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## Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 2 Life Mathematics Intext Questions

Exercise 2.1

Try These (Text book Page No. 33)

Classify the given examples as direct or inverse proportion:

(i) Weight of pulses to their cost.

Solution:

As weight increases cost also increases.

∴ Weight and cost are direct proportion.

(ii) Distance travelled by bus to the price of ticket.

Solution:

As the distance increases price to travel also increases.

∴ Distance and price are direct proportion.

(iii) Speed of the athelete to cover a certain distance.

Solution:

As the speed increases, the time to cover the distance become less.

So speed and time are in indirect proportion.

(iv) Number of workers employed to complete a construction in a specified time.

Solution:

As the number of workers increases, the amount of work become less, so they are in indirect proportion.

(v) Volume of water flown through a pipe to its pressure.

Solution:

As the pressure increases, volume also increases.

∴ They are direct proportions.

(vi) Area of a circle to its radius.

Solution:

If the radius of the circle increases its area also increases.

∴ Area and radius of circles are direct proportion.

Use the concept of direct and inverse proportions and try to answer the following questions:

Question 1.

A student can type 21 pages in 15 minutes. At the same rate, how long will it take the student to type 84 pages?

Solution:

Direct proportion

No. of minutes = x

k = \(\frac{21}{15}\)

\(\frac{21}{15}\) = \(\frac{84}{x}\)

Question 2.

The weight of an iron pipe varies directly with its length. If 8 feet of an iron pipe weighs 3.2 kg, find the proportionality constant k and determine the weight of a 36 feet iron pipe.

Solution:

Weight of 36 feet iron pipe = x

\(\frac{36}{x}\) = 2.5

Question 3.

A car covers a distance of 765 km in 51 litres of petrol. How much distance would it cover in 30 litres of petrol?

Solution:

Direct proportion

k = \(\frac{51}{765}\)

Distance cover = x km

\(\frac{30}{x}\) = \(\frac{51}{765}\)

Question 4.

If x and y vary inversely and x = 24 when y = 8, find x when y = 12.

Solution:

k = xy = 24 × 8 = 192

∴ 12 × x = 192

Question 5.

If 35 women can do a piece of work in 16 days, in how many days will 28 women do the same work?

Solution:

Inverse proportion

No. of days = x

k = 35 × 16

∴ 28 × x = 35 × 16

Question 6.

A farmer has food for 14 cows which can last for 39 days. How long would the food last, if 7 more cows join his cattle?

Solution:

Inverse variation

k = xy = 14 × 39

No. of cow = 14 + 7 = 21

No. of days = x

21 × x = 14 × 39

Question 7.

Identify the type of proportion and fill in the blank boxes:

Solution:

Direct proportion

\(\frac{x}{y}\) = k = \(\frac{1}{20}\)

(i) x = 2; y = ?

\(\frac{2}{y}\) = \(\frac{1}{20}\) ⇒ y = 2 × 20 = 40

(ii) x = ?; y = 60

\(\frac{x}{60}\) = \(\frac{1}{20}\) ⇒ x = \(\frac{60}{20}\) = 3

(iii) x = 4; y = ?

\(\frac{4}{y}\) = \(\frac{1}{20}\) ⇒ y = 80

(iv) x = 4; y = ?

\(\frac{8}{y}\) = \(\frac{1}{20}\) ⇒ y = 20 × 8 = 160

(v) x = ?; y = 180

\(\frac{x}{180}\) = \(\frac{1}{20}\)

x = \(\frac{180}{20}\) = 9

(vi) x = 12; y = ?

\(\frac{12}{y}\) = \(\frac{1}{20}\)

y = 12 × 20 = 240

(vii) x = ?; y = 360

\(\frac{x}{360}\) = \(\frac{1}{20}\) ⇒ x = \(\frac{360}{20}\) = 18

(viii) x = 24; y = ?

\(\frac{24}{y}\) = \(\frac{1}{20}\) ⇒ y = 24 × 20 = 480

Question 8.

Identify the type of proportion and fill in the blank boxes:

Solution:

Inverse proportion

k = xy = 1 × 144 = 144

(i) x = 2; y = ?

2y = 144

y = 72

(ii) X = ?; y = 48

48x = 144

x = \(\frac{144}{48}\) = 3

(iii) x = 4; y = ?

4y = 144

y = \(\frac{144}{4}\) = 36

(iv) x = 8; y = ?

8 y = 144

y = \(\frac{144}{8}\) = 18

(v) x = ?; y = 16

16x = 144

y = \(\frac{144}{16}\) = 9

(vi) x = 12; y = ?

12y = 144

y = \(\frac{144}{12}\) = 12

(vii) x = ?; y = 9

9x = 144

x = \(\frac{144}{9}\) = 16

(viii) x = 24; y = ?

24y = 144

y = \(\frac{144}{24}\) = 6

Try These (Text book Page No. 38)

Question 1.

When x = 5 and y = 5 find k, if x and y vary directly.

Solution:

If x and y vary directly then \(\frac{x}{y}\) = k

Here x = 5; y = 5

∴ k = \(\frac{5}{5}\)

k = 1

Question 2.

When x and y vary inversely, find the constant of variation when x = 64 and y = 0.75

Solution:

Given

x =64, y = 0.75

and also given x and y vary inversely.

∴ xy = k. the constant of variation.

∴ Constant = 64 × 0.75

Constant of variation = 48

Think (Text book Page No. 38)

(i) When x and y are in direct proportion and if y is doubled, then what happens to x?

Solution:

If x and y are in direct proportion \(\frac{x}{y}\) = k, constant.

if y is doubled, then \(\frac{x}{2}\) must be equal to k. So x also to be doubled.

(ii) if \(\frac{x}{y-x}\) = \(\frac{6}{7}\) What is \(\frac{x}{y}\)?

Solution:

if \(\frac{x}{y-x}\) = \(\frac{6}{7}\)

\(\frac{y-x}{x}\) = \(\frac{7}{6}\)

\(\frac{y}{x}\) – \(\frac{x}{x}\) = \(\frac{7}{6}\)

\(\frac{y}{x}\) = \(\frac{7}{6}\) + \(\frac{x}{x}\)

\(\frac{y}{x}\) = \(\frac{7}{6}\) + 1

\(\frac{y}{x}\) = \(\frac{7+6}{6}\)

\(\frac{y}{x}\) = \(\frac{13}{6}\)

\(\frac{x}{y}\) = \(\frac{6}{13}\)

Try These (Text book Page No. 40)

Identify the different variations present in the following questions:

Question 1.

24 men can make 48 articles in 12 days. Then, 6 men can make …………. articles in 6 days.

Solution:

Let the required no. of articles be x

(i) Mens and days are Indirect variables.

(ii) Men and Articles are direct variables

(iii) Days and articles are also direct variables using formula.

Let P_{1} = 24, D_{1} = 12, W_{1} = 48

P_{2} = 6, D_{2} = 6, W_{2} = x

Question 2.

15 workers can lay a road of length 4 km In 4 hours. Then, …………. workers can lay a road of length 8 km in 8 hours.

Solution:

Let the required number of workers be x

Length and workers are direct variable as more length need more workers.

The proportion is 4 : 8 : : 15 : x ……….(1)

Hours and workers are indirect variables as more working hours need less men.

∴ The proportion is 8 : 4 : : 15 : x ………..(2)

Combining (1) and (2)

Product of the extremes = product of the means

4 × 8 × x = 8 × 4 × 15

x = \(\frac{8×4×15}{4×8}\)

x = 15 workers

Question 3.

25 women working 12 hours a day can complete a work in 36 days. Then, 20 women must ……….. work hours to complete the same work in 30 days.

Solution:

Let the required hours be x

As women increases hours to work decreases

∴ It is an inverse proportion.

∴ Multiplying factor is \(\frac{25}{20}\)

As days increases hours needed become less

∴ It is also an indirect variation.

∴ Multiplying factor is \(\frac{36}{30}\)

∴ x = 12 × \(\frac{25}{20}\) × \(\frac{36}{30}\)

x = 18 hours

Question 4.

In a camp, there are 420 kg of rice sufficient for 98 persons for 45 days. The number of days that 60 kg of rice will last for 42 persons is…………

Solution:

Let the required number of days be x.

If amount of rice is more it will last for more days.

∴ It is Direct Proportion.

∴ Multiplying factor is \(\frac{60}{420}\)

If men increases number of days the rice lasts decreases

∴ It is an inverse proportion.

∴ Multiplying factor is \(\frac{98}{42}\)

x = 45 × \(\frac{60}{420}\) × \(\frac{98}{42}\)

x = 15 days

Try These (Text book Page No. 44)

Question 1.

Vikram can do one-third of work in p days. He can do \(\frac{3}{4}\)th of work in ………… days.

Solution:

\(\frac{1}{3}\) of the work will be done in p days

∴ Full work will be completed in 3p days

\(\frac{3}{4}\)th of the work will be done in = 3p × \(\frac{3}{4}\) = \(\frac{9}{4}\)p = 2\(\frac{1}{4}\)p days

Question 2.

If m persons can complete a work in n days, then 4m persons can complete the same work in ……….. days and \(\frac{m}{4}\) persons can complete the same work in…….. days

Solution:

Given m persons complete a work in n days

(i) Then work measured in terms of Man days = mn

4 m men do the work it will be completed in \(\frac{mn}{4m}\) days = \(\frac{n}{4}\) days.

(ii) \(\frac{m}{4}\) persons can complete the same work in \(\frac{mn}{\frac{m}{4}}\) days = \(\frac{4mn}{m}\) = 4n days