Time And Work - Aptitude MCQ Questions and Solutions with Explanations

If number of days at work is same
$$\eqalign{ & {\text{So, we can equate directly}} \cr & {\text{16 men}} = {\text{20 women}} \cr & {\text{4 men}} = {\text{5 women}} \cr & {\text{5 women}} = {\text{4 men}} \cr & {\text{1 women}} = \frac{4}{5}{\text{men}} \cr & {\text{15 women}} = \frac{4}{5} \times {\text{15}} \cr & {\text{15 women}} = {\text{12 men}} \cr} $$
⇒ 16 men complete a work in 25 days
1 men complete a work in 400 days
$$\eqalign{ & 28{\text{ men}} + {\text{15 women}} = ? \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \cr & 28{\text{ men}} + {\text{12 men}} = {\text{40 men}} \cr & {\text{work in }} = \frac{{25 \times 16}}{{40}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{10 days}} \cr & \cr & {\bf{Alternate:}} \cr & {\text{16 m or 20 w}} \to {\text{25 days}} \cr & {\text{28 m and 15 w}} \to {\text{?}} \cr & \frac{{16 \times 20 \times 25}}{{\left( {16 \times 15} \right) + \left( {20 \times 28} \right)}} \cr & = 10{\text{ days}} \cr} $$

Jyoti does $$\frac{3}{4}$$ unit of work in 12 days
Jyoti does 1 unit of work in
$$\eqalign{ & = 12 \times \frac{4}{3} \cr & = 16{\text{ days}} \cr} $$
According to the question,

	Mala	:	Jyoti
Efficiency →	2	:	1
Time →	1	:	2
	↓×8		←↓×8
	8 days		16 days

Ratio of digging speeds of man and boy = 8 : 5
Ratio of times taken by man and boy = 5 : 8
Suppose the man takes 5x days while the takes 8x
Then,
$$\eqalign{ & \Rightarrow \frac{1}{{5x}} + \frac{1}{{8x}} = \frac{1}{{40}} \cr & \Rightarrow \frac{{13}}{{40x}} = \frac{1}{{40}} \cr & \Rightarrow x = 13 \cr} $$
Hence, time taken by the boy to complete the work alone
$$\eqalign{ & {\text{ = }}\left( {8 \times 13} \right){\text{days}} \cr & {\text{ = 104 days}} \cr} $$

$$\eqalign{ & {\text{C's 1 day's work}} \cr & = \frac{1}{7} \cr & {\text{B's 1 day's work}} \cr & = \left( {\frac{1}{2} \times \frac{1}{7}} \right) \cr & = \frac{1}{{14}} \cr & {\text{A's 1 day's work}} \cr & = \left( {\frac{1}{2} \times \frac{1}{{14}}} \right) \cr & = \frac{1}{{28}} \cr & \left( {{\text{A}} + {\text{B}} + {\text{C}}} \right){\text{'s 1 day's work}} \cr & = \left( {\frac{1}{{28}} + \frac{1}{{14}} + \frac{1}{7}} \right) \cr & = \frac{7}{{28}} \cr} $$
Hence, A, B and C together can complete the work in 4 days.

Number of rosogollas eaten by Rosa in 1 minute = $$\frac{{32}}{{60}}$$
Number of rosogollas eaten by Lila in 1 minute =$$\frac{{32}}{{180}}$$
Number of rosogollas eaten by Rosa and Lila together in 1 minute
$$\eqalign{ & {\text{ = }}\left( {\frac{{32}}{{60}} + \frac{{32}}{{180}}} \right) \cr & {\text{ = }}\frac{{128}}{{180}} \cr & \therefore {\text{Required time}} \cr & = \left( {32 \div \frac{{128}}{{180}}} \right){\text{minutes}} \cr & = \left( {\frac{{32 \times 180}}{{128}}} \right){\text{minutes}} \cr & = {\text{ 45 minutes}} \cr} $$

	Tapas	:	Mihir
Efficiency units/day	2	:	1

$$\eqalign{ & {\text{T}} + {\text{M complete in 12 days}} \cr & {\text{Total work }} \cr & = 12 \times \left( {2 + 1} \right) \cr & = 36{\text{units}} \cr} $$
Tapas alone complete the whole work in
$$\eqalign{ & = \frac{{36}}{2} \cr & = 18{\text{ days}} \cr} $$

According to the question,
$$\eqalign{ & {\text{2m}} + {\text{3w}} = {\text{4m}} \cr & {\text{3w}} = {\text{4m}} - {\text{2m }} \cr & {\text{3w}} = {\text{2m}} \cr & {\text{3m}} + {\text{3w}} = {\text{3m}} + {\text{2m}} \cr & {\text{3m}} + {\text{3w}} = {\text{5m}} \cr} $$
4 men can do work in 20 days
1 men can do work in 20 × 4 days
5 men can do work in $$\frac{{{\text{20}} \times {\text{4}}}}{5}$$  = 16 days

$$\eqalign{ & {\bf{Alternate:}} \cr & \left( {2{\text{m}} + {\text{3w}}} \right) \times {\text{20}} = {\text{4m}} \times {\text{20}} \cr & \frac{{\text{m}}}{{\text{w}}} = \frac{3}{2} \cr & {\text{Total work }} \cr & {\text{ = }}\left( {{\text{2}} \times {\text{3}} + {\text{3}} \times {\text{2}}} \right) \times 20 \cr & = 240\,{\text{units}} \cr & {\text{5 men efficiency}} \cr & = 5 \times 3 \cr & = 15 \cr & {\text{Required number of days}} \cr & = \frac{{240}}{{15}} \cr & = 16{\text{ days}} \cr} $$

$$\eqalign{ & {\text{20 w}} \times {\text{16}} = {\text{16 m}} \times {\text{15}} \cr & {\text{20 w}} = {\text{15 m}} \cr & {\text{4 w}} = {\text{3 m}} \cr & \frac{{\text{m}}}{{\text{w}}} = \frac{4}{3} \cr & \therefore {\text{Man}}:{\text{Women}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,4:3 \cr} $$

Baggage delivered by first belt in 1 minute
$$ = \left( {\frac{3}{5}} \right){\text{tons}}$$
Baggage delivered by second belt in 1 minute
$$ = \left( {\frac{1}{2}} \right){\text{tons}}$$
Baggage delivered by both belt in 1 minute
$$\eqalign{ & = \left( {\frac{3}{5} + \frac{1}{2}} \right){\text{tons}} \cr & = \frac{{11}}{{10}}{\text{ tons}} \cr & \therefore {\text{Required time}} \cr & = \left( {33 \div \frac{{11}}{{10}}} \right){\text{ minutes}} \cr & = \left( {33 \times \frac{{10}}{{11}}} \right){\text{minutes}} \cr & = {\text{30 minutes}} \cr} $$

Number of envelopes addressed by first machine in 1 minute
$$ = \frac{{500}}{8}$$
Number of envelopes addressed by second machine in 1 minute
$$ = \frac{{500}}{x}$$
Number of envelopes addressed by both machine in 1 minute
$$\eqalign{ & {\text{ = }}\frac{{500}}{2} \cr & \therefore \frac{{500}}{8} + \frac{{500}}{x} = \frac{{500}}{2} \cr & \Rightarrow \frac{1}{8} + \frac{1}{x} = \frac{1}{2} \cr} $$