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

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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.

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