Time And Work - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & \left( {{\text{A}} + {\text{B}} + {\text{C}}} \right){\text{'s 1 day's work}} = \frac{1}{6} \cr & \left( {{\text{A}} + {\text{B}}} \right){\text{'s 1 day's work}} = \frac{1}{8} \cr & \left( {{\text{B}} + {\text{C}}} \right){\text{'s 1 day's work}} = \frac{1}{{12}}{\text{ }} \cr & \therefore \left( {{\text{A}} + {\text{C}}} \right){\text{'s 1 day's work}} \cr & = \left( {2 \times \frac{1}{6}} \right) - \left( {\frac{1}{8} + \frac{1}{{12}}} \right) \cr & = \left( {\frac{1}{3} - \frac{5}{{24}}} \right) \cr & = \frac{3}{{24}} \cr & = \frac{1}{8} \cr} $$
So, A and C together will do the work in 8 days.

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

$$\eqalign{ & \frac{{{{10}_{{\text{men}}}} \times {6_{{\text{hours}}}} \times {{18}_{{\text{days}}}}}}{{{1_{{\text{work}}}}}} = \frac{{{{15}_{{\text{men}}}} \times {{12}_{{\text{days}}}} \times {\text{H hour/day}}}}{{{1_{{\text{work}}}}}} \cr & \Leftrightarrow {\text{6 hours/day}} \cr} $$

Let total number of worker in beginning is N
$$\eqalign{ & {\text{According to the question,}} \cr & \frac{{{\text{N}} \times {{100}_{{\text{days}}}}}}{{{1_{{\text{work}}}}}} = \frac{{\left( {{\text{N}} - 10} \right) \times {{110}_{{\text{days}}}}}}{{{1_{{\text{work}}}}}} \cr & 100{\text{N}} = {\text{110N}} - {\text{1100}} \cr & \Rightarrow {\text{10N}} = {\text{1100}} \cr & \Rightarrow {\text{N}} = {\text{110}} \cr} $$

According to the question,
Total work
= 12 M × 12 D
= 144 units
Work done by 12 men in 6 days
= 12 × 6
= 72 units
Rest work
= 144 - 72
= 72 units
Required time for 6 men to complete the work
$$\eqalign{ & = \frac{{72}}{6} \cr & {\text{ = 12 days}} \cr & {\text{Hence,}} \cr & {\text{Total time}} = 12 + 6 = 18{\text{ days}} \cr & {\text{Extra time}} = 18 - 12 = 6{\text{ days}} \cr} $$

60 men work for 200 days.
They stops for 10 day due to bad weather.
So, the work is to complete in
= (50 - 10)
= 40 days
In order to complete in scheduled time i.e., 250 days.
Let 'n' number of more men is required
(60_men × 200_days) +{(60 + n)_men × 40_days} = 60_men × 250_days
⇒ 12000 + {(60 + n)_men × 40_days} = 15000
⇒ (60 + n)40_days = 3000
⇒ 60 + n = 75
⇒ n = 15

Alternate :
60 men can complete a work in 250 days.
But they work for 200 days.
Then remaining days = 50 days
So,
60 × 50 = 60 + x × 40
⇔ x = 15

$$\eqalign{ & \left( {{\text{A}} + {\text{B}}} \right){\text{'s 1 day's work}} = \frac{1}{{12}} \cr & \left( {{\text{B}} + {\text{C}}} \right){\text{'s 1 day's work}} = \frac{1}{8} \cr & \left( {{\text{A}} + {\text{C}}} \right){\text{'s 1 day's work}} = \frac{1}{{62}} \cr} $$
[ (A + B)'s 1 day's work + (B + C)'s 1 day's work ] - (A + C)'s 1 day's work
$$\eqalign{ & = \frac{1}{{12}} + \frac{1}{8} - \frac{1}{6} \cr & \Rightarrow 2\left( {{\text{B's 1 day's work}}} \right) = \frac{1}{{24}} \cr & \Rightarrow {\text{B's 1 day's work}} = \frac{1}{{48}} \cr} $$
Hence, B alone can do the work in 48 days.

$$\eqalign{ & \left( {{\text{A}} + {\text{B}}} \right){\text{'s 1 day's work}} = \frac{1}{{25}} \cr & {\text{C's 1 day's work}} = \frac{1}{{35}} \cr & \left( {{\text{A}} + {\text{B}} + {\text{C}}} \right){\text{'s 1 day's work}} \cr & = \left( {\frac{1}{{25}} + \frac{1}{{35}}} \right) \cr & = \frac{{12}}{{175}}.....({\text{i}}) \cr} $$
Also, A's 1 day's work = (B + C)'s 1 day's work.....(ii)
$$\eqalign{ & {\text{From (i) and (ii), we get : }} \cr & \Rightarrow {\text{2}} \times \left( {{\text{A's 1 day's work}}} \right) = \frac{{12}}{{175}} \cr & \Rightarrow {\text{A's 1 day's work}} = \frac{6}{{175}} \cr & \therefore {\text{B's 1 day's work}} \cr & = \left( {\frac{1}{{25}} - \frac{6}{{175}}} \right) \cr & = \frac{1}{{175}} \cr} $$

Suppose Uma takes x days to complete a work
Then, Madhu takes 2x days to complete the work
Uma's 1 day's work = $$\frac{1}{x}$$
Madhu's 1 day's work = $$\frac{1}{{2x}}$$
Rahul's 1 day's work = (Madhu + Uma)'s 1 day's work
$$\eqalign{ & = \frac{1}{{2x}} + \frac{1}{x} \cr & = \frac{3}{{3x}}{\text{ }} \cr} $$
(Madhu + Uma + Rahul)'s 1 day's work
$$\eqalign{ & = \frac{3}{{3x}} + \frac{1}{{2x}} + \frac{1}{x} = \frac{6}{{3x}} = \frac{3}{x} \cr & \therefore \frac{3}{x} = \frac{1}{6} \cr & \Rightarrow x = 18{\text{ }} \cr} $$
Hence, Madhu takes (2 × 18) = 36 days to complete the work

$$\eqalign{ & \frac{{{\text{28 M}} \times {\text{1 Week}}}}{{\frac{7}{8}}} = \frac{{x \times {\text{ 1 Week}}}}{{\frac{1}{8}}} \cr & \Leftrightarrow x = 4{\text{ men}} \cr} $$