Practice MCQ Questions and Answer on Time and Work

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: $$\eqalign{ & {\text{Given,}} \cr & {{\text{M}}_1} = 16{\text{ , }}{{\text{M}}_2} = ? \cr & {{\text{D}}_1} = 49{\text{ , }}{{\text{D}}_2} = 24 \cr & {{\text{W}}_1} = 1{\text{ ,}}{{\text{W}}_2} = ? \cr & {\text{According to the question,}} \cr & \frac{{{{\text{M}}_1}{{\text{D}}_1}}}{{{{\text{W}}_1}}}{\text{ = }}\frac{{{{\text{M}}_2}{{\text{D}}_2}}}{{{{\text{W}}_2}}} \cr & \Rightarrow \frac{{16 \times 49}}{1} = \frac{{14 \times 8}}{{{{\text{W}}_2}}} \cr & \Rightarrow {{\text{W}}_2} = \frac{{14 \times 8}}{{16 \times 49}} \cr & \Rightarrow {{\text{W}}_2} = \frac{1}{7} \cr & {\text{Remaining work}} \cr & = \left( {1 - \frac{1}{7}} \right) \cr & = \frac{6}{7} \cr & {\text{Again,}}\frac{{{{\text{M}}_1}{{\text{D}}_1}}}{{{{\text{W}}_1}}}{\text{ = }}\frac{{{{\text{M}}_2}{{\text{D}}_2}}}{{{{\text{W}}_2}}} \cr & \Rightarrow \frac{{16 \times 49}}{1} = \frac{{{{\text{M}}_2} \times 24}}{{\frac{6}{7}}} \cr & \Rightarrow 16 \times 49 = \frac{{{{\text{M}}_2} \times 24 \times 7}}{6} \cr & \Rightarrow 16 \times 49 = {{\text{M}}_2} \times 4 \times 7 \cr & \Rightarrow {{\text{M}}_2} = \frac{{16 \times 49}}{{4 \times 7}} \cr & \Rightarrow {{\text{M}}_2} = 28 \cr & \therefore {\text{Number of additional men}} \cr & = \left( {28 - 14} \right) \cr & = 14 \cr} $$

Solution: Ratio of time, taken by A and B = 160 : 100 = 8 : 5 If A takes 8 days, B takes 5 days If A takes 12 days, B takes $${\text{ = }}\frac{5}{8} \times 12 = \frac{{15}}{2} = 7\frac{1}{2}$$     days

Solution: Length of cloth produced by A and B in 10 hours $$ = 300000{\text{ metres}}$$ Length of cloth produced by B in 10 hours $$\eqalign{ & = \left( {\frac{{300000}}{{15}} \times 10} \right) \cr & = 200000{\text{ metres}} \cr} $$ ∴ Length of cloth produced by A in 10 hours $$\eqalign{ & = \left( {300000 - 200000} \right) \cr & = 100000{\text{ metres}} \cr} $$

Solution: Since we know efficiency and time are inversely proportional to each other. $$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{P}}:{\text{Q}} \cr & {\text{Efficiency }}3:4 \cr & {\text{Time }}\,\,\,\,\,\,\,\,\,\,{\text{ }}4:3 \cr} $$

Solution: Let to work Total Work = 36 One day work of A = $$\frac{{36}}{{36}}$$ = 1 unit/day One day work of B = $$\frac{{36}}{{18}}$$ = 2 unit/day One day work of C = $$\frac{{36}}{{12}}$$ = 3 unit/day In 3 days cycle total work done is A, A + B, A + C = 1 + (1 + 2) + (1 + 3) = 8 unit/Cycle ∴ 32 units of the work completed in 4 cycle and reminder 4 units of works in next two days. 1 cycle = 3 days ∴ 4 cycle = 12 days And remaining 4 unit work done in next two days. In days 13, A will work 1 unit and in day 14, A and B will work 3 units. Total number the day required to complete the work in the given condition is 14 days

Solution: Let the work be completed in x days. C work for x days then A works for (x - 8) days and B works for (x - 12) days. According to the question, $$\eqalign{ & {\frac{{ {x - 8} }}{{36}} + \frac{{ {x - 12} }}{{54}} + \frac{x}{{72}}} = 1 \cr & {\frac{{ {6x - 48 + 4x - 48 + 3x} }}{{216}}} = 1 \cr & 13x - 96 = 216 \cr & 13x = 216 + 96 = 312 \cr & x = \frac{{312}}{{13}} \cr & x = 24\,{\text{days}} \cr} $$

Solution: Let time taken by B and C = x days ∴ Time taken by A = 3x days ∴ Part of work done by A, B and C in 1 day $$\eqalign{ & = \frac{1}{{\text{x}}} + \frac{1}{{3{\text{x}}}} = \frac{{3 + 1}}{{3{\text{x}}}} = \frac{4}{{3{\text{x}}}} \cr & \therefore \frac{4}{{3{\text{x}}}} = \frac{1}{{24}} \cr & \Rightarrow 3{\text{x}} = 4 \times 24 \cr & \Rightarrow {\text{x}} = \frac{{4 \times 24}}{3} = 32\,{\text{days}} \cr} $$ ∴ Time taken by A = 32 × 3 = 96 days. Alternet : Work done by all of them together in $$\frac{1}{{24}}$$ Efficiency of A : Efficiency of (B + C) = 1 : 3 Work done by A in 1 day = $$\frac{1}{{24}}$$ × $$\frac{1}{{4}}$$ = $$\frac{1}{{96}}$$ i.e., A alone can finish the job in 96 days. Alternet : Suppose A can complete the job in 3x days and (B + C) can complete the job in x days. $$\eqalign{ & \frac{{3{\text{x}} \times {\text{x}}}}{{3{\text{x}} + {\text{x}}}} = 24 \cr & \Rightarrow \frac{{3{\text{x}}}}{4} = 24 \cr & \Rightarrow {\text{x}} = 32 \cr & 3{\text{x}} = 96\,{\text{days.}} \cr} $$

Solution: $$\frac{7}{{10}}$$ part of work has been completed by A in 15 days. Then, Rest work = 1 - $$\frac{7}{{10}}$$ = $$\frac{3}{{10}}$$ part Given, That $$\frac{3}{{10}}$$ part of the work is completed by A and B together in 4 days. Means, (A + B) completed the $$\frac{3}{{10}}$$ of work in 4 days So, (A + B)'s 1 day's work = $$\frac{3}{{10 \times 4}}$$  = $$\frac{3}{{40}}$$ Hence, (A + B) can complete the work in $$\frac{{40}}{3}$$ = $$13\frac{1}{3}$$ days

Solution: According to the question, Let the remaining work is complete in D days $${\text{4}}{{\text{5}}_{{\text{men}}}} \times {\text{1}}{{\text{6}}_{{\text{days}}}}$$    = $$\left( {{\text{4}}{{\text{5}}_{{\text{men}}}} \times {{\text{4}}_{{\text{days}}}}} \right)$$    + $${\left( {{\text{45}} + {\text{36}}} \right)_{{\text{men}}}}$$   $$ \times {\text{D}}$$ $$\eqalign{ & \Rightarrow \frac{{540}}{{81}} = {\text{D}} \cr & \Rightarrow {\text{D}} = 6\frac{2}{3}{\text{ days}} \cr} $$