Time And Work - Aptitude MCQ Questions and Solutions with Explanations

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

$$\eqalign{ & {\text{Dinesh's one hour work}} \cr & = \frac{{32}}{6} \cr & = \frac{{16}}{3}{\text{ pages/hour}} \cr & {\text{Rakesh's one hour work}} \cr & = \frac{{40}}{5} \cr & = 8{\text{ pages/hour}} \cr & {\text{Dinesh's and Rakesh's one hour work}} \cr & = \frac{{16}}{3} + 8 \cr & = \frac{{40}}{3}{\text{ pages/hour}} \cr & {\text{They will finish the work together}} \cr & \frac{{{\text{Total work}}}}{{{\text{Efficiency}}}} \cr & = \frac{{110}}{{\frac{{40}}{3}}} \cr & = 8\frac{1}{4} \cr & = {\text{8 hours, 15 minutes}} \cr} $$

9 hours 36 minutes
$$\eqalign{ & = 9 + \frac{{36}}{{60}}\,{\text{hours}} \cr & = 9\frac{3}{5} = \frac{{48}}{5}{\text{hours}} \cr} $$
(A + B)’s 1 hour’s work = $$\frac{5}{{48}}$$
C’s 1 hour’s work = $$\frac{1}{{48}}$$
(A + B + C)’s 1 hour’s work = $$\frac{5}{{48}} + \frac{1}{{48}}$$   = $$\frac{1}{8}$$ . . . . . . .(i)
A’s 1 hour’s work = (B + C)’s 1 hour’s work . . . . . . . . (ii)
From equation (i) and (ii),
2 × (A’s 1 hour’s work) = $$\frac{1}{8}$$
A’s 1 hour’s work = $$\frac{1}{{16}}$$
∴ B’s 1 hour’s work
$$\eqalign{ & = \frac{5}{{48}} - \frac{1}{{16}} \cr & = \frac{{5 - 3}}{{48}} \cr & = \frac{1}{{24}} \cr} $$
∴ B alone will finish the work in 24 hours.

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
$$\eqalign{ & = \left( {\frac{2}{9} \times \frac{{45}}{2}} \right){\text{men}} \cr & = 5{\text{ men}} \cr} $$

$$\eqalign{ & {\text{Let 1 men's 1 day's work}} = x \cr & {\text{and 1 boy's 1 day's work}} = y \cr & {\text{Then, }} \cr & \Rightarrow {\text{6}}\left( {25x + 10y} \right) = 5\left( {21x + 30y} \right) \cr & \Rightarrow 150x + 60y = 105x + 150y \cr & \Rightarrow 45x = 90y \cr & \Rightarrow x = 2y \cr & {\text{Let,}} \cr & {\text{The required number of boys be }}z \cr & {\text{Then,}} \cr & \Rightarrow {\text{4}}\left( {40x + zy} \right) = 6\left( {25x + 10y} \right) \cr & \Rightarrow 4\left( {80y + zy} \right) = 6\left( {50y + 10y} \right) \cr & \Rightarrow 80 + z = \frac{{6 \times 60}}{4} = 90 \cr & \Rightarrow z = 10 \cr} $$

Work done by 40 men in 5 days = $$\frac{1}{3}$$
(As if whole work is completed in 15 days then in 5 days $${{{\frac{1}{3}}^{{\text{rd}}}}}$$ of the work will be finished)
$$\eqalign{ & {\text{Remaining work}} = 1 - \frac{1}{3} = \frac{2}{3} \cr & \because 40{\text{ men do 1 work in 15 days}}{\text{.}} \cr & {\text{60 men can do }}\frac{2}{3}{\text{work in }}x{\text{ day}} \cr & \frac{{{{\text{M}}_1}{{\text{D}}_1}}}{{{{\text{W}}_1}}}{\text{ = }}\frac{{{{\text{M}}_2}{{\text{D}}_2}}}{{{{\text{W}}_2}}} \cr & {{\text{M}}_1} = 40{\text{ , }}{{\text{M}}_2} = 60 \cr & {{\text{D}}_1} = 15{\text{ , }}{{\text{D}}_2} = x \cr & {{\text{W}}_1} = 1{\text{ ,}}{{\text{W}}_2} = \frac{2}{3} \cr & \Rightarrow \frac{{40 \times 15}}{1} = \frac{{60 \times x}}{2} \cr & \Rightarrow \frac{2}{3}\left( {40 \times 15} \right) = 60x \cr & \Rightarrow 2 \times 40 \times 5 = 60x \cr & \Rightarrow x = \frac{{20}}{3} \cr & \Rightarrow x = 6\frac{2}{3}{\text{ days}} \cr} $$

L.C.M. of Total Work = 45
One day work of A = $$\frac{{45}}{{9}}$$ = 5 unit/day
One day work of B = $$\frac{{45}}{{15}}$$ = 3 unit/day
$$\eqalign{ & \left( {{\text{A}} + {\text{B}}} \right){\text{'s 2 days work }} \cr & = 5 + 3 \cr & = 8{\text{ units}} \cr & {\text{They will do in }} \cr & = \frac{{40}}{8} \times 2 \cr & = \left( {5 \times 2} \right) \cr & = 10{\text{ days}} \cr & \therefore {\text{Work left}} \cr & = 45 - 40 \cr & = {\text{5 units}} \cr & {\text{Now,}} \cr & {\text{A's turn and he will complete in}} \cr & = \frac{5}{5} \cr & = 1{\text{ days}} \cr & {\text{Then total work completed in}} \cr & = 10 + 1 \cr & = 11{\text{ days}} \cr} $$

$$\eqalign{ & {\text{Let the required time}} = {\text{T}} \cr & \Rightarrow \frac{{{{\text{m}}_{\text{1}}} \times {{\text{d}}_{\text{1}}} \times {{\text{t}}_{\text{1}}}}}{{{{\text{w}}_{\text{1}}}}}{\text{ = }}\frac{{{{\text{m}}_{\text{2}}} \times {{\text{d}}_{\text{2}}} \times {{\text{t}}_{\text{2}}}}}{{{{\text{w}}_{\text{2}}}}} \cr & \Rightarrow \frac{{12 \times 12}}{{12}} = \frac{{4 \times {\text{time}}}}{4} \cr & \Rightarrow {\text{Time}} = 12\operatorname{minutes} \cr} $$

$$\eqalign{ & {\text{Let,}} \cr & {\text{a}} = {\text{8h}} \cr & {\text{b}} = {\text{4}}\frac{1}{2}{\text{h}} = \frac{9}{2}{\text{h}} \cr} $$
Time required to finish the work together
$$\eqalign{ & = \sqrt {{\text{ab}}} \cr & = \sqrt {8 \times \frac{9}{2}} \cr & = 6{\text{ h}} \cr} $$

$$\eqalign{ & \Rightarrow {{\text{R}}_{\text{x}}} \cr & = \frac{{80}}{{20}} \cr & {\text{ = 4 pages/hour}} \cr & \Rightarrow {{\text{R}}_{{\text{x + y}}}} \cr & = \frac{{135}}{{27}} \cr & {\text{ = 5 pages/hour}} \cr & \Rightarrow {{\text{R}}_{\text{y}}} \cr & = {{\text{R}}_{{\text{x + y}}}} - {{\text{R}}_{\text{x}}} \cr & = 5 - 4 \cr & = 1{\text{ pages/hour}} \cr & \therefore {\text{y can copy 20 pages in}} \cr & = \frac{{{\text{20p}}}}{{{\text{1p/h}}}} \cr & = {\text{ }}20{\text{ hours}} \cr} $$