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

Answer:

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

Answer:

256 is smaller, so it is the 1st divisor

∴ HCF = 8

(iii) 6765 and 610

Answer:

∴ HCF = 5

(iv) 184, 230 and 276

Answer:

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

Answer:

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

Answer:

28 – 8 = 20

20 – 8 = 12

12 – 8 = 4

8 = 4 = 4

now m = n = 4

∴ HCF is 4

(iii) 280 and 420

Answer:

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

Answer:

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

Answer:

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

Answer:

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)

Answer:

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

Answer:

(c) 89

Hint:

∴ 11^{th} 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)

Answer:

(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 3^{rd} number of the Fibonacci sequence is a multiple of_______

(a) 2

(b) 3

(c) 5

(d) 8

Answer:

(a) 2

Hint:

Every 3^{rd} number in Fibonacci sequence is a multiple of 2

Question 8.

Every _____ number of the Fibonacci sequence is a multiple of 8

(a) 2^{th}

(b) 4^{th}

(c) 6^{th}

(d) 8^{th}

Answer:

(c) 6^{th}

Question 9.

The difference between the 18^{th} and 17^{th} Fibonacci number is

(a) 233

(b) 377

(c) 610

(d) 987

Answer:

(d) 987

Hint:

F(18) = F(17) + F(16)

F(18) – F(17) = F(16) = F(15) + F(14)

= 610 + 377 = 987

Question 10.

Common prime factors of 30 and 250 are

(a) 2 × 5

(b) 3 × 5

(c) 2 × 3 × 5

(d) 5 × 5

Answer:

(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

Answer:

(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

Answer:

(d) 11