Time And Work - Aptitude MCQ Questions and Solutions with Explanations

Number of units processed by computer A in 1 minute
$$ = \frac{1}{3}$$
Number of units processed by computer B in 1 minute
$$ = \frac{1}{5}{\text{ }}$$
Number of units processed by computer A, B and C in 1 minute
$$\eqalign{ & = \frac{{14 \times 3}}{{60}} \cr & {\text{ = }}\frac{7}{{10}} \cr} $$
Number of units processed by computer C in 1 minute
$$\eqalign{ & = \frac{7}{{10}} - \left( {\frac{1}{3} + \frac{1}{5}} \right) \cr & = \frac{7}{{10}} - \frac{8}{{15}} \cr & = \frac{5}{{30}} \cr & = \frac{1}{6} \cr} $$
Hence, computer C takes 6 minutes to process one input alone.

$$\frac{{{\text{5 men}} + {\text{2 women}}}}{{4{\text{work}}}}$$     = $$\left( {1{\text{ men}} + {\text{1 women}}} \right)$$
$$5{\text{ men}} + {\text{2 women}}$$     = $${\text{4 men}} + {\text{4 women}}$$
$$\eqalign{ & {\text{1 men}} = {\text{2 women}} \cr & \frac{{{\text{Men}}}}{{{\text{Women}}}} = \frac{2}{1} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{M}}:{\text{W}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2:1 \cr} $$

40 men = 60 women = 80 children
2 men = 3 women = 4 children
2 men = 3 women
1 women = $$\frac{2}{3}$$ men → 10 women
$$ \to \frac{2}{3} \times 10 = \frac{{20}}{3}{\text{ men}}$$
Similarly
2 men = 4 children
1 children = $$\frac{1}{2}$$ men → 10 children
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{10}}{2} = {\text{5 men}}$$
10 men = 10 women = 10 children
$$\eqalign{ & {\text{10 men}} + \frac{{20}}{3} + 5 \cr & \Rightarrow \frac{{30 + 20 + 15}}{3} \cr} $$
10 men + 10 women + 10 children = $$\frac{{65}}{3}$$ men
40 men can do a piece of work in 6 months
1 men can do a piece of work in 6 × 40
$$\frac{{65}}{3}$$ men can do a piece of work in
$$\eqalign{ & = \frac{{6 \times 40}}{{\frac{{65}}{3}}} \cr & = 11\frac{1}{{13}}{\text{ months}} \cr} $$

Alternate :
40 men = 60 women = 80 children
2 men = 3 women = 4 children
men : women : children = 6 : 4 : 3 (efficiency)
∴ Total work
$$\eqalign{ & = 40 \times 6 \times 6 \cr & = 1440\,{\text{units}} \cr} $$
Total time taken by (40 men + 60 women + 80 children)
$$\eqalign{ & = \frac{{{\text{Total work}}}}{{{\text{Efficiency}}}} \cr & = \frac{{1440}}{{130}} \cr & = 11\frac{1}{{13}}{\text{months}} \cr} $$

L.C.M. of Total Work = 15
One day work of A + B = $$\frac{{15}}{5}$$ = 3 unit/day
One day work of (2A + B) = $$\frac{{15}}{3}$$ = 5 unit/day
Now,
Assume A's efficiency is 2 units, B's is 1 unit.
So, it satisfies the equation of both cases
So, actual efficiency of A is 2 units/day
A alone can complete the work in
$$\eqalign{ & = \frac{{{\text{Total work}}}}{{{\text{Efficiency}}}} \cr & = \frac{{15}}{2} \cr & = 7\frac{1}{2}{\text{days}} \cr} $$

(3 men + 7 women) × 5 days = (4 men + 6 women) × 4 days
1 men = 11 women
∴ 3 men + 7 women
(3 × 11) women + 7 women
= 40 women
40 women can do a work in 5 days
1 women can do a work in (5 × 40) days
10 women can do a work in = $$\frac{{5 \times 40}}{{10}}$$ = 20 days

Let the number of pages typed in one hour by P, Q and R be x, y and z respectively
Then,
$$\eqalign{ & \Rightarrow x + y + z = \frac{{216}}{4} \cr & \Rightarrow x + y + z = 54z.....{\text{(i)}} \cr & {\text{ }}z - y = y - x \cr & \Rightarrow 2y = x + z.....{\text{(ii)}} \cr & {\text{ }}5z = 7x \cr & \Rightarrow x = \frac{5}{7}z......{\text{(iii)}} \cr} $$
Solving (i), (ii) and (iii), we get
$$\eqalign{ & x = 15, \cr & y = 18,{\text{ }} \cr & z = 21 \cr} $$

Number of pages typed by Ronald in 1 hour
$$\eqalign{ & = \frac{{32}}{6} \cr & = \frac{{16}}{3} \cr} $$
Number of pages typed by Elan in 1 hour
$$\eqalign{ & = \frac{{40}}{5} \cr & = 8 \cr} $$
Number of pages typed by both in 1 hour
$$\eqalign{ & = \left( {\frac{{16}}{3} + 8} \right) \cr & = \frac{{40}}{3} \cr} $$
∴Time taken by both to type 110 pages
$$\eqalign{ & {\text{ = }}\left( {100 \times \frac{3}{{40}}} \right){\text{hours}} \cr & = 8\frac{1}{4}{\text{hours}} \cr & = {\text{8 hours 15 minutes}} \cr} $$

Length of the cloth produced in 1 hour
$$\eqalign{ & {\text{ = }}\left( {\frac{{{\text{pq}}}}{{\text{r}}} \times 60} \right){\text{ m }} \cr & = \left( {\frac{{60{\text{pq}}}}{{\text{r}}}} \right){\text{ m}} \cr & \therefore {\text{Required time}} \cr & = \left( {20000 \div \frac{{60{\text{pq}}}}{{\text{r}}}} \right){\text{ hours }} \cr & {\text{ = }}\left( {\frac{{20000{\text{r}}}}{{60{\text{pq}}}}} \right){\text{ hours}} \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & {\text{1 man}} = {\text{2 women}} = {\text{3 boys}} \cr & {\text{1 man}} = {\text{2 women}} \cr & {\text{1 man}} = {\text{3 boys}} \cr & \frac{1}{2}{\text{ man}} = 1{\text{ women}} \cr & \frac{1}{3}{\text{ man}} = {\text{1 boys}} \cr & = {\text{1 man}} = 1{\text{ woman}} = 1{\text{ boys}} \cr & = {\text{1 man}} = \frac{1}{2}{\text{man}} = \frac{1}{3}{\text{ man}} \cr & = \frac{{11}}{6}{\text{ man}} \cr} $$
1 man can complete a work in 88 days
$$\eqalign{ & \frac{{11}}{6}{\text{man can complete a work in}} \cr & = \frac{{88}}{{\frac{{11}}{6}}}\, = 48{\text{ days }} \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & {\text{15 men}} = {\text{20 days}} \cr & {\text{300 men}} = 1{\text{ days}}.....{\text{(i)}} \cr & {\text{24 women}} = {\text{20 days}} \cr & {\text{480 men}} = 1{\text{ days}}......{\text{(ii)}} \cr & {\text{Compare equation (i) and (ii)}} \cr & {\text{300 men}} = 480{\text{ women}} \cr & {\text{5 men}} = 8{\text{ women}}.....{\text{(iii)}} \cr & {\text{10 men}} + 8{\text{ women}} = ? \cr & {\text{10 men}} + {\text{5 men}} = ? \cr & 15\,{\text{men}} = ? \cr} $$
$${\text{15 men}} \times {\text{20 days}}$$     = $${\text{15 men}}$$  $$ \times $$ $$x{\text{ days}}$$
$$x$$ = 20 days
Alternate
$$\eqalign{ & {\text{15m}} \times {\text{20 days}} = 24{\text{w}} \times 20{\text{ days}} \cr & \frac{{\text{m}}}{{\text{w}}} = \frac{8}{5} \cr} $$
So, 1 man work 8 units work in one day
and 1 woman work 5 units work in one day
Total work = 15 × 8 × 20
Hence, (10 men + 8 women) work whole in D days
$$\eqalign{ & \left( {{\text{10m}} + {\text{8w}}} \right) \times {\text{D}} = 15 \times 8 \times 20 \cr & \left( {{\text{10}} \times {\text{8}} + {\text{8}} \times {\text{5}}} \right) \times {\text{D}} = 15 \times 8 \times 20 \cr & \left( {{\text{80}} + 40} \right) \times {\text{D}} = 15 \times 8 \times 20 \cr & {\text{D}} = 20{\text{ days}} \cr} $$