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.

Short questions By selling a radio with 25% profit, the profit amount is Rs 60 for how much price was it sold? If a carpet that cost Rs 4500 is sold at a profit of 30%, what is the selling price ? A watch was bought for Rs 3405.50. At what price should it be sold to gain Rs 120. Bijay bought an old house for Rs 220,000 and spend Rs 83,500 for its repair and decoration. If he sold the house for Rs 300,000 find his loss percentage. A radio is sold for Rs 2700, there is a loss of 10% at what price should it be sold to gain 712%. What percent of discount should be given in a doll costing Rs. 180 such that a customer has to buy it for Rs 160? At 18% loss the selling price of an articles is Rs 164. Calculate the cost price of the article. How much should you have to pay to buy the volleyball shown in the figure?

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?
8 days
20 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?
13 days
9 days
18 days
20 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
16 days
14 days
12 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
22 days
24 days
18 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?
12 days
18 days
22 days
20 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?
15 days
16 days
23 days
18 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?
45 days
60 days
40 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
18 days
9 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 ?
17 days
21 days
15 days
10 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?
50,115 days
51,116 days
55,117 days
52,117 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?
30 days
45 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
8days
11 days
5 days

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?
4 men
5 men
6 men
3 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?
8 days
10 days
5 days
3 days

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AX 2=4 
Jan 12, 2017 
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Sundar KhatiwadaA can do a work in 12 days and B can destroy it in 3 days.If A does the work for 10 days and then B destroy the work up to 2 days. In how many days the work will be finished? 
Jan 12, 2017 
0 Replies View Replies 

krishnaA contracter undrtook to finish the work in 20 days.He employed 40 men to do the work.After 16 days,it was found that 2/3 of the work was completed.How many extra men should be employed so that the work should be finished in time? 
Jan 10, 2017 
1 Replies View Replies 

A contracter undrtook to finish the work in 20 days.He employed 40 men to do the work.After 16 days,it was found that 2/3 of the work was completed.How many extra men should be employed so that the work should be finished in time? 
Jan 10, 2017 
0 Replies View Replies 