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 ________ .

Answer:

9

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

Answer:

48

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

Answer:

5

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

Answer:

3

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

Answer:

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.

Answer:

True

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

Answer:

True

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

Answer:

False

(iv) The square of 75 is 4925.

Answer:

False

(v) The square root of 225 is 15.

Answer:

True

Question 3.

Find the square of the following numbers.

(i) 17

(ii) 203

(iii) 1098

Answer:

(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

Answer:

(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

Question 5.

Find the square root by prime factorisation method.

(i) 144

(ii) 256

(iii) 784

(iv) 1156

(v) 4761

(vi) 9025

Answer:

(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

Question 6.

Find the square root by long division method.

(i) 1764

(ii) 6889

(iii) 11025

(iv) 17956

(v) 418609

Answer:

(i) 1764

√1764 = 42

(ii) 6889

√6889 = 83

(iii) 11025

√11025 = 105

(iv) 17956

√17956 = 134

(v) 418609

√418609 = 647

Question 7.

Estimate the value of the following square roots to the nearest whole number:

(i) √440

(ii) √800

(iii) √1020

Answer:

(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

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

Answer:

(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.

Answer:

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 = 60

Objective Type Questions

Question 11.

The square of 43 ends with the digit .

(A) 9

(B) 6

(C) 4

(D) 3

Answer:

(A) 9

Hint:

Ones digit = 3 × 3 = 9

Question 12.

_______ is added to 24^{2} to get 25^{2}.

(A) 4^{2}

(B) 5^{2}

(C) 6^{2}

(D) 7^{2}

Answer:

(D) 7^{2}

Hint:

25^{2} = 25 × 25 = 625

24^{2} = 24 × 24 = 576

Question 13.

√48 is approximately equal to .

(A) 5

(B) 6

(C) 7

(D) 8

Answer:

(C) 7

Hint:

√49 = 7

Question 14.

\(\sqrt{128}-\sqrt{98}+\sqrt{18}\)

(A) √2

(B) √8

(C) √48

(D) √32

Answer:

(D) √32

Question 15.

The number of digits in the square root of 123454321 is ______.

(A) 4

(B) 5

(C) 6

(D) 7

Answer:

(B) 5

Hint:

\(\frac{n+1}{2}=\frac{10}{2}=5\)