Practice MCQ Questions and Answer on Time and Work

Solution: L.C.M. of Total Work = 120 One day work of A + B + C = $$\frac{{120}}{{8}}$$ = 15 unit/day One day work of A + B = $$\frac{{120}}{{\frac{{40}}{3}}}$$ = 9 unit/day One day work of A + C = $$\frac{{120}}{{12}}$$ = 10 unit/day $$\eqalign{ & {\text{C's efficiency}} = 15 - 9 = 6 \cr & {\text{B's efficiency}} = 15 - 10 = 5 \cr & {\text{A's efficiency}} = 15 - \left( {6 + 5} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 15 - 11 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4 \cr & {\text{So, A}}:{\text{B}}:{\text{C}} = 4:5:6 \cr} $$

Solution: $$\eqalign{ & {\text{According to the question,}} \cr & {\text{15 men}} = {\text{20 days}} \cr & {\text{300 men}} = 1{\text{ days}}.....{\text{(i)}} \cr & {\text{24 women}} = {\text{20 days}} \cr & {\text{480 men}} = 1{\text{ days}}......{\text{(ii)}} \cr & {\text{Compare equation (i) and (ii)}} \cr & {\text{300 men}} = 480{\text{ women}} \cr & {\text{5 men}} = 8{\text{ women}}.....{\text{(iii)}} \cr & {\text{10 men}} + 8{\text{ women}} = ? \cr & {\text{10 men}} + {\text{5 men}} = ? \cr & 15\,{\text{men}} = ? \cr} $$ $${\text{15 men}} \times {\text{20 days}}$$     = $${\text{15 men}}$$  $$ \times $$ $$x{\text{ days}}$$ $$x$$ = 20 days Alternate $$\eqalign{ & {\text{15m}} \times {\text{20 days}} = 24{\text{w}} \times 20{\text{ days}} \cr & \frac{{\text{m}}}{{\text{w}}} = \frac{8}{5} \cr} $$ So, 1 man work 8 units work in one day and 1 woman work 5 units work in one day Total work = 15 × 8 × 20 Hence, (10 men + 8 women) work whole in D days $$\eqalign{ & \left( {{\text{10m}} + {\text{8w}}} \right) \times {\text{D}} = 15 \times 8 \times 20 \cr & \left( {{\text{10}} \times {\text{8}} + {\text{8}} \times {\text{5}}} \right) \times {\text{D}} = 15 \times 8 \times 20 \cr & \left( {{\text{80}} + 40} \right) \times {\text{D}} = 15 \times 8 \times 20 \cr & {\text{D}} = 20{\text{ days}} \cr} $$

Solution: A, B and C together can fill 100% empty tank in 10 minutes Work rate of (A + B + C) = $$\frac{{100}}{{10}}$$ = 10% per minute A alone can fill the tank in 30 minutes Work rate of A = $$\frac{{100}}{{30}}$$ = 3.33% per minute B alone can fill the tank in 40 minutes Work rate of B = $$\frac{{100}}{{40}}$$ = 2.5% Work rate of (A + B) = 3.33 + 2.5 = 5.83% per minute Work rate of C, = Work rate of (A + B + C) - (A + B) = 10 - 5.83 = 4.17% per minute So, C takes = $$\frac{{100}}{{4.17}}$$ ≈ 24 minutes to fill the tank

Solution: Ratio of times taken by A and B $$\eqalign{ & = 100:80 \cr & = 5:4 \cr} $$ Suppose B takes x hours to do the work $$\eqalign{ & \Rightarrow 5:4::\frac{{15}}{2}:x \cr & \Rightarrow x = \left( {\frac{{4 \times 15}}{{2 \times 5}}} \right) \cr & \Rightarrow x = 6{\text{ hours}}{\text{}} \cr} $$

Solution: ⇒ A can do $$\frac{1}{3}$$ of a work in 30 day ⇒ A completed work in 90 days ⇒ B can do $$\frac{2}{5}$$ of the same work in 24 days ⇒ B can completed work in 60 days ⇒ Let the total work = LCM (90, 60) = 180 unit ⇒ Efficiency of A = 3 unit/day ⇒ Efficiency of B = 2 unit/day ⇒ They both worked for 20 days, work done in 20 days = 20 × 5 = 100 unit ⇒ Remaining work = 180 - 100 = 80 unit ⇒ Remaining work done by C in 8 days ⇒ Efficiency of C = 10 unit/day ⇒ Efficiency of (A + B + C) = 15 unit/day ⇒ Work completed = $$\frac{{180}}{{15}}$$ = 12 days ∴ If all worked together, the work complete in 12 days.

Solution:   Tapas   :   Mihir Efficiencyunits/day   2 : 1 $$\eqalign{ & {\text{T}} + {\text{M complete in 12 days}} \cr & {\text{Total work }} \cr & = 12 \times \left( {2 + 1} \right) \cr & = 36{\text{units}} \cr} $$ Tapas alone complete the whole work in $$\eqalign{ & = \frac{{36}}{2} \cr & = 18{\text{ days}} \cr} $$

Solution: (Ramesh & Rahman)'s 1 day's work $$ = \frac{1}{{20}} + \frac{1}{{25}} = \frac{{5 + 4}}{{100}} = \frac{9}{{100}}$$ ∴ Their 10 day's work $$ = \frac{{90}}{{100}} = \frac{9}{{10}}$$ ∴ Remaining work $$ = 1 - \frac{9}{{10}} = \frac{1}{{10}}$$ ∵ Suresh does $$\frac{1}{{10}}$$ work in 3 days ∴ Time taken by Suresh in doing 1 work = 3 × 10 = 30 days

Solution: 450 men × 20 = x men × 30 x = 300 men

Solution: According to the question, X → 24 days ⇒ Work done by X in 4 days alone = 4 × 1 = 4 units ⇒ Remaining work = 24 - 4 = 20 units ⇒ 20 units done by both together in (16 - 4 days) = 12 days ⇒ Then efficiencies of (X + Y) $$ = \frac{{{\text{work}}}}{{{\text{days}}}} = \frac{{20}}{{12}} = \frac{5}{3} = 1 + \frac{2}{3}$$ ⇒ Efficiency of Y = $$\frac{2}{3}$$ ⇒ Time taken by Y alone to complete the total work $$ = \frac{{24}}{{\frac{2}{3}}} = 36{\text{ days}}$$ Alternate solution: $$\eqalign{ & X \times 20 = \left( {X + Y} \right) \times 12 \cr & \frac{X}{{X + Y}} = \frac{{12}}{{20}} \cr & = \frac{{3 \to {\text{Efficiency of }}X}}{{5 \to {\text{Efficiency of }}\left( {X + Y} \right)}} \cr} $$ Efficiency of Y = 5 - 3 = 2 units/day Total work = 24 × 3 = 72 units Total time taken by Y = $$\frac{{72}}{2}$$ = 36 days

Solution: $$\eqalign{ & \left( {{\text{A}} + {\text{B}}} \right){\text{'s 20 day's work}}{\text{.}} \cr & = \left( {\frac{1}{{30}} \times 20} \right) \cr & = \frac{2}{3} \cr & {\text{Remaining work }} \cr & = \left( {1 - \frac{2}{3}} \right) \cr & = \frac{1}{3}{\text{ }} \cr} $$ Now, $$\frac{1}{3}$$ work is done by A in 20 days Whole work will be done by A in (20 × 3) = 60 days.