Time And Work - Aptitude MCQ Questions and Solutions with Explanations

12 men can do a piece of work in 24 days
$$ \Rightarrow {{\text{M}}_1} = 12{\text{ and }}{{\text{D}}_1} = 24$$
8 men can do this work in $${{\text{D}}_2}$$ days
$$\eqalign{ & \Rightarrow {{\text{M}}_2} = 8{\text{ }} \cr & \,\,\,\,\,\,\,\,\,{{\text{M}}_1}{{\text{D}}_1} = {{\text{M}}_{\text{2}}}{{\text{D}}_{\text{2}}} \cr & \Rightarrow 12 \times 24 = 8 \times {{\text{D}}_2} \cr & \Rightarrow {{\text{D}}_2} = \frac{{12 \times 24}}{8} \cr & \Rightarrow {{\text{D}}_2} = 36{\text{ days}} \cr} $$

Let time taken by A alone in doing work be x days
∴ Time taken by B alone = 3x days
$$\eqalign{ & {\text{A's 1 day's work}} = \frac{1}{x} \cr & {\text{B's 1 days work}} = \frac{1}{{3x}} \cr & \because {\text{A and B together finish}} \cr & = \frac{2}{5}{\text{work in 9 days}}{\text{.}} \cr} $$
∴ Time taken by A and B in doing whole work
$$\eqalign{ & = \frac{{9 \times 5}}{2} \cr & = \frac{{45}}{2}{\text{ days}} \cr} $$
According to given information we get
$$\eqalign{ & \therefore \frac{1}{x} + \frac{1}{{3x}} = \frac{2}{{45}} \cr & \Rightarrow \frac{{3 + 1}}{{3x}} = \frac{2}{{45}} \cr & \Rightarrow \frac{4}{{3x}} = \frac{2}{{45}} \cr & {\text{By cross - multiply we get }} \cr & \Rightarrow 2 \times 3x = 4 \times 45 \cr & \Rightarrow x = \frac{{4 \times 45}}{{2 \times 3}} \cr & \Rightarrow x = 30{\text{ days}} \cr & {\text{Time taken by A}} \cr & = x{\text{ days}} \cr & = {\text{30 days}} \cr & \therefore {\text{Time taken by B}} \cr & = 3x{\text{ days}} \cr & = 3 \times 30 \cr & = 90{\text{ days}} \cr} $$

$$\eqalign{ & {\text{A's 1 day's work}} = \frac{1}{{24}} \cr & {\text{B's 1 day's work}} = \frac{1}{5} \cr & {\text{C's 1 day's work}} = \frac{1}{{12}} \cr & \therefore \left( {{\text{A}} + {\text{B}} + {\text{C}}} \right){\text{'s 1 day's work}} \cr & = \frac{1}{{24}} + \frac{1}{5} + \frac{1}{{12}} \cr & {\text{L}}{\text{.C}}{\text{.M of 24, 5 and 12}} \cr} $$

2	24 - 5 - 12
2	12 - 5 - 6
3	6 - 5 - 3
	2 - 5 - 1

$$\eqalign{ & {\text{2}} \times 2 \times 3 \times 2 \times 5 = 120 \cr & = \frac{{5 + 24 + 10}}{{120}} \cr & = \frac{{39}}{{120}} \cr & = \frac{{13}}{{40}} \cr} $$
Time taken by A, B and C to complete the work, working together
$$\eqalign{ & = \frac{{40}}{{13}} \cr & = 3\frac{1}{{13}}{\text{days}} \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & \Rightarrow {\text{4m}} + {\text{6w}} = 8{\text{ days}}.....{\text{(i)}} \cr & {\text{or}} \cr & \Rightarrow {\text{32m}} + 48{\text{w}} = 1{\text{ days}} \cr & \Rightarrow {\text{3m}} + 7{\text{w}} = 10{\text{ days}}.....{\text{(ii)}} \cr & {\text{or}} \cr & \Rightarrow {\text{30m}} + 70{\text{w}} = 1{\text{ day}} \cr & \therefore {\text{32m}} + {\text{48w}} = 30{\text{m}} + 70{\text{w}} \cr & \Rightarrow {\text{2m}} = 22{\text{w}} \cr & \Rightarrow {\text{m}} = 11{\text{w}} \cr & {\text{From (i)}} \cr & \Rightarrow {\text{4m}} = 44{\text{w}} \cr & \therefore \left( {{\text{44w}} + {\text{6w}}} \right) \times 8 = 10{\text{w}} \times x \cr & \Rightarrow {\text{50w}} \times 8 = 10{\text{w}} \times x \cr & \Rightarrow x = 40{\text{ days}} \cr} $$

Days	Eff.	Total work
A - 12	2
		24
B - 12	1
	3

A and B together can finish the work
$$\eqalign{ & = \frac{{24}}{3} \cr & = 8{\text{ days}} \cr} $$

	Amit	Bhawana	Chandan
Eff. →	5x	x	2x

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

12 man × 15 = 20 women × 12 = Total work
3 man = 4 women
$$\eqalign{ & \frac{{{\text{Men}}}}{{{\text{Women}}}} = \frac{4}{3} \to {\text{Efficiency}} \cr & {\text{Total work}} = 12 \times 4 \times 15 = 720 \cr & \left( {5{\text{ man}} + 5{\text{ women}}} \right) \times {\text{D}} = 720 \cr & {\text{D}}\left( {5 \times 4 + 5 \times 3} \right) = 720 \cr & {\text{D}} = \frac{{720}}{{35}} = 20\frac{4}{7}{\text{ days}} \cr} $$

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Alternate:
According to the question,
12 men can do the 1 day work in $$\frac{1}{{15}}$$ days
So, 1 men can do the 1 day work in $$\frac{1}{{15 \times 12}}$$  = $$\frac{1}{{180}}$$ days
5 men can do the 1 day work in $$\frac{1}{{180}} \times 5$$   = $$\frac{5}{{180}}$$ days
Similarly 5 women do the 1 dys work $$\frac{5}{{240}}$$ days
∴ 5 men & 5 women together work in 1 day
$$\eqalign{ & = \frac{5}{{180}} + \frac{5}{{240}} \cr & = \frac{{20 + 15}}{{720}} \cr & = \frac{{35}}{{720}} \cr} $$
Hence, they complete the work $$\frac{{720}}{{35}}$$ = $$20\frac{4}{{35}}$$ days

$$\eqalign{ & {\text{X's 1 day's work}} = \frac{1}{{24}} \cr & {\text{X's 16 day's work}} = \frac{{16}}{{24}} \cr & {\text{Let,}} \cr} $$
Y alone complete the work in x days
$${\text{Y's 12 day's work}} = \frac{{12}}{x}$$
According to the question,
Complete work done by X and Y = 1
X's 16 day's work + Y's 12 day's work = 1
$$\eqalign{ & \Rightarrow \frac{{16}}{{24}} + \frac{{12}}{x} = 1 \cr & \Rightarrow \frac{2}{3} + \frac{{12}}{x} = 1 \cr & \Rightarrow \frac{{12}}{x} = 1 - \frac{2}{3} = \frac{1}{3} \cr & \Rightarrow x = 12 \times 3 \cr & \Rightarrow x = 36{\text{ days}} \cr} $$

6 men will complete the work in 12 days
1 men will complete the work in (6 × 12) = 72 days
8 women will complete two work in 18 days
1 women will complete the work in (8 × 18) = 144 days
18 children will complete the work in 10 days
1 children will complete the work in (18 × 10) = 180 days
$$\eqalign{ & {\text{1 men's 1 day's work}} = \frac{1}{{72}} \cr & {\text{1 women's 1 day's work}} = \frac{1}{{144}} \cr & {\text{1 children's 1 day's work}} = \frac{1}{{180}} \cr} $$
(4 men + 12 women + 20 children)'s 2 day's work
$$\eqalign{ & = 2\left( {\frac{4}{{72}} + \frac{{12}}{{144}} + \frac{{20}}{{180}}} \right) \cr & = 2\left( {\frac{1}{{18}} + \frac{1}{{12}} + \frac{1}{9}} \right) \cr & {\text{L}}{\text{.C}}{\text{.M of 18, 12 and 9}} = {\text{36}} \cr & {\text{ = }}\frac{{2\left( {2 + 3 + 4} \right)}}{{36}} \cr & = \frac{1}{2} \cr & \therefore {\text{Remaining work}} = \frac{1}{2} \cr & \therefore {\text{Required number of men}} \cr & = 72 \times \frac{1}{2} \cr & = 36 \cr} $$

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