## Time and Work

Subject: Compulsory Mathematics

#### Overview

The unitary method is a method or technique in an algebra for solving a class of problems in variation.Time and work are also related to the unitary method. Here, we calculate the part of a work in unit time and aslo calculate the time taken to do a work.

##### Time and Work

The unitary method is a method or technique in an algebra for solving a class of problems in variation.Time and work are also related to the unitary method. Here, we calculate the part of a work in unit time and also calculate the time taken to do a work.

• Work from Days:
If A can do a piece of work in n days, then A's 1 day's work =$\frac{1}{n}$

• Days from Work:
If A's 1 day's work = $\frac{1}{n}$,then A can finish the work in n days.

• Ratio:
If A is thrice as good workman as B, then:
The ratio of work done by A and B = 3: 1.
The ratio of times taken by A and B to finish a work = 1: 3.

• No. of days = total work / work done in 1 day

• Relationship between Men and Work
More men —can do → More work
Less men —can do →Less work

• Relationship between Work and Time
More work —takes →More Time
Less work —takes →Less Time

• Relationship between Men and Time
More men —can do in →Less Time
Less men —can do in →More Time

##### Things to remember
• A complete work is considered as 1.
• Work done by A in unit time added to the work done by B in unit time, gives the time work done by A and B together in unit time.
• The work done by A in unit time is subtracted from the work done by A and B together works in unit time gives the work done by B in unit time.
• Total work is done in different steps is always 1.
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.
##### Work, Rates, Time To Complete a Task

Solution:

Ram can do 1 work in 6 days
Ram can do $\frac{1}{6}$ work in 1 day

Ramesh can do 1 work in 9 days
Ramesh can do $\frac{1}{9}$ work in 1 day

Ram + Ramesh can do $\frac{1}{6} + \frac{1}{9}$ work in 1 day
They can do $\frac{3 + 2}{18} = \frac{5}{18}$ work in 1 day
They can do $\frac{5}{18} \times 3$ work in 3 days
They can do $\frac{5}{6}$ work in 3 days

Remaining work $= 1 - \frac {5}{6} = \frac {6 - 5}{6} = \frac{1}{6}$ work.
Ram can do $\frac{1}{6}$ work in 1 day
Total days = (3 + 1)days = 4 days Ans.

Solution:

Bhanu can do 1work in 20 days
Bhanu can do $\frac{1}{20}$ in 1 day

Hari can do 1 work in 25days
Hari can do$\frac{1}{25}$ work in 1 day

Bhanu + Hari can do $\frac{1}{20} + \frac{1}{25}$ work in 1 day
Bhanu + Hari can do$\frac{5 + 4}{100} = \frac{9}{100}$work in 1 day

Bhanu + Hari can do $\frac{9}{100} \times 5$work in 5 days
They can do$\frac{9}{20}$ work in 5 days.

Remaining work$= 1 - \frac{9}{20}= \frac{20 - 9}{20}= \frac{11}{20}$work.
Bhanu can do 1 work in 20 days.
Bhanu can do$\frac{11}{20}$ work in$\frac{20 \times 11}{20}$ days = 11 days Ans.

Solution:

Madan can do$\frac{2}{5}$ work in 9 days.
Madan can do$\frac{2}{5} \times \frac{1}{9} = \frac{2}{45}$work in 1 day

Remaining work =$1 - \frac{2}{5}= \frac{5 - 2}{5}= \frac{3}{5}$work.

Madan + Amar can do$\frac{3}{5}$work in 6 days
They can do$\frac{3}{5}\times \frac{1}{6}= \frac{1}{10}$work in 1 day.

Amar can do$\frac{1}{10} - \frac{2}{45}= \frac{9 - 4}{90}$work in 1 day

He can do $\frac{5}{90}= \frac{1}{18}$work in 1 day.

Solution:

Let, the work completed days be 'x'

A can do 1 work in 24 days.
A can do$\frac{1}{24}$work in 1 day.
A can do$\frac{4}{24} = \frac{1}{6}$work in 4 days.

B can do 1 work in 32 days.
B can do $\frac{1}{32}$ work in 1 day.
B can do $\frac{x - 6}{32}$ work in (x-6) day.

C can do 1 work in 48 days.
C can do $\frac{1}{48}$work in 1 days.
C can do$\frac{x}{48}$work in x days.

From Question,
\begin{align*} \frac{1}{6} +\frac{x - 6}{32} +\frac{x}{48} &= 1\\ or, \: \frac{16 + 3x -18 + 2x}{96} &=1\\ or, \: 5x - 2 &= 96 \\ or, \: 5x &= 98 \\ \therefore x &= \frac{98}{5} &= \frac {3}{5}19 \: days \end{align*}

$\therefore \text {The work will be completed in }\frac {3}{5}19 \: days. \: \: Ans.$

Solution:

Let, the work in completed in x days.
A work for 5 days, B worked for x days and C worked for (x - 5) days

A can do 1 work 10 days.
A can do $\frac{1}{10}$ work 1 days.
A can do $\frac {5}{10} = \frac {1}{2}$ work in 5 days.

B can do 1 work in 20 days.
B can do $\frac{1}{20}$ work in 1 day.
B can do $\frac{x}{20}$ work in xdays.

C can 1 work in 30 days.
C can do $\frac{1}{30}$ work in 1 day.
C can do $\frac{x - 5}{30}$ work in x - 5 days.

From Question,

\begin{align*} \frac{1}{2} +\frac{x}{20} +\frac{x - 5}{30} &= 1 \\ or, \: \frac{30 + 3x +2x -10}{60}&= 1\\or, 5x &= 60 - 20\\ or, x &= \frac{40}{5}\\ \therefore x &= 8\: days\end{align*}

$\therefore$ the work completed in 8 days.

Solution:

A can do 1 work in 30 days.
A can do $\frac{1}{30}$ work in 1 day.
A can do$\frac{14}{30} = \frac{7}{15}$ work in 14 days.

B can do 1 work in 40 days.
B can do$\frac{1}{40}$work in 1 day.
B can do$\frac{10}{40}= \frac{1}{4}$work in 10 days.

Remaining work$= 1 - \frac{7}{15} - \frac{1}{4}= \frac{60 - 28 - 15}{60} = \frac{17}{60}$ work.

C can do 1 work in 60 days.
C can do$\frac{17}{60}$ work in $60 \times \frac{17}{60}$work.

$\therefore$ C can do this work in 17 days.

Solution:

A + B can do 1 work in 10 days.
A + B can do $\frac{1}{10}$ work in 1 day.

B + C can do 1 work in 15 days.
B + C can do $\frac{1}{15}$ in 1 day.

A + C can do 1 work in 25 days.
A + C can do $\frac{1}{25}$ work in 1 day.

A + B + B + C + A + C = 2A + 2B + 2C can do$\frac{1}{10}+\frac{1}{15} + \frac{1}{25}$ work in 1 day.

A + B + C can do$\frac{1}{2} \left(\frac{30 + 20 + 12}{300} \right) = \frac{31}{300}$ work in 1 day.

A + B + C can do 1 work in$\frac{300}{31}$ days.

A + B + C can do 2 work in$\frac{600}{31}$ days. Ans.

Solution:

X can do 1 work in 20 days.
X can do $\frac{1}{20}$ work in 1 day.

Y can do 1 work in 30 days.
Y can do $\frac{1}{30}$work in 1 day.

Z can do 1 work in 40 days.
Z can do $\frac{1}{40}$work in 1 day.

X + Y + Z can do $\frac{1}{20} + \frac{1}{30} + \frac{1}{40}$ work in 1 day.
X + Y + Z can do$\frac{6 + 4 + 3}{120}= \frac{13}{120}$work in 1 day.
In 5 days X + Y + Z can do$\frac{13}{120} \times 5 = \frac{13}{24}$work.

Remaining work$= 1 - \frac{13}{24} = \frac{11}{24}$work.

Y + Z can do$\frac{1 }{30} + \frac{1}{40}$ work in 1 day.
Y + Z can do$\frac{4 + 3}{120} = \frac{7}{120}$work in 1 day.

Y + Z can do 1 work in$\frac{120}{7}$days.
Y + Z can do$\frac{11}{24}$ work in$\frac{120}{7} \times \frac{11}{24}$day $= \frac {55}{7} days.$

$\therefore$ the remaining work is completed in $7\frac{6}{6}$

Solution:

Bipin can do 1 work in x days.
Bipin can do $\frac{1}{x}$ work in 1 day.
Bipin can do$\frac{5}{x}$work in 5 day.

Gaurav can do$\frac{5}{x}$work in 4 days.
Gaurav can do$\frac{5}{4x}$work in 1 day.

Bipin + Gaurav can do$\frac{1}{x} + \frac{5}{4x}$work in 1 day.
Bipin + Gaurav can do$\frac{4 + 5}{4x} = \frac{9}{4x}$work in 1 day.
Bipin + Gaurav can do 1 work in$\frac{4x}{9}$ days.

From Question,

\begin{align*} \frac{4x}{9} &= 20 \\ or, x &= \frac{20 \times 9}{4}\\ &= 45 \: days \end{align*}

Bipin can do$\frac{1}{45}$work in 1 days.

Bipin + Gaurav can do$\frac{1}{20}$work in 1 day.

Gaurav can do$\frac{1}{20} - \frac{1}{45}$work in 1 day.
Gaurav can do$\frac{9 -4 }{180} = \frac{5}{180}= \frac{1}{36}$work in a day

Gaurav can do 1 work in 36 days.
Bipin takes 45 days for same work.

Solution:

A can do 1 work in 8 days.
A can do $\frac{1}{8}$work in 1 day.

B can do 1 work in 12 days.
B can do $\frac{1}{12}$work in 1 day.

A + B can do$\frac{1}{8} + \frac{1}{12}= \frac{3 + 2}{24}= \frac{5}{24}$ work in 1 day.
A + B can do $\frac{5}{24} \times 3$ work in 3 days.
A + B can do$\frac{5}{8}$ work in 3 days.

Remaining work = $1 - \frac{5}{8} = \frac{8 - 5}{8} = \frac{3}{8}$ work.

B + C can do $\frac{3}{8}$ work in 4 days.
B + C can do $\frac{3}{8} \times \frac{1}{4} = \frac{3}{32}$ work in 1 day.

C can do $\frac{3}{32} - \frac{1}{12}$ work in 1 day.
C can do $\frac{9 - 8}{96}= \frac{1}{96}$ work in 1 day.

$\therefore$ C can do 1 work in 96 days.

Solution:

Ajanta can do 1 work in 15 days.
Let, Amita can do 1 work in x days
From question,

\begin{align*} x - x \times 25\% &= 15 \\ or, x - x \times \frac{25}{100} &= 15\\or, x - \frac{x}{4} &= 15\\ or, \frac{4x - x}{4} &= 15 \\ or, \frac{3x}{4} &= 15\\ or, x &= \frac{15 \times 4}{3}\\ &= 20 \: days \end{align*}

Ajanta can do$\frac{1}{15}$ work in 1 day.

Amita can do 1 work in 1 day.
Amita can do $\frac{1}{20}$ work in 1 day.

Ajanta + Amita can do$\frac{1}{15} + \frac{1}{20}$ work in 1 day.
Ajanta + Amita can do$\frac{4 + 3}{60} = \frac{7}{60}$ work in 1 day
Ajanta + Amita can od 1 work in$\frac{60}{7}$ days.

Solution:

Let, A for x and B for y days to complete the work.

A can do 1 work in x days.
A can do$\frac{1}{x}$ work in 1 day.
A can do$\frac{2}{x}$ work in 2 days.
A can do$\frac{3}{x}$ work in 3 days.

B can do 1 work in y days.
B can do$\frac{1}{y}$ work in 1 day.
B can do$\frac{9}{y}$ work in 9 days.
B can do$\frac{6}{y}$ work in 6 days.

From first condition,
$\frac{2}{x} + \frac{9}{y} = 1 \: \: ......... (1)$

From Second condition,
$\frac{3}{x} + \frac{6}{y}= 1 \: \: \: ........(2)$

Eqn (1) is multiply by 3 andEqn (2) is multiply by 2 and subtract.

\begin{array}{rrrr} \frac{6}{x} &+ \frac {27}{y} &= 3\\ \frac{6}{x} &+ \frac {12}{y} &= 2\\ -&-&-\\ \hline\\ &\frac{15}{y}&=&1\\\end{array}

\begin{align*} y&= 15\\ \end{align*}

Putting the value of y in equation (1)

\begin{align*}\frac{2}{x} + \frac {9}{y} &= 1\\ or, \frac{2}{x} + \frac{9}{15} &= 1\\ or, \frac{2}{x}&= 1 - \frac{9}{15}\\ or, \frac{2}{x} &= \frac{15 - 9}{15} \\ or, \frac{2}{x}&= \frac{6}{15}\\ or, \frac{2}{x}&= \frac{2}{5}\\ \frac{2 \times 5}{2} &= x \\ \therefore x &= 5\end{align*}

A complete work in 5 days.
B complete work in 15 days.

Solution:

Let, a number of men employed in the beginning = x

 Men Days Work x 20 $\frac{1}{2}$ x + 60 (30 - 20)= 10 $\frac{1}{2}$

The relation between men and work are in direct variation, men and days are in indirect variation.

\begin{align*} \frac{x}{x + 60} &= \frac{10}{20} \times \frac{\frac{1}{2}}{\frac{1}{2}}\\or, \frac{x}{x + 60} &= \frac{1}{2}\\ or, 2x &= x + 60\\or, 2x - x &= 60 \\ \therefore x &= 60 \end{align*}

$\therefore$ the no. of men employed at first = 60.Ans.

Solution:

Remaining days = 60 - 40 = 20 days
Remaining work $= 1 - \frac{1}{2} =\frac{1}{2} \: work$

 Men Days Work 60 40 $\frac{1}{2}$ 60 + x 20 $\frac{1}{2}$

Men & work are in direct variation and men and days are in indirect variation.

\begin{align*}\frac{x + 60}{60} &= \frac{40}{20} \times \frac{\frac{1}{2}}{\frac{1}{2}} \\ or, \frac{x + 60}{60} &= 2 \\ or, x + 60 &= 120\\ or, x &= 120 - 60 &= 60 \end{align*}

$\therefore$ The no. of added people= 60 Ans.

Solution:

Let, the no. of labour of first be 'x'
Remaining days = 90 - 60 = 30 days
Remaining work = $1 - \frac {3}{5} = \frac{5 - 3}{5} = \frac{2}{5}$ work

 Men Days Work x 60 $\frac{3}{5}$ x + 15 30 $\frac{2}{5}$

Men and work are in direct proportion and men and days are in indirect proportion.

\begin{align*} \frac{x }{x + 15} &= \frac{30}{60} \times \frac{3}{5} \times \frac{5}{2}\\ or, \frac{x}{x + 15} &= \frac{3}{4} \\ or, 4x &= 3x + 45\\ or, 4x - 3x &= 45\\ \therefore x &= 45 \end{align*}

$\therefore$ The number of labours at first = 45Ans.

Solution:

Let, 4 men are added after x days.

 Men Days 20 x 20 + 4 =20 24 - 3 - x =21 - x

20 men can do 1 work in 24 days.

20 men can do $\frac{1}{24}$ work in 1 day.

20 men can do $\frac{x}{24}$ work in x days.

20 men can do $\frac{1}{24}$ work in 1 day.
1 men can do$\frac{1}{20 \times 24}$ work in 1 day.

24 men can do$\frac{1}{20} \times (21 - x)$ work in (21 - x ) days.

From above,

\begin{align*} \frac{x}{24} + \frac{21 - x}{20} &= 1 \\ or, \frac{5x + 126 - 6x}{120} &= 1\\ or, 126 - x &= 120\\ or, x &= 126 - 120 \\ \therefore x &= 6 \: days \end{align*}

$\therefore$ After 6 days men were added.Ans.

Solution:

A can fill 1 tank in 8 minutes.
A can fill $\frac{1}{8}$ tank in 1 minute.

B can fill 1 tank in 12 minutes.
B can fill $\frac{1}{12}$ tank in 1 minute.

C can empty 1 tank in 16 minutes.
C can empty$\frac{1}{16}$ tank in 1 minute.

Three taps fills$\frac{1}{8} + \frac{1}{12} - \frac{1}{16}$tank in 1 minute.

Three taps fills$\frac{6 + 4 - 3}{48}$ tank$= \frac{7}{48}$ tank in 1 minute.

Three tap fill 1 tank in $\frac{48}{7}$ $=6 \frac{6}{7}$ minutesAns.

Solution:

20 men can do 1 work in 24 days.
20 men can do $\frac{1}{24}$ work in 1 day.
2o men can do$\frac{6}{24} = \frac{1}{4}$ work in 6 days.

Remaining work =$1 - \frac{1}{4} = \frac{4 -1}{4} = \frac{3}{4}$ work.

Remaining days after added men = 21 - 6 = 15 days

In 24 days 1 work done by 20 men.
In 1 day 1 work done $20 \times 24$ men.

In 15 day 1 work done by$\frac{20 \times 24}{15}$ men.

In 15 day $\frac{3}{4}$ work done by$\frac{20 \times 24}{15} \times \frac{3}{4}$ men $= 24 \: men$

$\therefore$ Added men = 24 - 20 = 4 men

Solution:

A can do 1 work in 8 day.
A can do $\frac{1}{8}$ work in 1 day.
A can do$\frac{2}{8} = \frac{1}{4}$ work in 2 days.

B can do 1 work in 12 days.
B can do$\frac{1}{12}$ work in 1 day.
B can do$\frac{3}{12} = \frac{1}{4}$ work in 3 days.

Remaining work$= 1 - \frac{1}{4} - \frac{1}{4} = \frac{4 - 1 -1}{4} = \frac{2}{4} = \frac{1}{2}$ work.

A + B can do$\frac{1}{8} + \frac{1}{12}$ work in 1 day.
A + B can do$\frac{3 + 2}{24} = \frac{5}{24}$ work in 1 day.
A + B can do 1 work in$\frac{24}{5}$ days.
A + B can do $\frac{1}{2}$ work in $\frac{24}{5} \times \frac {1}{2} = \frac{12}{5} = 2\frac{2}{5}$ days.