Practice MCQ Questions and Answer on Time and Work

Solution: Mango Packing rate of Ram Lal = $$\frac{{70}}{7}$$ = 10 per hour Guavas Packing rate of Ram Lal = $$\frac{{56}}{7}$$ = 8 per hour Mango Packing rate of Ram Lal's Wife = $$\frac{{30}}{6}$$ = 5 per hour Guavas Packing rate of Ram Lal's Wife = $$\frac{{24}}{6}$$ = 4 per hour On comparing, 8 Guavas = 10 Mangoes 1 Guavas = $$\frac{{10}}{8}$$ Mangoes 2400 Guavas = $$\frac{{10 \times 2400}}{8}$$   = 3000 Mangoes This means that they need to pack (6300 + 3000) Mangoes Combined Mango Packing rate of Both = 10 + 5 = 15 per hour We will take 2 days = 1 time unit. Working 10 hours a day they will pack 150 Mangoes in 1 time unit Total time taken to pack 6300 Mangoes = $$\frac{{6300}}{{150}}$$  = 42 time unit So, total time taken = 2 × 42 = 84 days

Solution: $$\eqalign{ & {\text{According to the question,}} \cr & {\text{X}} \to {\text{24 days}} \cr & \Rightarrow {\text{Work done X in 4 days alone}} \cr & = 4 \times 1 \cr & = 4{\text{ units}} \cr & \Rightarrow {\text{Remaining work}} \cr & = 24 - 4 = 20{\text{ units}} \cr & \Rightarrow {\text{20 units done by both together in}} \cr & = \left( {16 - 4{\text{ days}}} \right) \cr & = 12{\text{ days}} \cr & {\text{Then efficiencies of }}\left( {{\text{X}} + {\text{Y}}} \right) \cr & = \frac{{{\text{Work}}}}{{{\text{Days}}}} \cr & = \frac{{20}}{{12}} \cr & = \frac{5}{3} \cr & = 1 + \frac{2}{3} \cr & \Rightarrow {\text{Efficiency of Y}} = \frac{2}{3} \cr} $$ Time taken by Y alone to complete the total work $$\eqalign{ & = \frac{{24}}{{\frac{2}{3}}} \cr & = 36{\text{ days}} \cr} $$ Alternate : $$\eqalign{ & {\text{X}} \times {\text{20}} = \left( {{\text{X}} + {\text{Y}}} \right) \times {\text{12}} \cr & \frac{{\text{X}}}{{{\text{X + Y}}}} \cr & = \frac{{12}}{{15}} \cr & = \frac{{3 \to {\text{Efficiency of X}}}}{{5 \to {\text{Efficiency of }}\left( {{\text{X}} + {\text{Y}}} \right)}} \cr & {\text{Efficiency of Y}} \cr & = 5 - 3 \cr & = 2{\text{units/day}} \cr & {\text{Total work}} \cr & = 24 \times 3 \cr & = 72{\text{units}} \cr & {\text{Total time taken by Y}} \cr & = \frac{{72}}{2} \cr & = {\text{36 days}} \cr} $$

Solution: $$\eqalign{ & {\text{From, }} \cr & \frac{{{{\text{m}}_1} \times {{\text{d}}_1} \times {{\text{t}}_1}}}{{{{\text{w}}_1}}} = \frac{{{{\text{m}}_2} \times {{\text{d}}_2} \times {{\text{t}}_2}}}{{{{\text{w}}_2}}} \cr & {\text{Let extra workers be x}} \cr & \Rightarrow \frac{{20 \times 12}}{{\frac{5}{8}}} = \frac{{\left( {20 + x} \right) \times 4}}{{\frac{3}{8}}} \cr & \Rightarrow 4 \times 12 = \frac{{\left( {20 + x} \right) \times 4}}{3} \cr & \Rightarrow 36 = 20 + x \cr & \Rightarrow x = 16 \cr & \Rightarrow {\text{Extra workers }} = {\text{16}} \cr} $$

Solution: $$\eqalign{ & \left( {{\text{A + B + C}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{6} \cr & \left( {{\text{A + B}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{8} \cr & \left( {{\text{B + C}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{12}} \cr & \therefore \left( {{\text{A + C}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = \left( {2 \times \frac{1}{6}} \right) - \left( {\frac{1}{8} + \frac{1}{{12}}} \right) \cr & = {\frac{1}{3} - \frac{5}{{24}}} \cr & = \frac{3}{{24}} \cr & = \frac{1}{8} \cr }$$ So, A and C together will do the work in 8 days

Solution: $$\frac{{{\text{5 men}} + {\text{2 women}}}}{{4{\text{work}}}}$$     = $$\left( {1{\text{ men}} + {\text{1 women}}} \right)$$ $$5{\text{ men}} + {\text{2 women}}$$     = $${\text{4 men}} + {\text{4 women}}$$ $$\eqalign{ & {\text{1 men}} = {\text{2 women}} \cr & \frac{{{\text{Men}}}}{{{\text{Women}}}} = \frac{2}{1} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{M}}:{\text{W}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2:1 \cr} $$

Solution: $$\eqalign{ & \frac{1}{3}{\text{work in 5 days}} \cr & {\text{then a complete work in}} \cr & = 5 \times 3 = 15{\text{ days}} \cr & {\text{B}} \to \frac{2}{5}{\text{work in 10 days}} \cr & {\text{then B completes work in}} \cr & = 10 \times \frac{5}{2} = {\text{25 days}} \cr} $$ L.C.M. of Total Work =75 One day work of A = $$\frac{{75}}{{15}}$$ = 5 unit/day One day work of B = $$\frac{{75}}{{25}}$$ = 3 unit/day $$\eqalign{ & \left( {{\text{A}} + {\text{B}}} \right){\text{ can do work}} \cr & = \frac{{75}}{8} = 9\frac{3}{8}{\text{ days}} \cr} $$

Solution: A : C = 2 : 1 A = 4 unit C = 2 unit B = 1 unit B = $$\frac{{90}}{1}$$ = 90 days

Solution: Ratio of 1 day's work of man and boy = 3 : 1 $$\eqalign{ & {\text{Total wages of the boy}} \cr & = {\text{Rs}}{\text{.}}\left( {800 \times \frac{1}{4}} \right) \cr & = {\text{Rs}}{\text{. 200}} \cr & \therefore {\text{Daily wages of the boy}} \cr & = {\text{Rs}}{\text{.}}\left( {\frac{{200}}{5}} \right) \cr & = {\text{Rs}}{\text{. 40}} \cr} $$

Solution: Let D days required to complete the remaining work $${40_{{\text{men}}}} \times {18_{{\text{days}}}}$$    = $$\left( {{{40}_{{\text{men}}}} \times {8_{{\text{days}}}}} \right)$$    + $$\left( {{{50}_{{\text{men}}}} \times {\text{D}}} \right)$$ $$\eqalign{ & \Rightarrow 720 - 320 = 50{\text{D}} \cr & \Rightarrow {\text{D}} = 8 \cr} $$

Solution: Working efficiency of both typist together, = $$\frac{{100}}{6}$$ = 16.66% per minute Now, let work efficiency of first typist be x and then second typist will be (16.66 - x) First typist typed alone for 4 minutes and second typed alone for 6 minutes and they left with $$\frac{1}{5}$$ (i.e 20%) of job, means they have completed 80% job Now, First Typist typed in 4 minute + Second typed in 6 minutes = 80% 4 × x + 6 × (16.66 - x) = 80% 4x + 100% - 6x = 80% x = 10% First Typist typed 10% per minutes. Then second typed (16.66 - 10) = 6.66% per minute Then, Second typist complete the whole job in $$\frac{{100}}{{6.66}}$$ = 15.01 = 15 minutes.