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## Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 2 Life Mathematics Ex 2.1

Question 1.

Fill in the blanks:

(i) A can finish a job in 3 days whereas B finishes it in 6 days. The time taken to complete the job together is………..days.

(ii) If 5 persons can do 5 jobs in 5 days, then 50 persons can do 50 jobs in………..days.

(iii) A can do a work in 24 days. A and B together can finish the work in 6 days. Then B alone can finish the work in…………days.

(iv) A alone can do a piece of work in 35 days. If B is 40% more efficient than A, then B will finish the work in………..days.

(v) A alone can do a work in 10 days and B alone in 15 days. They undertook the work for Rs 200000. The amount that A will get is………

Solution:

(i) 2 days

(ii) 5

(iii) 8

(iv) 25

(v) Rs 1,20,000

Question 2.

210 men working 12 hours a day can finish a job in 18 days. How many men are required to finish the job in 20 days working 14 hours a day?

Solution:

Let the required number of men be x.

More working hours ⇒ less men required.

∴ It is inverse proportion.

∴ Multiplying factor is \(\frac{12}{14}\)

Also more number of days ⇒ less men

∴ It is an inverse proportion.

∴ Multiplying factor is \(\frac{18}{20}\)

∴ x = 210 × \(\frac{12}{14}\) × \(\frac{18}{20}\)= 162 men

162 men are required.

Question 3.

A cement factory makes 7000 cement bags in 12 days with the help of 36 machines. How many bags can be made in 18 days using 24 machines?

Solution:

Let the required number of cement bags be x.

Number of days more ⇒ More cement bags.

∴ It is direct variation.

∴ The multiplying factor = \(\frac{18}{12}\)

Number of machines more ⇒ More cement bags.

∴ It is direct variation.

∴ The multiplying factor = \(\frac{24}{36}\)

∴ x = 7000 × \(\frac{18}{12}\) × \(\frac{24}{36}\)

x = 7000 cement bags

7000 cement bags can be made.

Question 4.

A soap factory produces 9600 soaps in 6 days working 15 hours a day. In how many days will it produce 14400 soaps working 3 hours more a day?

Solution:

Let the required number of days be x.

To produce more soaps more days required.

∴ It is direct proportion.

∴ Multiplying factor = \(\frac{14400}{9600}\)

If more hours spend, less days required.

∴ It is indirect proportion.

∴ Multiplying factor = \(\frac{15}{18}\)

∴ x = 6 × \(\frac{14400}{9600}\) × \(\frac{15}{18}\)

x = \(\frac{15}{2}\)

\(\frac{15}{2}\) days will be needed.

Question 5.

If 6 container lorries transport 135 tonnes of goods in 5 days, how many more lorries are required to transport 180 tonnes of goods in 4 days?

Solution:

Let the number of lorries required more = x.

As the goods are more ⇒ More lorries are needed to transport.

∴ It is direct proportion.

∴ Multiplying factor = \(\frac{180}{135}\)

Again if more days ⇒ less number of lorries enough.

∴ It is direct proportion.

∴ Multiplying factor = \(\frac{5}{4}\)

∴ 6 + x = 6 × \(\frac{180}{135}\) × \(\frac{5}{4}\)

6 + x = 10

x = 10 – 6

x = 4

∴ 4 more lorries are required.

Question 6.

A can do a piece of work in 12 hours, B and C can do it 3 hours whereas A and C can do it in 6 hours. How long will B alone take to do the same work?

Solution:

Time taken by A to complete the work =12 hrs.

∴ A’s 1 hr work = \(\frac{1}{12}\)…………(1)

(B + C) complete the work in 3 hrs.

∴ (B + C)’s 1 hour work = \(\frac{1}{3}\)…………(2)

(1) + (2) ⇒

∴ (A + B + C)’s 1 hour work = \(\frac{1}{12}\) + \(\frac{1}{3}\) = \(\frac{1+4}{12}\) = \(\frac{5}{12}\)

Now (A + C) complete the work in 6 hrs.

∴(A + C)’s 1 hour work = \(\frac{1}{6}\)

∴ B’s 1 hour work = (A + B + C)’s 1 hour work – (A + C)’s 1 hour work

= \(\frac{5}{12}\) – \(\frac{1}{6}\) = \(\frac{5-2}{12}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\)

∴ B alone take 4 days to complete the work.

Question 7.

A and B can do a piece of work in 12 days, while B and C can do it in 15 days whereas A and C can do it in 20 days. How long would each take to do the same work?

Solution:

(A + B) complete the work in 12 days.

∴ (A + B)’s 1 day work = \(\frac{1}{12}\)……….(1)

(B + C) complete the work in 15 days

∴ (B + C)’s 1 day work = \(\frac{1}{15}\)……….(2)

(A + C) complete the work in 20 days

∴ (A + C)’s 1 day work = \(\frac{1}{20}\)……….(3)

Now (1) + (2) + (3) ⇒

[(A + B) + (B + C) + (A + C)]’s 1 day work = \(\frac{1}{12}\) + \(\frac{1}{15}\) + \(\frac{1}{20}\)

(2A + 2B + 2C)’s 1 day work = \(\frac{5}{60}\) + \(\frac{4}{60}\) + \(\frac{3}{60}\)

2(A + B + C)’s 1 day work = \(\frac{5+4+3}{60}\)

(A + B + C)’s 1 day work = \(\frac{12}{60×2}\)

(A + B + C)’s 1 day work = \(\frac{1}{10}\)

Now A’s 1 day’s work = (A + B + C)’s 1 day work – (B + C)’s 1 day work

\(\frac{1}{10}\) – \(\frac{1}{15}\) = \(\frac{3}{30}\) – \(\frac{2}{30}\) = \(\frac{1}{30}\)

∴ A takes 30 days to complete the work.

B’s 1 day work = (A + B + C)’s 1 day’s work – (A + C)’s 1 day’s work

= \(\frac{1}{10}\) – \(\frac{1}{20}\) = \(\frac{6}{60}\) – \(\frac{3}{60}\)

\(\frac{6-3}{60}\) = \(\frac{3}{60}\) = \(\frac{1}{20}\)

B takes 20 days to complete the work.

C’s 1 day work = (A + B + C)’s 1 day work – (A + B)’s 1 day work

\(\frac{1}{10}\) – \(\frac{1}{12}\) = \(\frac{6}{60}\) – \(\frac{5}{60}\) = \(\frac{6-5}{60}\) = \(\frac{1}{60}\)

∴ C takes 60 days to complete the work.

Question 8.

Carpenter A takes 15 minutes to fit the parts of a chair while Carpenter B takes 3 more minutes than A to do the same work. Working together, how long will it take for them to fit the parts for 22 chairs?

Solution:

Time taken by A to fit a chair = 15 minutes

Time taken by B = 3 minutes more than A

= 15 + 3 = 18 minutes

∴ A’s 1 minute work = \(\frac{1}{15}\)

B’s 1 minute work = \(\frac{1}{18}\)

(A + B)’s 1 minutes work = \(\frac{1}{15}\) + \(\frac{1}{18}\)

\(\frac{12}{180}\) + \(\frac{22}{180}\) = \(\frac{22}{180}\) = \(\frac{11}{90}\)

∴ Time taken by (A + B) to fit a chair

= \(\frac{1}{\frac{11}{90}}\) = \(\frac{90}{11}\) minutes

∴ Time taken by (A + B) to fit 22 chairs

= \(\frac{90}{11}\) × 22 = 180 minutes

= \(\frac{180}{60}\) = 3 hours

Question 9.

A man takes 10 days to finish a job where as a woman takes 6 days to finish the same job. Together they worked for 3 days and then the woman left. In how many days will the man complete the remaining job?

Solution:

Man can finish the work in 10 days and women can finish the same work in 6 days.

∴ Man’s 1 day work = \(\frac{1}{10}\)

Woman’s 1 day work = \(\frac{1}{6}\)

(Man + Woman)s 1 day work = \(\frac{1}{10}\) + \(\frac{1}{6}\) = \(\frac{6}{60}\) + \(\frac{10}{60}\) = \(\frac{16}{60}\)

(Man + Woman)s 3 days work

In 3 days \(\frac{4}{5}\) th of the whole work is completed.

Remaining work = 1 – \(\frac{4}{5}\) = \(\frac{5}{5}\) – \(\frac{4}{5}\) = \(\frac{1}{5}\)

Complete work is done by the man by 10 days

∴ \(\frac{1}{5}\) of the work is done by man in \(\frac{1}{5}\) × 10 = 2 days.

Question 10.

A is thrice as fast as B. If B can do a piece of work in 24 days then, find the number of days they will take to complete the work together.

Solution:

If B does the work in 3 days, A will do it in 1 day.

B complete the work in 24 days.

∴ A complete the same work in \(\frac{24}{3}\) = 8 days.

∴ (A + B) complete the work in \(\frac{ab}{a+b}\) days = \(\frac{24×8}{24+8}\) days = \(\frac{24×8}{32}\)days = 6 days

They together complete the work in 6 days.