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 andTime
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.
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 (x6) 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,
\(\frac{3}{x} + \frac{6}{y}= 1 \: \: \: ........(2)\)
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.
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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A and B can do a piece of work in 15 and 20 days respectively. After working 4 days together, B left and A complete the remaining works. How any days the remaining work completed?
20 days
8 days
2 days
15 days

A and B can finish a piece of work in 16 and 24 days respectively.They worked together for 6 days and A left the job. In how many days B can complete the remaining work?
18 days
20 days
13 days
9 days

Sakar can do a work in 24 days and Bivas can do the same work in 30 days. They worked together for 8 days then Sakar goes away. In how many days will Bivas finish the remaining work ?
10 days
12 days
16 days
14 days

Shiva and Bishnu can do a piece of work in 12 days. Shiva alone can do it in 18 days. Both start the work together but Shiva leaves after 6 days. How long Bishnu work to finish the remaining work ?
26 days
18 days
22 days
24 days

Beena and Meena together can do a piece of work in 12 days. Meena alone can finish the work in 18 days. If she leaves after working alone for 6 days, in how many days will Beena finish the remaining work?
18 days
20 days
22 days
12 days

Bimila and Sharda can do a piece of work in 8 days. Bimala alone can do that work in 12 days. If Bimila worked alone for 4 days and stopped to do, how long will Sharda take to complete the remaining work?
16 days
15 days
18 days
23 days

A and B can finish a piece of work in 40 days. after they worked for 25 days ,B left.If a can finish the remaining work in 45 days, how long will B alone take to finish the whole work?
40 days
60 days
45 days
50 days

A can do a piece of work in 18 days and B in 24 days. They both worked together and after some days A left. If B finished the remaining work in 3 days, for how long did A work together?
15 days
9 days
18 days
12 days

A and B can do a piece of work in 30 and 45 days respectively. They work it together for some days then B leaves. If B leaves 5 days before the completion of the work, for how many days did they work together ?
10 days
17 days
21 days
15 days

A and B together can finish a work in 36 days. If A can do as much in 4 days as B can do it 9 days, how long would each take to do work alone?
52,117 days
55,117 days
51,116 days
50,115 days

X, Y and Z can do a piece of work in 20 days working together.Y can do the same work in 60 days and Z can do it in 120 days. In how many days X can do the whole work alone?
45 days
30 days
40 days
35 days

A, B and C can do a piece of work in 30, 40, and 60 days respectively. All of them started the work together but B left after working for 8 days. In how many days A and C can complete the remaining work?
10 days
11 days
5 days
8days

A man undertakes to do a certain job in 32 days. He hires 12 men for the job. At the end of 8 days, only onefifth of the work was completed. How many extra men must be employed in order to complete the work in time?
5 men
3 men
4 men
6 men

10 workers can do a work in 20 days, when they start the work, after how many days should 5 workers leave the work so that the work would be finished in 35 days?
3 days
10 days
8 days
5 days

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A and B can do a piece of work in 12 and 18 days respectively. both started the work together, but A left the work after some days and B completes the remaining work in 6 days. find in how many days A worked together?nothing 
Feb 23, 2017 
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arun paneru6 men and 8 boys can do some part of work in 10 days . 26 men and 48 boys can do same work in 2 days . how many days will 15 men and 20 boys can do same work ? 
Feb 22, 2017 
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MD KAMAL REZA 
Feb 22, 2017 
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why is time madewhy bis time made 
Feb 05, 2017 
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