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## Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 1 Numbers InText Questions

Recap Exercise (Text Book Page No. 3)

Question 1.
The simplest form of $$\frac{125}{200}$$ is
$$\frac{125}{200}=\frac{125 \div 25}{200 \div 25}=\frac{5}{8}$$
= $$\frac{5}{8}$$

Question 2.
Which of the following is not an equivalent fraction of $$\frac{8}{12}$$ ?
(A) $$\frac{2}{3}$$
(B) $$\frac{16}{24}$$
(C) $$\frac{32}{60}$$
(D) $$\frac{24}{36}$$
(C) $$\frac{32}{60}$$
$$\frac{8}{12}=\frac{8+4}{12 \div 4}=\frac{2}{3}$$
$$\frac{8}{12}=\frac{8 \times 2}{12 \times 2}=\frac{16}{24}$$
$$\frac{8}{12}=\frac{8 \times 3}{12 \times 3}=\frac{24}{36}$$
But $$\frac{32}{60}=\frac{32 \div 5}{60 \div 5}=\frac{6.4}{12}$$
∴ $$\frac{32}{60}$$ is not equivalent fraction of $$\frac{8}{12}$$

Question 3.
Which is bigger $$\frac{8}{9}$$ or $$\frac{4}{5}$$ ?
LCM of 5 and 9 = 45

$$\frac{8}{9}$$ is bigger than $$\frac{4}{5}$$

The LCM of 12 and 16 is 48.

Question 4.
Add the fractions : $$\frac{3}{5}+\frac{5}{8}+\frac{7}{10}$$.
LCM of 5, 8, 10 = 5 × 2 × 4
= 40

Question 5.
Simplify: $$\frac{1}{8}-\left(\frac{1}{6}-\frac{1}{4}\right)$$

Question 6.
Multiply: $$2 \frac{3}{5}$$ and $$1 \frac{4}{7}$$.
$$2 \frac{3}{5} \times 1 \frac{4}{7}=\frac{13}{5} \times \frac{11}{7}=\frac{143}{35}=4 \frac{3}{35}$$

Question 7.
Divide $$\frac{7}{36}$$ by $$\frac{35}{81}$$.
$$\frac{7}{36}+\frac{35}{81}=\frac{7}{36} \times \frac{81}{35}=\frac{9}{20}$$

Question 8.

Question 9.
In a city $$\frac{7}{20}$$ of the population are women and $$\frac{1}{4}$$ are children. Find the fraction of the population of men.
Let the total population = 1
Population of men = Total population – Women – Children

∴ Population of men = $$\frac{2}{5}$$

Question 10.
Represent $$\left(\frac{1}{2}+\frac{1}{4}\right)$$ by a diagram.

Try These (Text Book Page No. 3)

Question 1.
Is the number -7 a rational number? Why?
A rational number, Because – 7 = $$\frac{-14}{2}=\frac{p}{q}$$

Question 2.
Write any 6 rational numbers between 0 and 1.
$$\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \frac{1}{7}$$

Try These (Text Book Page No. 4)

That’s literally all there is to it! 1/8 as a decimal is 0.125.

Write the decimal forms of the following rational numbers:

Question 1.
$$\frac{4}{5}$$
$$\frac{4}{5}=\frac{4 \times 20}{5 \times 20}=\frac{80}{100}=0.80$$

Question 2.
$$\frac{6}{25}$$
$$\frac{6}{25}=\frac{6 \times 4}{25 \times 4}=\frac{24}{100}=0.24$$

Question 3.
$$\frac{486}{1000}$$
$$\frac{486}{1000}$$ = 0.486

Question 4.
$$\frac{1}{9}$$

$$\frac{1}{9}$$ = 0.11….

Question 5.
$$3 \frac{1}{4}$$

$$3 \frac{1}{4}$$ = $$\frac{13}{4}$$ = 3.25

That’s literally all there is to it! 5/8 as a decimal is 0.625.

Question 6.
$$-2 \frac{3}{5}$$

$$-2 \frac{3}{5}$$ = $$\frac{-13}{5}$$ = -2.6

Try These (Text Book Page No. 6)

Question 1.
$$\frac{7}{3}=\frac{?}{9}=\frac{49}{?}=\frac{-21}{?}$$

Question 2.
$$\frac{-2}{5}=\frac{?}{10}=\frac{6}{?}=\frac{-8}{?}$$

Try These (Text Book Page No. 7)

Question 1.
Which of the following pairs represents equivalent rational numbers?
(i) $$\frac{-6}{4}, \frac{18}{-12}$$
(ii) $$\frac{-4}{-20}, \frac{1}{-5}$$
(iii) $$\frac{-12}{-17}, \frac{60}{85}$$
(i) $$\frac{-6}{4}, \frac{18}{-12}$$
$$\frac{-6}{4}=\frac{-6 \times 3}{4 \times 3}=\frac{-18}{12}$$
∴ $$\frac{-6}{4}$$ equivalent to $$\frac{-18}{12}$$

(ii) $$\frac{-4}{-20}, \frac{1}{-5}$$
$$\frac{-4}{-20}=\frac{-4 \div(-4)}{-20 \div(-4)}=\frac{1}{5} \neq-\frac{1}{5}$$
∴ $$\frac{-4}{-20}$$ equivalent to $$\frac{1}{-5}$$

(iii) $$\frac{-12}{-17}, \frac{60}{85}$$
$$\frac{-12}{-17}=\frac{-12 x-5}{-17 x-5}=\frac{60}{85}$$
∴ $$\frac{-12}{-17}$$ equivalent to $$\frac{60}{85}$$

Question 2.
Find the standard form of
(i) $$\frac{36}{-96}$$
(ii) $$\frac{-56}{-72}$$
(iii) $$\frac{27}{18}$$
(i) $$\frac{36}{-96}$$
= $$\frac{-36 \div 12}{96 \div 12}=\frac{-3}{8}$$

(ii) $$\frac{-56}{-72}$$
= $$\frac{-56 \div(-8)}{-72 \div(-8)}=\frac{7}{9}$$

(iii) $$\frac{27}{18}$$
= $$1 \frac{9}{18}=1 \frac{1}{2}$$

Question 3.
Mark the following rational numbers on a number line.
(i) $$\frac{-2}{3}$$
$$\frac{-2}{3}$$ lies betveen 0 and -1.
T?ìe unit part between O and —lis divided into 3 equal parts and second part is taken.

(ii) $$\frac{-8}{-5}$$
$$\frac{-8}{-5}$$ = $$1 \frac{3}{5}$$
$$1 \frac{3}{5}$$ lies between I and 2, The unit part between I and 2 is divided into 5 equal parts and the third part is taken.

(iii) $$\frac{5}{-4}$$
$$\frac{5}{-4}$$ = $$-\frac{5}{4}$$ = $$-1 \frac{1}{4}$$
$$-1 \frac{1}{4}$$ lies between -1 and -2. The unit part between -1 and -2 is divided into four equal parts and the first part is taken.

Think (Text Book Page No. 15)

Is zero a rational number? If so, what is its additive inverse
Yes zero a rational number Additive inverse of zero is zero.

Think (Text Book Page No. 16)

What is the multiplicative inverse of 1 and -1?
Multiplicative inverse of 1 is 1 and -1 is -1.

Try These (Text Book Page No. 16)

Divide
(i) $$\frac{-7}{3}$$ by 5
(ii) 5 by $$\frac{-7}{3}$$
(iii) $$\frac{-7}{3}$$ by $$\frac{35}{6}$$
(i) $$\frac{-7}{3}$$ by 5

(ii) 5 by $$\frac{-7}{3}$$

(iii) $$\frac{-7}{3}$$ by $$\frac{35}{6}$$

Try These (Text Book Page No. 20)

The closure property on integers holds for subtraction and not for division. What about rational numbers? Verify.
Let 0 and $$\frac{1}{2}$$ te two rational numbers 0 – $$\frac{1}{2}$$ = –$$\frac{1}{2}$$ is a rational numter
∴ Closure property for subtraction holds for rational numbers.
But consider the two rational number $$\frac{5}{2}$$ and 0.
$$\frac{5}{2}$$ + 0 = $$\frac{5}{2 \times 0}=\frac{5}{0}$$
Here denominator = 0 and it is not a rational number.
∴ Closure property is not true for division of rational numbers.

Try These (Text Book Page No. 22)

(i) Is $$\frac{3}{5}-\frac{7}{8}=\frac{7}{8}-\frac{3}{5}$$ ?

LHS ≠ RHS
∴ $$\frac{3}{5} \div \frac{7}{8}$$ ≠ $$\frac{7}{8}-\frac{3}{5}$$
∴ Subtraction of rational numbers is not commutative.

(ii) $$\frac{3}{5} \div \frac{7}{8}=\frac{7}{8} \div \frac{5}{3}$$? So, what do you conclude?

LHS ≠ RHS
∴ $$\frac{3}{5} \div \frac{7}{8}$$ ≠ $$\frac{7}{8} \div \frac{5}{3}$$
∴ Commutative property not hold good br division of rational numbers.

Try This (Text Book Page No. 22)

Check whether associative property holds for subtraction and division.
Consider for rational numbers $$\frac{2}{3}, \frac{1}{2}$$ and $$\frac{3}{4}$$

∴ Associative property not holds for subraction of rational numbers

∴ Associative property not holds for division of rational numbers

Try This (Text Book Page No. 25)

Question 1.
Observe that,
$$\frac{1}{1.2}+\frac{1}{2.3}=\frac{2}{3}$$
$$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}=\frac{3}{4}$$
$$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}=\frac{4}{5}$$
Use your reasoning skills, to find the sum of the first 7 numbers in the pattern given above.

Think (Text Book Page No. 26)

Question 1.
Is the square of a prime number, prime?
No, the square of a prime number ‘P’ has at Least 3 divisors 1, P and P2. But a prime number is a number which has only two divisors, 1 and the number itself. So square of a prime number is not prime.

Question 2.
Will the sum of two perfect squares always be a perfect square? What about their difference and their product?
The sum of two perfect squares, need not be always a perfect square. Also the difference of two perfect squares need not be always a perfect square. Bu the product of two perfect square is a perfect square.

Try These (Text Book Page No. 26)

Question 1.
Which among 256, 576, 960, 1025,4096 are perfect square numbers? (Hint: Try to extend the table of squares already seen).
256 = 162
576 = 242
4096 = 642
∴ 256, 576, and 4096 are perfect squares

Question 2.
One can judge just by look that each of the following numbers 82, 113, 1972, 2057, 8888, 24353 is not a perfect square. Explain why?
Because the unit digit ola perfect square will be 0, 1,4, 5, 6, 9. But the given numbers have unit digits 2, 3, 7, 8. So they are not perfect squares.

Think (Text Book Page No. 27)

Consider the claim: “Between the squares of the consecutive numbers n and (n + 1), there are 2n non-square numbers’ Can it be true? FInd how many non-square numbers are there
(i) between 4 and 9 ?
(ii) between 49 and 64? and Verify the claim.

Therefore we conclude that there are 2n non-square numbers between two consecutive square numbers.

Think (Text Book Page No. 30)

In this quick guide we’ll describe what the factors of 96 are, how you find them and list out the factor pairs of 96 for you to prove the calculation works.

In this case, if we want to find the smallest factor with which we can multiply or divide 108 to get a square number, what should we do?
108 = 2 × 2 × 3 × 3 = 22 × 32 × 3
If we multiply the factors by 2, then we get
22 × 32 × 3 × 3 = 22 × 32 × 32 = (2 × 3 × 3)2
Which is perfect square.
∴ Again if we divide by 3 then we get 22 × 32 ⇒ (2 × 3)2, a perfect square.
∴ We have to multiply or divide 108 by 3 to get a perfect square.

Try These (Text Book Page No. 32)

Find the square root by long division method.
Question 1.
400

√400 = 20

Question 2.
1764

√1764 = 42

Question 3.
9801

√9801 = 66

Try These (Text Book Page No. 32)

without calculating the square root, guess the number of digits in the square root of the following numbers:
Question 1.
14400
$$\sqrt{14400}$$ = $$\sqrt{144 \times 100}$$
= $$\sqrt{144} \times \sqrt{100}$$
= 12 × 10
= 120

Question 2.
390625
$$\sqrt{390625}$$ = $$\sqrt{25 \times 25 \times 25 \times 25}$$
= $$\sqrt{25 \times 25} \times \sqrt{25 \times 25}$$
= 25 × 25
= 625

Question 3.
100000000
$$\sqrt{100000000}$$ = $$\sqrt{10000 \times 10000}$$
= $$\sqrt{10000} \times \sqrt{10000}$$
= 100 × 100
= 10,000

Try These (Text Book Page No. 33)

Find the square root of
Question 1.
5.4756

Question 2.
19.36

Question 3.
116.64

Think (Text Book Page No. 33)

Try to fill in the blanks using √ab = √a × √b

Try These (Text Book Page No. 34)

Using this method, find the square root of the numbers 1.2321 and 11.9025.
(i) 1.2321
√1.2321 = $$\sqrt{\frac{12321}{10000}}$$
= $$\frac{111}{100}$$ = 1.11

(ii) 11.9025
√11.9025 = $$\frac{\sqrt{119025}}{\sqrt{10000}}$$
= $$\frac{345}{100}$$ = 3.45

Try These (Text Book Page No. 34)

Write the numbers in ascending order.
Question 1.
4, √14, 5
Squaring all the numbers we get 42, (√14)2, 52 ⇒ 16, 14, 25
∴ Ascending order: 14, 16, 25
Ascending order: √14, 4, 5

Question 2.
7, √65, 8
Squaring 7, √65 and 8 we get 72, (√65)2, 82 ⇒ 49, 65, 64
Ascending order : 49, 64, 65
Ascending order: 7, 8, √65

Try These (Text Book Page No. 37)

Find the ones digit in the cubes of each of the following numbers.
(i) 12
(ii) 27
(iii) 38
(iv) 53
(v) 71
(vi) 84
(i) 12
12 ends with 2, so its cube ends with 8 i.e. ones digit in 123 is 8.

(ii) 27
27 ends with 7, so its cube end with 3. i.e., ones digit in 273 is 3.

(iii) 38
38 ends with 8, so its cube ends with 2 i.e. ones digit in 383 is 2.

(iv) 53
53 ends with 3, so its cube ends with 7. i.e, ones digit in 533 is 7.

(v) 71
71 ends with 1, so its cube ends with 1. i.e. ones digit in 713 is 1

(vi) 84
84 ends with 4, so its cube ends with 4. i.e, ones digit in 843 is 4.

Try These (Text Book Page No. 41)

Expand the following numbers using exponents:
(i) 8120
(ii) 20,305
(iii) 3652.01
(iv) 9426.521
(i) 8120
8120 = (8 × 1000) + (1 × 100) + (2 x×10) + 0 × 1
= (8 × 103) + (1 × 102) + (2 × 101)

(ii) 20,305
20305 = (2 × 10000) + (0 × 1000) + (3 × 100) + (0 × 10) + (5 × 1)
= (2 × 104) + (3 × 102) + 5

(iii) 3652.01
3652.01 = 3000 + 600 + 50 + 2 + $$\frac{0}{10}+\frac{1}{100}$$
= (3 × 1000) + (6 × 100) + (5 × 10) + (2 × 1) + (1 × $$\frac{1}{100}$$)
= (3 × 103) + (6 × 102) + (5 × 101) + 2 + (1 × 10-2)

(iv) 9426.521
= (9 × 1000) + (4 × 100) + (2 × 10) + (6 × 1) + $$\left(\frac{5}{10}\right)+\left(\frac{2}{100}\right)+\left(\frac{1}{1000}\right)$$
= (9 × 103) + (4 × 102) + (2 × 101) + 6 + (5 × 10-1) + (2 × 10-2) + (1 × 10-3)

Try These (Text Book Page No. 42)

Verify the following rules (as we did above). Here, a,b are non-zero integers and are any integers.
1. Product of same powers to power of product rule: am × bm = (ab)m
2. Quotient of same powers to power of quotient rule: $$\frac{a^{m}}{b^{m}}=\left(\frac{a}{b}\right)^{m}$$
3. Zero exponent rule: a0 = 1.
let a = 2; b = 3; m = 2
1. am × bm = 22 × 32
= 4 × 9 = 36 = (2 × 3)2

2. $$\frac{a^{m}}{b^{m}}=\frac{2^{2}}{3^{2}}=\frac{4}{9}=\left(\frac{2}{3}\right)^{2}$$

3. a0 = 20 = 1.

Try These (Text Book Page No. 44)

Question 1.
Write in standard form: Mass of planet Uranus is 8.68 × 1025 kg.
Mass of Planet Uranus = 86800000000000000000000000 kg
[23 zeros after 88]

Question 2.
Write in scientific notation:
(i) 0.000012005
0.000012005 = 1.2005 × 10-5

(ii) 43 12.345
43 12.345 = 4.312345 × 103

(iii) 0.10524
0.10524 = 1.0524 × 10-1

(iv) The distance between the Sun and the planet Saturn 1.4335 × 1012 miles.

## Samacheer Kalvi 8th Maths Guide Chapter 2 Measurements Ex 2.4

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 2 Measurements Ex 2.4 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 2 Measurements Ex 2.4

Question 1.
Two gates are fitted at the entrance of a library. To open the gates easily, a wheel is fixed at 6 feet distance from the wall to which the gate is fixed. If one of the gates is opened to 90°, find the distance moved by the wheel (π = 3.14).

Amswer:
Let A be the position of the wall AC be the gate in initial position and AB be position when it is moved 90°.
Now the arc length BC gives the distance moved by the wheel.

Length of the arc = $$\frac{\theta}{360^{\circ}}$$ × 2πr units
= $$\frac{90^{\circ}}{360^{\circ}}$$ × 2 × 3.14 × 6 feets
= 3.14 × 3 feets
= 9.42 feets
∴ Distance moved by the wheel = 9.42 feets.

Question 2.
With his usual speed, if a person covers a circular track of radius 150 m in 9 minutes, find the distance that he covers in 3 minutes (π = 3.14).
Radius of the circular track = 150m
Distance covers in 9 minutes = Perimeter of the circle = 2 × π × r units
Distance covered in 9 min = 2 × 3.14 × 150m
Distance covered in 1 min = $$\frac{2 \times 3.14 \times 150}{9} \mathrm{m}$$

Distance he covers in 3min = 314m

Question 3.
Find the area of the house drawing given in the figure.

Area of the house = Area of a square of side 6 cm + Area of a rectangle with l = 8cm, b = 6 cm + Area of a ∆ with b = 6 cm and h = 4 cm + Area of a parallelogram with b = 8 cm, h = 4 cm
= (side × side) + (l × b) + ($$\frac { 1 }{ 2 }$$ × b × h) + bh cm2
= (6 × 6) + (8 × 6) + ($$\frac { 1 }{ 2 }$$ × 6 × 4) + (8 × 4) cm2
= 36 + 48 + 12 + 32 cm2
= 128 cm2
Required Area = 128 cm2

Question 4.
Draw the top, front and side view of the following solid shapes
(i)

(a) Top view

(b) Front view

(c) Side view

(ii)

(a) Top view

(b) Front view

(c) Side view

Challenging Problems

Expert Guide to Buying and Selling Feet on Feet Finder ·

Question 5.
Guna has fixed a single door of width 3 feet in his room where as Nathan has fixed a double door, each of width 1 $$\frac { 1 }{ 2 }$$ feet in his room. From the closed position, if each of the single and double doors can open up to 120°, whose door takes a minimum area?
Width of the door that Guna fixed = 3 feet.
When the door is open the radius of the sector = 3 feet

Angle covered = 120°
∴ Area required to open the door

= 3π feet2

(b) Width of the double doors that Nathan fixed = 1 $$\frac { 1 }{ 2 }$$ feet.
Angle described to open = 120°
Area required to open = 2 × Area of the sector
z
= $$\frac { 1 }{ 2 }$$ (3π) feet2
∴ The double door requires the minimum area.

Question 6.
In a rectangular field which measures 15 m x 8m, cows are tied with a rope of length 3m at four corners of the field and also at the centre. Find the area of the field where none of the cow can graze. (π = 3.14).
Area of the field where none of the cow can graze = Area of the rectangle – [Area of 4 quadrant circles] – Area of a circle

Area of the rectangle = l × b units2
= 15 × 8m2 = 120m2
Area of 4 quadrant circles = 4 × $$\frac { 1 }{ 4 }$$ πr2 units
Area of 4 quadrant circles = 4 × $$\frac { 1 }{ 4 }$$ × 3.14 × 3 × 3 = 28.26 m2
Area of the circle at the middle = πr2 units
= 3.14 × 3 × 3m2 = 28.26m2
∴ Area where none of the cows can graze
= [120 – 28.26 – 28.26]m2 = 120 – 56.52 m2 = 63.48 m2

Question 7.
Three identical coins, each of diameter 6 cm are placed as shown. Find the area of the shaded region between the coins. (π = 3.14) (√3 = 1.732)

Given diameter of the coins = 6 cm
∴ Radius of the coins = $$\frac { 6 }{ 2 }$$ = 3 cm
Area of the shaded region = Area of equilateral triangle – Area of 3 sectors of angle 60°

Area of the equilateral triangle = $$\frac{\sqrt{3}}{4}$$ a2 units2 = $$\frac{\sqrt{3}}{4}$$ × 6 × 6 cm2
= $$\frac{1.732}{4}$$ × 6 × 6 cm2 = 15.588 cm2
Area of 3 sectors = 3 × $$\frac{\theta}{360^{\circ}}$$ × πr2 sq.units
= 3 × $$\frac{60^{\circ}}{360^{\circ}}$$ × 3.14 × 3 × 3 cm2 = 1.458 cm2
∴ Area of the shaded region = 15.588 – 14.13 cm2 = 1.458 cm2
Required area = 1.458 cm2 (approximately)

Question 8.
Using Euler’s formula, find the unknowns.

Euler’s formula is given by F + V – E = 2
(i) V = 6, E = 14
By Euler’s formula = F + 6 – 14 = 2
F = 2 + 14 – 6
F = 10

(ii) F = 8,E = 10
By Euler’s formula = 8 + V – 10 = 2
V = 2 – 8 + 10
V = 4

(iii) F = 20, V = 10
By Euler’s formula = 20 + 10 — E = 2
30 – E = 2
E = 30 – 2.
E = 28
Tabulating the required unknowns

## Samacheer Kalvi 8th Maths Guide Chapter 1 Numbers Ex 1.4

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 1 Numbers Ex 1.4 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 1 Numbers Ex 1.4

Question 1.
Fill in the blanks:
(i) The ones digit in the square of 77 is ________ .
9

(ii) The number of non-square numbers between 242 and 252 is ________ .
48

(iii) The number of perfect square numbers between 300 and 500 is ________ .
5

(iv) If a number has 5 or 6 digits in it, then its square root will have ________ digits.
3

(v) The value of Jii lies between integers ______ and ________ .
13, 14

Question 2.
Say True or False:
(i) When a square number ends in 6, its square root will have 6 in the unit’s place.
True

(ii) A square number will not have odd number of zeros at the end.
True

(iii) The number of zeros in the square of 91000 is 9.
False

(iv) The square of 75 is 4925.
False

(v) The square root of 225 is 15.
True

Question 3.
Find the square of the following numbers.
(i) 17
(ii) 203
(iii) 1098
(i) 17

(ii) 203

(iii) 1098

Question 4.
Examine if each of the following is a perfect square.
(i) 725
(ii) 190
(iii) 841
(iv) 1089
(i) 725
725 = 5 × 5 × 29 = 52 × 29
Here the second prime factor 29 does not have a pair.
Hence 725 is not a perfect square number.

(ii) 190
190 = 2 × 5 × 19
Here the factors 2, 5 and 9 does not have pairs.
Hence 190 is not a perfect square number.

(iii) 841
841 = 29 × 29
Hence 841 is a perfect square

(vi) 1089
1089 = 3 × 3 × 11 × 11 = 33 × 33
Hence 1089 is a perfect square

The factors of 144 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144.

Question 5.
Find the square root by prime factorisation method.
(i) 144
(ii) 256
(iii) 784
(iv) 1156
(v) 4761
(vi) 9025
(i) 144
144 = 2 × 2 × 2 × 2 × 3 × 3
√144 = 2 × 2 × 3 = 12

(ii) 256
256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
√256 = 2 × 2 × 2 × 2 = 16

(iii) 784
784 = 2 × 2 × 2 × 2 × 7 × 7
√784 = 2 × 2 × 2 × 2 × 7 × 7 = 28

(iv) 1156
1156 = 2 × 2 × 17 × 17
1156 = 22 × 172
1156 = (2 × 17)2
∴ $$\sqrt{1156}$$ = $$\sqrt{(2 \times 17)^{2}}$$ = 2 × 17 = 34
∴ $$\sqrt{1156}$$ = 34

(v) 4761
4761 = 3 × 3 × 23 × 23
4761 = 32 × 232
4761 = (3 × 23)2
√4761 = $$\sqrt{(3 \times 23)^{2}}$$
√4761 = 3 × 23
√4761 = 69

(vi) 9025
9025 = 5 × 5 × 19 × 19
9025 = 52 × 192
9025 = (5 × 19)2
√925 = $$\sqrt{(5 \times 19)^{2}}$$ = 5 × 19 = 95

Question 6.
Find the square root by long division method.
(i) 1764
(ii) 6889
(iii) 11025
(iv) 17956
(v) 418609
(i) 1764

√1764 = 42

(ii) 6889

√6889 = 83

(iii) 11025

√11025 = 105

(iv) 17956

√17956 = 134

(v) 418609

√418609 = 647

Roots Calculator is a free online tool that displays the roots of the given quadratic equation.

Question 7.
Estimate the value of the following square roots to the nearest whole number:
(i) √440
(ii) √800
(iii) √1020
(i) √440
we have 202 = 400
212= 441
∴ √440 ≃ 21

(ii) √800
we have 282 = 784
292= 841
∴ √800 ≃ 28

(iii) √1020
we have 312 = 961
322= 1024
∴ √1020 ≃ 32

Question 8.
Find the square root of the following decimal numbers and fractions.
(i) 2.89
(ii) 67.24
(iii) 2.0164
(iv) $$\frac{144}{225}$$
(v) $$7 \frac{18}{49}$$
(i) 2.89

√2.89 = 1.7

(ii) 67.24

√67.24 = 8.2

(iii) 2.0164

√2.0164 = 1.42

(iv) $$\frac{144}{225}$$

(v) $$7 \frac{18}{49}$$

$$\sqrt{7 \frac{18}{49}}=2 \frac{5}{7}$$

Question 9.
Find the least number that must be subtracted to 6666 so that it becomes a perfect square. Also, find the square root of the perfect square thus obtained.
Let us work out the process of finding the square root of 6666 by long division method.

The remainder in the last step is 105. Is if 105 be subtracted from the given number the
remainder will be zero and the new number will be a perfect square.
∴ The required number is 105. The square number is 6666 – 105 = 6561.

Question 10.
Find the least number by which 1800 should be multiplied so that it becomes a perfect square. Also, find the square root of the perfect square thus obtained.
We find 1800 = 2 × 2 × 3 × 3 × 5 × 5 × 2
= 22 × 32 × 52 × 2
Here the last factor 2 has no pair. So if we multiply 1800 by 2, then the number becomes a perfect square.

∴ 1800 × 2 = 3600 is the required perfect square number.
∴ 3600 = 1800 × 2
3600 = 22 × 32 × 52 × 2 × 2
3600 = 22 × 32 × 52 × 22
= (2 × 3 × 5 × 2)2
$$\sqrt{3600}=\sqrt{(2 \times 3 \times 5 \times 2)^{2}}$$
= 2 × 3 × 5 × 2 = 60
∴ √3600 = 60

Objective Type Questions

Question 11.
The square of 43 ends with the digit .
(A) 9
(B) 6
(C) 4
(D) 3
(A) 9
Hint:
Ones digit = 3 × 3 = 9

Question 12.
_______ is added to 242 to get 252.
(A) 42
(B) 52
(C) 62
(D) 72
(D) 72
Hint:
252 = 25 × 25 = 625
242 = 24 × 24 = 576

Question 13.
√48 is approximately equal to .
(A) 5
(B) 6
(C) 7
(D) 8
(C) 7
Hint:
√49 = 7

Question 14.
$$\sqrt{128}-\sqrt{98}+\sqrt{18}$$
(A) √2
(B) √8
(C) √48
(D) √32
(D) √32

Question 15.
The number of digits in the square root of 123454321 is ______.
(A) 4
(B) 5
(C) 6
(D) 7
(B) 5
Hint:
$$\frac{n+1}{2}=\frac{10}{2}=5$$

## Samacheer Kalvi 8th Maths Guide Chapter 7 Information processing Ex 7.2

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 7 Information processing Ex 7.2 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 7 Information processing Ex 7.2

Question 1.
Using repeated division method, find the HCF of the following:
(i) 455 and 26

Step 1: The larger number should be dividend 455 & smaller number should be divisor = 26
Step 2: After 1st division, the remainder becomes new divisor & the previous divisor becomes next dividend.
Step 3: This is done till remainder is zero.
Step 4: The last divisor is the HCF L.
∴ Ans: HCF is 13.

(ii) 392 and 256
256 is smaller, so it is the 1st divisor

∴ HCF = 8

(iii) 6765 and 610

∴ HCF = 5

(iv) 184, 230 and 276
First let us take 184 & 230

∴ 46 is the HCF of 184 and 230
Now the HCF of the first two numbers is the dividend for the third number.

∴ Ans: HCF of 184, 230 & 276 is 46

Question 2.
Using repeated subtraction method, find the HCF of the following:
(i) 42 and 70
Let number be m & n m > n
We do, m – n & the result of subtraction becomes new ‘m’. if m becomes less than n,
we do n – m and then assign the result as n. We should do this till m n. When m = n then ‘m’ is the HCF.
42 and 70
m = 70 n = 42
70 – 42 = 28

now m = 42, n = 28
42 – 28 = 14.

now m = 28, n = 14
28 – 14 = 14.

now m = 14. n = 14;
we stop here as m = n
∴ HCF of 42 & 70 is 14

(ii) 36 and 80

28 – 8 = 20
20 – 8 = 12
12 – 8 = 4
8 = 4 = 4
now m = n = 4
∴ HCF is 4

(iii) 280 and 420
Let m = 420, n = 280
m – n = 420 – 280 = 140
now m = 280, n = 140
m – n = 280 – 140 = 140
now m = n = 140
∴ HCF is 140

(iv) 1014 and 654
Let m = 1014, n = 654
m – n = 1014 – 654 = 360
now m = 654, n = 360
m – n = 654 – 360 = 294
now m = 360, n = 294
m – n = 360 – 294 = 66
now m = 294 n = 66
m – n = 294 – 66 = 228
now m = 66, n = 228
n – m = 228 – 66 = 162
now m = 162, n = 66
= 162 – 66 = 96
n – m = 96 – 66 = 30
Similarly, 66 – 30 = 36
36 – 30 = 6
30 – 6 = 24
24 – 6 = 18
18 – 6 = 12
12 – 6 = 6 now m = n
∴ HCF of 1014 and 654 is 6

Question 3.
Do the given problems by repeated subtraction method and verify the result.
(i) 56 and 12

Let n = 56 & n = 12
m – n = 56 – 12 = 44
now m = 44, n = 12
m – n = 44 – 12 = 32
m – n = 32 – 12 = 20
m – n = 20 – 12 = 8
n – m = 12 – 84
m – n = 8 – 4 = 4. now m = n
∴ HCF of 56 & 12 is 4

(ii) 320, 120 and 95

Let us take 320 & 120 first m = 320, n = 120
m – n = 320 – 120 = 200
m = 200, n = 120
∴ m – n = 200 – 120 = 80
120 – 80 = 40
80 – 40 = 40
∴ m = n = 40 → HCF of 320, 120
Now let us find HCF of 40 & 95
m = 95, n = 40
∴ m – n = 95 – 40 = 55
55 – 40 = 15
40 – 15 = 25
25 – 15 = 10
15 – 10 = 5
HCF of 40 & 95 is 5
10 – 5 = 5
∴ HCF of 320 120 & 95 is 5

Question 4.
Kalai wants to cut identical squares as big as she can, from a piece of paper measuring 168mm and by 196mm. What is the length of the side of the biggest square? (To find HCF using repeated subtraction method)
Sides are 168 & 196
To find HCF of 168 & 196, we are to use repeated subtraction method.
∴ m = 196, n = 168
m – n = 196 – 168 = 28

now n = 28, m = 168
m – n = 168 – 28 = 140

now m = 140, n = 28
m – n = 140 – 28 = 112

now m = 112, n = 28
m – n = 112 – 28 = 84

now m = 84, n = 28
m – n = 84 – 28 = 56

now m = 56, n = 28
m – n = 56 – 28 = 28
∴ HCF is 28
∴ Length of biggest square is 28

Objective Type Questions

Question 5.
What is the eleventh Fibonacci number?
(a) 55
(b) 77
(c) 89
(d) 144
(c) 89
Hint:

∴ 11th Fibonacci number is 89

Question 6.
If F(n) is a Fibonacci number and n = 8, which of the following is true?
(a) F(8) = F(9) + F( 10)
(b) F(8) = F(7) + F(6)
(c) F(8) = F(10) × F(9)
(d) F(8) = F(7) – F(6)
(b) F(8) = F(7) + F(6)
Hint:
Given F(n) is a Fibonacci number & n = 8
∴ F(8) = F(7) + F (6) as any term in Fibonacci series is the sum of preceding 2 terms

Question 7.
Every 3rd number of the Fibonacci sequence is a multiple of_______
(a) 2
(b) 3
(c) 5
(d) 8
(a) 2
Hint:
Every 3rd number in Fibonacci sequence is a multiple of 2

Question 8.
Every _____ number of the Fibonacci sequence is a multiple of 8
(a) 2th
(b) 4th
(c) 6th
(d) 8th
(c) 6th

Question 9.
The difference between the 18th and 17th Fibonacci number is
(a) 233
(b) 377
(c) 610
(d) 987
(d) 987
Hint:
F(18) = F(17) + F(16)
F(18) – F(17) = F(16) = F(15) + F(14)
= 610 + 377 = 987

Factors of 70 are the list of integers that we can split evenly into 70.

Question 10.
Common prime factors of 30 and 250 are
(a) 2 × 5
(b) 3 × 5
(c) 2 × 3 × 5
(d) 5 × 5
(a) 2 × 5
Prime factors of 30 are 2 × 3 × 5
Prime factors of 250 are 5 × 5 × 5 × 2
∴ Common prime factors are 2 × 5

Question 11.
Common prime factors of 36,60 and 72 are
(a) 2 × 2
(b) 2 × 3
(c) 3 × 3
(d) 3 × 2 × 2
(d) 3 × 2 × 2
Hint:
Prime factors of 36 are 2 × 2 × 3 × 3
Prime factors of 60 are 2 × 2 × 3 × 5
Prime factors of 72 are 2 × 2 × 2 × 3 × 3
∴ Common prime factors are 2 × 2 × 3

Question 12.
Two numbers are said to be co-prime numbers if their HCF is
(a) 2
(b) 3
(c) 0
(d) 1
(d) 11

## Samacheer Kalvi 8th Social Science Guide Geography Chapter 5 Hazards

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Social Science Guide Pdf Geography Chapter 5 v Text Book Back Questions and Answers, Important Questions, Notes.

## Tamilnadu Samacheer Kalvi 8th Social Science Solutions Geography Chapter 5 Hazards

### Samacheer Kalvi 8th Social Science Hazards Text Book Back Questions and Answers

Percentage Off Calculator is a free online tool that displays the percentage off of the product.

1. ………………. percentage of nitrogen is present in the air.
a) 78.09%
b) 74.08%
c) 80.07%
d) 76.63%
a) 78.09%

2. Tsunami in the Indian Ocean took place in the year ………………. .
a) 1990
b) 2004
c) 2005
d) 2008
b) 2004

3. The word tsunami is derived from ………………. language.
a) Hindi
b) French
c) Japanese
d) German
c) Japanese

4. The example of surface water is
a) Artesian well
b) Groundwater
c) Subsurface water
d) Lake
d) Lake

5. Event that occurs due to the failure of monsoons.
a) Condensation
b) Drought
c) Evaporation
d) Precipitation
b) Drought

II. Fill in the blanks.

1. Hazards may lead to …………….
affect the community most severely

2. Landslide is an example of …………… hazard.
geologic

3. On the basis of origin, the hazard can be grouped into …………… categories.
8 (or) eight

4. Terrorism is an example of …………… hazard.
human-induced

5. Oxides of nitrogen are …………… pollutants which affect human beings.
primary

6. Chernobyl nuclear accident took place in
26th April 1986

III. Match the following.

 List I List II 1. Primary pollutant a. Terrorism 2. Hazardous waste b. Tsunami 3. Earthquake c. Outdated drugs 4. Meteorological drought d. Oxides of sulphur 5. Human induced hazard e. Reduction in rainfall

 List I List II 1. Primary pollutant d. Oxides of sulphur 2. Hazardous waste c. Outdated drugs 3. Earthquake b. Tsunami 4. Meteorological drought e. Reduction in rainfall 5. Human induced hazard a. Terrorism

1. Define ‘hazard’?
‘Hazards are defined as a thing, person, event or factor that poses a threat to people, structures or economic assets and which may cause a disaster.’

2. What are the major types of hazards?

• Natural hazards
• Socio-natural hazards

3. Write a brief note on hazardous wastes.
The wastes that may or tend to cause adverse health effects on the ecosystem and human beings are called hazardous wastes.

4. List out the major flood-prone areas of our country.
Punjab, West Bengal, Odisha are the region which is also prone to flood often.

5. Mention the types of drought.
Drought could be classified into three types. They are:

1. Meteorological drought
2. Hydrological drought
3. Agricultural drought

6. Why should not we construct houses in foothill areas?
Because of the presence of steep slopes and heavy rainfall we should not construct houses at the foothills areas.

V. Distinguish the following.
1. Hazards and disasters.

 Hazards Disasters Hazards occur frequently and threaten people. It is a hazardous event that occurs over a limited time span. Ex: Natural hazards, Earthquake. Ex: Flood

2. Natural hazards and human-made hazards.

 Natural hazard Human-made hazard Man has no role to play in such hazards. These are caused by the undesirable activities of humans. Ex: Flood, drought, etc. Ex: Pollution of air and water

3. Flood and drought.

 Flood Drought It occurs through heavy rainfall. It occurs through no rainfall. It is often held in the coastal Andhra Pradesh & Odisha. The dry region has seen in the leeward side of Western ghats.

4. Earthquake and tsunami.

 Earthquake Tsunami The violent tremor of the earth’s crust is called an Earthquake. Tsunami refers to huge ocean waves caused by landslides. The study of earthquakes is called Seismograph. The study of the Tsunami is called Oceanography.

1. Write an essay on air pollution.
Pollution of Air:

1. Air is a mixture of several gases.
2. The main gases are nitrogen (78.09%) for forming products such as fertilizers for plants and for making the air inert, oxygen (20.95%) for breathing and carbon dioxide (0.03%) for photosynthesis.
3. Some other gases like argon, neon, helium, krypton, hydrogen, zenon and methane are also present.
4. Air pollution is the contamination of indoor or outdoor air by a range of gases. Air pollution can be categorized into primary and secondary pollutants.
5. A primary pollutant is an air pollutant emitted directly from a source.
6. A secondary pollutant is not directly emitted.
7. Primary pollutants are as follows:
• Oxides of Sulphur
• Oxides of Nitrogen
• Oxides of Carbon
• Particulate Matter and
• Other primary pollutants
8. Secondary pollutants are as follows:
• Ground Level Ozone
• Smog

2. Define earthquake and list out its effects.
Earthquakes:

• Earthquake is a violent tremor in the earth’s crust, sending out a series of shock waves in all directions from its place of origin.
• Earthquake-prone regions of the country have been identified on the basis of scientific inputs relating to seismicity, earthquakes that occurred in the past, and the tectonic setup of the region.
• Based on these inputs, the Bureau of Indian Standards has grouped the country into four seismic zones: Zone II, Zone III, Zone IV, and Zone V (No area of India is classified as Zone I).
 Seismic Zones Level of Risk Zone V Very High Zone IV High Zone III Moderate Zone II Low

3. Give a detailed explanation of the causes of landslides.

1. Landslide is a rapid downward movement of rock, soil, and vegetation down the slope under the influence of gravity.

2. The causes of landslides are wide-ranging, They have two aspects in common.

3. Force of gravity and

4. Failure of Soil

5. Landslides are considered of two types. They are:

• Naturally occurring disaster.
• Human-induced changes in the environment.

6. Natural causes of landslides are:

• Climatic changes
• Seismic activities
• Weathering
• Soil erosion
• Forest fires
• Gravity and
• Volcanic eruption

7. Human causes of landslides include deforestation mining, construction of roads, and railways over the mountain.

4. Elaborately discuss the effects of water pollution.

• It may cause a harmful effect on any living thing that drinks or uses or lives in it.
• It may be defined as alternation in the physical, chemical, and biological characteristics of water, which may cause harmful effects in human and aquatic life.

The major causes of water pollution in India are:

• Urbanisation
• Industrial effluents
• Sewages etc.
• Agricultural runoff and improper agricultural practices
• Seawater intrusion
• Solid wastes

### Samacheer Kalvi 8th Social Science Hazards Additional Important Questions and Answers

1. The content of Oxygen in the air is …………….
a) 20.95%
b) 21.95%
c) 22.95%
d) 23.95%
a) 20.95%

2. The word ‘hazard’ Originated from the ……………. language.
a) French
b) Telugu
c) Malayalam
a) French

3. National Institute of Disaster Management located in ……………..
a) Britain
b) New Delhi
c) France
d) Japan
b) New Delhi

4. In Tamil Nadu ……………. coastal districts are frequently affected by Storm Surges.
a) 13
b) 14
c) 15
d) 16
a) 13

5. The areas that receive an annual rainfall of less than ……………. are the drought zone regions of India.
a) 60 cm
b) 50 cm
c) 40 cm
d) 45 cm
a) 60 cm

6. The meaning of Tsunami is ……………..
a) Wave
b) Harbour wave
c) Storm
d) Storm wave
b) Harbour wave

7. In ……………. the Ukraine part of Exclusion Zone was declared as a radiogical and environmental biosphere.
a) 2013
b) 2014
c) 2015
d)2016
d) 2016

8. In India about ……………. of the land mass is prone to landslide hazard.
a) 15%
b) 18%
c) 19%
d) 20%
a) 15%

9. In Trophical Cyclones the wind Speed may reach upto ……………. per hour.
a) 200 km
b) 100 km
c) 300 km
d) 400 km
a) 200 km

10. High-Pressure Zone is mentioned in the Geographical map as ……………. letter.
a) L
b) P
c) H
d) S
c) H

II. Fill in the blanks.
1. The meaning of hazard is ………………..
a game of dice

2. ………………. may be worsened by the destruction of mangroves.
Storm Surge hazard

3. ………………. is a violent tremor in the earth’s crust.
Earthquake

4. ………………. is a strong wind circulating around a low-pressure area in the atmosphere.
Cyclonic Storm

5. During Cyclonic the wind speed may reach upto …………. km and rainfall may record up to …………… cm.
200, 50

6. A sudden rise of sea water due to trophical cyclone is called ………………..
Storm Surge

7. ………………. and ………………. are the major causes of landslides.
Steep slope, heavy rainfall

8. In Tamil Nadu ………………. and ………………. are frequently affected by landslides.
Kodaikanal, Ooty

9. The gas used for Photosynthesis is ………………..
Carbon dioxide

10. A ………………. pollutant is an air pollutant emitted directly from a Source.
Primary

11. ………………. may be defined by alteration is the physical, chemical, and biological characteristics of water.
Water Pollution

12. In India ………………. is perhaps the most critical components in Managing Disasters.
prevention

13. Example for Biological hazard is ………………..
Chicken Box

14. ………………., ………………. are Human-made hazards.
Explosions, hazardous waste

15. Primary Pollutant example ………………..
oxides of Sulphur

III. Match the following.

 1. Geological hazard a) Floods, Coastal erosion 2. Hydrologic hazard b) Eruptions and Lava flows 3. Volcanic hazard c) Chickenpox, Smallpox 4. Environmental hazard d) Earthquakes, Tsunami 5. Biological hazard e) Soil, air

 1. Geological hazard d) Earthquakes, Tsunami 2. Hydrologic hazard a) Floods, Coastal erosion 3. Volcanic hazard b) Eruptions and Lava flows 4. Environmental hazard e) Soil, air 5. Biological hazard c) Chickenpox, Smallpox

IV. State whether the following statements are true or false.
1. Tsunami means Stormy Surge.
False

2. A Secondary Pollutant is not directly emitted.
True

V. Assertion and Reasons.
1. Assertion: Landslides are generally Sudden and infrequent.
Reason: The presence of Sleep Slope and heavy rainfall are the major causes of landslides.
a) A is correct but R is incorrect
b) Both A and R are incorrect
c) A is incorrect and R is correct
d) Both A and R are correct
d) Both A and R are correct

1. Define Catastrophe.
A catastrophe is a massive disaster that requires a significant expenditure of money and a long time for recovery.

2. What are the major causes of floods?
The major causes of floods are:
Meteorological factors

• Heavy rainfall
• Tropical cyclones
• Cloudburst Physical factors
• Large catchment area

Human factors

• Deforestation
• Siltation
• Faulty agricultural practices
• Faulty irrigation practices
• Collapse of dams
• Accelerated urbanisation.

3. Write short notes on Indian Ocean Tsunami?

• On December 26, 2004, at 7:59 a.m. local time, an undersea earthquake with a magnitude of 9.1 struck off the coast of the Indonesian island of Sumatra.
• The tsunami killed at least 2,25,000 people across a dozen countries, with Indonesia, Sri Lanka, India, Thailand, Somalia and Maldives, sustaining massive damage.

4. Explain Prevention Measures.
Prevention is defined as the activities taken to prevent a natural calamity or potential hazard from having harmful effects on either people or economic assets.

VII. Distinguish the following.
1. Geologic hazard and Environmental hazard.

 Geologic hazard Environmental hazard Earthquakes, Tsunami, Landslide and Land subsidence. Pollution of soil/ air/water, Desertification, Global warming and Deforestation.

1. Explain about Socio – natural hazards.
Socio-natural hazards:
These are caused by the combined effect of natural forces and misdeeds of human. Some of the examples are:

• The frequency and intensity of floods and droughts may increase due to indiscriminate felling of trees, particularly in the catchment areas of the rivers.
• Landslides are caused by natural forces and their frequency, and impact may be aggravated as a result of construction of roads, houses etc., in mountainous areas, excavating tunnels and by mining and quarrying.
• Storm surge hazards may be worsened by the destruction of mangroves.
• Smog is a serious problem in most big urban areas. The emissions from vehicles and industries, combustion of wood and coal together combined with fog leads to smog.

2. Explain the Hazards based on their origin.
Hazards can be grouped into eight categories
1. Atmospheric hazard:
Tropical storms, Thunderstorms, Lightning, Tornadoes, Avalanches, Heat waves, Fog and Forest fire.

2. Geologic/Seismic hazard:
Earthquakes, Tsunami, Landslide and Land subsidence.

3. Hydrologic hazard:
Floods, Droughts, Coastal erosion and Storm surges.

4. Volcanic hazard:
Eruptions and Lava flows.

5. Environmental hazard:
Pollution of soil/ air/water, Desertification, Global warming and Deforestation.

6. Biological hazard:
Chickenpox, Smallpox, AIDS [HIV] and Killer bees.

7. Technological hazard:
Hazardous material incidents, Fires, Infrastructure failures [Bridges, Tunnels, Dams, Nuclear and Radiological accidents].

8. Human-induced hazard:
Terrorism, Bomb blast, War, Transportation accidents and Civil disorder.

## Samacheer Kalvi 8th Maths Guide Chapter 4 Life Mathematics Ex 4.1

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 4 Life Mathematics Ex 4.1 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 4 Life Mathematics Ex 4.1

Question 1.
Fill in the blanks:
(i) If 30 % of x is 150, then x is _______ .
500
Hint:
Given 30% of x is 150
i.e $$\frac{30}{100}$$ × x = 150

∴ x = 500

(ii) 2 minutes is _______ % to an hour.
3$$\frac{1}{3}$$%
Hint:
Let 2 min be x% of an hour
and 1 hr = 60mm
x% = $$\frac{2}{60} \times 100=\frac{200}{60}=\frac{10}{3}=3 \frac{1}{3}$$
x = 3$$\frac{1}{3}$$%

(iii) If x% of x = 25, then x = _______ .
50
Hint:
Given that x% of x is 25
∴ $$\frac{x}{100}$$ × x = 25
∴ x2 = 25 × 100 = 2500
∴ x = √2500 = 50

(iv) In a school of 1400 students, there are 420 girls. The percentage of boys in the school is _______ .
70
Hint:
Given total number of students in school = 1400
Number of girls in school = 420
∴ Number of boys in school = 1400 – 420 = 980

= $$\frac{980}{14}$$ = 70
% of boys = 70%

(v) 0.5252 is _______ %.
52.52%
Hint:
Given a number, and to express as a percentage, we need to multiply by 100
∴ to express 0.5252 as percentage, we should multiply by 100
∴ 0.5252 × 100 = 52.52%

percent difference Calculator can be found by first finding the difference between the numbers, then finding the average of the numbers, and then dividing.

Question 2.
Rewrite each underlined part using percentage language.
(i) One half of the cake is distributed to the children.
50% of the cake is distributed to the children
Hint:
One half is nothing but $$\frac { 1 }{ 2 }$$
as percentage, we need to multiply by 100
∴ $$\frac { 1 }{ 2 }$$ × 100 = 50%

(ii) Aparna scored 7.5 points out of 10 in a competition.
Aparna scored 75% in a competition
Hint:
7.5 points out of 10 is $$\frac{7.5}{10}$$ = 0.75
For percentage, we need to multiply by 100
We get 0.75 × 100 = 75%

(iii) The statue was made of pure silver.
The statue was made of 100% pure silver
Hint:
Pure silver means there are no other metals
so, 100 out of 100 parts is made of silver = $$\frac{100}{100}$$
∴ to express as percentage, $$\frac{100}{100}$$ × 100% = 100%

(iv) 48 out of 50 students participated in sports.
96% students participated in sports.
Hint:
48 out of 50 students in fraction form is $$\frac{48}{50}$$
As a percentage, we need to multiply by 100

(v) Only 2 persons out of 3 will be selected in the interview.
Only 66$$\frac{2}{3}$$% will be selected in the interview.
Hint:
2 out of 3 in fraction form is $$\frac{2}{3}$$
to express as percentage, we need to multiply by 100
$$\frac{2}{3} \times 100=\frac{200}{3}=66 \frac{2}{3} \%$$

To convert CGPA to Percentage, we need to multiply CGPA by 9.5, which will give us the percentage.

Question 3.
48 is 32% of which number?
Let the number required to be found be ‘x’
Given that 32% of x is 48
i.e., $$\frac{32}{100}$$ × x = 48

∴ x = 150

Question 4.
What is 25% of 30% of 400?
Required to find 25% of 30% of 400

The percentage decrease calculator determines the change from one amount to a lesser amount in terms of percent decrease.

Question 5.
If a car is sold for ₹ 2,00,000 from its original price of ₹ 3,00,000, then find the percentage of decrease in the value of the car.

original price of car = ₹ 3,00,000
actual selling price of car = ₹ 2,00,000
Decrease in amount from original = 3,00,000 – 2,00,000 = 1,00,000

Question 6.
If the difference between 75% of a number and 60% of the same number is 82.5, then find 20% of that number.
Given that 75% of number less 60% of number is 82.5
Let the number be ‘x’
∴ $$\frac{75}{100}$$ x x – $$\frac{60}{100}$$ x x = 82.5
∴ 0.75 x – 0.60 x = 82.5
∴ 0.15 x = 82.5
∴ x = $$\frac{82.5}{0.15}=\frac{8250}{15}$$ = 550
Required to find 20% of number ie 20% of x.

Question 7.
A number when increased by 18% gives 236. Find the number.
Let the number be x. Given that when it is increased by 18%, we get 236.

Question 8.
A number when decreased by 20% gives 80. Find the number.
Let the number be x. Given that when it is increased by 20% we get 80.

x = 100

It is important to learn to convert CGPA into percentage because both the systems are interconnected.

Question 9.
A number is increased by 25% and then decreased by 20%. Find the percentage change in that number.
Method 1.
Let the number be x.
First it is increased by 25%
∴ It becomes x + $$\frac{25}{100}$$ × x = $$\frac{125}{100}$$
Secondary it is decreased by 20%
$$\frac{125 x}{100}-\frac{20}{100} \times \frac{125}{100} x=\frac{125}{100} x \times \frac{80}{100}=x$$
Now we get back x, therefore there is no change.
Hence percentage change in that number is 0%

Method 2.
[to understand, let us assume that number is 100]
So, first when we increase by 25%, we get

Now this 125 is decreased by 20%, we get
125 – $$\frac{25}{100}$$ × 125 = 125 – 25 = 100
∴ We get back 100 ⇒ No change
Hence percentage change in that number is 0%

Question 10.
The ratio of boys and girls in a class is 5:3. If 16% of boys and 8% of girls failed in an examination, then find the percentage of passed students.
Let number of boys be ‘b’ and number of girls be ‘g’
Ratio of boys and girls is given as 5:3
b:g = 5:3 ⇒ $$\frac{b}{g}=\frac{5}{3}$$ …… (A)
Failure in boys = 16% = $$\frac{16}{100}$$ × b = $$\frac{16b}{100}$$
Failure in girls = 8% = $$\frac{8}{100}$$ × g = $$\frac{8g}{100}$$
Pass in boys = 100 – 16% = 84% = $$\frac{84}{100} b$$ …… (1)
Pass in girls = 100 – 8% = 92% = $$\frac{92}{100} g$$ ……. (2)
From A, we have $$\frac{b}{g}=\frac{5}{3}$$ , adding I on both sides, we get
$$\frac{b}{g}$$ + 1 = $$\frac{5}{3}$$ + 1
$$\frac{b+g}{g}=\frac{5+3}{3}=\frac{8}{3}$$
∴ g = $$\frac{3}{8}$$(b + g) ……. (3)
Similarly b = $$\frac{5}{8}$$ (b + g) ……. (4)
Total pass = pass in girls + pass in boys
= (1) + (2) = $$\frac{84}{100} b+\frac{92}{100} g$$

Objective Type Questions

Question 11.
12% of 250 litre is the same as ________ of 150 litre.
(A) 10%
(B) 15%
(C) 20%
(D) 30%
(C) 20%
Hint:
12% of 250 = $$\frac{12}{100}$$ × 250 = 30 lit.
Percentage: $$\frac{30}{150}$$ × 100 = 20%

Question 12.
If three candidates A, B and C in a school election got 153,245 and 102 votes respectively, then the percentage of votes got by the winner is ________ .
(A) 48%
(B) 49%
(C) 50%
(D) 45%
(B) 49%
Hint:
Candidate 1: 153
Candidate 2: 245 – winner [as maximum votes)
Candidate 3: 102
Total votes = 1 + 2 + 3 = 153 + 245 + 102 = 500

= $$\frac{245}{500}$$ × 100 = 49%

Question 13.
15% of 25% of 10000 = ________ .
(A) 375
(B) 400
(C) 425
(D) 475
(A) 375
Hint:
15% of 25% of 10000 is
First let us do 25% of 10,000, which is

Next 15% of the above is $$\frac{15}{100}$$ × 2500 = 375

Question 14.
When 60 is subtracted from 60% of a number to give 60, the number is
(A) 60
(B) 100
(C) 150
(D) 200
(D) 200
Hint:
Let the number be ‘X’
60% of the number is $$\frac{60}{100}$$ × x = $$\frac{60x}{100}$$
Given that when 60 is subtracted from 60%, we get 60
i.e $$\frac{60}{100}$$ x – 60 = 60
∴ $$\frac{60}{100}$$ x 60 + 60 = 120
∴ x = $$\frac{120 \times 100}{60}$$ = 200

Question 15.
If 48% of 48 = 64% of x, then x =
(A) 64
(B) 56
(C) 42
(D) 36
(D) 36
Hint:
Given that 48% of 48 = 64% of x

x = 36

## Samacheer Kalvi 8th Maths Guide Chapter 4 Life Mathematics InText Questions

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

## Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 4 Life Mathematics InText Questions

Try These (Text Book Page No. 122)

Find the indicated percentage value of the given numbers.

Try These (Text Book Page No. 124)

Question 1.
What percentage ola day is 10 hours?
In a day, there are 24 hours .
∴ 10 hrs out of 24 hrs is $$\frac{10}{24}$$
As a percentage, we need to multiply by 100
∴ Percentage = $$\frac{10}{24}$$ × 100 = 41.67%

Question 2.
Divide ₹ 350 among P, Q and R such that P gets 50% of what Q gets and Q gets 50% of what R gets.
Let R get x, Q gets 50% of what R gets
∴ Q gets = $$\frac{50}{100} \times x=\frac{x}{2}$$
P gets 50% of what Q gets .
∴ P gets = $$\frac{50}{100} \times \frac{x}{2}=\frac{x}{4}$$
Since 350 is divided among the three
∴ 350 = $$x+\frac{x}{2}+\frac{x}{4}$$
350 = $$\frac{4 x+2 x+x}{4}=\frac{7 x}{4}$$ = 350
x = $$\frac{350 \times 4}{7}$$
Q gets = $$\frac{x}{2}=\frac{200}{2}$$ = 100,
P gets = $$\frac{x}{4}=\frac{200}{4}$$ = 50
∴ p = 50, Q = 100, R = 200

Think (Text Book Page No. 124)

With a lot of pride, the traffic police commissioner of a city reported that the accidents had decreased by 200% in one year. He came up with this number by stating that the increase in accidents from 200 to 600 is clearly a 200% rise and now that it had gone down from 600 last year to 200 this year should be a 200% fall. Is this decrease from 600 to 200, the same 200% as reported by him? Justify.

Increase from original value 200 to 600

Decrease from original value 600 to 200

here original value is 600
% decrease = $$\frac{600-200}{600}$$ × 100 = $$\frac{400}{600}$$ × 100 = 66.67 % decrease
Increase from 200 → 600 and % decrease from 600 → 200 are not the same

Try These (Text Book Page No. 126)

Question 1.
If the selling price of an article is less than the cost price of the article, then there is a ________ .
Loss

Percentage difference calculator tool makes the calculation faster, and it displays the difference in the percentage in a fraction of seconds.

Question 2.
An article costing 5000 is sold for ₹ 4850. Is there a profit or loss? What percentage is it?
Loss
Percentage of Loss

Question 3.
If the ratio of cost price and the selling price of an article is 5:7, then the profit / gain is ________ %.
C.P = 5x
S.P = 7x
Profit = 7x – 5x = 2x

Think (Text Book Page No. 129)

A shopkeeper marks the price of a marker board 15% above the cost price and then allows a discount of 15% on the marked price. Does he gain or lose in the transaction?
Let cost price of marker board be 100
CP = 100 Marks it 15% above CP
∴ Marked price MP = $$\frac{15}{100}$$ × CP + CP
= $$\frac{15}{100}$$ × 100 + 100 = 15 + 100 = 115
Discount % = 15 %

∴ He sells it 97.75 which is less than his cost price. Therefore he loses
Loss = 97.75 – 100 = – 2.25

Try These (Text Book Page No. 129)

Question 1.
The formula to find the simple interest for a given principal is ________ .
$$\frac{\mathrm{PNR}}{100}$$

Question 2.
Find the simple interest on ₹ 900 for 73 days at 8% p.a.

Question 3.
In how many years will ₹ 2000 become ₹ 3600 at 10% p.a simple interest?
I = 3600 – 2000 = 1600

Try These (Text Book Page No. 141)

Question 1.
Classify the given examples as direct or inverse proportion:
(i) Weight of pulses to their cost.
As weight increases cost also increases.
∴ Weight and cost are direct proportion.

(ii) Distance travelled by bus to the price of ticket.
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.
As the speed increases, the time to cover the distance become less.
So speed and üme are in indirect proportion.

(iv) Number of workers employed to complete a construction in a specified time.
As the number of workers increases, the amount of work become less, so they are in indirect proportion.

(v) Area of a circle to its radius.
If the radius of the circle increases its area also increases.
∴ Area and radius of circles are direct proportion.

Question 2.
A student can type 21 pages in 15 minutes. At the same rate, how long will it take student to type 84 pages?
Direct proportion
No. of minutes = x
k = $$\frac{21}{15}$$
$$\frac{21}{15}=\frac{84}{x}$$

Question 3.
If 35 women can do a piece of work in 16 days, In how many days will 28 women do the same work?
Inverse proportion
No. of days = x
k = 35 × 16
∴ 28 × x = 35 × 16

Try These (Text Book Page No. 145)

Question 1.
If x and y vary directly, find k when x = y = 5.
I If x andy vary directly then $$\frac{x}{y}$$ = k.
Here x = 5; y = 5
∴ k = $$\frac{5}{5}$$
k = 1

Question 2.
If x and y vary inversely, find the constant of proportionality when x = 64 and y = 0.75
Gìven x = 64, y = 0.75
and also given x andy vary inversely.
∴ xy = k. the constant of variation.
∴ Constant = 64 × 0.75
Constant of variation = 48

Activity (Text Book Page No. 145)

Draw a circle of a given radius. Then, draw its radii in such a way that the angles between any two consecutive pair of radii are equal. Start drawing 3 radii and end with drawing 12 radii in the circle. List and prepare a table for the number of radii to the angle between a pair of consecutive radii and check whether they are in inverse proportion. What is the proportionality constant?

As the number of radii increases angle decreases.
Hence they are in inverse proportion
∴ xy = 4 proportional constant
3 × 120° = 360° = k = 360°

Try These (Text Book Page No. 147)

The percentage difference calculator is here to help you compare two numbers.

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.
Let the required no. of articles be x

 Men (P) Days (D) Articles (W) 24 12 48 6 6 x

(i) Mens and days are Indirect variables.
(ii) Men and Articles are direct vanables
(iii) Days and articles are also direct variables using formula.
Let
P1 = 24
P2 = 6

D1 = 12
D2 = 6

W1 = 48
W1 = x

x = 6 Articles

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.
Let the required no. of workers be x

 Length (work) Hours Workers 4 km 4 hrs 15 8 km 8 hrs x

(i) Length and workers are direct variable as more length need more workers.
The proportion is 4 : 8 : : 15 : x ——– (1)
(ü) Hours and workers are indirect variables as more working hours need less men.
∴ The proportion is : 4 : : 15 : x ——– (2)
Combining (1) and (2)

Product of the extremes = Product of the mean
4 × 8 × x = 8 × 4 × 15
x = $$\frac{8 \times 4 \times 15}{4 \times 8}$$
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 a day to complete the same work in 30 days.
Let the required hours be x.

 Women Days Hours 25 36 12 20 30 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 = $$12 \times \frac{25}{20} \times \frac{36}{30}$$
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 .
Let the required number of days be x.

 Rice (kg) Men Days 420 98 45 60 42 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 = $$45 \times \frac{60}{420} \times \frac{98}{42}$$

x = 15 days

Try These (Text Book Page No. 150)

Question 1.
Vikram can do one-third of work in p days. He can do th of work In ________ days.
$$\frac{1}{3}$$ of the work will be done mp days.
$$\frac{3}{4}$$ th of the work will be done in = 3p x $$\frac{3}{4}$$
= $$\frac{9}{4}$$p = 2$$\frac{1}{4}$$ p days.
work in ______ days and $$\frac{m}{4}$$ persons can complete the same work in ______ days.
4 m men do the work it will be completed in $$\frac{m n}{4 m}$$ days = $$\frac{m}{4}$$ days.