Practice MCQ Questions and Answer on Time and Work

Solution: Let the time taken by the three friends together to do the work be x hours Then, time taken by Anne alone = (x + 6) hours Time taken by Bob alone = (x + 1) hours Time taken by Bob alone = 2x hours $$\therefore \frac{1}{{x + 6}} + \frac{1}{{x + 1}} + \frac{1}{{2x}} = \frac{1}{x}$$ $$ \Rightarrow $$ $$\frac{{2x\left( {x + 1} \right) + 2x\left( {x + 6} \right) + \left( {x + 1} \right)\left( {x + 6} \right)}}{{2x\left( {x + 6} \right)\left( {x + 1} \right)}}$$         $$ = $$ $$\frac{1}{x}$$ $$\eqalign{ & \Rightarrow 5{x^2} + 21x + 6 = 2\left( {{x^2} + 7x + 6} \right) \cr & \Rightarrow 3{x^2} + 7x - 6 = 0 \cr & \Rightarrow \left( {x + 3} \right)\left( {3x - 2} \right) = 0 \cr & \Rightarrow x = \frac{2}{3}{\text{ }}\left[ {\because x \ne - 3} \right] \cr & \therefore {\text{Required times}} \cr & = \frac{2}{3}{\text{ hours}} \cr & = \left( {\frac{2}{3} \times 60} \right){\text{hours}} \cr & = 40{\text{ minutes}} \cr} $$

Solution: \[\begin{array}{*{20}{c}} {}&{\text{P}}&{}&{\text{Q}} \\ {{\text{Efficiency}}}&3&:&1 \\ {{\text{Time}}}&1&:&3 \end{array}\] 3 unit → 24 days 1 unit → 8 days Total work = 24 P + Q can complete the same work = $$\frac{{24}}{4}$$ = 6 days

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: (A's 1 day's work) : (B's 1 day's work) $$\eqalign{ & = 2:1 \cr & \left( {{\text{A}} + {\text{B}}} \right){\text{'s 1 day's work}} = \frac{1}{{18}} \cr & \therefore {\text{A's 1 day's work}} \cr & = \left( {\frac{1}{{18}} \times \frac{2}{3}} \right) \cr & = \frac{1}{{27}} \cr} $$ Hence, A alone can finish the work in 27 days.

Solution: Applying formula Let work done by 'n' men and women $$\frac{{{{\text{p}}_{{\text{men}}}} \times {{\text{p}}_{{\text{hours}}}} \times {{\text{p}}_{{\text{days}}}}}}{{{{\text{p}}_{{\text{units}}}}}} = $$     $$\frac{{{{\text{n}}_{{\text{men}}}} \times {{\text{n}}_{{\text{hours}}}} \times {{\text{n}}_{{\text{days}}}}}}{{{{\text{W}}_{{\text{units}}}}}}$$ $$\eqalign{ & \Leftrightarrow {{\text{p}}^{\text{2}}} = \frac{{{{\text{n}}^{\text{3}}}}}{{\text{w}}} \cr & \Leftrightarrow {\text{W}} = \frac{{{{\text{n}}^{\text{3}}}}}{{{{\text{p}}^{\text{2}}}}} \cr} $$

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: $$\eqalign{ & {\text{Let}}\,{\text{A's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = x\,{\text{and}} \cr & {\text{B's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = y \cr & {\text{Then}},\,x + y = \frac{1}{{30}} \cr & 16x + 44y = 1 \cr & {\text{Solving}}\,{\text{these}}\,{\text{two}}\,{\text{equations,}}\,{\text{we}}\,{\text{get}} \cr & x = \frac{1}{{60}}\,{\text{and}}\,y = \frac{1}{{60}} \cr & \therefore {\text{B's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{60}} \cr & {\text{Hence,}}\,{\text{B}}\,{\text{alone}}\,{\text{shall}}\,{\text{finish}}\,{\text{the}} \cr & {\text{whole}}\,{\text{work}}\,{\text{in}}\,{\text{60}}\,{\text{days}} \cr} $$

Solution: \[40\% \to \frac{{ - 2}}{5}\begin{array}{*{20}{c}} {} \\ { \to \,{\text{Men}}} \end{array}\] $$\eqalign{ & 9 \times 34 \times 5 = 3 \times 51 \times x \cr & x = 10{\text{ hours}} \cr} $$

Solution: Acreage reaped by 5 men and 3 women in 1 day $$\eqalign{ & = \frac{{18}}{4} \cr & = \frac{9}{2} \cr} $$ Acreage reaped by 3 men and 2 women in 1 day $$\eqalign{ & = \frac{{22}}{8} \cr & = \frac{{11}}{4} \cr} $$ Suppose 1 man can reap x acres in 1 day and 1 women can reap y acres in 1 day $$\eqalign{ & \therefore 5x + 3y = \frac{9}{2} \cr & \Rightarrow 10x + 6y = 9\,.....{\text{(i)}} \cr & 3x + 2y = \frac{{11}}{4} \cr & \Rightarrow 9x + 6y = \frac{{33}}{4}\,.....{\text{(ii)}} \cr & {\text{Subtracting (ii) from (i),}} \cr & {\text{We get}}:x = 9 - \frac{{33}}{4} = \frac{3}{4} \cr & {\text{Putting x}} = \frac{3}{4}{\text{ in (i), we get}} \cr & \Rightarrow 6y = 9 - \frac{{15}}{2} \cr & \Rightarrow 6y = \frac{3}{2} \cr & \Rightarrow y = \frac{1}{4} \cr} $$ Acreage reaped by 21 women in 6 days $$\eqalign{ & = \left( {\frac{1}{4} \times 21 \times 6} \right) \cr & = \frac{{63}}{2} \cr} $$ Remaining acreage to be reaped $$\eqalign{ & = \left( {54 - \frac{{63}}{2}} \right) \cr & = \frac{{45}}{2} \cr} $$ Acreage reaped by 1 men in 6 days $$\eqalign{ & = \left( {\frac{3}{4} \times 6} \right) \cr & = \frac{9}{2} \cr} $$ In 6 days, $$\frac{9}{2}$$ acre is reaped by 1 man ∴ In 6 days, $$\frac{{45}}{2}$$ acre is reaped by $$\eqalign{ & = \left( {\frac{2}{9} \times \frac{{45}}{2}} \right){\text{men}} \cr & = 5{\text{ men}} \cr} $$

Solution: $$\eqalign{ & {\text{According to the question,}} \cr & \Rightarrow 2{\text{A}} = {\text{3B}} \cr & \Rightarrow \frac{{\text{A}}}{{\text{B}}} = \frac{3}{2} \cr & \Rightarrow {\text{Then efficiency ratio }} \cr & {\text{A}}:{\text{B}} = 3:2 \cr} $$ ⇒ We know that time is inversely proportional to efficiency ⇒ Then time taken by them in ratio $$\eqalign{ & {\text{A}}:{\text{B}} = \mathop {\mathop {{\text{ }}2}\limits_{4 \times \downarrow } {\text{ }}}\limits_{8{\text{days}}} :\mathop {\mathop 3\limits_{{\text{ }} \downarrow \times 4} }\limits_{12{\text{days}}} \cr & \because {\text{A can do the work in 8 days}} \cr & \Rightarrow {\text{i}}{\text{.e, 2 units}} \to {\text{8}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{1 unit}} \to {\text{4}} \cr & \Rightarrow {\text{Time taken by B}} \to {\text{3 units}} \cr & = 3 \times 4 \cr & = 12{\text{ days}} \cr} $$