Time And Work - Aptitude MCQ Questions and Solutions with Explanations

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

$$\eqalign{ & {\text{C's 3 day's work}} \cr & = \left( {\frac{1}{{15}} \times 3} \right) \cr & = \frac{1}{5} \cr & \left( {{\text{B}} + {\text{C}}} \right){\text{'s 2 day's work}} \cr & = \left[ {\left( {\frac{1}{{12}} + \frac{1}{{15}}} \right) \times 2} \right] \cr & = \left( {\frac{3}{{20}} \times 2} \right) \cr & = \frac{3}{{10}} \cr & \therefore {\text{Remaining work}} \cr & = \left[ {1 - \left( {\frac{1}{5} + \frac{3}{{10}}} \right)} \right] \cr & = \left( {1 - \frac{1}{2}} \right) \cr & = \frac{1}{2} \cr & \left( {{\text{A}} + {\text{B}} + {\text{C}}} \right){\text{'s 1 day's work}} \cr & = \left( {\frac{1}{{10}} + \frac{1}{{12}} + \frac{1}{{15}}} \right) \cr & = \frac{{15}}{{60}} \cr & = \frac{1}{4} \cr} $$
$$\frac{1}{4}$$ work is done by A, B ans C in 1 day.
∴ $$\frac{1}{2}$$ work is done by A, B and C in
$$\eqalign{ & = \left( {4 \times \frac{1}{2}} \right) \cr & = {\text{2 days}} \cr & {\text{Total number of days}} \cr & = \left( {3 + 2 + 2} \right) \cr & = 7 \cr} $$

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

$$\eqalign{ & {\text{Boy's 1 day's work}} \cr & = \frac{1}{3} - \left( {\frac{1}{7} + \frac{1}{8}} \right) \cr & = \left( {\frac{1}{3} - \frac{{15}}{{56}}} \right) \cr & = \frac{{11}}{{168}} \cr} $$
∴ Ratio of wages of the first man, second man and boy
$$\eqalign{ & = \frac{1}{7}:\frac{1}{8}:\frac{{11}}{{168}} \cr & = 24:21:11 \cr & {\text{First man's share}} \cr & = {\text{Rs}}{\text{.}}\left( {\frac{{24}}{{56}} \times 1400} \right) \cr & = {\text{Rs}}{\text{.600}} \cr & {\text{Second man's share}} \cr & = {\text{Rs}}{\text{.}}\left( {\frac{{21}}{{56}} \times 1400} \right) \cr & = {\text{Rs}}{\text{.525}} \cr & {\text{Boy's man's share}} \cr & = {\text{Rs}}{\text{.}}\left[ {1400 - \left( {600 + 525} \right)} \right] \cr & = {\text{Rs}}{\text{.275}} \cr} $$

20 men → 18 days
⇒ Work done by 20 men working
Together = 1 work
⇒ Work done by them in 3 days working
Together = 1 × 3 = 3 work
⇒ Remaining work = 18 - 3 = 15 work
⇒ 15 work is to be done by (20 + 5) = 25 men
$$\eqalign{ & \therefore {\text{Efficiency of 1 man}} = \frac{1}{{20}} \cr & \Rightarrow {\text{Efficincy of 5 men}} \cr & = \frac{5}{{20}} \cr & = \frac{1}{4} \cr & \Rightarrow {\text{So, efficiency of }}\left( {20 + 5} \right) \cr & \Rightarrow 25{\text{ men}} = 1 + \frac{1}{4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{5}{4}{\text{ working days}} \cr & {\text{Required time}} \cr & = \frac{{{\text{Work}}}}{{{\text{Efficiency}}}} \cr & = \frac{{15}}{{\frac{5}{4}}} \cr & = 12{\text{ days}} \cr} $$
Therefore, 12 more days will be taken to finish the remaining work

Alternate : 20 men can do 18 days
So, total work = 18 × 20 = 360
20 men 3 days work = 20 × 3 = 60
Remaining work = 360 - 60 = 300
After joining 5 men total men = 20 + 5 = 25
So, finish the remaining work in = $$\frac{{300}}{{25}} = 12{\text{ days}}$$

L.C.M. of total work = 60
One day work of A = $$\frac{{60}}{{10}}$$ = 6 unit/day
One day work of B = $$\frac{{60}}{{12}}$$ = 5 unit/day
One day work of C = $$\frac{{60}}{{15}}$$ = 4 unit/day
$$\eqalign{ & {\text{A leaves before 5 days}} \cr & = 6 \times 5 \cr & = 30 \cr & {\text{B leaves before 3 days}} \cr & = 3 \times 5 \cr & = 15 \cr & {\text{Then total work}} \cr & = 60 + 30 + 15 \cr & = 105 \cr & {\text{Total efficiency}} \cr & = 6 + 5 + 4 \cr & = 15 \cr & {\text{Total days requierd}} \cr & = \frac{{105}}{{15}} \cr & = 7{\text{ days}} \cr} $$

Let total work = 40 units
$$\eqalign{ & {\text{X's 1 day work}} = 1{\text{ unit}} \cr & {\text{X's 8 days work is}} \cr & = 8 \times 1 \cr & = 8{\text{ units}} \cr & {\text{Work left}} = 40 - 8 = 32 \cr & {\text{Y's 1 day work}} = 2{\text{ unit}} \cr & {\text{X's 8 days work is}} = 1{\text{ units}} \cr} $$
X + Y complete the whole work together in
$$\eqalign{ & = \frac{{40}}{{2 + 1}} \cr & = 13\frac{1}{3}\,{\text{days}} \cr} $$

L.C.M. of Total Work =60
One day work of A + B = $$\frac{{60}}{{30}}$$ = 2 unit/day
One day work of B + C = $$\frac{{60}}{{20}}$$ = 3 unit/day
$$\eqalign{ & {\text{According to the question,}} \cr & \Rightarrow \left( {{\text{A}} + {\text{B}}} \right).....{\text{30 days}} \cr & \Rightarrow \left( {{\text{B}} + {\text{C}}{\text{.}}} \right).....{\text{20 days}} \cr & \Rightarrow {\text{A}}.....{\text{5}} \cr & \Rightarrow {\text{B}}.....{\text{5}} + {\text{10}} \cr & \Rightarrow {\text{C}}.....{\text{10}} + {\text{8 days}} \cr} $$
Work done by (A +B) in 5 days
= 2 × 5 = 10 units
⇒ Work done by (B + C) in 10 days
= 10 × 3 = 30 units
⇒ Total work (finished till now).....40 units
⇒ Remaining work
= 60 - 40
= 20 units
⇒ Here we find that C does remaining 20 units in 8 days
⇒ C's efficiency = $$\frac{{{\text{Work}}}}{{{\text{Day}}}}$$
$$ = \frac{{{\text{20 Work}}}}{{{\text{8 Days}}}} = \frac{5}{2}$$
⇒ Therefore, time taken by C alone to complete the work
$$ = \frac{{60}}{5} \times 2 = 24{\text{ days}}$$

$$\eqalign{ & {\text{Let total money be Rs}}{\text{.}}x \cr & {\text{A's 1day's wages}} = {\text{Rs}}{\text{.}}\frac{x}{{21}} \cr & {\text{B's 1day's wages}} = {\text{Rs}}{\text{.}}\frac{x}{{28}} \cr & \therefore \left( {{\text{A}} + {\text{B}}} \right){\text{'s 1 day's wages}} \cr & = {\text{Rs}}{\text{.}}\left( {\frac{x}{{21}} + \frac{x}{{28}}} \right) \cr & = {\text{Rs}}{\text{.}}\frac{x}{{12}} \cr} $$
∴ Money is sufficient to pay the wages of both for 12 days.

$$\eqalign{ & {\text{Part of the done by A}} \cr & = \left( {\frac{1}{{10}} \times 5} \right) \cr & = \frac{1}{2} \cr & {\text{Part of the done by B}} \cr & = \left( {\frac{1}{{15}} \times 5} \right) \cr & = \frac{1}{3} \cr & {\text{Part of the done by C}} \cr & = 1 - \left( {\frac{1}{2} + \frac{1}{3}} \right) \cr & = \frac{1}{6} \cr} $$
So, (A's share) : (B's share) : (C's share)
$$\eqalign{ & = \frac{1}{2}:\frac{1}{3}:\frac{1}{6} \cr & = 3:2:1 \cr & \therefore {\text{A's share}} \cr & = {\text{Rs}}{\text{.}}\left( {\frac{3}{6} \times 1500} \right) \cr & = {\text{Rs}}.750 \cr & {\text{B's share}} \cr & = {\text{Rs}}{\text{.}}\left( {\frac{2}{6} \times 1500} \right) \cr & = {\text{Rs}}.500 \cr & {\text{C's share}} \cr & = {\text{Rs}}{\text{.}}\left( {\frac{1}{6} \times 1500} \right) \cr & = {\text{Rs}}.250 \cr & {\text{A's daily wages}} \cr & = {\text{Rs}}{\text{.}}\left( {\frac{{750}}{5}} \right) \cr & = {\text{Rs}}.150 \cr & {\text{B's daily wages}} \cr & = {\text{Rs}}{\text{.}}\left( {\frac{{500}}{5}} \right) \cr & = {\text{Rs}}.100 \cr & {\text{C's daily wages}} \cr & = {\text{Rs}}{\text{.}}\left( {\frac{{250}}{2}} \right) \cr & = {\text{Rs}}.125 \cr & \therefore {\text{Daily wages of B and C}} \cr & = {\text{Rs}}{\text{.}}\left( {100 + 125} \right) \cr & = {\text{Rs}}{\text{.225}} \cr} $$