Samacheer Kalvi Class 8 Maths Solutions Chapter 7 Information processing Exercise 7.2

 

Question 1. Using repeated division method, find the HCF of the following:
(i) 455 and 26
Answer: We use the repeated division method to find the HCF.
First, divide 455 by 26:
\( 455 = 26 \times 17 + 13 \)
Next, the remainder 13 becomes the new divisor, and the previous divisor 26 becomes the new dividend:
\( 26 = 13 \times 2 + 0 \)
Since the remainder is now 0, the last divisor, which is 13, is the HCF.
In simple words: To find the HCF, we divide the bigger number by the smaller number. Then, we divide the first divisor by the remainder. We keep doing this until the remainder is zero. The last number we divided by (the last divisor) is the HCF.

🎯 Exam Tip: The repeated division method, also known as the Euclidean algorithm, is efficient for finding the HCF of larger numbers by continually reducing them to smaller remainders.

 

Question 1. Using repeated division method, find the HCF of the following:
(ii) 392 and 256
Answer: We use the repeated division method to find the HCF.
First, divide 392 by 256:
\( 392 = 256 \times 1 + 136 \)
Next, the remainder 136 becomes the new divisor, and the previous divisor 256 becomes the new dividend:
\( 256 = 136 \times 1 + 120 \)
Again, the remainder 120 becomes the new divisor, and 136 becomes the new dividend:
\( 136 = 120 \times 1 + 16 \)
The remainder 16 becomes the new divisor, and 120 becomes the new dividend:
\( 120 = 16 \times 7 + 8 \)
Finally, the remainder 8 becomes the new divisor, and 16 becomes the new dividend:
\( 16 = 8 \times 2 + 0 \)
Since the remainder is 0, the last divisor, which is 8, is the HCF.
In simple words: We find the HCF by dividing repeatedly. The remainder from each step becomes the next divisor. We stop when there is no remainder, and the last divisor is our HCF.

🎯 Exam Tip: When using the repeated division method, ensure each step uses the previous divisor as the new dividend and the previous remainder as the new divisor.

 

Question 1. Using repeated division method, find the HCF of the following:
(iii) 6765 and 610
Answer: We use the repeated division method to find the HCF.
First, divide 6765 by 610:
\( 6765 = 610 \times 11 + 55 \)
Next, the remainder 55 becomes the new divisor, and the previous divisor 610 becomes the new dividend:
\( 610 = 55 \times 11 + 5 \)
Finally, the remainder 5 becomes the new divisor, and 55 becomes the new dividend:
\( 55 = 5 \times 11 + 0 \)
Since the remainder is 0, the last divisor, which is 5, is the HCF.
In simple words: We keep dividing the bigger number by the smaller one. Then, we take the remainder and divide the old divisor by it. We repeat this until the remainder is zero. The last number used as a divisor is the HCF.

🎯 Exam Tip: The Euclidean algorithm works for any two positive integers; it's a fundamental concept in number theory.

 

Question 1. Using repeated division method, find the HCF of the following:
(iv) 184, 230 and 276
Answer: We find the HCF of three numbers by finding the HCF of two numbers first, and then finding the HCF of that result with the third number.
**Step 1: Find the HCF of 184 and 230**
Divide 230 by 184:
\( 230 = 184 \times 1 + 46 \)
Now, divide 184 by the remainder 46:
\( 184 = 46 \times 4 + 0 \)
The HCF of 184 and 230 is 46.

**Step 2: Find the HCF of 46 (the result from Step 1) and 276**
Divide 276 by 46:
\( 276 = 46 \times 6 + 0 \)
The HCF of 46 and 276 is 46.

Therefore, the HCF of 184, 230 and 276 is 46.
In simple words: To find the HCF of three numbers, first find the HCF of any two numbers. Then, find the HCF of that answer and the third number. This final HCF is the answer for all three.

🎯 Exam Tip: When finding the HCF of more than two numbers, always break it down into finding the HCF of two numbers at a time.

 

Question 2. Using repeated subtraction method, find the HCF of the following:
(i) 42 and 70
Answer: In the repeated subtraction method, we keep subtracting the smaller number from the larger number until one of the numbers becomes zero or both become equal. The non-zero number is the HCF.
Let \( m = 70 \) and \( n = 42 \). Since \( m > n \), we subtract \( n \) from \( m \).
\( 70 - 42 = 28 \)
Now, the new \( m = 42 \) and \( n = 28 \). Again, \( m > n \), so we subtract \( n \) from \( m \).
\( 42 - 28 = 14 \)
Now, the new \( m = 28 \) and \( n = 14 \). Again, \( m > n \), so we subtract \( n \) from \( m \).
\( 28 - 14 = 14 \)
Now, the new \( m = 14 \) and \( n = 14 \). Since \( m = n \), we stop here. The common value is the HCF.
Therefore, the HCF of 42 and 70 is 14.
In simple words: To find the HCF using repeated subtraction, take the larger number and subtract the smaller one. Then, use the smaller number and the answer you just got, and repeat the subtraction. Keep going until both numbers are the same; that number is the HCF.

🎯 Exam Tip: The repeated subtraction method is most intuitive for smaller numbers, helping visualize the concept of common factors.

 

Question 2. Using repeated subtraction method, find the HCF of the following:
(ii) 36 and 80
Answer: We use the repeated subtraction method. We subtract the smaller number from the larger number repeatedly.
Let \( m = 80 \) and \( n = 36 \). Since \( m > n \):
\( 80 - 36 = 44 \)
Now, \( m = 44 \) and \( n = 36 \). Since \( m > n \):
\( 44 - 36 = 8 \)
Now, \( m = 36 \) and \( n = 8 \). Since \( m > n \):
\( 36 - 8 = 28 \)
Now, \( m = 28 \) and \( n = 8 \). Since \( m > n \):
\( 28 - 8 = 20 \)
Now, \( m = 20 \) and \( n = 8 \). Since \( m > n \):
\( 20 - 8 = 12 \)
Now, \( m = 12 \) and \( n = 8 \). Since \( m > n \):
\( 12 - 8 = 4 \)
Now, \( m = 8 \) and \( n = 4 \). Since \( m > n \):
\( 8 - 4 = 4 \)
Now, \( m = 4 \) and \( n = 4 \). Since \( m = n \), we stop.
Therefore, the HCF is 4.
In simple words: For HCF by subtraction, always subtract the smaller number from the bigger one. The answer and the smaller number become the new pair. Keep going until both numbers are equal, and that number is the HCF.

🎯 Exam Tip: This method shows how the HCF is the largest number that divides both original numbers without a remainder, as it's the result of their continuous difference.

 

Question 2. Using repeated subtraction method, find the HCF of the following:
(iii) 420 and 280
Answer: We use the repeated subtraction method to find the HCF.
Let \( m = 420 \) and \( n = 280 \). Since \( m > n \):
\( m - n = 420 - 280 = 140 \)
Now, the new \( m = 280 \) and \( n = 140 \). Since \( m > n \):
\( m - n = 280 - 140 = 140 \)
Now, the new \( m = 140 \) and \( n = 140 \). Since \( m = n \), we stop.
Therefore, the HCF is 140.
In simple words: Start with two numbers. Subtract the smaller from the larger. Then, take the smaller number and the result of your subtraction and do it again. Keep doing this until both numbers are the same; that number is the HCF.

🎯 Exam Tip: This method helps understand that the HCF is also the HCF of the original numbers' difference and the smaller number, a key property of the Euclidean algorithm.

 

Question 2. Using repeated subtraction method, find the HCF of the following:
(iv) 1014 and 654
Answer: We use the repeated subtraction method to find the HCF.
Let \( m = 1014 \) and \( n = 654 \). Since \( m > n \):
\( m - n = 1014 - 654 = 360 \)
Now, the new \( m = 654 \) and \( n = 360 \). Since \( m > n \):
\( m - n = 654 - 360 = 294 \)
Now, the new \( m = 360 \) and \( n = 294 \). Since \( m > n \):
\( m - n = 360 - 294 = 66 \)
Now, the new \( m = 294 \) and \( n = 66 \). Since \( m > n \):
\( m - n = 294 - 66 = 228 \)
Now, the new \( m = 228 \) and \( n = 66 \). Since \( m > n \):
\( m - n = 228 - 66 = 162 \)
Now, the new \( m = 162 \) and \( n = 66 \). Since \( m > n \):
\( m - n = 162 - 66 = 96 \)
Now, the new \( m = 96 \) and \( n = 66 \). Since \( m > n \):
\( m - n = 96 - 66 = 30 \)
Now, the new \( m = 66 \) and \( n = 30 \). Since \( m > n \):
\( m - n = 66 - 30 = 36 \)
Now, the new \( m = 36 \) and \( n = 30 \). Since \( m > n \):
\( m - n = 36 - 30 = 6 \)
Now, the new \( m = 30 \) and \( n = 6 \). Since \( m > n \):
\( m - n = 30 - 6 = 24 \)
Now, the new \( m = 24 \) and \( n = 6 \). Since \( m > n \):
\( m - n = 24 - 6 = 18 \)
Now, the new \( m = 18 \) and \( n = 6 \). Since \( m > n \):
\( m - n = 18 - 6 = 12 \)
Now, the new \( m = 12 \) and \( n = 6 \). Since \( m > n \):
\( m - n = 12 - 6 = 6 \)
Now, the new \( m = 6 \) and \( n = 6 \). Since \( m = n \), we stop.
Therefore, the HCF of 1014 and 654 is 6.
In simple words: Keep subtracting the smaller number from the larger one until both numbers become equal. That common number is the HCF. It might take many steps, but the process stays the same.

🎯 Exam Tip: For larger numbers, the repeated subtraction method can be lengthy; make sure to do each subtraction carefully to avoid errors.

 

Question 3. Do the given problems by repeated subtraction method and verify the result.
(i) 56 and 12
Answer: We use the repeated subtraction method to find the HCF of 56 and 12.
Let \( m = 56 \) and \( n = 12 \). Since \( m > n \):
\( m - n = 56 - 12 = 44 \)
Now, the new \( m = 44 \) and \( n = 12 \). Since \( m > n \):
\( m - n = 44 - 12 = 32 \)
Now, the new \( m = 32 \) and \( n = 12 \). Since \( m > n \):
\( m - n = 32 - 12 = 20 \)
Now, the new \( m = 20 \) and \( n = 12 \). Since \( m > n \):
\( m - n = 20 - 12 = 8 \)
Now, the new \( m = 12 \) and \( n = 8 \). Since \( m > n \):
\( m - n = 12 - 8 = 4 \)
Now, the new \( m = 8 \) and \( n = 4 \). Since \( m > n \):
\( m - n = 8 - 4 = 4 \)
Now, the new \( m = 4 \) and \( n = 4 \). Since \( m = n \), we stop.
The HCF of 56 and 12 is 4.

**Verification using repeated division:**
Divide 56 by 12:
\( 56 = 12 \times 4 + 8 \)
Divide 12 by the remainder 8:
\( 12 = 8 \times 1 + 4 \)
Divide 8 by the remainder 4:
\( 8 = 4 \times 2 + 0 \)
The HCF is 4, which matches the result from repeated subtraction.
In simple words: We find the HCF by repeatedly taking away the smaller number from the larger one until they are the same. Then, we check this answer using a different method, like division, to be sure it is correct. Both methods give the same HCF.

🎯 Exam Tip: When asked to verify, choose a different method (like prime factorization or division) to confirm your answer from the initial method.

 

Question 3. Do the given problems by repeated subtraction method and verify the result.
(ii) 320, 120 and 95
Answer: We find the HCF of three numbers using repeated subtraction by finding the HCF of two numbers first, and then finding the HCF of that result with the third number.
**Step 1: Find the HCF of 320 and 120 using repeated subtraction**
Let \( m = 320 \) and \( n = 120 \). Since \( m > n \):
\( 320 - 120 = 200 \)
Now, \( m = 200 \) and \( n = 120 \). Since \( m > n \):
\( 200 - 120 = 80 \)
Now, \( m = 120 \) and \( n = 80 \). Since \( m > n \):
\( 120 - 80 = 40 \)
Now, \( m = 80 \) and \( n = 40 \). Since \( m > n \):
\( 80 - 40 = 40 \)
Now, \( m = 40 \) and \( n = 40 \). Since \( m = n \), we stop.
The HCF of 320 and 120 is 40.

**Step 2: Find the HCF of 40 (the result from Step 1) and 95 using repeated subtraction**
Let \( m = 95 \) and \( n = 40 \). Since \( m > n \):
\( 95 - 40 = 55 \)
Now, \( m = 55 \) and \( n = 40 \). Since \( m > n \):
\( 55 - 40 = 15 \)
Now, \( m = 40 \) and \( n = 15 \). Since \( m > n \):
\( 40 - 15 = 25 \)
Now, \( m = 25 \) and \( n = 15 \). Since \( m > n \):
\( 25 - 15 = 10 \)
Now, \( m = 15 \) and \( n = 10 \). Since \( m > n \):
\( 15 - 10 = 5 \)
Now, \( m = 10 \) and \( n = 5 \). Since \( m > n \):
\( 10 - 5 = 5 \)
Now, \( m = 5 \) and \( n = 5 \). Since \( m = n \), we stop.
The HCF of 40 and 95 is 5.

Therefore, the HCF of 320, 120 and 95 is 5.

**Verification using repeated division:**
First, HCF of 320 and 120:
\( 320 = 120 \times 2 + 80 \)
\( 120 = 80 \times 1 + 40 \)
\( 80 = 40 \times 2 + 0 \)
HCF(320, 120) = 40.

Now, HCF of 40 and 95:
\( 95 = 40 \times 2 + 15 \)
\( 40 = 15 \times 2 + 10 \)
\( 15 = 10 \times 1 + 5 \)
\( 10 = 5 \times 2 + 0 \)
HCF(40, 95) = 5. This matches the result from repeated subtraction.
In simple words: To find the HCF of three numbers using subtraction, first find the HCF of the first two. Then, use that HCF and the third number to find their HCF. This final number is the HCF for all three. We then check our answer using the division method.

🎯 Exam Tip: Keep the calculations organized when dealing with three numbers. Clearly separate the steps for finding the HCF of each pair of numbers.

 

Question 4. Valai wants to cut identical squares big as she can, from a piece of paper measuring 168mm and 196mm. What is the length of the side of the biggest square? (To find HCF using repeated subtraction method)
Answer: To find the side length of the biggest identical square that can be cut from a rectangular piece of paper, we need to find the HCF (Highest Common Factor) of its length and width. Here, the dimensions are 168 mm and 196 mm. We will use the repeated subtraction method.
Let \( m = 196 \) and \( n = 168 \). Since \( m > n \):
\( m - n = 196 - 168 = 28 \)
Now, the new \( m = 168 \) and \( n = 28 \). Since \( m > n \):
\( m - n = 168 - 28 = 140 \)
Now, the new \( m = 140 \) and \( n = 28 \). Since \( m > n \):
\( m - n = 140 - 28 = 112 \)
Now, the new \( m = 112 \) and \( n = 28 \). Since \( m > n \):
\( m - n = 112 - 28 = 84 \)
Now, the new \( m = 84 \) and \( n = 28 \). Since \( m > n \):
\( m - n = 84 - 28 = 56 \)
Now, the new \( m = 56 \) and \( n = 28 \). Since \( m > n \):
\( m - n = 56 - 28 = 28 \)
Now, the new \( m = 28 \) and \( n = 28 \). Since \( m = n \), we stop.
The HCF of 168 and 196 is 28.
Therefore, the length of the side of the biggest square Valai can cut is 28 mm.
In simple words: To find the largest square that fits perfectly, we need to find the biggest number that divides both the length and width of the paper. We use repeated subtraction to find this number, and the answer is 28 mm.

🎯 Exam Tip: Word problems involving "biggest square," "maximum length," or "largest number of items" usually require finding the HCF of the given quantities.

 

Objective Type Questions

 

Question 5. What is the eleventh Fibonacci number?
(a) 77
(b) 89
(c) 144
(d) 110
Answer: (b) 89
Hint: The Fibonacci sequence starts with 0 and 1 (or 1 and 1), and each next number is the sum of the two before it. Let's list the terms to find the 11th one.

From the table, the 11th Fibonacci number is 89.
In simple words: The Fibonacci sequence is like a special list of numbers where each number is found by adding the two numbers before it. If we count up to the 11th number in this list, we find it is 89.

🎯 Exam Tip: Remember that the Fibonacci sequence often starts with \( F_0 = 0, F_1 = 1 \) or \( F_1 = 1, F_2 = 1 \). Be careful to count the terms from the correct starting point based on the problem's context.

 

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) \( F(8) = F(10) \times F(9) \)
(d) \( F(8) = F(7) - F(6) \)
Answer: (b) F(8) = F(7) + F(6)
Hint: By the definition of a Fibonacci number, any term in the Fibonacci series is the sum of its two preceding terms. This is a core rule for Fibonacci numbers.
So, for \( n=8 \), \( F(8) \) would be the sum of \( F(8-1) \) and \( F(8-2) \), which means \( F(8) = F(7) + F(6) \).
In simple words: Fibonacci numbers follow a rule: any number in the series is made by adding the two numbers right before it. So, the 8th Fibonacci number is just the 7th number plus the 6th number.

🎯 Exam Tip: Understand the fundamental recursive definition of the Fibonacci sequence: \( F(n) = F(n-1) + F(n-2) \). This is crucial for solving problems related to its properties.

 

Question 7. Every 3rd number of the Fibonacci sequence is a multiple of _____
(a) 2
(b) 3
Answer: (a) 2
Hint: Let's list the Fibonacci sequence and check the 3rd, 6th, 9th terms etc.
Sequence: 1, 1, **2**, 3, 5, **8**, 13, 21, **34**, 55, 89, **144** ...
The 3rd term is 2.
The 6th term is 8.
The 9th term is 34.
The 12th term is 144.
All these numbers (2, 8, 34, 144) are even numbers, meaning they are multiples of 2. This pattern continues throughout the sequence.
In simple words: If you look at every third number in the Fibonacci list, you will notice they are all even numbers. This means they can all be divided by 2 without any remainder.

🎯 Exam Tip: To discover patterns in sequences like Fibonacci, list out the first several terms and observe the properties of specific positions (e.g., every 3rd, 4th, or 6th term).

 

Question 8. Every _______ number of the Fibonacci sequence is a multiple of 8
(a) 2th
(b) 4th
(c) 6th
(d) 8th
Answer: (c) 6th
Hint: Let's list the Fibonacci sequence and check which terms are multiples of 8.
Sequence: 1, 1, 2, 3, 5, **8**, 13, 21, 34, 55, 89, **144**, ...
The 6th term is 8 (which is \( 8 \times 1 \)).
The 12th term is 144 (which is \( 8 \times 18 \)).
This pattern shows that every 6th number in the Fibonacci sequence is a multiple of 8.
In simple words: When you look at the Fibonacci numbers, every 6th number in the list can be divided by 8 perfectly. For example, the 6th number is 8, and the 12th number is 144, both of which are multiples of 8.

🎯 Exam Tip: Recognizing divisibility patterns in sequences is a useful skill. Continue the sequence sufficiently to confirm a pattern before choosing an answer.

 

Question 9. The difference between the 18th and 17th Fibonacci number is
(a) 233
(b) 377
(c) 610
(d) 987
Answer: (d) 987
Hint: We know the basic rule for Fibonacci numbers: \( F(n) = F(n-1) + F(n-2) \).
Using this rule, we can say that \( F(18) = F(17) + F(16) \).
If we rearrange this equation, we get \( F(18) - F(17) = F(16) \).
Now we need to find \( F(16) \). We can find this by looking at a Fibonacci sequence table or calculating it.

From the table, \( F(15) = 610 \) and \( F(14) = 377 \).
So, \( F(16) = F(15) + F(14) = 610 + 377 = 987 \).
Therefore, \( F(18) - F(17) = 987 \).
In simple words: In the Fibonacci series, any number is the sum of the two numbers before it. So, if you subtract the 17th number from the 18th number, you will get the 16th Fibonacci number. By adding the 14th and 15th Fibonacci numbers, which are 377 and 610, we get 987.

🎯 Exam Tip: Remember the basic recurrence relation \( F_n = F_{n-1} + F_{n-2} \). This allows you to find relationships like \( F_n - F_{n-1} = F_{n-2} \) quickly without calculating the actual numbers.

 

Question 10. Common prime factors of 30 and 250 are
(a) \( 2 \times 5 \)
(b) \( 3 \times 5 \)
(c) \( 2 \times 3 \times 5 \)
(d) \( 5 \times 5 \)
Answer: (a) \( 2 \times 5 \)
Hint: First, we find the prime factors for each number.
Prime factors of 30: \( 30 = 2 \times 3 \times 5 \)
Prime factors of 250: \( 250 = 25 \times 10 = (5 \times 5) \times (2 \times 5) = 2 \times 5 \times 5 \times 5 \)
Now, we look for the factors that are common to both lists.
Common prime factors are 2 and 5.
So, their product is \( 2 \times 5 \).
In simple words: To find common prime factors, first break each number down into its smallest building blocks (prime numbers). Then, find the prime numbers that appear in both lists. Multiply these common prime numbers together for the answer.

🎯 Exam Tip: Remember that prime factorization is unique for every number; list all prime factors for each number before identifying the common ones. The HCF is the product of these common prime factors.

 

Question 11. Common prime factors of 36, 60 and 72 are
(a) \( 2 \times 2 \)
(b) \( 2 \times 3 \)
(c) \( 3 \times 3 \)
(d) \( 3 \times 2 \times 2 \)
Answer: (d) \( 3 \times 2 \times 2 \)
Hint: First, we find the prime factors for each number.
Prime factors of 36: \( 36 = 2 \times 2 \times 3 \times 3 \)
Prime factors of 60: \( 60 = 2 \times 2 \times 3 \times 5 \)
Prime factors of 72: \( 72 = 2 \times 2 \times 2 \times 3 \times 3 \)
Now, we identify the prime factors that are common to all three lists.
From 36: \( \textbf{2} \times \textbf{2} \times \textbf{3} \times 3 \)
From 60: \( \textbf{2} \times \textbf{2} \times \textbf{3} \times 5 \)
From 72: \( \textbf{2} \times \textbf{2} \times 2 \times \textbf{3} \times 3 \)
The common prime factors are two 2s and one 3. So, their product is \( 2 \times 2 \times 3 = 12 \).
The option \( 3 \times 2 \times 2 \) also results in 12, just in a different order.
In simple words: Break down each of the three numbers into its prime factors. Then, find all the prime factors that appear in the lists for all three numbers. Multiply these common prime factors together to get the HCF.

🎯 Exam Tip: When finding common factors for three numbers, a factor must be present in the prime factorization of ALL three numbers to be considered common.

 

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) 1
Hint: Co-prime numbers (or relatively prime numbers) are a pair of integers that have no common factors other than 1. This means their highest common factor (HCF) must be 1. For example, 7 and 10 are co-prime because their only common factor is 1.
In simple words: If two numbers are co-prime, it means the biggest number that can divide both of them exactly is 1. They don't share any other common factors apart from 1.

🎯 Exam Tip: A common misconception is that co-prime numbers must be prime themselves, but this is false; for example, 4 and 9 are co-prime even though neither is prime.