Samacheer Kalvi Class 6 Maths Solutions Term 3 Chapter 5 Information Processing Exercise 5.1

Question 1. Study and complete the following pattern.
(i) \( 1 \times 1 = 1 \)
\( 11 \times 11 = 121 \)
\( 111 \times 111 = 12321 \)
\( 1111 \times 1111 = ? \)
(ii)

Answer:
(i) The pattern shows that when you multiply a number made of only 1s by itself, the result goes up and then down, like 1, 2, 3, then back down 3, 2, 1. For `1111 x 1111`, there are four '1's, so the number will go up to 4 and then come back down. So, \( 1111 \times 1111 = 1234321 \). The next number in this sequence, \( 11111 \times 11111 \), would be \( 123454321 \).
(ii) This is a number pyramid where each brick's value is the sum of the two bricks directly below it. The completed numbers in the pyramid, from top to bottom and left to right, are: `144, 60, 84, 36, 48, 15, 27`. These numbers are all derived by adding the two numbers below them to find the number above them in the pyramid.
In simple words: For (i), the number \( 1111 \times 1111 \) is \( 1234321 \). For (ii), the numbers that fill the pyramid are 144, 60, 84, 36, 48, 15, 27, by adding the two numbers below to get the one above.

🎯 Exam Tip: When working with number patterns, try to find the rule of addition, subtraction, multiplication, or division between consecutive terms. For geometric patterns, look for how shapes rotate, reflect, or change their internal elements.

 

Question 2. Find next three numbers in the following number patterns.
(i) 50, 51, 53, 56, 60......
(ii) 77, 69, 61, 53, ......
(iii) 10, 20, 40, 80,..
(iv) \( \frac{21}{33}, \frac{321}{444}, \frac{4321}{555} \)
Answer:
(i) The pattern is made by adding 1, then 2, then 3, then 4, and so on. So, after 60, you will add 5, then 6, then 7.
\( 50 \xrightarrow{+1} 51 \xrightarrow{+2} 53 \xrightarrow{+3} 56 \xrightarrow{+4} 60 \xrightarrow{+5} 65 \xrightarrow{+6} 71 \xrightarrow{+7} 78 \)
The next three numbers are 65, 71, 78.
(ii) The pattern is made by subtracting 8 each time. So, after 53, you will subtract 8, then 8 again, then 8 one more time.
\( 77 \xrightarrow{-8} 69 \xrightarrow{-8} 61 \xrightarrow{-8} 53 \xrightarrow{-8} 45 \xrightarrow{-8} 37 \xrightarrow{-8} 29 \)
The next three numbers are 45, 37, 29.
(iii) The pattern is made by multiplying the previous number by 2 each time. So, after 80, you will multiply by 2, then by 2 again, then by 2 one more time. This is also called a geometric progression.
\( 10 \xrightarrow{\times 2} 20 \xrightarrow{\times 2} 40 \xrightarrow{\times 2} 80 \xrightarrow{\times 2} 160 \xrightarrow{\times 2} 320 \xrightarrow{\times 2} 640 \)
The next three numbers are 160, 320, 640.
(iv) In this pattern, the numerator (top number) adds one digit (2, then 3, then 4) to the left side in each step, and the denominator (bottom number) adds a 1 to the right side of the repeating digit. This means the numerator increases its digit sequence, while the denominator keeps the same digit repeating more times. Following this logic:
The given fractions are: \( \frac{21}{33}, \frac{321}{444}, \frac{4321}{555} \)
The next three fractions will be: \( \frac{54321}{66666}, \frac{654321}{777777}, \frac{7654321}{8888888} \)
In simple words: For (i) add 1, 2, 3... The next numbers are 65, 71, 78. For (ii) subtract 8. The next numbers are 45, 37, 29. For (iii) multiply by 2. The next numbers are 160, 320, 640. For (iv), the top numbers grow by adding a digit on the left (2, 3, 4, 5, 6, 7), and the bottom numbers add another repeating digit (3, 4, 5, 6, 7, 8).

🎯 Exam Tip: Always look for the difference or ratio between consecutive numbers to identify arithmetic or geometric progressions. Sometimes the difference itself forms a pattern.

 

Question 3. Consider the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55,.... Observe and complete the following table by understanding the number patterns? followed. After filling the table discuss the pattern followed in addition and subtraction, of the numbers of the sequence?
Answer:
The Fibonacci sequence is 1, 1, 2, 3, 5, 8, 13, 21, 34, 55.
The table is completed by matching sums and differences with specific terms in the sequence.

The pattern followed is that the sum of the first 'n' Fibonacci numbers (skipping the initial '1') is always equal to the (n+2)th Fibonacci number minus 1. For example, in (ii), \( 1+3+8 \) is the sum of three numbers (excluding the first 1). The 5th Fibonacci number (1, 1, 2, 3, **5**, 8, **13**) is 5, so the (3+2)th term is the 5th, which is 5. Wait, this interpretation of the question solution mapping is a bit off. Let's re-read the solution to Q3: (i) 12, 13 – 1 = 12 (This means for 1+3+8, the answer is 12. And 13-1 is 12. This seems to be the answer for *step ii* of the table.) (ii) 33, 34-1=33 (This seems to be the answer for *step iii* of the table.) (iii) 1 + 3 + 8 + 21 + 55 = 88, 89 − 1 = 88 (This seems to be the answer for *step iv* of the table.) The solution labels (i), (ii), (iii) do not match the table labels (i), (ii), (iii), (iv) for the *missing parts*. Let's assume the solution provides the answers for the question marks in the table in order. Pattern 1: i) 1+3 = 4 (given) ii) 1+3+8 = 12 iii) 1+3+8+21 = 33 iv) 1+3+8+21+55 = 88 (using the next number 55 from sequence) Pattern 2: i) 5-1 = 4 (given) ii) 13-1 = 12 (13 is the 7th term in 1,1,2,3,5,8,13) iii) 34-1 = 33 (34 is the 9th term) iv) 89-1 = 88 (89 is the 11th term) This means the pattern for sums is: Sum of first `n` terms (starting from `F_2 = 1`) is `F_{n+2} - 1`. Sum \( (F_2 + F_3 + \dots + F_{n+1}) = F_{n+3} - 1 \). For (ii): Sum \( F_2+F_3+F_4 = 1+2+3 = 6 \). This is not matching `1+3+8=12`. The sequence is 1, 1, 2, 3, 5, 8, 13, 21, 34, 55. Pattern 1 is adding certain numbers from the sequence: i) \( F_1 + F_3 = 1+2 = 3 \) (no, source says 1+3, if F_1=1, F_2=1, F_3=2, F_4=3). Let's assume the problem means the numbers 1, 3, 8, 21, 55 are elements from the Fibonacci sequence, but not necessarily consecutive. If the Fibonacci sequence is \( F_0=0, F_1=1, F_2=1, F_3=2, F_4=3, F_5=5, F_6=8, F_7=13, F_8=21, F_9=34, F_{10}=55 \). Pattern 1 uses \( F_1=1, F_4=3, F_6=8, F_8=21, F_{10}=55 \). So: i) \( 1+3 = 4 \). (Here 4 is \( F_5-1 \)). ii) \( 1+3+8 = 12 \). (Here 12 is \( F_7-1 \)). iii) \( 1+3+8+21 = 33 \). (Here 33 is \( F_9-1 \)). iv) \( 1+3+8+21+55 = 88 \). (Here 88 is \( F_{11}-1 \)). Pattern 2 is \( F_n - 1 \). i) \( F_5-1 = 5-1 = 4 \). ii) \( F_7-1 = 13-1 = 12 \). iii) \( F_9-1 = 34-1 = 33 \). iv) \( F_{11}-1 = 89-1 = 88 \). (Note: 89 is not listed in the Fibonacci sequence up to 55, it's the next term after 55+34). So, the pattern is: Sum of selected odd-indexed Fibonacci terms \( (F_1 + F_4 + F_6 + \dots) \) equals the next term in the Fibonacci sequence after the last added term, minus 1. Or more precisely: \( F_1 + F_3 + F_5 + \dots + F_{2n-1} = F_{2n}-1 \). Here, the terms added are \( F_1, F_4, F_6, F_8, F_{10} \). This isn't strictly odd-indexed. The pattern in the first column is: The sum of numbers \( F_1, F_4, F_6, F_8, F_{10} \) (starting at \( F_1 \), then adding terms whose indices are \( 3, 2, 2, 2 \) greater than the previous index, or perhaps \( F_1 \), then adding Fibonacci numbers at indices \( 2, 4, 6, 8, 10 \) for \( F_1, F_3, F_5, F_7, F_9 \)). The question states `1, 3, 8, 21, 55` are from the Fibonacci sequence `1, 1, 2, 3, 5, 8, 13, 21, 34, 55`. This implies \( F_1=1, F_4=3, F_6=8, F_8=21, F_{10}=55 \). The sums are: \( 1 \rightarrow 1 \) \( 1+3 \rightarrow 4 \) \( 1+3+8 \rightarrow 12 \) \( 1+3+8+21 \rightarrow 33 \) \( 1+3+8+21+55 \rightarrow 88 \) The corresponding results in Pattern 2 are \( F_5-1=4 \), \( F_7-1=12 \), \( F_9-1=33 \), \( F_{11}-1=88 \). The pattern is that the sum in Pattern 1 (which adds terms from the Fibonacci sequence) equals the result of taking the Fibonacci number that is one step beyond the last term *used in the sum* (for instance, after using 3, the next term is 5; after using 8, the next term is 13), and then subtracting 1. More accurately, if the last term in the sum is \( F_k \), the result is \( F_{k+1} - 1 \). No, that's not right. Let's use the sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89 (where \( F_1=1, F_2=1 \), etc. or \( F_0=1, F_1=1 \)). Let's use the source's implied sequence from 1,1,2,3... Fibonacci: 1 (F1), 1 (F2), 2 (F3), 3 (F4), 5 (F5), 8 (F6), 13 (F7), 21 (F8), 34 (F9), 55 (F10), 89 (F11) Pattern 1: i) 1 (F1) + 3 (F4) = 4 ii) 1 (F1) + 3 (F4) + 8 (F6) = 12 iii) 1 (F1) + 3 (F4) + 8 (F6) + 21 (F8) = 33 iv) 1 (F1) + 3 (F4) + 8 (F6) + 21 (F8) + 55 (F10) = 88 Pattern 2: i) 5 (F5) - 1 = 4 ii) 13 (F7) - 1 = 12 iii) 34 (F9) - 1 = 33 iv) 89 (F11) - 1 = 88 The pattern is: the sum of the sequence \( F_1 + F_4 + F_6 + \dots + F_{2n} \) is equal to \( F_{2n+1} - 1 \). For example, in step ii), \( 1+3+8 \). The last term is 8 (\( F_6 \)). The next term after \( F_6 \) is \( F_7=13 \). The pattern for sums here is \( F_7-1=12 \). For step iii), \( 1+3+8+21 \). The last term is 21 (\( F_8 \)). The next term after \( F_8 \) is \( F_9=34 \). The pattern for sums here is \( F_9-1=33 \). This is the relation: Sum of terms \( F_1 + F_4 + F_6 + \dots + F_{2n} = F_{2n+1}-1 \). I will reword this in simple language.
In simple words: The table shows two patterns using the Fibonacci sequence. In Pattern 1, specific numbers from the sequence are added together. In Pattern 2, a Fibonacci number is picked, and 1 is subtracted from it. Both patterns result in the same answer for each step. For example, adding 1, 3, and 8 gives 12, which is also what you get by taking 13 (the next number after 8 in the sequence, then skipping one) and subtracting 1.

🎯 Exam Tip: When dealing with sequence patterns, look for how terms relate to previous terms (addition, subtraction, multiplication, division), or how their position affects their value. Fibonacci sequences often have interesting sum properties.

 

Question 4. Complete the following patterns.
(i)

(ii)(iii)Answer:
(i) This pattern rotates letters. The letter A rotates clockwise by 90 degrees in each step. A becomes V (upside down A), V becomes A (right-side up A), and A becomes rotated A (like a triangle on its side). Then it repeats. The completed pattern is:

(ii) This pattern involves a small black circle moving inside the squares. It moves from top-left, to top-right, to bottom-right, to bottom-left, then back to top-left. It's a cyclic movement through the corners. The completed pattern is:

(iii) This pattern shows arrows or diagonal lines inside squares. Each arrow points to a different corner. The pattern follows a sequence of rotations of these lines. The missing part in the bottom right will have diagonal lines pointing to the bottom-right and top-left corners, similar to the top-left square, but rotated.
The completed pattern is:

In simple words: For (i), the letter 'A' shape turns around. For (ii), the small black dot moves in a circle inside the squares. For (iii), the diagonal lines in the square rotate their position.

🎯 Exam Tip: For visual patterns, look for rotations, reflections, changes in size, number of elements, or the movement of a specific part. Try to identify the core rule that governs the change from one step to the next.

 

Question 5. Find the HCF of the following pair of numbers by Euclid's game
(i) 25 and 35
(ii) 36 and 12
(iii) 15 and 29
Answer:
(i) For 25 and 35:
We find the Highest Common Factor (HCF) by using Euclid's game, which involves finding the HCF of the smaller number and the difference between the two numbers.
HCF of (25, 35) \( = \) HCF of (25, \( 35 - 25 \))
\( = \) HCF of (25, 10)
Now, let's find the factors of 25 and 10.
\( 25 = 5 \times 5 \)
\( 10 = 2 \times 5 \)
The common factor is 5. So, the HCF of (25, 10) is 5.
Therefore, the HCF of 25 and 35 is 5.

(ii) For 36 and 12:
Using Euclid's game, we find the HCF of the smaller number and the difference.
HCF of (36, 12) \( = \) HCF of (36, \( 36 - 12 \))
\( = \) HCF of (36, 24)
Now, let's find the factors of 36 and 24.
\( 36 = 2 \times 2 \times 3 \times 3 \)
\( 24 = 2 \times 2 \times 2 \times 3 \)
The common factors are \( 2 \times 2 \times 3 = 12 \). So, the HCF of (36, 24) is 12.
Therefore, the HCF of 36 and 12 is 12.

(iii) For 15 and 29:
Using Euclid's game, we find the HCF of the smaller number and the difference.
HCF of (15, 29) \( = \) HCF of (15, \( 29 - 15 \))
\( = \) HCF of (15, 14)
Now, let's find the factors of 15 and 14.
\( 15 = 3 \times 5 \times 1 \)
\( 14 = 2 \times 7 \times 1 \)
The only common factor is 1. So, the HCF of (15, 14) is 1.
Therefore, the HCF of 15 and 29 is 1.
In simple words: To find the HCF using Euclid's game, replace the larger number with the difference between the two numbers, and keep repeating until you find a common factor. The HCF for (i) is 5, for (ii) is 12, and for (iii) is 1.

🎯 Exam Tip: Euclid's algorithm for HCF relies on the property that HCF(a, b) = HCF(a, b-a) (if b > a). This method is very efficient for larger numbers too.

 

Question 6. Find HCF of 48 and 28. Also find the HCF of 48 and the number obtained by finding their difference.
Answer:
First, let's find the HCF of 48 and 28.
Factors of 48: \( 2 \times 2 \times 2 \times 2 \times 3 \)
Factors of 28: \( 2 \times 2 \times 7 \)
The common factors are \( 2 \times 2 \). So, HCF of (48, 28) \( = 4 \).

Next, we find the difference between 48 and 28:
Difference \( = 48 - 28 = 20 \).
Now, we need to find the HCF of 48 and this difference, which is 20.
Factors of 48: \( 2 \times 2 \times 2 \times 2 \times 3 \)
Factors of 20: \( 2 \times 2 \times 5 \)
The common factors are \( 2 \times 2 \). So, HCF of (48, 20) \( = 4 \).
Notice that the HCF of (48, 28) is the same as the HCF of (48, 20). This shows a property of HCF where HCF(a, b) = HCF(a, b-a).
In simple words: The HCF of 48 and 28 is 4. When we find the difference between 48 and 28 (which is 20), and then find the HCF of 48 and this new number 20, the HCF is still 4.

🎯 Exam Tip: This question demonstrates a key property of HCF: HCF(a, b) = HCF(a, b-a). Using this can simplify finding the HCF of two numbers, especially larger ones.

 

Question 7. Give instructions to fill in a bank withdrawal form issued in a bank.
Answer:
To fill in a bank withdrawal form, follow these steps:

• First, write your full name in capital letters from left to right in the designated space.
• Next, write the correct date of withdrawal in the top right corner of the form.
• Then, clearly write the amount of money you want to withdraw, both in words, in the space provided.
• After that, write the same amount of money in numbers (figures) in the box provided.
• Finally, sign your signature at the right bottom, exactly above the label 'signature of the depositor'. Make sure your signature matches the one held by the bank.

In simple words: To fill a bank form, write your name in capital letters, put the date, write how much money you want (both in words and numbers), and then sign it.

🎯 Exam Tip: Always write clearly and neatly on bank forms. Double-check all details, especially the date, amount, and signature, as mistakes can cause delays.

 

Question 8. Arrange the name of your classmates alphabetically.
Answer:
Here is a list of names arranged alphabetically:

• Ajay. S
• Anbu. T
• Balamurugan. M
• Darshan. S
• Elizabeth. N
• Franklin. P
• Godwin. A
• Harsha Varthan. M
• Immanuel. S
• Jothipriya. B
• Kannan. L
• Lakshmi. S
• Muthu. N
• Nagaraj. A
• Pattu. R
• Sethu. M
• Thangam. R
• Velu. S

In simple words: The names are put in alphabetical order, just like in a dictionary, from A to Z.

🎯 Exam Tip: To alphabetize names with initials, first sort by the first name. If first names are the same, then sort by the initial of the last name.

 

Question 9. Follow and execute the instructions given below.
(i) Write the number 10 in the place common to the three figures
(ii) Write the number 5 in the place common for square and circle only.
(iii) Write the number 7 in the place common for triangle and circle only.
(iv) Write the number 2 in the place common for triangle and square only.
(v) Write the numbers 12, 14, and 8 only in square, circle, and triangle respectively.
Answer:
The numbers are placed in the diagram according to the given instructions:

In simple words: The numbers are put into the drawing where the shapes overlap. 10 goes where all three shapes meet. 5 goes where only the square and circle meet. 7 goes where only the triangle and circle meet. 2 goes where only the triangle and square meet. 12 is only in the square, 14 is only in the circle, and 8 is only in the triangle.

🎯 Exam Tip: Carefully read each instruction and identify the specific overlapping region for placement. It helps to lightly shade or mark regions as you go to avoid confusion.

 

Question 10. Fill in the following information
Answer:
The form is filled with example data. Here's how each field is completed:
1. **Candidate's EMIS No:** 1456893210325814
2. **Name of the candidate:** RAJA P
3. **Class:** 6
4. **Date of Birth:** 16 / 04 / 1996
5. **Father's name:** PAULRAJ A
6. **Mother's Name:** MEENA K
7. **Sex:** Male (indicated with a tick mark)
8. **Area to which candidates resides:** Urban (indicated with a tick mark)
In simple words: The form is filled out with a student's information, including their EMIS number, name, class, birth date, parents' names, gender, and where they live.

🎯 Exam Tip: When filling out official forms, always use capital letters for names and ensure all numbers (like dates and EMIS) are written clearly in their correct boxes.

 

Question 11. The next term in the sequence 15, 17, 20, 22, 25, ... is
(a) 28
(b) 29
(c) 27
(d) None of the options
Answer: (c) 27
In simple words: The numbers in the sequence go up by 2, then by 3, then by 2, then by 3, and so on. So, after 25, you add 2 to get 27.

🎯 Exam Tip: When a sequence doesn't follow a single addition or multiplication rule, look for alternating rules, such as adding different numbers in a repeating pattern.

 

Question 12. What will be the 25th letter in the pattern? ABC AABBCCC AAA BBBCCC,...
(a) B
(b) C
(c) D
(d) A
Answer: (a) B
In simple words: The pattern is one A, then two Bs, then three Cs, then repeat (A, B, C) with increasing counts. The 25th letter will be 'B' because the sequence repeats as 1-A, 2-B, 3-C, then 1-A, 2-B, 3-C. After three full cycles, you'll reach the 18th letter (6 letters per cycle). The 25th letter is the 7th letter of the next cycle, which would be 'B'.

🎯 Exam Tip: For repeating letter patterns, identify the length of one full cycle and the number of repetitions. Then use division and remainder to find the letter at a specific position.

 

Question 13. The difference between 6th term add 5th term in the Fibonacci sequence is ___.
(a) 6
(b) 8
(c) 5
(d) 3
Answer: (d) 3
In simple words: The Fibonacci sequence starts 1, 1, 2, 3, 5, 8, 13... The 5th term is 5, and the 6th term is 8. The difference between them is \( 8 - 5 = 3 \).

🎯 Exam Tip: Always write down the first few terms of the Fibonacci sequence to avoid errors in identifying the terms. Remember it starts with two 1s.

 

Question 14. The 11th term in the Lucas sequence 1, 3, 4, 7, is
(a) 199
(b) 76
(c) 123
(d) 47
Answer: (a) 199
In simple words: The Lucas sequence is like Fibonacci, where each number is the sum of the two before it, but it starts with 2 and 1 (or 1 and 3 as given here). The terms are 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199. So, the 11th term is 199.

🎯 Exam Tip: Like the Fibonacci sequence, the Lucas sequence is defined by adding the two preceding terms. Make sure to generate enough terms to reach the required position.

 

Question 15. If the Highest Common Factor of 26 and 54 is 2, then HCF of 54 and 28 is .
(a) 26
(b) 2
(c) 54
(d) 1
Answer: (b) 2
In simple words: The HCF of 54 and 28 is 2. This is because 28 and 54 are both even numbers, and their factors are \( 28 = 2 \times 2 \times 7 \) and \( 54 = 2 \times 3 \times 3 \times 3 \). The only common factor is 2.

🎯 Exam Tip: The HCF of two numbers is the largest number that divides both of them without leaving a remainder. Prime factorization is a reliable way to find it.