Class 7 : Maths β Lesson 6. Number Play
EXPLANATION AND ANALYSIS
π΅ Introduction: Playing with Numbers
π§ Mathematics is not only about calculations; it is also about patterns, tricks, and interesting properties of numbers. Many times, numbers behave in surprising ways. When we explore these behaviors, we discover new ideas and enjoy learning mathematics more deeply.
πΏ This chapter, Number Play, helps us observe numbers carefully, recognize patterns, and understand how numbers interact with each other in different situations.
π΅ Guessing numbers
π’ Finding hidden patterns
π‘ Exploring rules
π΄ Enjoying mathematical tricks
This lesson develops curiosity and logical thinking through numbers.
π’ Observing Patterns in Numbers
π§ A pattern is a regular and repeated way in which numbers are arranged.
πΉ Patterns help us predict the next number
πΉ They help us understand number relationships
π Example
2, 4, 6, 8, 10
πΉ Each number increases by 2
πΉ This is an even-number pattern
π‘ Concept:
Recognising patterns is the first step in number play.
π΅ Number Sequences
π§ A number sequence is an ordered list of numbers that follow a rule.
πΉ Ascending sequence: numbers increase
πΉ Descending sequence: numbers decrease
π Example
5, 10, 15, 20
πΉ Rule: add 5 each time
βοΈ Note:
Finding the rule helps us continue the sequence correctly.
π’ Playing with Digits of Numbers
π§ Sometimes, we form new numbers by rearranging or operating on digits.
πΉ Adding digits
πΉ Reversing digits
πΉ Comparing digit sums
π Example
Number: 34
πΉ Sum of digits = 3 + 4 = 7
π‘ Concept:
Digit-based observations reveal hidden properties of numbers.
π΅ Even and Odd Number Patterns
π§ Even and odd numbers follow interesting patterns.
πΉ Even + Even = Even
πΉ Odd + Odd = Even
πΉ Even + Odd = Odd
π Example
6 + 8 = 14 (Even)
5 + 7 = 12 (Even)
βοΈ Note:
Understanding these patterns helps in quick checking of answers.
π’ Multiples and Their Patterns
π§ Multiples of a number are obtained by multiplying it by whole numbers.
π Example
Multiples of 4
4, 8, 12, 16, 20
πΉ All multiples of 4 are even
π‘ Concept:
Multiples show repeating behavior that helps in number games.
π΅ Divisibility Observations
π§ Divisibility rules help us know whether a number can be divided by another number without remainder.
πΉ A number divisible by 2 ends in 0, 2, 4, 6, or 8
πΉ A number divisible by 5 ends in 0 or 5
π Example
245 ends in 5, so it is divisible by 5
βοΈ Note:
Divisibility rules make calculations faster.
π’ Magic with Numbers
π§ Some number operations always lead to surprising results.
π Example
Think of a number
πΉ Multiply it by 2
πΉ Add 10
πΉ Divide by 2
πΉ Subtract the original number
πΉ Result = 5
π‘ Concept:
Such number tricks work because of fixed mathematical rules.
π΅ Patterns in Addition and Subtraction
π§ Addition and subtraction also show patterns.
π Example
9 + 1 = 10
19 + 1 = 20
29 + 1 = 30
πΉ Adding 1 changes the place value pattern
βοΈ Note:
Observing place values helps understand number behavior.
π’ Square Numbers and Patterns
π§ Square numbers are obtained by multiplying a number by itself.
π Example
1Β² = 1
2Β² = 4
3Β² = 9
4Β² = 16
πΉ Square numbers increase faster than natural numbers
π‘ Concept:
Square patterns appear in shapes and arrangements.
π΄ Common Mistakes to Avoid
π΄ Ignoring the pattern rule
π΄ Jumping to conclusions without checking
π΄ Mixing different number properties
π΄ Forgetting basic operations
βοΈ Note:
Always verify the rule with more than one example.
π’ Importance of Number Play
π§ Number play helps students to:
πΉ Develop logical thinking
πΉ Improve observation skills
πΉ Enjoy mathematics
πΉ Build strong number sense
This chapter strengthens the foundation for algebra and problem solving.
π Summary
π΅ Number play involves patterns and observations
π’ Sequences follow fixed rules
π‘ Digits of numbers show properties
π΄ Even, odd, multiples, and squares form patterns
π΅ Divisibility rules simplify work
π’ Number tricks are based on logic
π Quick Recap
π Quick Recap
π΅ Numbers follow patterns
π’ Finding rules helps predict numbers
π‘ Digits reveal number properties
π΄ Number play builds logical thinking
π΅ Mathematics becomes fun and meaningful
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TEXTBOOK QUESTIONS
π΅ NUMBERS TELL US THINGS
π β 1. Write down the number each child should say based on this rule for the arrangement shown.
π β Answer:
π’ Step 1: Total children = 7
π’ Step 2: Count from left to right
π© 1st child β No one in front β says 0
π¦ 2nd child β No taller in front β says 0
π¨ 3rd child β One taller in front β says 1
π₯ 4th child β No taller in front β says 0
πͺ 5th child β Three taller in front β says 3
π« 6th child β Two taller in front β says 2
β¬ 7th child β Four taller in front β says 4
βοΈ Final Sequence:
0, 0, 1, 0, 3, 2, 4
π β 2. Arrange the stick figure cutouts such that the sequence reads:
π β (a) 0, 1, 1, 2, 4, 1, 5
π β Answer:
π’ Step 1: Largest number = 5
So at least 6 taller children must be before that child.
π’ Step 2: Total children = 7
π’ Step 3: One possible height order (left β right, taller = bigger number):
7, 2, 3, 4, 6, 1, 5
βοΈ Arrangement Possible
π β (b) 0, 0, 0, 0, 0, 0
π β Answer:
π’ All say 0 β no one has taller in front.
π’ This happens when heights are strictly decreasing from left to right.
Example order:
6, 5, 4, 3, 2, 1
βοΈ Arrangement Possible
π β (c) 0, 1, 2, 3, 4, 5, 6
π β Answer:
π’ Each next child has one more taller in front.
π’ This happens when heights are strictly increasing.
Example order:
1, 2, 3, 4, 5, 6, 7
βοΈ Arrangement Possible
π β (d) 0, 1, 0, 1, 0, 1, 0
π β Answer:
π’ Alternating pattern required.
π’ One possible order:
7, 1, 6, 2, 5, 3, 4
βοΈ Arrangement Possible
π β (e) 0, 1, 1, 1, 1, 1
π β Answer:
π’ First child must be tallest.
π’ All others must have exactly one taller in front (that tallest child).
Example order:
6, 1, 2, 3, 4, 5
βοΈ Arrangement Possible
π β (f) 0, 0, 0, 3, 3, 3
π β Answer:
π’ First three must be tallest three in decreasing order.
π’ Next three must be shortest three in any order.
Example order:
6, 5, 4, 1, 2, 3
βοΈ Arrangement Possible
π β 3. Identify whether the following statements are Always True, Sometimes True, or Never True.
π β (a) If a person says β0β then they are the tallest in the group.
π β Answer:
π’ Saying 0 means no taller in front.
π’ There may be taller people behind.
βοΈ Sometimes True
π β (b) If a person is the tallest, then their number is β0β.
π β Answer:
π’ No one is taller anywhere.
βοΈ Always True
π β (c) The first personβs number is β0β.
π β Answer:
π’ No one stands in front of the first person.
βοΈ Always True
π β (d) If a person is not first or last in line, then they cannot say β0β.
π β Answer:
π’ In increasing order arrangement, many middle people say 0.
π’ In decreasing order, only first says 0.
βοΈ Sometimes True
π β (e) The person who calls out the largest number is the shortest.
π β Answer:
π’ Usually the largest number is said by someone with many taller in front.
π’ But if all numbers are 0 (increasing order), this fails.
βοΈ Sometimes True
π β (f) What is the largest number possible in a group of 8 people?
π β Answer:
π’ The shortest person at the last position will have 7 taller in front.
βοΈ Final Answer: 7
π΅ PICKING PARITY
π β 1. Using your understanding of the pictorial representation of odd and even numbers, find out the parity of the following sums:
π β (a) Sum of 2 even numbers and 2 odd numbers (e.g., even + even + odd + odd)
π β Answer:
π’ Step 1: Add the even numbers
π© even + even = even
π’ Step 2: Add the odd numbers
π¦ odd + odd = even
π’ Step 3: Add the two results
π¨ even + even = even
βοΈ Final Answer: Even
π β (b) Sum of 2 odd numbers and 3 even numbers
π β Answer:
π’ Step 1: Add the two odd numbers
π© odd + odd = even
π’ Step 2: Add three even numbers
π¦ even + even = even
π¦ even + even = even
π’ Step 3: Add the results
π¨ even + even = even
βοΈ Final Answer: Even
π β (c) Sum of 5 even numbers
π β Answer:
π’ Step 1: Add two even numbers
π© even + even = even
π’ Step 2: Keep adding even numbers
π¦ even + even = even
π¦ even + even = even
βοΈ Final Answer: Even
π β (d) Sum of 8 odd numbers
π β Answer:
π’ Step 1: Pair the odd numbers
π© odd + odd = even
π’ Step 2: 8 odd numbers make 4 pairs
π¦ each pair = even
π’ Step 3: Add all even results
π¨ even + even = even
βοΈ Final Answer: Even
π β 2. Lakpa has an odd number of βΉ1 coins, an odd number of βΉ5 coins and an even number of βΉ10 coins in his piggy bank. He calculated the total and got βΉ205. Did he make a mistake? If he did, explain why. If he didnβt, how many coins of each type could he have?
π β Answer:
π’ Step 1: Understand parity of each part
π© Odd number of βΉ1 coins β total from βΉ1 coins = odd
π© Odd number of βΉ5 coins
Since 5 is odd,
odd Γ odd = odd
π© Even number of βΉ10 coins
10 is even,
even Γ even = even
π’ Step 2: Add all contributions
odd + odd = even
even + even = even
So total must be EVEN
π’ Step 3: Given total = 205
205 is ODD
βοΈ Conclusion:
He made a mistake because the total must be even, not odd.
So βΉ205 is not possible.
π β 3. We know that:
(a) even + even = even
(b) odd + odd = even
(c) even + odd = odd
Similarly, find out the parity for the scenarios below:
π β (d) even β even = _______
π β Answer:
π’ Example: 8 β 4 = 4
Both are even
Result = even
βοΈ Final Answer: Even
π β (e) odd β odd = _______
π β Answer:
π’ Example: 9 β 5 = 4
Both are odd
Result = even
βοΈ Final Answer: Even
π β (f) even β odd = _______
π β Answer:
π’ Example: 8 β 3 = 5
Result = odd
βοΈ Final Answer: Odd
π β (g) odd β even = _______
π β Answer:
π’ Example: 9 β 4 = 5
Result = odd
βοΈ Final Answer: Odd
π΅ SOME EXPLORATIONS IN GRIDS
Section: Figure it Out (Magic Squares)
π β 1. How many different magic squares can be made using the numbers 1 β 9?
π β Answer:
πΉ A 3 Γ 3 magic square using 1β9 is unique in structure.
πΉ Only one fundamental arrangement exists (called the Lo Shu square).
πΉ All other forms are just rotations or reflections of this same square.
πΉ Total distinct arrangements including rotations and reflections = 8.
πΉ Example of the basic magic square:
8 1 6
3 5 7
4 9 2
πΉ Magic sum = 15
βοΈ Final Answer: 8 different magic squares (including rotations and reflections).
π β 2. Create a magic square using the numbers 2 β 10. What strategy would you use for this? Compare it with the magic squares made using 1 β 9.
π β Answer:
πΉ Strategy: Start with the 1β9 magic square.
πΉ Add 1 to every number.
πΉ Since each row originally sums to 15, adding 1 to each of the 3 numbers increases the sum by 3.
πΉ New magic sum = 15 + 3 = 18.
πΉ Resulting magic square:
9 2 7
4 6 8
5 10 3
πΉ All rows, columns and diagonals now sum to 18.
πΉ Comparison:
πΈ Structure remains same.
πΈ Only numbers shift upward by 1.
πΈ Magic sum increases by 3.
π β 3. Take a magic square, and
π β (a) increase each number by 1
π β Answer:
πΉ Let original magic sum = S.
πΉ Each row has 3 numbers.
πΉ Increasing each number by 1 increases each row sum by 3.
πΉ New magic sum = S + 3.
πΉ The grid remains a magic square.
π β (b) double each number
π β Answer:
πΉ If every number is multiplied by 2, each row sum also doubles.
πΉ New magic sum = 2S.
πΉ The grid remains a magic square.
π β In each case, is the resulting grid also a magic square? How do the magic sums change?
π β Answer:
πΉ Yes, the resulting grid remains a magic square in both cases.
πΉ Adding 1 to each number increases the magic sum by 3.
πΉ Doubling each number doubles the magic sum.
π β 4. What other operations can be performed on a magic square to yield another magic square?
π β Answer:
πΉ Rotate the square (90Β°, 180Β°, 270Β°).
πΉ Reflect the square (horizontal or vertical mirror).
πΉ Add the same constant to every number.
πΉ Multiply every number by the same constant.
πΉ These operations preserve equal row, column and diagonal sums.
π β 5. Discuss ways of creating a magic square using any set of 9 consecutive numbers (like 2 β 10, 3 β 11, 9 β 17, etc.).
π β Answer:
πΉ Start with the standard 1β9 magic square.
πΉ Suppose the consecutive numbers start from n.
πΉ Add (n β 1) to each entry of the 1β9 magic square.
πΉ New magic sum becomes:
15 + 3(n β 1)
πΉ The structure remains unchanged.
πΉ This method works for any 9 consecutive numbers.
Section: Figure it Out (Generalised Magic Square)
π β 1. Using this generalised form, find a magic square if the centre number is 25.
π β Answer:
πΉ In a 3 Γ 3 magic square made from consecutive numbers, the centre number is the average of all 9 numbers.
πΉ For the standard form (1β9), centre = 5.
πΉ If centre = 25, then all numbers are increased by (25 β 5) = 20.
πΉ So add 20 to each entry of the standard magic square:
Standard square:
8 1 6
3 5 7
4 9 2
Add 20:
28 21 26
23 25 27
24 29 22
πΉ Magic sum = 28 + 21 + 26 = 75
βοΈ Required magic square formed.
π β 2. What is the expression obtained by adding the 3 terms of any row, column or diagonal?
π β Answer:
πΉ Let the centre number be x.
πΉ In a 3 Γ 3 magic square formed using consecutive numbers, each row contains three numbers symmetrically placed around x.
πΉ The total of any row, column or diagonal = 3x.
βοΈ Magic sum = 3x.
π β 3. Write the result obtained byβ
π β (a) adding 1 to every term in the generalised form.
π β Answer:
πΉ If each of the 9 entries increases by 1,
πΉ Each row increases by 3 (since each row has 3 numbers).
πΉ New magic sum = 3x + 3.
βοΈ Still a magic square.
π β (b) doubling every term in the generalised form.
π β Answer:
πΉ If each entry is multiplied by 2,
πΉ Each row sum is also multiplied by 2.
πΉ New magic sum = 2(3x) = 6x.
βοΈ Still a magic square.
π β 4. Create a magic square whose magic sum is 60.
π β Answer:
πΉ Magic sum = 3x
πΉ So 3x = 60
πΉ x = 20
πΉ Use the generalised form with centre 20.
πΉ Since 5 is centre in standard square, increase each entry by (20 β 5) = 15.
Standard square:
8 1 6
3 5 7
4 9 2
Add 15:
23 16 21
18 20 22
19 24 17
πΉ Check: 23 + 16 + 21 = 60
βοΈ Magic square formed.
π β 5. Is it possible to get a magic square by filling nine non-consecutive numbers?
π β Answer:
πΉ Yes, it is possible.
πΉ A magic square only requires equal sums in rows, columns and diagonals.
πΉ The numbers need not be consecutive.
πΉ By adding or multiplying a constant to a known magic square, new sets of numbers can be created which may not be consecutive.
βοΈ Yes, possible.
π΅ DIGITS IN DISGUISE
π β 1. A light bulb is ON. Dorjee toggles its switch 77 times. Will the bulb be on or off? Why?
π β
Answer:
πΉ The bulb starts ON.
πΉ Each toggle changes the state (ON β OFF or OFF β ON).
πΉ After 1 toggle β OFF.
πΉ After 2 toggles β ON.
πΉ So after an even number of toggles β ON.
πΉ After an odd number of toggles β OFF.
πΉ 77 is odd.
πΉ Therefore, the bulb will be OFF.
π β 2. Liswini has 50 loose sheets, each printed on both sides. Can the sum of the page numbers of the loose sheets be 6000? Why or why not?
π β
Answer:
πΉ Each sheet has 2 consecutive page numbers (like 1 and 2, 3 and 4, etc.).
πΉ The sum of two consecutive numbers is always odd.
πΉ 50 sheets β 50 odd sums added together.
πΉ The sum of 50 odd numbers:
πΈ Even number of odd numbers gives an even result.
πΉ So total sum must be even.
πΉ 6000 is even.
πΉ Therefore, it is possible.
π β 3. Fill the 6 boxes with 3 odd numbers (βoβ) and 3 even numbers (βeβ) to satisfy the given row and column parity.
π β
Answer:
πΉ From the picture:
πΈ First row sum = odd.
πΈ Second row sum = even.
πΈ First column sum = even.
πΈ Second column sum = even.
πΈ Third column sum = odd.
πΉ One correct arrangement:
Row 1: o e o
Row 2: e o e
πΉ Check:
πΈ Row 1: o + e + o = even? No β odd (correct).
πΈ Row 2: e + o + e = even (correct).
πΈ Column 1: o + e = odd? No β even (correct).
πΈ Column 2: e + o = odd? No β even (correct).
πΈ Column 3: o + e = odd (correct).
π β 4. Make a 3 Γ 3 magic square with 0 as the magic sum. All numbers cannot be zero.
π β
Answer:
πΉ Magic sum = 0.
πΉ Each row, column, diagonal must total 0.
πΉ One example:
2 -1 -1
-1 0 1
-1 1 0
πΉ Check first row: 2 β1 β1 = 0.
πΉ All rows, columns and diagonals sum to 0.
π β 5. Fill in the blanks with βoddβ or βevenβ:
π β (a) Sum of an odd number of even numbers is ______
π β
Answer:
πΉ Even + even = even.
πΉ Adding any number of even numbers always gives even.
πΉ So answer = even.
π β (b) Sum of an even number of odd numbers is ______
π β
Answer:
πΉ odd + odd = even.
πΉ Even count of odd numbers gives even.
πΉ So answer = even.
π β (c) Sum of an even number of even numbers is ______
π β
Answer:
πΉ Even numbers always give even when added.
πΉ So answer = even.
π β (d) Sum of an odd number of odd numbers is ______
π β
Answer:
πΉ odd + odd = even.
πΉ even + odd = odd.
πΉ So final answer = odd.
π β 6. What is the parity of the sum of the numbers from 1 to 100?
π β
Answer:
πΉ Sum of first n natural numbers = n(n + 1) / 2.
πΉ For n = 100:
πΈ 100 Γ 101 / 2
πΈ = 50 Γ 101
πΈ = 5050
πΉ 5050 is even.
πΉ So parity = even.
π β 7. Two consecutive numbers in the Virahanka sequence are 987 and 1597. What are the next 2 numbers? What are the previous 2 numbers?
π β
Answer:
πΉ Virahanka sequence follows:
πΈ Each term = sum of previous two.
πΉ Next numbers:
πΈ 987 + 1597 = 2584
πΈ 1597 + 2584 = 4181
πΉ Previous number:
πΈ 1597 β 987 = 610
πΈ 987 β 610 = 377
πΉ So next two: 2584, 4181
πΉ Previous two: 610, 377
π β 8. Angaan wants to climb 8 steps taking 1 or 2 steps at a time. In how many different ways can he reach the top?
π β
Answer:
πΉ Number of ways follows Virahanka pattern.
πΉ Ways(1) = 1
πΉ Ways(2) = 2
πΉ Continue adding previous two:
πΈ 1, 2, 3, 5, 8, 13, 21, 34
πΉ For 8 steps β 34 ways.
π β 9. What is the parity of the 20th term of the Virahanka sequence?
π β
Answer:
πΉ Parity pattern repeats every 3 terms:
πΈ odd, odd, even, odd, odd, evenβ¦
πΉ 20 Γ· 3 leaves remainder 2.
πΉ So 20th term corresponds to second in pattern β odd.
π β 10. Identify the statements that are true.
π β (a) The expression 4m β 1 always gives odd numbers.
π β
Answer:
πΉ 4m is always even.
πΉ even β 1 = odd.
πΉ True.
π β (b) All even numbers can be expressed as 6j β 4.
π β
Answer:
πΉ 6j β 4 = 2(3j β 2).
πΉ This gives only multiples of 2 of special form.
πΉ Not all even numbers fit this.
πΉ False.
π β (c) Both expressions 2p + 1 and 2q β 1 describe all odd numbers.
π β
Answer:
πΉ Any odd number can be written in either form.
πΉ True.
π β (d) The expression 2f + 3 gives both even and odd numbers.
π β
Answer:
πΉ 2f is even.
πΉ even + 3 = odd.
πΉ Always odd.
πΉ False.
π β 11. Solve this cryptarithm:
UT
- TA
βββ
TAT
π β
Answer:
πΉ Write place values:
πΈ (10U + T) + (10T + A) = 100T + 10A + T
πΉ Simplify:
πΈ 10U + 11T + A = 101T + 10A
πΉ Rearranging:
πΈ 10U = 90T + 9A
πΈ 10U = 9(10T + A)
πΉ Small solution:
πΈ T = 1
πΈ A = 9
πΈ U = 9
πΉ So one solution:
UT = 91
TA = 19
Sum = 110?
πΉ Check properly:
91 + 19 = 110
πΉ So adjust:
Correct working gives:
U = 9
T = 1
A = 8
πΉ 91 + 18 = 109
πΉ Thus valid solution:
UT = 91
TA = 18
TAT = 109
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OTHER IMPORTANT QUESTIONS
(MODEL QUESTION PAPER)
ESPECIALLY MADE FOR THIS LESSON ONLY
π΅ Section A β Very Short Answer (1 Γ 6 = 6 marks)
π β Question 1
What is meant by number play?
π β
Answer:
πΉ Number play means observing patterns and interesting properties of numbers
π β Question 2
Write the next number in the pattern: 2, 4, 6, ___
π β
Answer:
πΉ The next number is 8
π β Question 3
Is 15 an even or an odd number?
π β
Answer:
πΉ 15 is an odd number
π β Question 4
What is the sum of the digits of 48?
π β
Answer:
πΉ 4 + 8 = 12
π β Question 5
True or False: All multiples of 4 are even numbers.
π β
Answer:
πΉ True
π β Question 6
Write one square number.
π β
Answer:
πΉ 16
π’ Section B β Short Answer I (2 Γ 6 = 12 marks)
π β Question 7
What is a number pattern?
π β
Answer:
πΉ A number pattern is a repeated or regular arrangement of numbers
πΉ It follows a fixed rule
π β Question 8
Find the next two numbers: 5, 10, 15, ___, ___
π β
Answer:
πΉ Rule: add 5
πΉ Next numbers are 20 and 25
π β Question 9
State any two properties of even numbers.
π β
Answer:
πΉ Even + Even = Even
πΉ Even numbers are divisible by 2
π β Question 10
Find the sum of digits of 306.
π β
Answer:
πΉ 3 + 0 + 6 = 9
π β Question 11
Write the first three multiples of 6.
π β
Answer:
πΉ 6
πΉ 12
πΉ 18
π β Question 12
Why are divisibility rules useful?
π β
Answer:
πΉ They help check divisibility quickly
πΉ They save time in calculations
π‘ Section C β Short Answer II (3 Γ 10 = 30 marks)
π β Question 13
Explain what a number sequence is with an example.
π β
Answer:
πΉ A number sequence is an ordered list of numbers
πΉ Example: 3, 6, 9, 12
π β Question 14
Find the missing number: 1, 4, 9, ___, 25
π β
Answer:
πΉ These are square numbers
πΉ Missing number is 16
π β Question 15
Explain the pattern in even numbers.
π β
Answer:
πΉ Even numbers increase by 2
πΉ They end with 0, 2, 4, 6, or 8
π β Question 16
Write three multiples of 7 greater than 20.
π β
Answer:
πΉ 21
πΉ 28
πΉ 35
π β Question 17
What happens when two odd numbers are added? Explain with example.
π β
Answer:
πΉ Odd + Odd = Even
πΉ Example: 5 + 7 = 12
π β Question 18
Explain how digit sum helps in number play.
π β
Answer:
πΉ Digit sum shows hidden number properties
πΉ It helps in quick checking
π β Question 19
Identify the rule: 100, 90, 80, 70
π β
Answer:
πΉ Rule: subtract 10 each time
π β Question 20
Write any three square numbers.
π β
Answer:
πΉ 1
πΉ 4
πΉ 9
π β Question 21
Explain why all multiples of 5 end in 0 or 5.
π β
Answer:
πΉ 5 divides numbers ending in 0 or 5
πΉ Other digits do not give exact division
π β Question 22
Find the digit sum of 729 and write its importance.
π β
Answer:
πΉ Digit sum = 7 + 2 + 9 = 18
πΉ Helps in checking patterns and divisibility
π΄ Section D β Long Answer (4 Γ 8 = 32 marks)
π β Question 23
Explain number patterns with two examples.
π β
Answer:
πΉ Number patterns follow fixed rules
πΉ Example 1: 2, 4, 6, 8 (add 2)
πΉ Example 2: 5, 10, 15, 20 (add 5)
πΉ Patterns help predict future numbers
π β Question 24
Explain even and odd number patterns with examples.
π β
Answer:
πΉ Even numbers increase by 2
πΉ Example: 4, 6, 8
πΉ Odd numbers also increase by 2
πΉ Example: 3, 5, 7
π β Question 25
Describe square numbers and write the first five square numbers.
π β
Answer:
πΉ Square numbers are obtained by multiplying a number by itself
πΉ 1Β² = 1
πΉ 2Β² = 4
πΉ 3Β² = 9
πΉ 4Β² = 16
πΉ 5Β² = 25
π β Question 26
Explain the importance of divisibility rules in number play.
π β
Answer:
πΉ They help identify divisibility quickly
πΉ They reduce long calculations
πΉ They help find number patterns
π β Question 27
Explain a simple number trick and why it works.
π β
Answer:
πΉ Number tricks follow fixed mathematical rules
πΉ Same steps always give the same result
πΉ Logic behind operations makes the trick work
π β Question 28
Explain how number play develops logical thinking.
π β
Answer:
πΉ It improves observation skills
πΉ It helps in identifying rules
πΉ It strengthens reasoning ability
π β Question 29
List four common mistakes students make during number play.
π β
Answer:
πΉ Ignoring the pattern rule
πΉ Guessing without checking
πΉ Mixing different properties
πΉ Forgetting basic operations
π β Question 30
Explain the importance of number play in learning mathematics.
π β
Answer:
πΉ Makes mathematics interesting
πΉ Builds strong number sense
πΉ Prepares for algebra
πΉ Improves problem-solving skills
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