Practice MCQ Questions and Answer on Time and Work

Solution: $$\eqalign{ & {\text{From }}{{\text{M}}_1}{{\text{D}}_1} = {{\text{M}}_2}{{\text{D}}_2} \cr & \Rightarrow \frac{{{{\text{M}}_1}{{\text{D}}_1}}}{{{{\text{M}}_2}{{\text{D}}_2}}} = \frac{5}{6} \cr & \Rightarrow \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = \frac{5}{6} \cr & \Rightarrow \frac{{\left( {x - 1} \right)}}{{\left( {x + 2} \right)}} = \frac{5}{6} \cr & \Rightarrow 6x - 6 = 5x + 10 \cr & \Rightarrow x = 16 \cr} $$

Solution:   Amit     Bhawana     Chandan Eff. →   5x x 2x $$\eqalign{ & {\text{Let total work}} = 1 \cr & {\text{Efficiency of }}\left( {{\text{A}} + {\text{B}} + {\text{C}}} \right) = 1 \cr & {\text{Then,}} \cr & \Leftrightarrow 5x + x + 2x = 1 \cr & \Leftrightarrow x = \frac{1}{8} \cr & {\text{Days taken by Amit}} \cr & = \frac{1}{{\frac{5}{8}}} \Rightarrow \frac{8}{5} \cr & {\text{Days taken by Chandan}} \cr & = \frac{1}{{\frac{2}{8}}} \Rightarrow 4 \cr & {\text{Difference of days}} \cr & = 4 - \frac{8}{5} \Rightarrow \frac{{20 - 8}}{5} \Rightarrow 2\frac{2}{5} \cr} $$

Solution: Total work = 25 × 60 = 1500 x × 25 + (80 - x) × 15 = 1500 25x - 15x + 1200 = 1500 10x = 300 x = 30

Solution: $$\eqalign{ & {\text{A's 1 hour's work}} = \frac{1}{4} \cr & \left( {{\text{B}} + {\text{C}}} \right){\text{'s 1 hour's work}} = \frac{1}{3} \cr & \left( {{\text{A}} + {\text{C}}} \right){\text{'s 1 hour's work}} = \frac{1}{2} \cr & \left( {{\text{A}} + {\text{B}} + {\text{C}}} \right){\text{'s 1 hour's work}} \cr & = \frac{1}{4} + \frac{1}{3} \cr & = \frac{7}{{12}} \cr & \therefore {\text{B's 1 hour's work}} \cr} $$ = (A + B + C)'s 1 hour's work - (A + C)'s 1 hour's work $$\eqalign{ & = \frac{7}{{12}} - \frac{1}{2} \cr & = \frac{1}{{12}} \cr} $$ So, B alone can complete the work in 12 hours.

Solution: Assume work to be done 100%. First case, A + B = $$\frac{{100}}{5}$$ = 20% work done per day -------- (1) Second case, 2A + $$\frac{{\text{B}}}{2}$$ = $$\frac{{100}}{4}$$ = 25% work done per day ------ (2) On solving equation (1) and (2), we get A = 10 days

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: Part of wall built by A in 1 day = $$\frac{1}{8}$$ Part of wall broken by B in 1 day = $$\frac{1}{3}$$ Part of wall built by A in 4 days $$\eqalign{ & = \left( {\frac{1}{8} \times 4} \right) \cr & = \frac{1}{2} \cr} $$ Part of wall broken by B and built by A in 2 days$$\eqalign{ & = 2\left( {\frac{1}{3} - \frac{1}{8}} \right) \cr & = \frac{5}{{12}} \cr} $$ $$\eqalign{ & {\text{Part of wall built in 6 days}} \cr & = \left( {\frac{1}{2} - \frac{5}{{12}}} \right) \cr & = \frac{1}{{12}} \cr & {\text{Remaining part to be built}} \cr & = \left( {1 - \frac{1}{{12}}} \right) \cr & = \frac{{11}}{{12}} \cr} $$ Now, $$\frac{1}{8}$$ part of wall built by A in 1 day $$\eqalign{ & \therefore \frac{{11}}{{12}}{\text{ part of wall built by A in}} \cr & = \left( {8 \times \frac{{11}}{{12}}} \right) \cr & = \frac{{22}}{3} \cr & = 7\frac{1}{3}{\text{ day}} \cr} $$

Solution: $$\eqalign{ & {\text{A's 1 day's work}} = \frac{1}{{18}} \cr & {\text{And}} \cr & {\text{B's 1 day's work}} = \frac{1}{9} \cr & \therefore \left( {{\text{A}} + {\text{B}}} \right)'{\text{s 1 day's work}} \cr & = \left( {\frac{1}{{18}} + \frac{1}{9}} \right) \cr & = \frac{1}{6} \cr} $$

Solution: $$\eqalign{ & \left( {{\text{A}} + {\text{B}}} \right){\text{'s 1 day's work}} = \frac{1}{{25}} \cr & {\text{C's 1 day's work}} = \frac{1}{{35}} \cr & \left( {{\text{A}} + {\text{B}} + {\text{C}}} \right){\text{'s 1 day's work}} \cr & = \left( {\frac{1}{{25}} + \frac{1}{{35}}} \right) \cr & = \frac{{12}}{{175}}.....({\text{i}}) \cr} $$ Also, A's 1 day's work = (B + C)'s 1 day's work.....(ii) $$\eqalign{ & {\text{From (i) and (ii), we get : }} \cr & \Rightarrow {\text{2}} \times \left( {{\text{A's 1 day's work}}} \right) = \frac{{12}}{{175}} \cr & \Rightarrow {\text{A's 1 day's work}} = \frac{6}{{175}} \cr & \therefore {\text{B's 1 day's work}} \cr & = \left( {\frac{1}{{25}} - \frac{6}{{175}}} \right) \cr & = \frac{1}{{175}} \cr} $$

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 $$\eqalign{ & \left( {{\text{10m}} + {\text{8w}}} \right) \times {\text{D}} = 15 \times 8 \times 20 \cr & \left( {{\text{10}} \times {\text{8}} + {\text{8}} \times {\text{5}}} \right) \times {\text{D}} = 15 \times 8 \times 20 \cr & \left( {{\text{80}} + 40} \right) \times {\text{D}} = 15 \times 8 \times 20 \cr & {\text{D}} = 20{\text{ days}} \cr} $$