Time and Work

Subject: Compulsory Mathematics

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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

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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.
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Questions and Answers

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:

Let, Added men = x
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.

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