Samacheer Class 8 Maths - Numbers

(i) \(\frac{9}{4}\)
\(\frac{9}{4}=2 \frac{1}{4}\)
∴ \(\frac{9}{4}\) lies between 2 and 3(ii) \(\frac{-8}{3}\)
\(\frac{-8}{3}=-2 \frac{2}{3}\)
\(-2 \frac{2}{3}\) lies between -2 and 3(iii) \(\frac{-17}{-5}\)
\(\frac{-17}{-5}=3 \frac{2}{5}\)
\(3 \frac{2}{5}\) lies between 3 and 4 in the number line.(iv) \(\frac{15}{-4}\)
\(\frac{15}{-4}=-3 \frac{3}{4}\)
\(-3 \frac{3}{4}\) lies between -3 and -4

(i) 2 and 0
i.e., \(\frac{-2}{1}\) and \(\frac{0}{1}\)∴ Five rational number between \(\frac { -20 }{ 10 }\) (= -2) and \(\frac { 0 }{ 10 }\) (= 0) are(ii) \(\frac{-1}{2}\) and \(\frac{3}{5}\)
LCM of 2 and 5 = 2 × 5 = 10∴ Five rational number between(iii) \(\frac{1}{4}\) and \(\frac{7}{20}\)∴ Five rational number between(iv) \(\frac{-6}{4}\) and \(\frac{-23}{10}\)∴ Five rational number between

(i) \(\frac{-11}{5}, \frac{-21}{8}\)
LCM of 5, 8 is 40(ii) \(\frac{3}{-4}, \frac{-1}{2}\)
LCM of 4 and 2 = 4(iii) \(\frac{2}{3}, \frac{4}{5}\)
LCM of 3 and 5 is 15.

(i) \(\frac{-5}{12}, \frac{-11}{8}, \frac{-15}{24}, \frac{-7}{-9}, \frac{12}{36}\)
LCM of 12, 8, 24, 9, 36 is 4 × 3 × 2 × 3 = 72Now comparing the numerators – 30, – 99, -45, 56, 24 we get 56 > 24 > – 30 > – 45 > – 99(ii) \(\frac{-17}{10}, \frac{-7}{5}, 0, \frac{-2}{4}, \frac{-19}{20}\)
LCM of 10, 5, 4, 20 is 5 × 2 × 2 = 20Negative numbers are less than zero.
∴ Arranging the numerators we get
– 34 < – 28 < – 19 < – 10 < 0Objective Type Questions:

closure property for addition
Let a = \(\frac{-5}{7}\) and b = \(\frac{8}{9}\)∴ Closure property is true for addition of rational numbers.
Closure property for multiplication∴ Closure property is true for rnultiplìcation of rational numbers.

Let a = \(\frac{-10}{11}\) and \(\frac{-8{33}\) be the given rational numbers.From (1) and (2)
a + b = b + a and hence additionis commutative for rational numbersFrom (3) and (4) a × b = b × a
Hence multiplication is commutative for rational numbers.

From (1) and (2), (a + b) + c = a + (b + c) is true for rational numbers.From (1) and (2) (a × b) × c = (a × b) × c is true for rational numbers.
Thus associative property.

From (1) and (2) we have a × (b + c) = (a × b) + (a × c) is true
Hence multiplication is distributive over addition for rational numbers Q.

(i) 725
725 = 5 × 5 × 29 = 5 2 × 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.

(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 = 2 2 × 17 2
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 = 3 2 × 23 2
4761 = (3 × 23) 2
√4761 = \(\sqrt{(3 \times 23)^{2}}\)
√4761 = 3 × 23
√4761 = 69(vi) 9025
9025 = 5 × 5 × 19 × 19
9025 = 5 2 × 19 2
9025 = (5 × 19) 2
√925 = \(\sqrt{(5 \times 19)^{2}}\) = 5 × 19 = 95

(i) 1764√1764 = 42
(ii) 6889√6889 = 83(iii) 11025√11025 = 105
(iv) 17956√17956 = 134
(v) 418609√418609 = 647Roots Calculator is a free online tool that displays the roots of the given quadratic equation.

(i) √440
we have 20 2 = 400
21 2 = 441
∴ √440 ≃ 21
(ii) √800
we have 28 2 = 784
29 2 = 841
∴ √800 ≃ 28
(iii) √1020
we have 31 2 = 961
32 2 = 1024
∴ √1020 ≃ 32

We find 1800 = 2 × 2 × 3 × 3 × 5 × 5 × 2
= 2 2 × 3 2 × 5 2 × 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 = 2 2 × 3 2 × 5 2 × 2 × 2
3600 = 2 2 × 3 2 × 5 2 × 2 2
= (2 × 3 × 5 × 2) 2
\(\sqrt{3600}=\sqrt{(2 \times 3 \times 5 \times 2)^{2}}\)
= 2 × 3 × 5 × 2 = 60
∴ √3600 = 60Objective Type Questions

1944 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3= 2 3 × 3 3 × 3 × 3
There are two triplets to make further triplets we need one more 3.
∴ 1944 is not a perfect cube.

We have 10985 = 5 × 13 ×13 × 13
= 5 × 13 ×13 × 13
Here we have a triplet of 13 and we are left over with 5.
If we divide 10985 by 5, the new number will be a perfect cube.
∴ The required number is 5.

Grouping the prime factors of 200 as triplets, we are left with 5 × 5
We need one more 5 to make it a perfect cube.
So to make 200 a perfect cube multiply both sides by 5.1000 = 2 × 2 × 2 × 5 × 5 × 5 × 5 × 5
Now 1000 is a perfect cube.
∴ The required number is 5.

(729) 1/3 = 3 × 3 = 9
∴ The cube of 9 is 729.
9 = 3 × 3 [ie 3 is squared to get 9]We have to find out √3,
√3 = 1.732

Consider the numbers 2 2 and 4 2
The numbers are 4 and 16.
Their procluct 4 × 16 = 64
64 = 4 × 4 × 4
∴ The required square numbers are 4 and 16

(i) (3 2 ) 3 × (2 × 3 5 ) -2 × (18) 2(ii) \(\frac{9^{2} \times 7^{3} \times 2^{5}}{84^{3}}\)(iii) \(\frac{2^{8} \times 2187}{3^{5} \times 3^{2}}\)= 2 8-5 × 3 7-5
= 2 3 × 3 2
= 8 × 9
= 72

(i) 6054.321
6054.321 = (6 × 1000) + (0 × 100) + (5 × 10) + (4 × 10 0 ) + \(\frac{3}{10}+\frac{2}{100}+\frac{1}{1000}\)
= (6 × 10 3 ) + (5 × 10 1 ) + (4 × 10 0 ) + \(\frac{3}{10}+\frac{2}{100}+\frac{1}{1000}\)
= (6 × 10 3 ) + (5 × 10 1 ) + (4 × 10 0 ) + (3 × 10 -1 ) + (2 × 10 -2 ) + (1 × 10 -3 )
(ii) 897.14
= (8 × 100) + (9 × 10) + (7 × 10 0 ) + \(\frac{1}{10}+\frac{4}{100}\)
= (8 × 1o 2 ) +( 9 × 10 1 ) + (7 × 10 0 ) + \(\left(1 \times \frac{1}{10}\right)+\left(4 \times \frac{1}{100}\right)\)
= (8 × 10 3 ) + (9 × 10 3 ) + (7 × 10 0 ) + (1 × 10 -1 ) + (4 × 10 -2 )

(i) 8 × 10 4 + 7 × 10 3 + 6 × 10 2 + 5 × 10 1 + 2 × 1 + 4 × 10 -2 + 7 × 10 -4
= 8 × 10 4 + 7 × 10 3 + 6 × 10 2 + 5 × 10 1 + 2 × 1 + 4 × 10 -2 + 7 × 10 -4
= 8 × 10000 + 7 × 1000 + 6 × 100 + 5 × 10 + 2 × 1 + 4 × \(\frac{1}{100}\) + 7 × \(\frac{1}{10000}\)
= 80000 + 7000 + 600 + 50 + 2 + \(\frac{4}{100}\) + \(\frac{7}{10000}\)
= 87652.0407(ii) 5 × 10 3 + 5 × 10 1 + 5 × 10 -1 + 5 × 10 -3
= 5 × 10 3 + 5 × 10 1 + 5 × 10 -1 + 5 × 10 -3
= 5 × 1000 + 5 × 10 + 5 × \(\frac{1}{10}\) + 5 × \(\frac{1}{1000}\)
= 5000 + 50 + \(\frac{5}{10}+\frac{5}{1000}\) = 5050.505
(iii) The radius of a hydrogen atom is 2.5 10 -11 m
Radiys of a hydrogen atom = 2.5 × 10 -11 m
= \(2.5 \times \frac{1}{10^{11}} \mathrm{m}=\frac{2.5}{10^{11}} \mathrm{m}\)
= 0.000000000025 m

467800000000 = 4.678 × 10 11
(ii) 0.000001972
0.000001972 = 1.972 × 10 -6
(iii) 1642.398
1642.398 = 1.642398 × 10 3(iv) Earth’s volume is about 1,083,000,000,000 cubic kilometres
1,083,000,000,000
Earth’s volume = 1.083 110 × 10 2 cubic kilometres
(v) If you fill a bucket with dirt, the portion of the whole Earth that is in the bucket will be 0.00000000000000000000000 16 kg
Portion of earth in the bucket = 0.00000000000000000000000 16 kg
= 1.6 10 × 10 24 kg.
Objective Type Questions

Capacity of the small waterug = 3\(\frac{4}{5}\) litres.
Capacity of the big jug = \(2 \frac{2}{3}\) times the small one.
= \(2 \frac{2}{3} \times 3 \frac{4}{5}=\frac{8}{3} \times \frac{19}{5}=\frac{152}{15}\)
= \(\frac{2}{15}\) litres
Capacity of the large jug = \(\frac{2}{15}\) litres.

Length of the room × Breadth = Area of the room
Breadth of the room = \(2 \frac{11}{20}\) m
Area of the room = \(\frac{153}{10}\) sq.m
Length x \(2\frac{11}{20}\) = \(\frac{153}{10}\)Length of the room = 6 m

Area of the square = 4489 cm 2
(side)2 = 4489 cm 2
(side)2 = 67 × 67
side = 67 2
Length of a side = 67
Length of a side with liner = 67 + 2 + 2 cm
= 71 cmArea of the larger square = 71 × 71 cm 2
= 5041 cm 2
Area of the liner = Area of big square – Area of small square
= (5041 – 4489) cm 2
= 552 cm 2

Area of the greeting card = 90 cm 2
(side) 2 = 90 cm 2
(side) 2 = 2 × 5 × 3 × 3 = 2 × 5 × 3 2Side = 3\(\sqrt{2 \times 5}\)
side = 3√10 cm
side = 3 × 3.2cm
side = 9.6 cm
∴ Side lies between the whole numbers 9 and 10.

Area of one tile = 1 sq.decimeter
Area of 225 tiles = 225 sq.decimeter
225 square tiles exactly covers the square shaped verandah.
∴ Area of 225 tiles = Area of the verandah
Area of the verandah = 225 sq.decimeter
side × side = 15 × 15 sq.decimeter
side = 15 decimeters
Length of each side of verandah = 15 decimeters.

Heart beat per minute = 80 beats
(i) an hour
One hour = 60 minutes
Heart beat in an hour = 60 × 80
= 4800
= 4.8 × 10 3
(ii) In a day
One day = 24 hours = 24 × 60 minutes
∴ Heart beat in one day = 24 × 60 × 80 = 24 × 4800 = 115200
= 1.152 × 10 5
(iii) a year
One year = 365 days = 365 × 24 hours = 365 × 24 × 60 minutes
∴ Heart beats in a year = 365 × 24 × 60 × 80
= 42048000
= 4.2048 × 10 7(iv) 100 years
Heart beats in one year = 4.2048 × 10 7
heart beats in 100 years = 4.2048 × 10 7 × 100 = 4.2048 × 10 7 × 10 2
= 4.2048 × 10 9
Challenging Problems:

1 inch = 120 km
Distance between A and B = \(4\frac{1}{6}\)
Distance between A and C = \(3\frac{1}{3}\)
∴ Distance between B and C = \(4 \frac{1}{6}+3 \frac{1}{3}\) inches1 inch = 120km
∴ \(\frac{45}{6}\) inches = \(\frac{45}{6}\) × 120 km = 900 km
Distance between B and C = 900 km

let a = \(\frac{5}{6}\) and b = \(\frac{-4}{3}\) be two non zero rational numbers.∴ Collection of non-zero rational numbers are closed under division.
(ii) Subtraction is not commutative for rational numbers.
let a = \(\frac{1}{2}\) and b = \(-\frac{5}{6}\) be two rational numbers.a – b ≠ b – a
∴ Subtraction is not commutative for rational numbers.(iii) Division is not associative for rational numbers.
Let a = \(\frac{2}{5}\), b = \(\frac{6}{5}\), c = \(\frac{3}{5}\) be three rational numbers.a ÷ (b ÷ c) ≠ (a ÷ b) ÷ c
∴ Division is not associative for rational numbers.
(iv) Distributive property of multiplication over subtraction is true for rational numbers. That is, a (b – c) = ab – ac.
Let a = \(\frac{2}{9}\), b = \(\frac{3}{6}\), c = \(\frac{1}{3}\) be three rational numbers.
To prove a × (b – c) = ab – bc∴ From (1) and (2)
a × (b – c) = ab – bc
∴ Distributivity of multiplication over subtraction is true for rational numbers.(v) The mean of two rational numbers is rational and lies between them.
Let a = \(\frac{2}{11}\) and b = \(\frac{5}{6}\) be two rational numbers∴ The mean lies between the given rational numbers \(\frac{2}{11}\) and \(\frac{5}{6}\)

Let the weight of 1 ragi adai = x grams
given \(\frac { 1 }{ 4 }\) of x = 120gm
\(\frac { 1 }{ 4 }\) × x = 120
x = 120 × 4
x = 480gm
∴ \(\frac { 2 }{ 3 }\) of the adai = \(\frac { 2 }{ 3 }\) × 480 gm = 2 × 160 gm = 320gm
\(\frac { 2 }{ 3 }\) of the weight of adai = 320gm

Number of cadets to form square designThe numbers 2 and 3 are unpaired
∴ It is impossible to have the parade forming square design with 1536 cadets.39 × 39 = 1521
Also 40 × 40 = 1600
∴ We have to add (1600 – 1536) = 64 to make 1536 a perfect square.
∴ 64 more cadets would be required to form the square design.

16 25 = (2 4 ) 25 = 2 100
8 100 = (2 3 ) 100 = 2 300
4 400 = (2 2 ) 400 = 2 800
2 600 = 2 600
Comparing the powers we have.
2 100 < 2 300 < 2 600 < 2 800
∴ The required order: 16 25 , 8 100 , 3 500 , 4 400 , 2 600

\(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \frac{1}{7}\)
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:

\(\frac{-2}{3}\) 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 numbersTry This (Text Book Page No. 25)

No, the square of a prime number ‘P’ has at Least 3 divisors 1, P and P 2 . 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.

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)

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 = 2 2 × 3 2 × 3
If we multiply the factors by 2, then we get
2 2 × 3 2 × 3 × 3 = 2 2 × 3 2 × 3 2 = (2 × 3 × 3) 2
Which is perfect square.
∴ Again if we divide by 3 then we get 2 2 × 3 2 ⇒ (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.

0.000012005 = 1.2005 × 10 -5
(ii) 43 12.345
43 12.345 = 4.312345 × 10 3(iii) 0.10524
0.10524 = 1.0524 × 10 -1
(iv) The distance between the Sun and the planet Saturn 1.4335 × 10 12 miles.