WBBSE Solutions For Class 8 Maths Arithmetic Chapter 7 Time And Work

Arithmetic Chapter 7 Time And Work

1. If the number of workers is fixed, then the amount of work done is directly proportional to the time spent on doing the work.
2. If the allotted time for doing the work is constant, then the number of workers required varies directly with the amount of work to be done.
3. If the amount of work to be done is fixed, then the time taken to do a work varies inversely to the number of workers doing it.

Arithmetic Chapter 7 Time And Work Examples

Example 1. 4 tractors can plough 28 bighas in a day. Calculate how many bighas can be ploughed by 8 tractors in a day.

Solution: 4 tractors in a day can plough 28 bighas.

1 tractors in a day can plough \(\frac{28}{4}\) bighas.

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8 tractors in a day can plough \(\frac{28 \times 8}{4}\) bighas = 56 bighas

∴ 56 bighas can be ploughed by 8 tractors in a day.

Example 2. In a factory 120 parts of machine are made in 5 days. Calculate how many parts of the machine will be made in 9 days.

Solution: Let the number of parts of machine will be made in 9 days is x [x > 0].

In the mathematical language, the problem is:

Time (in days)
5
9

Number of parts
120
x

The number of parts of machine and time taken by a fixed number of labourers are directly proportional.

∴ 5: 9 : : 120: x

⇒ \(\frac{5}{9}\) = \(\frac{120}{x}\)

⇒ 5x = 9 × 120

⇒ x = \(\frac{9 \times 120}{5}\)

⇒ x = 216

∴ 216 parts of machine will be made in 9 days.

Example 3. If 45 men can do a piece of work in 5 days, then how many days will be needed to complete this work?

Solution: 45 men can do a piece of work in 5 days.

1 men can do this work in (5 x 45) days

15 men can do this work in \(\frac{5 \times 45}{15}\) = 15 days

∴ 15 days will be needed to complete this work.

Example 4. Mahim can do a piece of work in 10 days. Calculate the rest of the total work after 3 days.

Solution: Mahim completes the work in 10 days

In 1 day he can do the work \(\frac{1}{10}\) part

In 3 days he can do the work or part or (\(\frac{1 \times 3}{10}\)) part or \(\frac{3}{10}\)

∴ After 3 days the rest of the total work is (1-\(\frac{3}{10}\)) part or \(\frac{7}{10}\) part

∴ 3 days will be needed to complete this work.

Example 5. Aritra, Soumyadip, and Subhadip can do a piece of work separately in x days, y days, and z days respectively. If they do the work together, in how many days they together will complete the work?

Solution: Aritra alone does in 1 day the \(\frac{1}{x}\) part

Soumyadip alone does in 1 day the \(\frac{1}{y}\) part

Subhadip alone does in 1 day the \(\frac{1}{z}\) part

They together do in 1 day (\(\frac{1}{x}\) + \(\frac{1}{y}\) + \(\frac{1}{z}\)) part = \(\left(\frac{y z+x z+x y}{x y z}\right)\) part

∴ They do \(\left(\frac{y z+x z+x y}{x y z}\right)\) part of the work in 1 day.

∴ Total work i.e. 1 part is done in \(\left(1 \div \frac{x y+y z+z x}{x y z}\right)\) days = \(\frac{x y z}{x y+y z+z x}\)

∴ \(\frac{x y z}{x y+y z+z x}\) days they together will complete the work.

Example 6. B can do a piece of work in 12 days. If the efficiency of A is 60% more than the efficiency of B, then what time can A do it alone?

Solution: B does in 1 day \(\frac{1}{12}\) part

As the efficiency of A is 60% more than the efficiency of B

∴ A does in 1 day (\(\frac{1}{12}\)+\(\frac{1}{12}\)x\(\frac{60}{1200}\)) part = \(\left(\frac{1}{12}+\frac{1}{20}\right)=\left(\frac{5+3}{60}\right)\) part

= \(\frac{8}{60}\) part = \(\frac{2}{15}\) part

∴ A do \(\frac{2}{15}\) part of total work in 1 day

∴ Total work i.e. 1 part is done in (1 + \(\frac{2}{15}\)) days = 1 x \(\frac{15}{2}\) day = 7\(\frac{1}{2}\) day

∴ A can do it alone in 7\(\frac{1}{12}\) days.

Example 7. A does \(\frac{7}{10}\) part of a piece of work in 14 days and then with the help of B, A finish the work in 2 days. How long would B take to do the work alone?

Solution: A does the work in (14 + 2) days or 16 days.

A in 14 days do \(\frac{7}{10}\) part of the total work

A in 1 day do (\(\frac{7}{10 \times 14}\)) part of the total work

A in 16 days do (\(\left(\frac{7 \times 16}{10 \times 14}\right)\)) part or \(\frac{4}{5}\) part of the total work

∴ The remaining part is (1-\(\frac{4}{5}\)) part = \(\frac{1}{5}\) part

B do \(\frac{1}{5}\) part of the total work in 2 day.

B do 1 part in (2÷\(\frac{1}{5}\)) days = (2x\(\frac{5}{1}\)) days = 10 days

∴ B alone take 10 days to finish the works.

Example 8. The empty tank is filled with the first pipe in 45 minutes and the full tank is made empty with another pipe in 1 hour. If two pipes are opened together then find the time to fill the empty tank.

Solution: 1 hour 60 minutes

It takes 1 minute to fill up the \(\frac{1}{45}\) part of the empty tank by the first pipe

It takes 1 minute to make empty the \(\frac{1}{60}\) part of the tank by the second pipe

If the two pipes are open, it takes 1 minute to fill the (\(\frac{1}{45}\)–\(\frac{1}{60}\)) part = \(\frac{4-3}{180}\) part = \(\frac{1}{180}\) part

The \(\frac{1}{180}\) part is filled in in 1 minute 1 part is filled in (1÷\(\frac{1}{180}\)) minutes = (1 x 180) minutes = 3 hours

∴ If two pipe are open the full tank will be filled in 3 hours.

Example 9. A, B and C can do a piece of work in 10 days, 12 days, and 15 days respectively. If they do the work together then how many days they will complete the work? [Using proportion method]

Solution: A does in 1 day \(\frac{1}{10}\) part

B does in 1 day \(\frac{1}{12}\) part

C does in 1 day \(\frac{1}{15}\) part

Together they do in 1 day (\(\frac{1}{10}\)+\(\frac{1}{12}\)+\(\frac{1}{15}\)) part = \(\frac{6+5+4}{45}\) part = \(\frac{15}{60}\) part = \(\frac{1}{4}\) part

In the mathematical language, the problem is:

Amount of work (in part)
\(\frac{1}{4}\)
1

Time taken (in days)
1
? (x) (say)

The amount of work and time taken are directly proportional if the number of man is fixed.

∴ \(\frac{1}{4}\): 1 : : 1: x

⇒ \(\frac{\frac{1}{4}}{1}=\frac{1}{x}\)

⇒ \(\frac{x}{4}\) = 1

⇒ x = 4

∴ The required time to complete the work is 4 days.

Example 10. If 50 men working 8 hours a day can do a piece of work in 12 days, in how many days can 60 men working 10 hours a day do twice as much work?

Solution: 50 men working 8 hours a day do the work in 12 days

1 man working 8 hours a day do it in (12 x 50) days

1 man working 1 hour a day do it in (12 x 50 x 8) days

60 men working 1 hour a day do it in \(\frac{12 \times 50 \times 8}{60}\) days

60 men working 10 hours a day do it in \(\frac{12 \times 50 \times 8}{60 \times 10}\) = 8 days

∴ 60 men working 10 hours a day do twice as much work in (8 x 2) days or 16 days.

Example 11. A and B together can do a piece of work in 3 days, B and C can do a piece of work in 4 days and A and C can do a piece of work in 5 days. In how many days can each of them do it separately?

Solution: In 1 day (A + B) can do \(\frac{1}{3}\) part of the work

In 1 day (B + C) can do \(\frac{1}{4}\) part of the work

In 1 day (A + C) can do \(\frac{1}{5}\) part of the work

In 1 day twice the work of (A + B + C) together = \(\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right) \text { part }=\left(\frac{20+15+12}{60}\right)\) part = \(\frac{47}{60}\) part.

∴ In 1 day (A + B + C) can do (\(\frac{47}{60 \times 2}\)) or \(\frac{47}{120}\) part of the work

∴ In 1 day A can do (\(\frac{47}{120}\)–\(\frac{1}{4}\)) part = \(\frac{47-30}{120}\) part = \(\frac{17}{120}\) part

∴ A can do the whole work in (1÷\(\frac{17}{120}\)) days = \(\frac{1 \times 120}{17}\) days = 7\(\frac{1}{17}\) days

Similarly, in 1 day B can do the work (\(\frac{47}{120}\)–\(\frac{1}{5}\)) part = \(\frac{47-24}{120}\) part = \(\frac{23}{120}\) part

∴ B can do the whole work in \(\frac{120}{23}\) days or 5\(\frac{5}{23}\) days

In 1 day C can do (\(\frac{47}{120}\)–\(\frac{1}{3}\)) part = \(\frac{47-40}{120}\) part = \(\frac{7}{120}\) part

∴ C can do the whole work in \(\frac{120}{7}\) days or 17\(\frac{1}{7}\) days.

Example 12. X, Y and Z individually can complete a work in 10, 12 and 15 days respectively. They started doing the work together. After 3 days X left away. Calculate in how many days Y and Z will complete the remaining work.

Solution: X does in 1 day \(\frac{1}{10}\) part

Y does in 1 day \(\frac{1}{12}\) part

Z does in 1 day \(\frac{1}{15}\) Part

Together they do in 1 day (\(\frac{1}{110}\)+\(\frac{1}{12}\)+\(\frac{1}{15}\)) part = \(\frac{6+5+4}{60}\) part = \(\frac{1}{4}\) part

Together they do in 3 days \(\frac{3}{4}\) part

Remaining part of the work is (1-\(\frac{3}{4}\)) part = \(\frac{1}{4}\) part

After 3 days X left away

Y and Z together do in 1 day is (\(\frac{1}{12}\)+\(\frac{1}{15}\)) part = \(\frac{5+4}{60}\) part =\(\frac{9}{60}\) part = \(\frac{3}{20}\) part

In the mathematical language the problem is:

Amount of work (in part)
\(\frac{3}{20}\)
\(\frac{1}{4}\)

Time (in days)
1
? (x) [say]

If the number of men are fixed, the amount of work and time are in direct proportion

∴ \(\frac{3}{20}\): \(\frac{1}{4}\) : : 1: x

⇒ \(\frac{\frac{3}{20}}{\frac{1}{4}}=\frac{1}{x}\)

⇒ x x \(\frac{3}{20}\) = \(\frac{1}{4}\)

⇒ x = \(\frac{20}{3}\) x \(\frac{1}{4}\) = \(\frac{5}{3}\) = 1\(\frac{2}{3}\)

∴ Y and Z will complete the remaining work in 1\(\frac{2}{3}\) days.

Example 13. The inlet pipe of a roof tank can fill it in 30 minutes. If outlet pipe is opened to supply water to the house, the tank is completely emptied in 45 minutes. When the tank is half filled, if both the pipes are opened together then what time it take to fill the tank?

Solution: The inlet pipe of a roof tank can fill it in 1 minute \(\frac{1}{30}\) part of the tank.

The outlet pipe can empty the tank in 1 minute \(\frac{1}{45}\) part of the tank.

When both the pipes are opened together, then the tank filled in 1 minute is

(\(\frac{1}{30}\)–\(\frac{1}{45}\)) part = \(\frac{6-4}{180}\) part = \(\frac{2}{180}\) part = \(\frac{1}{90}\) part

Now, according to the problem, the tank is half filled.

So the part which is to be filled is (1-\(\frac{1}{2}\)) part = \(\frac{1}{2}\) part of the tank.

In the mathematical language, the problem is:

Part filled
\(\frac{1}{90}\)
\(\frac{1}{2}\)

Time taken (in minutes)
1
? (x) [say]

∴ Part filled and time is direct proportion

\(\frac{1}{90}\): \(\frac{1}{2}\) : : 1: x

⇒ \(\frac{\frac{1}{90}}{\frac{1}{2}}=\frac{1}{x}\)

⇒ \(\frac{x}{90}\) = \(\frac{1}{2}\)

⇒ x = \(\frac{90}{2}\) = 45

∴ Required time is 45 minutes.

Example 14. A tailor can make 12 shirts in 3 days. How long will he take to make 8 shirts?

Solution: In the mathematical language, the problem is:

No. of shirts made
12
8

Time taken (days)
3
?

So, we get in this problem is 12: 8: 3: Required time.

∴ Required time = \(\frac{8 \times 3}{12}\) = 2 days.

Example 15. Out of 100 trees, 20 wood-cutters cut 20 trees in 4 hours. After that 4 wood- cutters leave the job, then when will the whole job get finished?

Solution: 4 wood-cutters leave the job.

So number of wood-cutters left is (20-4) = 16.

Let the time taken by them be x hours.

According to the problem in mathematical language is:

No. of trees
20
(100 – 20) = 80

Number of Wood-cutters
20
(20-4) = 16

Time taken (hrs.)
4
x

So, we get that 80: 16 : : x: 4

or, \(\frac{80 \times 4}{16}\) = 20 hours

So that the total cutting is done after 4+20= 24 hours.

Example 16. Choose the correct answer:

1. In a factory 216 parts of machine are made in 3 days. The number of parts of the machine will be made in 7 days is

1. 72
2. 432
3. 648
4. 504

Solution: Let the number of parts of machine will be made in 7 days is x [x > 0]

In the mathematical language, the problem is:

Time (in days)
3
7

Number of parts
216
x

The number of parts time and time are directly proportional.

∴ 3: 7 : : 216: x ⇒ \(\frac{3}{7}\) = \(\frac{216}{x}\)

⇒ x = \(\frac{7 \times 216}{3}\) = 504

504 parts will be made in 7 days.

∴ The correct answer is 4. 504

The number of parts of the machine will be made in 7 days is 504.

2. If 3 tractors can plough 18 bighas in a day, then 7 tractors can plough in a day is

1. 42 bighas
2. \(\frac{54}{7}\) bighas
3. \(\frac{7}{6}\) bighas
4. None of these

Solution: 3 tractors in a day can plough 18 bighas

1 tractors in a day can plough \(\frac{18}{3}\) bighas

7 tractors in a day can plough \(\frac{18 \times 7}{3}\) bighas

∴ The correct answer is 1. 42 bighas.

7 tractors can plough in a day is 42 bighas.

Example 17. Write ‘True’ or ‘False’:

1. If 24 men need 12 days to cut a pond, then 36 men will be needed to did that pond in 8 days.

Solution: In 12 days to cut a pond required 24 men

In 1 day to cut that pond required (24 × 12) men

In 8 days to cut that pond required (\(\frac{24 \times 12}{8}\)) men = 36 men

∴ So the statement is true.

2. If two pipes can fill a cistern in 10 hours and 15 hours respectively, then in 5 hours they together fill it.

Solution: In 1 hour the 1st pipe fill \(\frac{1}{10}\) part and second pipe fill \(\frac{1}{15}\) part.

When both pipes are opened together it takes 1 hour to fill up the (\(\frac{1}{10}\)+\(\frac{1}{15}\)) part = \(\frac{6+4}{60}\) part = \(\frac{10}{60}\) part = \(\frac{1}{6}\) part

Thus \(\frac{1}{6}\) part is filled in 1 hour

1 part is filled in (1÷\(\frac{1}{6}\)) hours (1 x 6) hours = 6 hours

∴ So the statement is false.

Example 18. Fill in the blanks:

1. If x men can do a piece of work in y days, then z men can do it in _______ days.

Solution: x men can do the work in y days

1 man can do it in (x x y) days

z men can do it in \(\frac{xy}{z}\) days.

2. Amal and Kamal completes a work separately in x days and y days respectively. If they do the work together then they will completes the work in ______ days.

Solution: In 1 day Amal does \(\frac{1}{x}\) part

In 1 day Kamal does \(\frac{1}{y}\) part

Together they do in 1 day (\(\frac{1}{x}\)+\(\frac{1}{y}\)) part = \(\frac{y+x}{xy}\) part

∴ They do \(\frac{x+y}{xy}\) part of the total work in 1 day

Total work i.e. 1 part is done in (1÷\(\frac{x+y}{xy}\)) days

or 1 x \(\frac{xy}{x+y}\) days = \(\frac{xy}{x+y}\) days.

They do the work together then they will completes the work in 1 x \(\frac{xy}{x+y}\) days = \(\frac{xy}{x+y}\) days