Time And Work - Aptitude MCQ Questions and Solutions with Explanations

According to the question,
3 men = 5 women
As they complete the same work in same time
6 men + 5 women
= 6 men + 3 men
= 9 men
If, 3 men does a work in 12 days
1 men does a work in 12 × 3
9 men does a work in $$\frac{{{\text{12}} \times {\text{3}}}}{9}$$  = 4 days

Alternate :
$$\eqalign{ & 3{\text{ men}} \times {\text{12}} = {\text{5 women}} \times {\text{12}} \cr & \frac{{{\text{men}}}}{{{\text{women}}}} = \frac{5}{3} \cr & {\text{Total work}} \cr & = 3 \times 5 \times 12 \cr & = 15 \times 12 \cr & = 180{\text{ units}} \cr & {\text{Efficiency of 6 men}} + {\text{5 women}} \cr & = 6 \times 5 + 5 \times 3 \cr & = 45 \cr & {\text{Time taken by 6 men}} + {\text{5 women}} \cr & = \frac{{180}}{{45}} \cr & = 4{\text{ days}} \cr} $$

$$\eqalign{ & {\text{10 men}} \times {\text{12 days}} = {\text{10 women}} \times {\text{6 days}} \cr & {\text{2 men}} = {\text{1 woman}} \cr & \frac{{\text{M}}}{{\text{W}}} = \frac{1}{2}{\text{ }} \cr & {\text{1 man work}} = {\text{1 unit/day}} \cr & {\text{1 woman work}} = {\text{2 unit/day}} \cr & {\text{Total work}} = {\text{10 men}} \times {\text{12 days}} \cr & = 10 \times 1 \times 12 \cr & = 120{\text{ units}} \cr & {\text{Time required}}\left( {{\text{10 men}} + {\text{10 women}}} \right) \cr & = \frac{{{\text{Total work}}}}{{{\text{Efficiency}}}} \cr & = \frac{{120}}{{10 \times 1 + 10 \times 2}} \cr & = \frac{{120}}{{30}} \cr & = 4{\text{ days}} \cr} $$

Let A and B together take x hours to complete the work.
Then, A alone takes (x + 8) hours
And
B alone takes $$\left( {{\text{x}} + \frac{9}{2}} \right)$$   hours to complete the work
Then,
$$\eqalign{ & \Rightarrow \frac{1}{{\left( {x + 8} \right)}} + \frac{1}{{\left( {x + \frac{9}{2}} \right)}} = \frac{1}{x} \cr & \Rightarrow \frac{1}{{\left( {x + 8} \right)}} + \frac{2}{{\left( {2x + 9} \right)}} = \frac{1}{x} \cr & \Rightarrow x\left( {4x + 25} \right) = \left( {x + 8} \right)\left( {2x + 9} \right) \cr & \Rightarrow 2{x^2} = 72 \cr & \Rightarrow {x^2} = 36 \cr & \Rightarrow x = 6 \cr} $$

Let the time taken by the three friends together to do the work be x hours
Then, time taken by Anne alone = (x + 6) hours
Time taken by Bob alone = (x + 1) hours
Time taken by Bob alone = 2x hours
$$\therefore \frac{1}{{x + 6}} + \frac{1}{{x + 1}} + \frac{1}{{2x}} = \frac{1}{x}$$
$$ \Rightarrow $$ $$\frac{{2x\left( {x + 1} \right) + 2x\left( {x + 6} \right) + \left( {x + 1} \right)\left( {x + 6} \right)}}{{2x\left( {x + 6} \right)\left( {x + 1} \right)}}$$         $$ = $$ $$\frac{1}{x}$$
$$\eqalign{ & \Rightarrow 5{x^2} + 21x + 6 = 2\left( {{x^2} + 7x + 6} \right) \cr & \Rightarrow 3{x^2} + 7x - 6 = 0 \cr & \Rightarrow \left( {x + 3} \right)\left( {3x - 2} \right) = 0 \cr & \Rightarrow x = \frac{2}{3}{\text{ }}\left[ {\because x \ne - 3} \right] \cr & \therefore {\text{Required times}} \cr & = \frac{2}{3}{\text{ hours}} \cr & = \left( {\frac{2}{3} \times 60} \right){\text{hours}} \cr & = 40{\text{ minutes}} \cr} $$

$$\eqalign{ & \left( {{\text{A}} + {\text{B}}} \right){\text{'s 1 day's work}} = \frac{1}{{12}} \cr & \left( {{\text{B}} + {\text{C}}} \right){\text{'s 1 day's work}} = \frac{1}{8} \cr & \left( {{\text{A}} + {\text{C}}} \right){\text{'s 1 day's work}} = \frac{1}{{16}} \cr} $$
Adding, we get 2(A + B + C)'s 1 day's work
$$\eqalign{ & = \left( {\frac{1}{{12}} + \frac{1}{8} + \frac{1}{{16}}} \right) \cr & = \frac{{13}}{{96}} \cr} $$
So, A, B and C together can complete the work in
$$\eqalign{ & = \frac{{96}}{{13}} \cr & = 7\frac{5}{{13}}{\text{days}} \cr} $$

$$\eqalign{ & {\text{Let additional men be }}x \cr & \frac{{25 \times 24}}{{{\text{Work}} \to \frac{1}{3}}} = \frac{{\left( {25 + x} \right) \times 12}}{{\frac{2}{3}\left( {{\text{Remaining work 1}} - \frac{1}{3}} \right)}} \cr & \Rightarrow x = 75 \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & {\text{Let, D is number of days}} \cr & \frac{{639 \times 12 \times 5}}{{1{\text{ road}}}} = \frac{{30 \times 6 \times {\text{D}}}}{{1{\text{ road}}}} \cr & \Leftrightarrow {\text{D}} = {\text{213 days}} \cr} $$

$$\eqalign{ & {\text{Here work is 280 m length of wall }} \cr & {\text{And 100 m length of wall}} \cr & {\text{Let, M men will finish 100 m wall}}{\text{.}} \cr & \frac{{72 \times 21}}{{280}} = \frac{{{\text{M}} \times {\text{18}}}}{{100}} \cr & \Leftrightarrow {\text{M}} = 30 \cr} $$

$$\eqalign{ & \frac{{{4_{{\text{mat - wevers}}}} \times {4_{{\text{days}}}}}}{{{4_{{\text{mats}}}}}} = \frac{{{8_{{\text{mat - wevers}}}} \times {8_{{\text{days}}}}}}{{{{\text{N}}_{{\text{mats}}}}}} \cr & \Leftrightarrow {\text{N}} = {\text{16 mats}} \cr} $$

$$\eqalign{ & {\text{A's 1 hour's work}} = \frac{1}{4} \cr & \left( {{\text{B}} + {\text{C}}} \right){\text{'s 1 hour's work}} = \frac{1}{3} \cr & \left( {{\text{A}} + {\text{C}}} \right){\text{'s 1 hour's work}} = \frac{1}{2} \cr & \left( {{\text{A}} + {\text{B}} + {\text{C}}} \right){\text{'s 1 hour's work}} \cr & = \frac{1}{4} + \frac{1}{3} \cr & = \frac{7}{{12}} \cr & \therefore {\text{B's 1 hour's work}} \cr} $$
= (A + B + C)'s 1 hour's work - (A + C)'s 1 hour's work
$$\eqalign{ & = \frac{7}{{12}} - \frac{1}{2} \cr & = \frac{1}{{12}} \cr} $$
So, B alone can complete the work in 12 hours.