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## Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 Numbers Ex 1.1

Question 1.

Fill in the blanks:

i. The ones digit in the square of 77 is …………

ii. The number of non-square numbers between 24² and 25² is ……………….

iii. If a number ends with 5, its square ends with …………

iv. A square number will not end with numbers ………….

v. The number of perfect square numbers between 300 and 500 is …………

Solution:

i. 9

ii. 48

iii. 5

iv. 2, 3, 7, 8

v. 5

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.

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

iii. The number of zeros in the square of 961000 is 9.

iv. (7, 24, 25) is a Pythagorean triplet.

v. The square root of 221 is 21.

Solution:

i. True

ii. True

iii. False

iv. True

v. False

Question 3.

What will be the ones digit in the squares of the following numbers?

(i) 36

(ii) 252

(iii) 543

Solution:

(i) If a number ends with 6, its square ends with 6.

∴ Ones’ digit in the square of 36 is 6.

(ii) If a number ends with 2, its square ends with 4.

∴ Ones’ digit in the square of 252 is 4

(iii) If a number ends with 3, its square ends with 9.

∴ Ones’ digit in the square of 543 is 9.

Question 4.

Study the given numbers and justify why each of them obviously cannot be a perfect square.

(i) 1000

(ii) 34567

(iii) 408

Solution:

We know that the numbers end with odd number of zeros, 7 and 8 not perfect squares.

∴ 1000, 34567 and 408 cannot be perfect squares.

Question 5.

Find the sum without actually adding the following odd numbers:

(i) 1 + 3 + 5 + 7 +……..+ 35

(ii) The first 99 odd natural numbers.

Solution:

1 + 3 + 5 + 7 +……..+ 35.

Here there are 18 odd numbers from 1 to 35.

Sum of first n consecutive odd natural numbers = n²

∴ Sum of first 18 consecutive odd natural numbers = 18² = 18 × 18 = 324

(ii) The first 99 odd natural numbers.

Sum of first n consecutive natural numbers = n²

∴ Sum of 99 odd natural numbers = 99² = 99 × 99 = 9801

Question 6.

Express

(i) 15² and

(ii) 19²

as the sum of two consecutive positive integers.

Solution:

Question 7.

Write

(i) 10² and

(ii) 11²

as the sum of consecutive odd natural numbers.

Solution:

(i) 10²

10² = 100 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19

(ii) 11²

11² = 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

Question 8.

Find a Pythagorean triplet whose

(i) largest member is 65

(ii) smallest member is 10

Solution:

(i) Largest number is 65

Given largest number is 65.

We know that 2m, m² – 1, m² + 1 form a Pythagorean triplet.

let m² + 1 = 65

m² = 65 – 1

m² = 64

m² = 8 × 8

m = 8

∴ 2m = 2 × 8 = 16

m² – 1 = 64 – 1 = 63

∴ The required Pythagorean triplet is (16, 63, 65)

(ii) Smallest number is 10

We know that (2m, m² – 1, m² + 1) form a Pythagorean triplet.

Given smallest number is 10

let 2m = 10

m = \(\frac{10}{2}\)

m = 5

m² + 1 = 5² + 1 = 25 + 1 = 26

m² – 1 = 5² – 1 = 25 – 1 = 24

∴ The required Pythagorean triplet is (10, 24, 26)

Question 9.

Find the square root of the following by repeated subtraction method.

(i) 144

(ii) 256

(iii) 784

Solution:

(i) 144

Step 1: 144 – 1 = 143

Step 2: 143 – 3 = 140

Step 3: 140 – 5 = 135

Step 4: 135 – 7 = 128

Step 5: 128 – 9 = 119

Step 6: 119 – 11 = 108

Step 7: 108 – 13 = 95

Step 8: 95 – 15 = 80

Step 9: 80 – 17 = 63

Step 10: 63 – 19 = 44

Step 11: 44 – 21 = 23

Step 12: 23 – 23 = 0.

We get 0 in the 12th step.

∴ 144 is a perfect square and ⇒ \(\sqrt{144}\) = 12.

(ii) 256

Step 1: 256 – 1 = 255

Step 2: 255 – 3 = 252

Step 3: 252 – 5 = 247

Step 4: 247 – 7 = 240

Step 5: 240 – 9 = 231

Step 6: 231 – 11 = 220

Step 7: 220 – 13 = 207

Step 8: 207 – 15 = 192

Step 9: 192 – 17 = 175

Step 10: 175 – 19 = 156

Step 11: 156 – 21 = 135

Step 12: 135 – 23 = 112

Step 13: 112 – 25 = 87

Step 14: 87 – 27 = 60

Step 15: 60 – 29 = 31

Step 16: 31 – 31 = 0

We have subtracted odd numbers starting from 1 repeatedly from 256. We get 0 in the 8th step.

∴ 256 is a perfect square and \(\sqrt{256}\) = 16

(iii) 784

Step 1: 784 – 1 = 783

Step 2: 783 – 3 = 780

Step 3: 780 – 5 = 775

Step 4: 775 – 7 = 768

Step 5: 768 – 9 = 759

Step 6: 759 – 11 = 748

Step 7: 748 – 13 = 735

Step 8: 735 – 15 = 720

Step 9: 720 – 17 = 703

Step 10: 703 – 19 = 684

Step 11: 684 – 21 = 663

Step 12: 663 – 23 = 640

Step 13: 640 – 25 = 615

Step 14: 615 – 27 = 588

Step 15: 588 – 29 = 559

Step 16: 559 – 31 = 528

Step 17: 528 – 33 = 495

Step 18: 495 – 35 = 460

Step 19: 460 – 37 = 423

Step 20: 423 – 39 = 384

Step 21: 384 – 41 = 343

Step 22: 343 – 43 300

Step 23: 300 – 45 = 255

Step 24: 255 – 47 = 208

Step 25: 208 – 49 = 159

Step 26: 159 – 51 = 108

Step 27: 108 – 53 = 55

Step 28: 55 – 55 = 0

We have subtracted odd numbers starting from 1 repeatedly from 784, we get zero in the 28th step.

∴ 784 is a perfect square. \(\sqrt{784}\) = 28

Question 10.

Find the square root by prime factorisation method.

(i) 1156

(ii) 4761

(iii) 9025

Solution:

(i) 1156

1156 = 2 × 2 × 17 × 17

1156 = 2² × 17²

1156 = (2 × 17)²

∴ \(\sqrt{1156}\) = \((\sqrt{2×17})^{2}\) = 2 × 17 = 34

∴ \(\sqrt{1156}\) = 34

(ii) 4761

4761 = 3 × 3 × 23 × 23

4761 = 3² × 23²

4761 = (3 × 23)²

\(\sqrt{4761}\) = \((\sqrt{3 × 23})^{2}\)

\(\sqrt{4761}\) = 3 × 23

\(\sqrt{4761}\) = 69

(iii) 9025

9025 = 5 × 5 × 19 × 19

9025 = 5² × 19²

9025 = (5 × 19)²

\(\sqrt{925}\) = \((\sqrt{5 × 19})^{2}\) = 5 × 19 = 95

Question 11.

Examine if each of the following is a perfect square:

(i) 725

(ii) 190

(iii) 841

(iv) 1089

Solution:

(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

Question 12.

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.

Solution:

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

\(\sqrt{3600}\) = \((\sqrt{2 × 3 × 5 × 2})^{2}\) = 2 × 3 × 5 × 2 = 60

∴ \(\sqrt{3600}\) = 60.

Question 13.

Find the smallest number by which 10985 should be divided so that the quotient is a perfect square.

Solution:

We find 10985 = 5 × 13 × 13 × 13

= 5 × 13 × 13²

Here the prime factors 5 and 13 do not have pairs.

∴ We can divide 10985 by 65 (5 × 13) to get a perfect square

∴ When we divide 10985 by 65 we get quotient 169.

169 = 13 × 13

\(\sqrt{169}\) = 13

Question 14.

Is 2352 a perfect square? If not, find the smallest number by which 2352 must be multiplied so that the product is a perfect square. Find the square root of the new number.

Solution:

We find 2352 = 2 × 2 × 2 × 2 × 3 × 7 × 7

2352 = 2² × 2² × 3 × 7²

Here the factor 3 has no pair.

∴ 2352 is not a perfect square.

We have to multiply 2352 by 3 so that the product is a perfect square.

Img 13

∴ 2352 × 3 = 2² × 2² × 7² × 3 × 3

7056 = 2² × 2² × 7² × 3²

7056 = (2 × 2 × 7 × 3)²

∴ \(\sqrt{7056}\) = \((\sqrt{2 × 2 × 7 × 3})^{2}\)

= 2 × 2 × 7 × 3 = 84

\(\sqrt{7056}\) = 84

Question 15.

Find the least square number which is divisible by each of the numbers 8, 12 and 15.

Solution:

The least number divisible by each of the numbers 8, 12, 15 is their L.C.M

∴ LCM of 8, 12, 15 is (4 × 3 × 2 × 5) = 120

Resolving 120 into prime factors

We get 120 = 2 × 2 × 2 × 3 × 5

Grouping into pairs of equal factors

120 = (2 x 2) x 2 x 3 x 5

∴ The factors 2, 3 and 5 had no pairs.

Thus to make 120 a perfect square, we must multiply it by (2 x 3 x 5) = 30.

∴ The least square number divisible by 8,12 and 15 is 120 x 30 = 3600