9}}$$   = 31.75% per hour Time Taken to empty the tank = $$\frac{{100}}{{31.75}}$$  ≈ 3.15 hours So, Capacity of the tank = 3.15 × 180 = 567 liters

Solution: $$\eqalign{ & {\text{Part of the done by A}} \cr & = \left( {\frac{1}{{10}} \times 5} \right) \cr & = \frac{1}{2} \cr & {\text{Part of the done by B}} \cr & = \left( {\frac{1}{{15}} \times 5} \right) \cr & = \frac{1}{3} \cr & {\text{Part of the done by C}} \cr & = 1 - \left( {\frac{1}{2} + \frac{1}{3}} \right) \cr & = \frac{1}{6} \cr} $$ So, (A's share) : (B's share) : (C's share) $$\eqalign{ & = \frac{1}{2}:\frac{1}{3}:\frac{1}{6} \cr & = 3:2:1 \cr & \therefore {\text{A's share}} \cr & = {\text{Rs}}{\text{.}}\left( {\frac{3}{6} \times 1500} \right) \cr & = {\text{Rs}}.750 \cr & {\text{B's share}} \cr & = {\text{Rs}}{\text{.}}\left( {\frac{2}{6} \times 1500} \right) \cr & = {\text{Rs}}.500 \cr & {\text{C's share}} \cr & = {\text{Rs}}{\text{.}}\left( {\frac{1}{6} \times 1500} \right) \cr & = {\text{Rs}}.250 \cr & {\text{A's daily wages}} \cr & = {\text{Rs}}{\text{.}}\left( {\frac{{750}}{5}} \right) \cr & = {\text{Rs}}.150 \cr & {\text{B's daily wages}} \cr & = {\text{Rs}}{\text{.}}\left( {\frac{{500}}{5}} \right) \cr & = {\text{Rs}}.100 \cr & {\text{C's daily wages}} \cr & = {\text{Rs}}{\text{.}}\left( {\frac{{250}}{2}} \right) \cr & = {\text{Rs}}.125 \cr & \therefore {\text{Daily wages of B and C}} \cr & = {\text{Rs}}{\text{.}}\left( {100 + 125} \right) \cr & = {\text{Rs}}{\text{.225}} \cr} $$

Solution: Ratio of times taken by Sakshi and Tanya = 125 : 100 = 5 : 4 Suppose Tanya takes x days to do the work 5 : 4 :: 20 : x ⇒ $$x = {\frac{{4 \times 20}}{5}} $$ ⇒ x = 16 days Hence, Tanya takes 16 days to complete the work

Solution: $$\eqalign{ & {\text{Dinesh's one hour work}} \cr & = \frac{{32}}{6} \cr & = \frac{{16}}{3}{\text{ pages/hour}} \cr & {\text{Rakesh's one hour work}} \cr & = \frac{{40}}{5} \cr & = 8{\text{ pages/hour}} \cr & {\text{Dinesh's and Rakesh's one hour work}} \cr & = \frac{{16}}{3} + 8 \cr & = \frac{{40}}{3}{\text{ pages/hour}} \cr & {\text{They will finish the work together}} \cr & \frac{{{\text{Total work}}}}{{{\text{Efficiency}}}} \cr & = \frac{{110}}{{\frac{{40}}{3}}} \cr & = 8\frac{1}{4} \cr & = {\text{8 hours, 15 minutes}} \cr} $$

Solution: Work done by 40 men in 5 days = $$\frac{1}{3}$$ (As if whole work is completed in 15 days then in 5 days $${{{\frac{1}{3}}^{{\text{rd}}}}}$$ of the work will be finished) $$\eqalign{ & {\text{Remaining work}} = 1 - \frac{1}{3} = \frac{2}{3} \cr & \because 40{\text{ men do 1 work in 15 days}}{\text{.}} \cr & {\text{60 men can do }}\frac{2}{3}{\text{work in }}x{\text{ day}} \cr & \frac{{{{\text{M}}_1}{{\text{D}}_1}}}{{{{\text{W}}_1}}}{\text{ = }}\frac{{{{\text{M}}_2}{{\text{D}}_2}}}{{{{\text{W}}_2}}} \cr & {{\text{M}}_1} = 40{\text{ , }}{{\text{M}}_2} = 60 \cr & {{\text{D}}_1} = 15{\text{ , }}{{\text{D}}_2} = x \cr & {{\text{W}}_1} = 1{\text{ ,}}{{\text{W}}_2} = \frac{2}{3} \cr & \Rightarrow \frac{{40 \times 15}}{1} = \frac{{60 \times x}}{2} \cr & \Rightarrow \frac{2}{3}\left( {40 \times 15} \right) = 60x \cr & \Rightarrow 2 \times 40 \times 5 = 60x \cr & \Rightarrow x = \frac{{20}}{3} \cr & \Rightarrow x = 6\frac{2}{3}{\text{ days}} \cr} $$

Solution: $$\eqalign{ & \left( {{\text{A + B}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{10}} \cr & {\text{C's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{50}} \cr & \left( {{\text{A + B + C}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = {\frac{1}{{10}} + \frac{1}{{50}}} = \frac{6}{{50}} = \frac{3}{{25}}........\left( {\text{i}} \right) \cr & {\text{A's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = \left( {{\text{B + C}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}}\,........\left( {{\text{ii}}} \right) \cr & {\text{From}}\,\left( {\text{i}} \right)\,{\text{and}}\,\left( {{\text{ii}}} \right){\text{,we}}\,{\text{get}}:2 \times \left( {{\text{A's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}}} \right) \cr & = \frac{3}{{25}} \cr & \Rightarrow {\text{A's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{3}{{50}} \cr & \therefore {\text{B's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}}\left( {\frac{1}{{10}} - \frac{3}{{50}}} \right) \cr & = \frac{2}{{50}} = \frac{1}{{25}} \cr & {\text{So,}}\,{\text{B}}\,\,{\text{alone}}\,{\text{could}}\,{\text{do}}\,{\text{the}}\,{\text{work}}\,{\text{in}}\,{\text{25}}\,{\text{days}} \cr} $$

Solution: Jyoti does $$\frac{3}{4}$$ unit of work in 12 days Jyoti does 1 unit of work in $$\eqalign{ & = 12 \times \frac{4}{3} \cr & = 16{\text{ days}} \cr} $$ According to the question,   Mala   :   Jyoti Efficiency →   2   :   1 Time → 1 : 2   ↓×8   ←↓×8   8 days   16 days

Solution: L.C.M. of Total Work =60 One day work of A + B = $$\frac{{60}}{{30}}$$ = 2 unit/day One day work of B + C = $$\frac{{60}}{{20}}$$ = 3 unit/day $$\eqalign{ & {\text{According to the question,}} \cr & \Rightarrow \left( {{\text{A}} + {\text{B}}} \right).....{\text{30 days}} \cr & \Rightarrow \left( {{\text{B}} + {\text{C}}{\text{.}}} \right).....{\text{20 days}} \cr & \Rightarrow {\text{A}}.....{\text{5}} \cr & \Rightarrow {\text{B}}.....{\text{5}} + {\text{10}} \cr & \Rightarrow {\text{C}}.....{\text{10}} + {\text{8 days}} \cr} $$ Work done by (A +B) in 5 days = 2 × 5 = 10 units ⇒ Work done by (B + C) in 10 days = 10 × 3 = 30 units ⇒ Total work (finished till now).....40 units ⇒ Remaining work = 60 - 40 = 20 units ⇒ Here we find that C does remaining 20 units in 8 days ⇒ C's efficiency = $$\frac{{{\text{Work}}}}{{{\text{Day}}}}$$ $$ = \frac{{{\text{20 Work}}}}{{{\text{8 Days}}}} = \frac{5}{2}$$ ⇒ Therefore, time taken by C alone to complete the work $$ = \frac{{60}}{5} \times 2 = 24{\text{ days}}$$