Practice MCQ Questions and Answer on Time and Work

Solution: M1D1H1 Remaining work of three painters of (12 - 3) days is completed by two painters 3 × 9 × 6 = 2 × 6 × D2 D2 = $$13\frac{1}{2}$$ days

Solution: $$\eqalign{ & {\text{Suppose}}\,{\text{A,}}\,{\text{B}}\,{\text{and}}\,{\text{C}}\,{\text{take}} \cr & x,\,\frac{x}{2},\,\frac{x}{3}\,{\text{days}}\,{\text{respectively}}\,{\text{to}}\,{\text{finish}}\,{\text{the}}\,{\text{work}} \cr & {\text{Then}},\, {\frac{1}{x} + \frac{2}{x} + \frac{3}{x}} = \frac{1}{2} \cr & \Rightarrow \frac{6}{x} = \frac{1}{2} \cr & \Rightarrow x = 12 \cr & {\text{So,}}\,{\text{B}}\,{\text{takes}}\, {\frac{{12}}{2}} \cr & = 6\,{\text{days}}\,{\text{to}}\,{\text{finish}}\,{\text{the}}\,{\text{work}} \cr} $$

Solution: Answer & Solution Answer: Option E Solution: $$\eqalign{ & {\text{1 men's 1 day's work}} \cr & = \frac{1}{{21 \times 3}} \cr & = \frac{1}{{63}} \cr & {\text{1 boy's 1 day's work}} \cr & = \frac{1}{{21 \times 9}} \cr & = \frac{1}{{189}} \cr & \left( {{\text{5 men}} + {\text{6 boy}}} \right){\text{'s 1 day's work}} \cr & = \frac{5}{{63}} + \frac{6}{{189}} \cr & = \frac{5}{{63}} + \frac{2}{{63}} \cr & = \frac{7}{{63}} \cr & = \frac{1}{9} \cr} $$ Hence, 5 men's and 6 boy's together can do the work in 9 days.

Solution: If B worked 5 days more = 3 × 5 = 15 unit If C worked 10 days more = 2 × 10 = 20 unit Now, total work = 108 + 15 + 20 = 143 (A + B + C) = $$\frac{{143}}{{11}}$$ = 13 days

Solution: Let the capacity of the tank = q litres According to the question, Quantity of water emptied by the leak in 1 hour = $$\frac{\text{q}}{10}$$ litres Qunatity of water filled by the tap in 1 hour = 240 litres According to the question, $$\eqalign{ & \Rightarrow \frac{{\text{q}}}{{10}} - \frac{{\text{q}}}{{15}} = 240 \cr & \Rightarrow \frac{{3{\text{q}} - 2{\text{q}}}}{{30}} = 240 \cr & \Rightarrow \frac{9}{{30}} = 240 \cr & \Rightarrow {\text{q}} = 240 \times 30 \cr & \Rightarrow {\text{q}} = 7200\,{\text{litres}} \cr} $$

Solution: $$\eqalign{ & {\text{Required average,}} \cr & = \frac{{ {\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{{10}} + \frac{1}{{12}}} }}{5} \cr & = {\frac{{48}}{{60}} \times \frac{1}{5}} \cr & = \frac{4}{{25}} \cr & = 0.16 \cr} $$

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: $$\eqalign{ & 3 \times V = 2 \times K \cr & \frac{V}{K} = \frac{2}{3} = \frac{{10}}{{15}} \cr & 4 \times K = D \times 5 \cr & \frac{K}{D} = \frac{5}{4} = \frac{{15}}{{12}} \cr & {\text{Total work}} = 48 \times \left( {10 + 15 + 12} \right) \cr & {\text{Time of Dinesh}} = \left( {\frac{{48 \times 37}}{{12}}} \right) = 148{\text{ hr}} \cr} $$

Solution: Work done by X in 8 days = $$ {\frac{1}{{40}} \times 8} $$   = $$\frac{1}{5}$$ Remaining work = $$ {1 - \frac{1}{5}} $$   = $$\frac{4}{5}$$ Now, $$\frac{4}{5}$$ work is done by Y in 16 days Whole work will be done by Y in = $$ {16 \times \frac{5}{4}} $$   = 20 days ∴ X's 1 day's work = $$\frac{1}{{40}}$$ ∴ Y's 1 day's work = $$\frac{1}{{20}}$$ (X + Y)'s 1 day's work $$\eqalign{ & = {\frac{1}{{40}} + \frac{1}{{20}}} \cr & = \frac{3}{{40}} \cr} $$ Hence, X and Y will together complete the work in $$\eqalign{ & = {\frac{{{\text{40}}}}{{\text{3}}}} \cr & {\text{ = 13}}\frac{{\text{1}}}{{\text{3}}}\,\,{\text{days}} \cr} $$

Solution: $$\eqalign{ & {\text{Number}}\,{\text{of}}\,{\text{pages}}\,{\text{typed}}\,{\text{by}}\,{\text{Ravi}}\,{\text{in}}\,{\text{1}}\,{\text{hour}} \cr & = \frac{{32}}{6} = \frac{{16}}{3} \cr & {\text{Number}}\,{\text{of}}\,{\text{pages}}\,{\text{typed}}\,{\text{by}}\,{\text{Kumar}}\,{\text{in}}\,{\text{1}}\,{\text{hour}} \cr & = \frac{{40}}{5} = 8 \cr & {\text{Number}}\,{\text{of}}\,{\text{pages}}\,{\text{typed}}\,{\text{by}}\,{\text{both}}\,{\text{in}}\,{\text{1}}\,{\text{hour}} \cr & = {\frac{{16}}{3} + 8} = \frac{{40}}{3} \cr & \therefore {\text{Time}}\,{\text{taken}}\,{\text{by}}\,{\text{both}}\,{\text{to}}\,{\text{type}}\,{\text{110}}\,{\text{pages}} \cr & = {110 \times \frac{3}{{40}}} {\text{hours}} \cr & = 8\frac{1}{4}\,{\text{hours}}\,{\text{(or)}}\,{\text{8}}\,{\text{hours}}\,{\text{15}}\,{\text{minutes}} \cr} $$