Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 4 Life Mathematics Ex 4.4 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 4 Life Mathematics Ex 4.4

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 working together is __________days.
2 days

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

(iii) A can do a work in 24 days. If A and B together can finish the work in 6 days, then B alone can finish the work in ________ days.
8

(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.
25

(v) A alone can do a work in 10 days and B alone in 15 days. They undertook the work for ₹ 200000. The amount that A will get is .
₹ 1,20,000

Question 2.
210 men working 12 hours a day can finish ajob in 18 days. How many men are required to finish the job in 20 days working 14 hours a day?
Let the required number of men be x.

 Hours Day Men 12 18 210 14 20 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 = $$210 \times \frac{12}{14} \times \frac{18}{20}$$

x = 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?
Let he required number of cement bags be x.

 Days Machines Cement bags 12 36 7000 18 24 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 = $$7000 \times \frac{18}{12} \times \frac{24}{36}$$

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 more hours a day?
Let the required number of days be x.

 Soaps Hours Days 9600 15 6 14400 (15 + 3) = 18 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 = $$6 \times \frac{14400}{9600} \times \frac{15}{18}$$

x = $$\frac{15}{2}$$
$$\frac{15}{2}$$ days will be needed.

Question 5.
If 6 container lorries can transport 135 tonnes of goods in 5 days, how many more lorries are required to transport 180 tonnes of goods in 4 days?
Let the number of lorries required more = x.

 Container lorries Goods (tonnes) Days 6 135 5 6 + x 180 4

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 = $$6 \times \frac{180}{135} \times \frac{5}{4}$$

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?
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 hr 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?
(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 = $$\frac{5+4+3}{60}$$

LCM = 5 × 4 × 3 = 60
(A + B + C)’s 1 day work = $$\frac{12}{60 \times 2}$$
(A + B + C)’s 1 day work = $$\frac{1}{10}$$
Now A’s I 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 I day work – (A + B)’s I 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 minutes more than A to do the same work. Working together, how long will it take for them to fit the parts for 22 chairs?
Time taken by A to fit a chair = 15 minutes
Time taken by B = 3 minutes more than A
= 15 + 3 = 118 minutes
∴ As 1 minute work = $$\frac{1}{15}$$
B’s 1 minute work = $$\frac{1}{18}$$
(A+B)’s 1 minutes work = $$\frac{1}{15}+\frac{1}{18}$$
$$\frac{12}{180}+\frac{10}{180}=\frac{22}{180}=\frac{11}{90}$$
∴ Time taken by (A + B) to fit a chair

LCM = 3 × 5 × 6 = 180

∴ Time taken by (A + B) to fit a chair
= $$\frac{90}{11}$$ × 22 = 180 minutes
= $$\frac{180}{60}$$ = 3 hours

Question 9.
A can do a work In 45 days. He works at it for 15 days and then, B alone finishes the remaining work in 24 days. Find the time taken to complete 80% of the work, if they work together.
A completes the work in 45 days.
∴ A’s 1 day work = $$\frac{1}{45}$$

Remaining work = $$1-\frac{1}{3}=\frac{3-1}{3}=\frac{2}{3}$$
B finishes $$\frac{2}{3}$$ rd work in 24 days

Let x days required

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.
∴ A complete the same work in $$\frac{24}{3}$$ = 8 days.
∴ (A + B) complete the work in $$\frac{a b}{a+b}$$ days
= $$\frac{24 \times 8}{24+8}$$ days
= $$\frac{24 \times 8}{32}$$ days = 6 days