Subject: Compulsory Mathematics

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.

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

- 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.

Ram can do a work in 6 days and Ramesh can do the same work in 9 days. They work together for 3 days. Ram left the work then

(a) In how many days can Ram will do the remaining work along?

(b) How many days are needed to complete the whole work?

**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.

Bhanu and Hari can do a piece of work in 20 days and 25 days respectively. They work together for 5 days and Hari goes away. In how many days will Bhanu finish the remaining work?

**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.**

Madan can do \( \frac{2}{5}\) part of work in 9 days. He calls Amar to finish the remaining works, they work together and finish work in 6 days. How many days will be taken to finish work alone?

**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.

A, B and C can do a piece of work in 24, 32 and 48 days. Three of them started the work together. A left the work after the 4 days and B left it in 6 days before it completion. Find in how many days the work might have been finished.

**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*}

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

A can do a piece of work in 10 days, B can do it in 20 days and C can do it in 30 days. Three of them started the work together A left the work after 5 days and C left it in 5 days before it completion. Find in how many days the work might have been finished?

**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.

A, B and C can finish a piece of work in 30, 40 and 60 days respectively. 10 days after they started to work together B leaves. A leaves 4 days after B left, and C completes the remaining work. Find how many days C had worked together?

**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.

A and B can do a piece of work in 10 days, B and C in 15 days and A and C in 25 days. In how many days they can finished double of the same work working together.

**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.**

X, Y and Z can finish a piece of work in 20, 30 and 40 days respectively. If X left the work after working for 5 days, In how many days can Y and Z together complete the remaining work?

**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 \(7\frac{6}{6}\)

Gaurav does as much work work in 4 days as Bipin does in 5 days. If they together can do it in 20 days, how long will each work take. If they work along?

**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.

A and B can do a piece of work in 8 and 12 days respectively. They start the work together out A leaves after 3 days. Then B calls C and they completed the remaining work together in 4 days. In how many days C alone can do the whole 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.

Ajanta can do a piece of work in 15 days. Amita is found to be 25% less efficient than Ajanta. Find in how many day will

(a) Amita do the work alone.

(b) Ajanta and Amita do the same work together.

**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.

A is engaged to do a piece of work. After working for 2 days he leaves and B finished the remaining work in 9 days. A had left the work after working for 3 days B would have finished the work in 6 days. In how many days can each working along finished the whole work.

**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,

\(\frac{3}{x} + \frac{6}{y}= 1 \: \: \: ........(2)\)

Eq^{n} (1) is multiply by 3 andEq^{n} (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.

A contractor had to finish a work in 30 days and he employed some men to do work. They finished half of the work in 20 days. When 60 more man were added, the work was finished on the specific time. How many men were employed in the beginning?

**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.

A contractor had to finish a work in 60 days and he employed 60 labours to do the work. They finished half of the work in 40 days. How many more labours should be added to finish the work in specific time.

**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.

A contractor committed to finish a work in 90 days. He employed some labour upon it, but they could do only \(\frac{3}{5}\) of the work in 60 days. He employed 15 more labours and finished the work in time. How many labours had he employed first?

**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.

20 men can do a piece of work in 24 days. After working for few days, 4 men are added and the work was finished 3 days earlier. After how many days were 4 men added?

**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.

There are three water taps in a water tank. Tap A can fill the tank in 8 minutes, B can fill in 12 minutes and C can empty that tank in 16 minutes. If all three taps are opened together, how long will it take to fill that empty tank.

**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.

20 men can do a piece of work in 24 days. After working for 6 days and additional number of men is taken to finish the work in 21 days from the beginning. Find the number of additional men.

**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

A can do a piece of work in 8 days and B can do the same work in 12 days. A worked alone for 2 days and B for 3 days. In how many days will the remaining work be finished. If they work together? Find it.

**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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