Time And Work - Aptitude MCQ Questions and Solutions with Explanations

Number of page copied by (Subhash + Prakash) in 1 hour = $$\frac{{300}}{{40}}$$  = 7.5 pages;
Subhash copied pages in one hour = $$\frac{{50}}{{10}}$$ = 5 pages
Hence, Prakash copied pages in one hour = 7.5 - 5 = 2.5
Thus,
Prakash can copied 30 pages in = $$\frac{{30}}{{2.5}}$$ = 12 hour

Variation Method:
In 100 days only 2.5 km road i.e. 16.66 % of work has been completed

Men	Days	Road (km)
45	100↓	2.5
x↑	200	12.5↑

Arrows show the directions of variation of quantity with respect to each other
$$\frac{{\text{x}}}{{45}} = \frac{{100 \times 12.5}}{{200 \times 2.5}}$$
x = 113 men;
Required men to be increased,
= 113 - 45
= 68

Work equivalence method:
1200 × 45 = 1000 × x
Hence, x = 54 days

Variation Method:
After 15 days 200 soldiers leaved.
Soldiers              Food for days
1200 ↓                 45
1000                 ↑ x (let)
Arrows show the opposite variation to each other
$$\frac{{1200}}{{1000}} = \frac{{\text{x}}}{{45}}$$
Or, x = $$\frac{{1200 \times 45}}{{1000}}$$   = 54 days.

One Day's work of A and B together,
$$\frac{1}{{\text{A}}} + \frac{1}{{\text{B}}} = \frac{1}{5}\,........({\text{i}})$$

When A works with twice efficiency,Then,
$$\frac{2}{{\text{A}}} + \frac{1}{{3{\text{B}}}} = \frac{1}{3}\,........({\text{ii}})$$

on solving equations (i) and (ii), we get
$$A = \frac{{25}}{4} = 6\frac{1}{4}$$

Let the obstruction remain for x min.
Hence,
Part of cistern filled in X min + part of cistern filled in 3 min = full cistern
$$\left[ {\frac{{7{\text{x}}}}{{8 \times 12}} + \frac{{5{\text{x}}}}{{6 \times 16}}} \right]$$    $$ + $$ $$\left[ {\frac{3}{{12}} + \frac{3}{{16}}} \right]$$   = 1
$$\frac{{12{\text{x}}}}{{96}} + \frac{7}{{16}} = 1$$
Thus,
X = 4.5 min.

Let pipe C alone can empty the cistern in x min.
A fills cistern in 1 min = $$\frac{1}{{10}}$$
B fills cistern in 1 min = $$\frac{1}{{15}}$$
A and B together fill in 1 min $$ = \frac{{10 \times 15}}{{10 + 15}} = \frac{{150}}{{25}} = 6\,{\text{min}}$$
Since, waste pipe was left open for 6 min then,
6 min, $$\frac{6}{x}$$ part of cistern will be emptied by C
Now,
$$\frac{6}{x}$$ part of the cistern would be filled by A and B in 2 min.
Hence,
cistern will be filled in $$\frac{3}{x}$$ min. And
$$\frac{x}{3}$$ = 6
x = 18 min.

Let 1 boy's 1 day's work = x
And 1 girl's 1 day's work = y
Now,
(5 boys + 2 girls)'s work = 5x + 2y
Given ,
5x + 2y is equal to 4 times work done by a boy and a girl
Thus,
5x + 2y = 4(x + y)
5x + 2y = 4x + 4y
x = 2y
$$\frac{{\text{x}}}{{\text{y}}} = \frac{2}{1}$$
Hence, the required ratio is 2 : 1

Let x be the required number of days
Given,
12 men and 12 women can complete a work separately in 14 days and 21 days respectively
Then,
12 men's 1 day work = $$\frac{1}{{14}}$$
And,
12 women's 1 day work = $$\frac{1}{{21}}$$
Then ,
12 women's 3 days work = $$\frac{3}{{21}}$$ = $$\frac{1}{7}$$

The remaining work = $$1 - \frac{1}{7}$$  = $$\frac{6}{7}$$
Man's group leaves 3 days before the completion of work
That is, they were working together for x - 3 days
Thus, we have $$\frac{1}{7}$$ work left to be done in last 3 days by the women's group. This also means $$\frac{6}{7}$$ th of work has been done by both the groups (before men left)
Now, (12 men + 12 women)'s 1 day work = $$\frac{1}{{14}} + \frac{1}{{21}}$$   = $$\frac{5}{{42}}$$
i.e., $$\frac{5}{{42}}$$ work is done by 2 groups in 1 day.
So, $$\frac{6}{7}$$ of work is done by 2 groups together in $$\frac{{42}}{5} \times \frac{6}{7}$$   = $$\frac{{36}}{5}$$ days
Total time take to complete the work will be
= $$\frac{{36}}{5}$$ + 3 = $$\frac{{51}}{5}$$

Vimal's 1 day work = $$\frac{1}{{20}}$$
Since, Vimal and Kamal can together complete in 12 days
i.e. (Vimal + Kamal)'s 1 day work = $$\frac{1}{{12}}$$
Then,
Kamal's 1 day work,
$$ = \frac{1}{{12}} - \frac{1}{{20}} \Rightarrow \frac{2}{{60}} \Rightarrow \frac{1}{{30}}$$
If Kamal Works only for half a day daily, then his 1 day work becomes $$\frac{1}{2} \times \frac{1}{{30}}$$  = $$\frac{1}{{60}}$$
Therefore, 1 day work of both Vimal and Kamal,
$$ = \frac{1}{{20}} + \frac{1}{{60}} \Rightarrow \frac{4}{{60}} \Rightarrow \frac{1}{{15}}$$
Hence, the work will be completed in 15 days.

Combined efficiency of all the three boats = 60 passengers /trip
Now, consider option (A)
15 trips and 150 passengers means efficiency of A = 10 passengers per trip
A's efficiency = 10 passengers per trip
Then, (B + C) combined efficiency = 50 passengers per trip
Since, combined efficiency is 60 so option (A) is correct