Practice MCQ Questions and Answer on Time and Work

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.

Solution: Let B can alone finish the work = x days So, A can alone finish the work = (x - 10) days Now, one day work of A = $$\frac{1}{{{\text{x}} - 10}}$$ and one day work of B = $$\frac{1}{{\text{x}}}$$ Now, given (A + B) can finish the work = 12 day So, one day work of (A + B) = $$\frac{1}{{12}}$$ $$\eqalign{ & \Rightarrow \frac{1}{{{\text{x}} - 10}} + \frac{1}{{\text{x}}} = \frac{1}{{12}} \cr & \Rightarrow \frac{{x + x - 10}}{{x \times \left( {x - 10} \right)}} = \frac{1}{{12}} \cr & \Rightarrow \frac{{2x - 10}}{{{x^2} - 10x}} = \frac{1}{{12}} \cr & \Rightarrow 12\left( {2x - 10} \right) = {x^2} - 10x \cr & \Rightarrow 24x - 120 = {x^2} - 10x \cr & \Rightarrow {x^2} - 10x - 24x + 120 = 0 \cr & \Rightarrow {x^2} - 34x + 120 = 0 \cr & \Rightarrow {x^2} - 30x - 4x + 120 = 0 \cr & \Rightarrow x\left( {x - 30} \right) - 4\left( {x - 30} \right) = 0 \cr & \Rightarrow \left( {x - 30} \right) \times \left( {x - 4} \right) = 0 \cr & \Rightarrow x = 30,\,4 \cr} $$ if x = 4, then A alone can finish the work = 4 - 10 = -6, which is not possible. So, x = 30 Hence, B can alone finish the work = 30 days

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: $$\eqalign{ & {\text{Amit's 1 day's work }} \cr & = \left( {\frac{1}{4} - \frac{1}{6}} \right) \cr & = \frac{1}{{12}} \cr} $$ ∴ Amit alone can plough the field in 12 days.

Solution: $$\eqalign{ & \left( {{\text{A}} + {\text{B}}} \right){\text{'s 1 day's work}} = \frac{1}{{12}} \cr & \left( {{\text{B}} + {\text{C}}} \right){\text{'s 1 day's work}} = \frac{1}{8} \cr & \left( {{\text{A}} + {\text{C}}} \right){\text{'s 1 day's work}} = \frac{1}{{16}} \cr} $$ Adding, we get 2(A + B + C)'s 1 day's work $$\eqalign{ & = \left( {\frac{1}{{12}} + \frac{1}{8} + \frac{1}{{16}}} \right) \cr & = \frac{{13}}{{96}} \cr} $$ So, A, B and C together can complete the work in $$\eqalign{ & = \frac{{96}}{{13}} \cr & = 7\frac{5}{{13}}{\text{days}} \cr} $$

Solution: According to question, Ganga begins at 9 am and she does 3 units/hours Saraswati begins at 10 am and she does 2 units/hours So by 11 am they complete 5 units Time $$ = \frac{{{\text{T}}{\text{.W}}.}}{{3 + 2}} = \frac{{24}}{5}$$ (4 cycle of 2 hrs each + 4 units left) And now ganga will complete 3 unit out of 4 units in 1 hr Now, rest 1 unit work done by = $$\frac{1}{2}$$ hr Total time = 8 + 1 + $$\frac{1}{2}$$ = 9$$\frac{1}{2}$$ hr Hence, Work finished at = 9 am + 9$$\frac{1}{2}$$ hr = 6:30 PM Alternate Solution: Work done by Ganga in 1 hour = $$\frac{1}{8}$$ Work done by Saraswati in 1 hour = $$\frac{1}{12}$$ They are working alternatively with Ganga beginning the job. Work done in every two hours = $$\frac{1}{8}$$ + $$\frac{1}{12}$$ = $$\frac{5}{24}$$ Work done in 4 × 2 = 8 hours = $$\frac{5\times4}{24}$$ = $$\frac{5}{6}$$ Remaining work = 1 - $$\frac{5}{6}$$ = $$\frac{1}{6}$$ In 9th hour, Ganga starts the work and does $$\frac{1}{8}$$ of the work Work remaining = $$\frac{1}{6}$$ - $$\frac{1}{8}$$ = $$\frac{1}{24}$$ In 10th hour, Saraswati starts the work Time needed to finish the remaining work $$\eqalign{ & = \frac{{\frac{1}{{24}}}}{{\frac{1}{{12}}}} \cr & = \frac{1}{{24}} \times 12 \cr} $$   $$=$$ 0.5 hours   $$=$$ 30 minutes i.e., work will be completed in 9 hour 30 minutes, after 9 AM i.e., at 6:30 PM

Solution: $$\eqalign{ & 6W \times 3 = 9G \times 2 \cr & \frac{W}{G} = \frac{1}{1} \cr & {\text{Total work}} = 18 \cr & 18 = \left( {8W + 6G} \right) \times d \cr & 18 = 14 \times d \cr & d = \frac{9}{7}{\text{ days}} \cr} $$

Solution: Let capacity of the tank = x gallons; Part of the tank filled in 1 minute $$\eqalign{ & = {\frac{x}{{24}}} + {\frac{x}{{40}}} - 30 \cr & {\text{Or}}, {\frac{x}{{24}}} + {\frac{x}{{40}}} - 30 = \frac{x}{{60}} \cr & {\text{Or}}, \frac{x}{{24}} + \frac{x}{{40}} - \frac{x}{{60}} = 30 \cr & {\text{Or}}, {\frac{{ {10x + 6x - 4x} }}{{240}}} = 30 \cr & {\text{Or}},\,12x = 30 \times 240 \cr & {\text{Or}},\,x = 600\,{\text{gallons}} \cr} $$

Solution: $$\eqalign{ & \left( {{\text{A}} + {\text{B}}} \right){\text{'s 1 day's work}} = \frac{1}{2} \cr & \left( {{\text{B}} + {\text{C}}} \right){\text{'s 1 day's work}} = \frac{1}{4} \cr & \left( {{\text{A}} + {\text{C}}} \right){\text{'s 1 day's work}} = \frac{5}{{12}} \cr & {\text{Adding, we get: }} \cr & {\text{2}}\left( {{\text{A}} + {\text{B}} + {\text{C}}} \right){\text{'s 1 day's work}} \cr & = \left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{{12}}} \right) \cr & = \frac{{14}}{{12}} \cr & = \frac{7}{6} \cr & \Rightarrow \left( {{\text{A}} + {\text{B}} + {\text{C}}} \right){\text{'s 1 day's work}} = \frac{7}{{12}} \cr & {\text{So, A's 1 day's work}} \cr & = \left( {\frac{7}{{12}} - \frac{1}{4}} \right) \cr & = \frac{4}{{12}} \cr & = \frac{1}{3} \cr & \therefore {\text{A alone can do the work in 3 days}}{\text{.}} \cr} $$

Solution: Let X litres be the per day filling and L litres be the capacity of the reservoir, then 90X + L = 40000 × 90 ----------- (1) 60X + L = 32000 × 60 ----------- (2) Solving the equation, X = 56000 litres Thus, 56000 litres per day can be used without the failure of supply.