Sunday, December 7, 2025

Chapter 5: Number Play – Complete Question Bank

📘 Chapter 5: Number Play – Complete Question Bank

Grade 8 Mathematics

NCERT Ganita Prakash

Comprehensive Worksheet

Logical Reasoning Problem-Solving Analytical Thinking Pattern Recognition Algebraic Generalization

📝 Study Notes & Key Properties

🔢 Insert NCERT image from p. 84 here – Consecutive number sums

5.1 Sums of Consecutive Numbers: Every odd number = sum of two consecutive numbers.
5.2 Parity of Expressions: For any 4 consecutive numbers, all +/– expressions yield even numbers.
5.3 Divisibility by 4: Even numbers → multiples of 4 (remainder 0) or remainder 2.
5.4 Divisibility Rules: If a divides M & N → divides M+N and M–N.
5.5 Remainders: Numbers leaving remainder r when divided by n: nk + r.
5.6 Digital Roots: Sum digits repeatedly until single digit. Multiples of 9 → digital root 9.

🔢 Insert NCERT image from p. 86 here – Divisibility shortcuts table

🧮 Multiple Choice Questions (20)

1. Which of the following is NOT a sum of consecutive natural numbers?

(a) 7

(b) 10

(d) 15

Explanation: 11 cannot be expressed as a sum of consecutive natural numbers. Powers of 2 generally cannot, except 2 itself. 7=3+4, 10=1+2+3+4, 15=7+8=4+5+6.

2. For any 4 consecutive numbers, all expressions with ‘+’ and ‘–’ yield:

(a) Odd numbers

(b) Even numbers

(d) Multiples of 3

Explanation: Changing a sign changes the value by an even number. Starting expression (e.g., a+b-c-d) gives even result, so all 8 possibilities yield even numbers.

3. The sum of two even numbers is divisible by 4 if:

(a) Both are multiples of 4

(b) Both leave remainder 2

(d) None

Explanation: Even numbers are either multiples of 4 (4k) or 4k+2. Sum of two same type: 4k+4m=4(k+m) or (4k+2)+(4m+2)=4(k+m+1).

4. If a number is divisible by 8, it is also divisible by:

(a) 2

(b) 4

(d) Only 2

Explanation: 8 = 2³, so any multiple of 8 contains factor 2²=4 and 2¹=2.

5. The remainder when 427 is divided by 9 is:

(a) 4

(b) 5

(d) 7

Explanation: Digit sum: 4+2+7=13 → 1+3=4. For divisibility by 9, remainder = digital root (unless digital root is 9, remainder 0).

🔍 Assertion & Reasoning (20)

1. Assertion: All odd numbers can be expressed as sums of two consecutive numbers.
Reason: Odd numbers are of the form 2n+1, which equals n + (n+1).

(a) Both correct, Reason explains Assertion

(b) Both correct, Reason does not explain

(d) Both wrong

Explanation: Assertion is true (e.g., 7=3+4, 9=4+5). Reason correctly shows odd number 2n+1 = n + (n+1), which are consecutive integers.

2. Assertion: For 4 consecutive numbers, all ‘+’/‘–’ expressions give even results.
Reason: Changing a sign changes value by an even number.

(a) Both correct, Reason explains

(b) Both correct, no relation

(d) Both wrong

Explanation: Assertion true (verified). Reason: Switching +b to -b changes total by 2b (even), so parity remains same.

✅ True/False (10)

1. Every even number can be expressed as a sum of consecutive numbers.

True

False

Explanation: False. Some even numbers like 2, 8, 32 (powers of 2) cannot be expressed as sums of consecutive natural numbers.

2. Digital root of a multiple of 9 is always 9.

True

False

Explanation: True. Multiples of 9 have digit sums divisible by 9, and repeated summing yields 9 (except 0 which gives 9 as digital root).

📝 Short Answer I (2 Marks × 15)

1. Express 15 as sums of consecutive numbers in two ways.

Answer: 15 = 7 + 8
15 = 4 + 5 + 6
Also 15 = 1 + 2 + 3 + 4 + 5

2. Show that for 4 consecutive numbers, a ± b ± c ± d is always even.

Answer: Let numbers be n, n+1, n+2, n+3. Changing a sign changes total by even number (e.g., +b to -b changes by 2b). Starting expression n+(n+1)-(n+2)-(n+3) = -4 (even). All other expressions have same parity.

📝 Short Answer II (3 Marks × 10)

1. Prove: If a number is divisible by both 9 and 4, it is divisible by 36.

Proof: Let N be divisible by 9 → contains factor 3². Divisible by 4 → contains factor 2². Thus N contains 2²×3²=36 as factor.
Alternatively: LCM(9,4)=36. If divisible by both, divisible by LCM.

2. Solve cryptarithm: UT × 3 = PUT

Solution: Try T=7 → 7×3=21, carry 2. U×3+2 ends with U. Try U=5 → 5×3+2=17, doesn't end with 5. Try U=1 → 1×3+2=5, doesn't end with 1. Try U=7 → 7×3+2=23, ends with 3? No.
Actually: 17×3=51 → U=1, T=7, P=5 ✓.

📚 Long Answer (5 Marks × 10)

1. Explore and write which numbers can be expressed as sums of consecutive numbers in more than one way. Provide reasoning.

Answer:
• Numbers with odd divisors >1 can be expressed in multiple ways.
• Example: 15 (divisors: 1,3,5,15) → 3 ways.
• General: If N has an odd divisor d>1, it can be written as sum of d consecutive numbers centered at N/d.
• Powers of 2 have only one representation (as single number).
• Formula: Number of ways = number of odd divisors of N minus 1.
• Example: 45 has odd divisors 1,3,5,9,15,45 → 6-1=5 ways.

2. Prove that the sum of three consecutive even numbers is divisible by 6.

Proof: Let three consecutive even numbers be 2n, 2n+2, 2n+4.
Sum = 2n + (2n+2) + (2n+4) = 6n + 6 = 6(n+1).
Clearly divisible by 6 for all integer n.

🧩 Case-Based Questions

Case 1: Sums of Consecutive Numbers

Anshu is exploring sums of consecutive numbers. He writes: 7=3+4, 10=1+2+3+4, 12=3+4+5, 15=7+8=4+5+6=1+2+3+4+5. He wonders: Can every natural number be written as a sum of consecutive numbers?

1. Which of the following numbers CANNOT be expressed as a sum of consecutive natural numbers?

(a) 9

(b) 11

(d) 25

Explanation: 11 cannot be expressed. Powers of 2 generally cannot (except 2 itself). 9=4+5, 16= cannot, 25=12+13.

2. Which number can be expressed as a sum of consecutive numbers in the most ways?

(a) 15

(b) 21

(d) 45

Explanation: 45 can be written in 5 ways: 22+23, 14+15+16, 7+8+9+10+11, 5+6+7+8+9+10, 1+2+3+4+5+6+7+8+9.

3. Which statement is TRUE?

(a) All odd numbers can be expressed as sum of 2 consecutive numbers

(b) All even numbers can be expressed as sum of consecutive numbers

(d) Prime numbers cannot be expressed as sum of consecutive numbers

Explanation: (a) True: odd number n = (n/2 - 0.5) + (n/2 + 0.5). (b) False: e.g., 2,8. (c) False: needs negative numbers. (d) False: e.g., 5=2+3.

4. How many ways can 18 be expressed as sum of consecutive natural numbers?

(a) 1

(b) 2

(d) 4

Explanation: 18 = 5+6+7 = 3+4+5+6. Two ways only.

Case 2: Parity and 4 Consecutive Numbers

Take any 4 consecutive numbers, say 3,4,5,6. Place '+' and '–' signs between them in all 8 possible ways. All results are even numbers.

1. What is common about all results?

(a) All are positive

(b) All are negative

(d) All are multiples of 3

Explanation: Verified by evaluating all 8 expressions: 18,6,8,-4,10,-2,0,-12 → all even.

2. Why are all results even?

(a) Because 4 consecutive numbers contain 2 evens and 2 odds

(b) Because sum of 4 consecutive numbers is always even

(d) Both (a) and (c)

Explanation: Both reasons contribute. The parity remains same because sign changes alter by even amounts.

🔢 Insert NCERT image from p. 90 here – Digital root example

❓ "Figure It Out" Questions

1. The sum of four consecutive numbers is 34. What are these numbers?

Solution: Let numbers be n, n+1, n+2, n+3.
Sum = 4n + 6 = 34 → 4n = 28 → n=7.
Numbers: 7, 8, 9, 10.

2. "I hold some pebbles... When grouped by 3's, one remains... by 5's, one remains... by 7's, none remain. Less than 100. How many pebbles?"

Solution: N mod 3=1, mod 5=1, mod 7=0, N<100.
N=7k, try multiples of 7: 7,14,21,28,35,42,49,56,63,70,77,84,91,98.
Check odd (since mod 2=1): 7,21,35,49,63,77,91.
Check mod 5=1: 21,91.
Check mod 3=1: 91 ✓.
Answer: 91 pebbles.

Chapter 6: We Distribute, Yet Things Multiply - Complete with SVG Diagrams

Section E: Short Answer II (3 Marks × 10 Questions)

Show geometrically that \((a+b)^2 = a^2 + 2ab + b^2\) (describe with diagram).

Fig: Geometric proof of (a+b)² - Square divided into a², b², and two ab rectangles

Geometric Proof:
1. Draw a square of side length (a+b)
2. Divide horizontally at distance 'a' from left
3. Divide vertically at distance 'a' from top
4. This creates four regions:
- Top-left: Square of side a (area = a²)
- Top-right: Rectangle a×b (area = ab)
- Bottom-left: Rectangle b×a (area = ab)
- Bottom-right: Square of side b (area = b²)
5. Total area = a² + ab + ab + b² = a² + 2ab + b² = (a+b)²

The diagram visually proves the algebraic identity using area decomposition

Prove: \((m+n)^2 - 4mn = (n-m)^2\). Show geometrically.

Fig: (m+n)² square with four corner rectangles (mn each) removed, leaving center square (n-m)²

Geometric Proof:
1. Start with square of side (m+n), area = (m+n)²
2. Remove four corner rectangles, each of size m×n
3. Total area removed = 4mn
4. Remaining center square has side = (n-m)
5. Area of center square = (n-m)²
6. Therefore: (m+n)² - 4mn = (n-m)²

The diagram shows removing corner rectangles from large square leaves smaller center square

Section F: Long Answer Questions (5 Marks × 10 Questions)

Derive all three identities geometrically using diagrams.

Three Algebraic Identities - Geometric Proofs

Identity 1A: (a+b)²

Identity 1B: (a-b)²

Identity 1C: (a+b)(a-b) = a²-b²

Geometric Derivations:

1. (a+b)² = a² + 2ab + b²
• Draw square of side (a+b)
• Divide into a², b², and two ab rectangles
• Total area = sum of all parts

2. (a-b)² = a² - 2ab + b²
• Start with a² square
• Remove two ab rectangles (overlaps b² area)
• Add back b² (was subtracted twice)
• Remaining = a² - 2ab + b²

3. (a+b)(a-b) = a² - b²
• Start with a² square
• Remove b×a rectangle from side
• Also remove a×b rectangle from bottom
• b² area was removed twice, so add back once
• Result = a² - ab - ab + b² = a² - b²

All three identities can be proven geometrically using area decomposition and rearrangement

Prove algebraically that the diagonal products in a 2×2 calendar square differ by 7.

Fig: 2×2 Calendar square pattern - Diagonal products always differ by 7

Algebraic Proof:

In a 2×2 calendar square:
Top-left number = a
Top-right number = a+1
Bottom-left number = a+7 (next week same day)
Bottom-right number = a+8

Diagonal Products:
D₁ = a × (a+8) = a² + 8a
D₂ = (a+1) × (a+7) = a² + 8a + 7

Difference:
D₂ - D₁ = (a² + 8a + 7) - (a² + 8a)
= a² + 8a + 7 - a² - 8a
= 7

Therefore, the diagonal products always differ by 7, regardless of the starting number 'a'.

This works because calendar rows are 7 days apart, creating the constant difference

Section G: Case-Based Questions (5 Cases × 4 Sub-Questions Each)

Case 3: Geometric Proof of Identities

A square of side length \( (m+n) \) is drawn. Four rectangles of dimensions \( m \times n \) are removed from the corners, leaving a smaller shaded square in the center.

Fig: Interactive diagram - Hover over parts to see details

What is the area of the large square?

a) \( m^2 + n^2 \)

b) \( (m+n)^2 \)

c) \( m^2 + 2mn + n^2 \)

d) Both b and c

\((m+n)^2 = m^2+2mn+n^2\) (expand using identity)

What is the total area of the four removed rectangles?

a) \( mn \)

b) \( 2mn \)

c) \( 4mn \)

d) \( m^2n^2 \)

c) \( 4mn \)

Four rectangles each of area \( m \times n \)

The area of the shaded square is:

a) \( (m-n)^2 \)

b) \( (n-m)^2 \)

c) \( m^2 + n^2 - 2mn \)

d) All of these

All are equivalent expressions for the same area

If \( m = 3 \) and \( n = 7 \), what is the area of the shaded square?

a) 16

b) 25

c) 36

d) 49

a) 16

\((7-3)^2 = 4^2 = 16\)

Case 4: Pattern Diagrams

The number of circles in a pattern grows as shown:

Fig: Triangular number pattern - Each step adds one more row than previous

How many circles in Step 4?

a) 10

b) 15

c) 20

d) 25

b) 15

Step 4 = 1+2+3+4+5 = 15 circles (triangular number T₅)

Saturday, December 6, 2025

Chapter 6: We Distribute, Yet Things Multiply - Question Bank

Chapter 6: We Distribute, Yet Things Multiply - Complete Question Bank

Section A: Multiple Choice Questions (20 Questions)

What is the expanded form of \((x + 4)(x + 3)\)?

a) \(x^2 + 7x + 12\)

b) \(x^2 + 12x + 7\)

c) \(x^2 + 7x + 7\)

d) \(x^2 + 12x + 12\)

a) \(x^2 + 7x + 12\)

Using distributive property: \((x+4)(x+3) = x(x+3) + 4(x+3) = x^2 + 3x + 4x + 12 = x^2 + 7x + 12\)

The expression \((a - 7)^2\) equals:

a) \(a^2 - 49\)

b) \(a^2 - 14a + 49\)

c) \(a^2 + 14a + 49\)

d) \(a^2 - 7a + 49\)

b) \(a^2 - 14a + 49\)

Using identity: \((a-b)^2 = a^2 - 2ab + b^2\). Here \(b=7\), so \(a^2 - 2×a×7 + 49 = a^2 - 14a + 49\)

Which identity is used to find \(98 \times 102\) quickly?

a) \((a + b)^2\)

b) \((a - b)^2\)

c) \((a + b)(a - b)\)

d) \(a(b + c)\)

c) \((a + b)(a - b)\)

\(98 \times 102 = (100-2)(100+2) = 100^2 - 2^2\). This uses difference of squares identity.

The product \(45 \times 55\) can be written as:

a) \((50 - 5)(50 + 5)\)

b) \((40 + 5)(50 + 5)\)

c) \((50 + 5)^2\)

d) \((50 - 5)^2\)

a) \((50 - 5)(50 + 5)\)

\(45 = 50-5\) and \(55 = 50+5\), so \(45×55 = (50-5)(50+5)\)

If \(a\) and \(b\) are integers, \((a - b)^2\) is always equal to:

a) \((b - a)^2\)

b) \(-(b - a)^2\)

c) \(a^2 - b^2\)

d) \(b^2 - a^2\)

a) \((b - a)^2\)

Squaring removes the sign: \((a-b)^2 = [-(b-a)]^2 = (b-a)^2\)

The value of \(101^2\) using the identity is:

a) 10201

b) 10001

c) 10101

d) 11001

a) 10201

\(101^2 = (100+1)^2 = 100^2 + 2×100×1 + 1^2 = 10000 + 200 + 1 = 10201\)

The expression \(3p(2q - 5)\) expands to:

a) \(6pq - 15p\)

b) \(6pq - 5\)

c) \(3p - 15q\)

d) \(6pq + 15p\)

a) \(6pq - 15p\)

\(3p(2q-5) = 3p×2q - 3p×5 = 6pq - 15p\)

If the product of two numbers is \(ab\), and both are increased by 1, the new product is:

a) \(ab + 1\)

b) \(ab + a + b + 1\)

c) \(ab + a + b\)

d) \(ab + 2\)

b) \(ab + a + b + 1\)

\((a+1)(b+1) = ab + a + b + 1\)

The product \((x + 2)(x - 2)\) simplifies to:

a) \(x^2 - 4\)

b) \(x^2 + 4\)

c) \(x^2 - 2\)

d) \(x^2 + 2x - 4\)

a) \(x^2 - 4\)

Using difference of squares: \((x+2)(x-2) = x^2 - 2^2 = x^2 - 4\)

10.

Which of these is NOT an identity?

a) \(a(b + c) = ab + ac\)

b) \(a^2 - b^2 = (a + b)(a - b)\)

c) \(a + b = b + a\)

d) \(a^2 + b^2 = (a + b)^2\)

\((a+b)^2 = a^2 + 2ab + b^2\), not \(a^2 + b^2\)

11.

The increase in \(23 \times 27\) if 27 is increased by 1 is:

a) 23

b) 27

c) 1

d) 50

a) 23

If second number increases by 1, product increases by first number

12.

The expression \((2x + 5)^2\) equals:

a) \(4x^2 + 25\)

b) \(4x^2 + 10x + 25\)

c) \(4x^2 + 20x + 25\)

d) \(4x^2 + 20x + 10\)

c) \(4x^2 + 20x + 25\)

\((2x+5)^2 = (2x)^2 + 2×2x×5 + 5^2 = 4x^2 + 20x + 25\)

13.

Which pattern follows \(2(a^2 + b^2) = (a+b)^2 + (a-b)^2\)?

a) Pattern 1 from the chapter

b) Pattern 2 from the chapter

c) Pattern for 11, 101, 1001

d) Coin triangle pattern

a) Pattern 1 from the chapter

Pattern 1 shows \(2(a^2+b^2) = (a+b)^2 + (a-b)^2\)

14.

The product \(3874 \times 11\) using distributive property is:

a) 42614

b) 42514

c) 42624

d) 42714

a) 42614

\(3874×11 = 3874×(10+1) = 38740 + 3874 = 42614\)

15.

If \(m + n = 10\) and \(mn = 21\), then \(m^2 + n^2 = ?\)

a) 58

b) 100

c) 42

d) 79

a) 58

\(m^2+n^2 = (m+n)^2 - 2mn = 100 - 42 = 58\)

16.

Which is equivalent to \(k^2 + 2k\)?

a) \((k+1)^2 - 1\)

b) \(k(k+1) + k\)

c) \(k(k+2)\)

d) All of these

All simplify to \(k^2 + 2k\)

17.

The product \((a + 3)(a - 4)\) expands to:

a) \(a^2 - a - 12\)

b) \(a^2 + a - 12\)

c) \(a^2 - 12\)

d) \(a^2 + 7a - 12\)

a) \(a^2 - a - 12\)

\((a+3)(a-4) = a^2 - 4a + 3a - 12 = a^2 - a - 12\)

18.

To multiply by 101 quickly, we write the number as:

a) \(\times (100 + 1)\)

b) \(\times (10 + 1)\)

c) \(\times 100 + 1\)

d) \(\times 101 + 0\)

a) \(\times (100 + 1)\)

\(101 = 100 + 1\), so use distributive property

19.

The expression \((x - y)^2 + (x + y)^2\) simplifies to:

a) \(2x^2 + 2y^2\)

b) \(x^2 + y^2\)

c) \(4xy\)

d) \(2xy\)

a) \(2x^2 + 2y^2\)

\((x-y)^2 + (x+y)^2 = (x^2-2xy+y^2) + (x^2+2xy+y^2) = 2x^2 + 2y^2\)

20.

Which method is NOT used in "Mind the Mistake"?

a) Correcting sign errors

b) Expanding brackets

c) Using Pythagorean theorem

d) Combining like terms

c) Using Pythagorean theorem

"Mind the Mistake" deals with algebraic errors, not geometry theorems

Section B: Assertion & Reasoning Questions (20 Questions)

Assertion (A): \((a + b)^2 = a^2 + 2ab + b^2\) for all integers \(a, b\).
Reason (R): The distributive property holds for integers.

a) Both A and R are true and R explains A.

b) Both A and R are true but R does not explain A.

c) A is true but R is false.

d) A is false but R is true.

a) Both A and R are true and R explains A.

The distributive property is used to derive \((a+b)^2 = a^2 + 2ab + b^2\)

A: \((a - b)^2\) and \((b - a)^2\) are equal.
R: Squaring a negative gives a positive.

a) Both true, R explains A.

b) Both true, R does not explain A.

c) A true, R false.

d) A false, R true.

a) Both true, R explains A.

\((a-b)^2 = [-(b-a)]^2 = (b-a)^2\) because square of negative is positive

A: \(99 \times 101 = 9999\).
R: \(99 \times 101 = (100-1)(100+1) = 100^2 - 1^2\).

a) Both true, R explains A.

b) Both true, R does not explain A.

c) A true, R false.

d) A false, R true.

a) Both true, R explains A.

R shows the method using difference of squares identity

A: The product \(23 \times 27\) increases by 23 if 27 is increased by 1.
R: \(a(b+1) = ab + a\).

a) Both true, R explains A.

b) Both true, R does not explain A.

c) A true, R false.

d) A false, R true.

a) Both true, R explains A.

R gives the general rule that explains A

A: \(2(a^2 + b^2) = (a+b)^2 + (a-b)^2\) is an identity.
R: It follows from adding the two square identities.

a) Both true, R explains A.

b) Both true, R does not explain A.

c) A true, R false.

d) A false, R true.

a) Both true, R explains A.

Adding \((a+b)^2\) and \((a-b)^2\) gives \(2(a^2+b^2)\)

A: \(3874 \times 101 = 391374\).
R: \(3874 \times 101 = 387400 + 3874\).

a) Both true, R explains A.

b) Both true, R does not explain A.

c) A true, R false.

d) A false, R true.

a) Both true, R explains A.

R shows the calculation method using distributive property

A: \((x+2)(x+5) = x^2 + 7x + 10\).
R: Using distributive property: \(x(x+5) + 2(x+5)\).

a) Both true, R explains A.

b) Both true, R does not explain A.

c) A true, R false.

d) A false, R true.

a) Both true, R explains A.

R shows how to expand using distributive property to get A

A: \((a+b)^2\) is always greater than \(a^2 + b^2\).
R: Because \(2ab\) is always positive.

a) Both true, R explains A.

b) Both true, R does not explain A.

c) A true, R false.

d) A false, R true.

A is false because if a or b is negative, \(2ab\) can be negative. R is true about squares of negatives.

A: The pattern \(k^2 + 2k\) can be written as \(k(k+2)\).
R: Both represent the same algebraic expression.

a) Both true, R explains A.

b) Both true, R does not explain A.

c) A true, R false.

d) A false, R true.

a) Both true, R explains A.

R explains why they are equivalent expressions

10.

A: The product of two numbers remains same if one is increased by 2 and other decreased by 2.
R: \((a+2)(b-2) = ab + 2b - 2a - 4\).

a) Both true, R explains A.

b) Both true, R does not explain A.

c) A true, R false.

d) A false, R true.

A is false - product changes unless special case. R is true as expansion.

11.

A: \((a+b)(a-b) = a^2 - b^2\) for all real numbers.
R: This is the difference of squares identity.

a) Both true, R explains A.

R names the identity that A represents

12.

A: \(104^2 = 10816\).
R: \(104^2 = (100+4)^2 = 100^2 + 2×100×4 + 4^2\).

a) Both true, R explains A.

R shows the calculation using identity to get A

13.

A: The product decreases when one number increases by 1 and other decreases by 1 if \(b < a + 1\).
R: \((a+1)(b-1) = ab + b - a - 1\).

a) Both true, R explains A.

From R: increase is \(b-a-1\), so product decreases if \(b-a-1 < 0\) i.e., \(b < a+1\)

14.

A: \(97^2 = 9409\).
R: \(97^2 = (100-3)^2 = 100^2 - 2×100×3 + 3^2\).

a) Both true, R explains A.

R shows calculation using \((a-b)^2\) identity

15.

A: \(2(5^2 + 6^2) = (11)^2 + (1)^2\).
R: Using identity \(2(a^2+b^2) = (a+b)^2 + (a-b)^2\) with \(a=6,b=5\).

a) Both true, R explains A.

R applies the pattern identity to verify A

16.

A: \((a+b)^2 - (a-b)^2 = 4ab\).
R: Subtracting the two square identities gives this result.

a) Both true, R explains A.

R explains how to derive A from known identities

17.

A: \((x+1)^3 = x^3 + 3x^2 + 3x + 1\).
R: \((x+1)^3 = (x+1)(x+1)^2 = (x+1)(x^2+2x+1)\).

a) Both true, R explains A.

R shows the expansion steps to get A

18.

A: The distributive property can be used for subtraction: \(a(b-c) = ab - ac\).
R: Subtraction is adding the negative: \(a(b-c) = a[b + (-c)]\).

a) Both true, R explains A.

R explains why distributive property works for subtraction

19.

A: \(73^2 = 5329\).
R: \(73^2 = (70+3)^2 = 70^2 + 2×70×3 + 3^2\).

a) Both true, R explains A.

R shows the decomposition method to calculate square

20.

A: The product \(46 \times 54\) is 2484.
R: \(46 \times 54 = (50-4)(50+4) = 50^2 - 4^2\).

a) Both true, R explains A.

R uses difference of squares to calculate product efficiently

Section C: True/False Questions (10 Questions)

\((p + q)^2 = p^2 + q^2\)

False

Correct identity: \((p+q)^2 = p^2 + 2pq + q^2\)

\((a - b)^2 = (b - a)^2\)

True

Squaring removes the sign difference

\(a(b + c) = ab + ac\) for all real numbers

True

This is the distributive property

\(11 \times 12 = 132\) can be found using \(10 \times 12 + 1 \times 12\)

True

\(11 \times 12 = (10+1) \times 12 = 10 \times 12 + 1 \times 12 = 120 + 12 = 132\)

\((x + 3)(x - 3) = x^2 - 9\)

True

Difference of squares: \((x+3)(x-3) = x^2 - 3^2 = x^2 - 9\)

\(2(a^2 + b^2) = (a+b)^2 + (a-b)^2\)

True

Adding the two square identities gives this result

The product \(23 \times 27\) increases by 50 if both numbers increase by 1

False

Increase is \(a + b + 1 = 23 + 27 + 1 = 51\), not 50

\(101^2 = 10201\)

True

\((100+1)^2 = 10000 + 200 + 1 = 10201\)

\(k^2 + 2k = (k+1)^2 - 1\)

True

\((k+1)^2 - 1 = k^2 + 2k + 1 - 1 = k^2 + 2k\)

10.

\((a+b)(a-b) = a^2 + b^2\)

False

Should be \(a^2 - b^2\)

Section D: Short Answer I (2 Marks × 15 Questions)

Expand: \((3x + 2)(x + 5)\)

\(3x^2 + 17x + 10\)

\((3x+2)(x+5) = 3x(x+5) + 2(x+5) = 3x^2 + 15x + 2x + 10 = 3x^2 + 17x + 10\)

Find \(96^2\) using \((a - b)^2\) identity.

9216

\(96^2 = (100-4)^2 = 100^2 - 2×100×4 + 4^2 = 10000 - 800 + 16 = 9216\)

Simplify: \(4a(3b - 2) + 5a\)

\(12ab - 3a\)

\(4a(3b-2) + 5a = 12ab - 8a + 5a = 12ab - 3a\)

How much does \(23 \times 27\) increase if 23 is increased by 1?

When first number increases by 1, product increases by second number

Verify \(2(a^2 + b^2) = (a+b)^2 + (a-b)^2\) for \(a=5, b=3\).

LHS: \(2(25+9)=68\), RHS: \((8)^2 + (2)^2 = 64+4=68\) ✓

Both sides equal 68, identity verified

Multiply \(48 \times 52\) using a suitable identity.

2496

\(48×52 = (50-2)(50+2) = 50^2 - 2^2 = 2500 - 4 = 2496\)

Expand \((2m + 3n)^2\).

\(4m^2 + 12mn + 9n^2\)

\((2m+3n)^2 = (2m)^2 + 2×2m×3n + (3n)^2 = 4m^2 + 12mn + 9n^2\)

Find \(999^2\) using \((1000 - 1)^2\).

998001

\(999^2 = (1000-1)^2 = 1000000 - 2000 + 1 = 998001\)

Simplify: \((x+4)^2 - (x-4)^2\)

\(16x\)

\((x^2+8x+16) - (x^2-8x+16) = 16x\)

10.

Expand: \((a + b - c)(a + b + c)\)

\(a^2 + 2ab + b^2 - c^2\)

Consider \((a+b)\) as single term: \((a+b)^2 - c^2 = a^2+2ab+b^2-c^2\)

11.

Find the product: \(234 \times 11\) using distributive method.

2574

\(234×11 = 234×(10+1) = 2340 + 234 = 2574\)

12.

If \(a = 7, b = 3\), find \((a+b)^2 - (a-b)^2\).

\((10)^2 - (4)^2 = 100 - 16 = 84\) or using identity: \(4ab = 4×7×3 = 84\)

13.

Write \(k^2 + 2k\) in two other equivalent forms.

\((k+1)^2 - 1\) and \(k(k+2)\)

Both simplify to \(k^2 + 2k\)

14.

Expand: \((p - 8)(p + 8)\)

\(p^2 - 64\)

Difference of squares: \(p^2 - 8^2 = p^2 - 64\)

15.

Find \(73^2\) using \((70 + 3)^2\).

5329

\(73^2 = (70+3)^2 = 70^2 + 2×70×3 + 3^2 = 4900 + 420 + 9 = 5329\)

Section E: Short Answer II (3 Marks × 10 Questions)

Expand \((a + b)(a^2 + 2ab + b^2)\) and simplify.

\(a^3 + 3a^2b + 3ab^2 + b^3\)

\((a+b)(a^2+2ab+b^2) = a(a^2+2ab+b^2) + b(a^2+2ab+b^2) = a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3 = a^3 + 3a^2b + 3ab^2 + b^3\)

Multiply \(3874 \times 101\) using distributive property in one line.

391374

\(3874 \times 101 = 3874 \times (100+1) = 387400 + 3874 = 391374\)

Show geometrically that \((a+b)^2 = a^2 + 2ab + b^2\) (describe with diagram).

Draw square of side (a+b). It consists of: one square of side a (area a²), one square of side b (area b²), and two rectangles of sides a and b (each area ab). Total area = a² + b² + 2ab = (a+b)².

Geometric proof using area decomposition

If \(x = 8, y = 3\), find the area of the shaded region from page 95.

Area = \((n-m)^2\) where n=8, m=3 gives \((8-3)^2 = 5^2 = 25\)

Prove: \((m+n)^2 - 4mn = (n-m)^2\).

LHS: \(m^2+2mn+n^2 - 4mn = m^2 - 2mn + n^2 = (n-m)^2\)

Expanding and simplifying gives the identity

Find three examples where product decreases when one number is increased by 1 and the other decreased by 1.

1. (2,8)→(3,7): 16→21 (actually increases)
2. (5,5)→(6,4): 25→24 (decreases)
3. (10,1)→(11,0): 10→0 (decreases)

Product decreases when \(b < a+1\)

Verify Pattern 1: \(2(5^2 + 6^2) = (11)^2 + (1)^2\).

LHS: \(2(25+36)=122\), RHS: \(121+1=122\) ✓

Using identity \(2(a^2+b^2) = (a+b)^2 + (a-b)^2\) with a=6,b=5

Expand \((a - b)(a^3 + a^2b + ab^2 + b^3)\) and find the pattern.

\(a^4 - b^4\)

Pattern: \((a-b)(a^n + a^{n-1}b + ... + b^n) = a^{n+1} - b^{n+1}\)

Correct the mistake: \((5m + 6n)^2 = 25m^2 + 36n^2\).

Correct: \(25m^2 + 60mn + 36n^2\)

Missing middle term \(2×5m×6n = 60mn\)

10.

Use Identity 1C to find \(45 \times 55\).

2475

\(45×55 = (50-5)(50+5) = 50^2 - 5^2 = 2500 - 25 = 2475\)

Section F: Long Answer Questions (5 Marks × 10 Questions)

Derive all three identities: \((a+b)^2, (a-b)^2, (a+b)(a-b)\) using distributive property.

1. \((a+b)^2 = (a+b)(a+b) = a(a+b) + b(a+b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2\)
2. \((a-b)^2 = (a-b)(a-b) = a(a-b) - b(a-b) = a^2 - ab - ba + b^2 = a^2 - 2ab + b^2\)
3. \((a+b)(a-b) = a(a-b) + b(a-b) = a^2 - ab + ba - b^2 = a^2 - b^2\)

All derived using distributive property \(a(b+c) = ab + ac\)

A park has two square green plots each of side \(g\) and a walking path of width \(w\) around them. Write an expression for the tiled area.

Tiled area = \((2g+2w)^2 - 2g^2 = 4g^2 + 8gw + 4w^2 - 2g^2 = 2g^2 + 8gw + 4w^2\)

Total area minus green area gives tiled area

For the pattern in "This Way or That Way" (p. 96), show that all four expressions simplify to \(k^2 + 2k\).

1. \((k+1)^2 - 1 = k^2 + 2k + 1 - 1 = k^2 + 2k\)
2. \(k^2 + 2k\) (given)
3. \(k(k+1) + k = k^2 + k + k = k^2 + 2k\)
4. \(k(k+2) = k^2 + 2k\)

All four methods give same algebraic expression

Explain the method for fast multiplication by 11 and 101 with two examples each.

By 11: Multiply by (10+1): \(123×11 = 1230+123 = 1353\), \(456×11 = 4560+456 = 5016\)
By 101: Multiply by (100+1): \(123×101 = 12300+123 = 12423\), \(456×101 = 45600+456 = 46056\)

Use distributive property: \(a×11 = a×(10+1)\), \(a×101 = a×(100+1)\)

From "Mind the Mistake," identify and correct any 5 errors with explanations.

1. \(-3p(-5p+2q) = 15p^2 - 6pq\) (not \(-3p+5p-2q\))
2. \(2(x-1)+3(x+4) = 2x-2+3x+12 = 5x+10\) (not \(5x+3\))
3. \(y+2(y+2) = y+2y+4 = 3y+4\) (not \((y+2)^2\))
4. \((5m+6n)^2 = 25m^2 + 60mn + 36n^2\) (not \(25m^2+36n^2\))
5. \(5w^2 + 6w\) cannot combine (not \(11w^2\))

Common errors: forgetting to multiply all terms, incorrect expansion of squares, combining unlike terms

Prove algebraically that the diagonal products in a 2×2 calendar square differ by 7.

Numbers: \(a, a+1, a+7, a+8\)
Diagonal products: \(a(a+8) = a^2+8a\) and \((a+1)(a+7) = a^2+8a+7\)
Difference: \((a^2+8a+7) - (a^2+8a) = 7\)

The difference is always 7 regardless of \(a\)

If a number leaves remainder 3 when divided by 7 and another leaves remainder 5, find remainders for their sum, difference, and product when divided by 7.

Let \(A = 7x+3\), \(B = 7y+5\)
Sum: \(A+B = 7(x+y) + 8 = 7(x+y+1) + 1\) → remainder 1
Difference: \(A-B = 7(x-y) - 2 = 7(x-y-1) + 5\) → remainder 5
Product: \(AB = (7x+3)(7y+5) = 49xy + 35x + 21y + 15 = 7(7xy+5x+3y+2) + 1\) → remainder 1

Using algebraic representation of numbers with remainders

Show that \((6n+2)^2 - (4n+3)^2\) is 5 less than a perfect square.

\((6n+2)^2 - (4n+3)^2 = (10n+5)(2n-1) = 20n^2 - 5\)
Add 5: \(20n^2\) which is a perfect square when \(n\) is an integer

Using difference of squares and simplification

Find \(406^2, 72^2, 145^2\) using suitable identities.

\(406^2 = (400+6)^2 = 160000 + 4800 + 36 = 164836\)
\(72^2 = (70+2)^2 = 4900 + 280 + 4 = 5184\)
\(145^2 = (150-5)^2 = 22500 - 1500 + 25 = 21025\)

Decompose numbers into easy-to-square parts

10.

Explore the coin triangle flipping problem: Find minimum moves for 15 coins and generalize for triangular number \(T_n\).

For triangle with \(T_n = n(n+1)/2\) coins:
- 3 coins (n=2): 1 move
- 6 coins (n=3): 2 moves
- 10 coins (n=4): 3 moves
- 15 coins (n=5): 5 moves
Pattern: Minimum moves = ⌊n/2⌋ for n>1

Triangular number pattern with practical problem-solving

Section G: Case-Based Questions (5 Cases × 4 Sub-Questions Each)

Case 1: Fast Multiplication Tricks

Rahul learns that the distributive property can be used to multiply numbers quickly. He sees the example: \( 3874 \times 11 = 3874 \times (10 + 1) = 38740 + 3874 = 42614 \). He also learns that for a 4-digit number \( dcba \): \( dcba \times 101 = dcba \times (100 + 1) = dcba00 + dcba \).

Which property is Rahul using here?

a) Commutative property

b) Distributive property

c) Associative property

d) Identity property

b) Distributive property

Using \(a(b+c) = ab + ac\)

What is \(495 \times 11\) using this method?

a) 5445

b) 5440

c) 4945

d) 4950

a) 5445

\(495 \times 11 = 495 \times (10+1) = 4950 + 495 = 5445\)

Using the rule for multiplying by 101, what is \(3874 \times 101\)?

a) 391374

b) 387400

c) 391474

d) 387474

a) 391374

\(3874 \times 101 = 3874 \times (100+1) = 387400 + 3874 = 391374\)

Which is NOT true about multiplying by 11?

a) You add the number to itself shifted left by one digit

b) It works for any number of digits

c) It only works for 2-digit numbers

d) It uses \( \times (10 + 1) \)

c) It only works for 2-digit numbers

The method works for numbers with any number of digits

Case 2: Calendar Number Patterns

In a calendar, Priya marks a 2×2 square of numbers. She labels the top-left number as \( a \), so the square becomes:

| \( a \) | \( a+1 \) |
| \( a+7 \) | \( a+8 \) |

What is the difference between the diagonal products?

a) 1

b) 7

c) 8

d) 0

b) 7

\(a(a+8) - (a+1)(a+7) = a^2+8a - (a^2+8a+7) = -7\), absolute difference = 7

For \( a = 9 \), what are the diagonal products?

a) 153 and 160

b) 144 and 153

c) 135 and 144

d) 126 and 135

a) 153 and 160

\(9×17=153\), \(10×16=160\)

The algebraic expression for the difference of diagonal products is:

a) \((a+8)(a) - (a+1)(a+7)\)

b) \((a+1)(a+7) - a(a+8)\)

c) \(a(a+8) - (a+1)(a+7)\)

d) \((a+7)(a+8) - a(a+1)\)

c) \(a(a+8) - (a+1)(a+7)\)

This represents product of top-left & bottom-right minus product of top-right & bottom-left

When simplified, this difference always equals:

a) 5

b) 7

c) 10

d) 14

b) 7

\(a(a+8) - (a+1)(a+7) = a^2+8a - (a^2+8a+7) = -7\), so absolute difference = 7

Case 3: Geometric Proof of Identities

A square of side length \( (m+n) \) is drawn. Four rectangles of dimensions \( m \times n \) are removed from the corners, leaving a smaller shaded square in the center.

What is the area of the large square?

a) \( m^2 + n^2 \)

b) \( (m+n)^2 \)

c) \( m^2 + 2mn + n^2 \)

d) Both b and c

\((m+n)^2 = m^2+2mn+n^2\)

What is the total area of the four removed rectangles?

a) \( mn \)

b) \( 2mn \)

c) \( 4mn \)

d) \( m^2n^2 \)

c) \( 4mn \)

Four rectangles each of area \( mn \)

The area of the shaded square is:

a) \( (m-n)^2 \)

b) \( (n-m)^2 \)

c) \( m^2 + n^2 - 2mn \)

d) All of these

All are equivalent expressions

If \( m = 3 \) and \( n = 7 \), what is the area of the shaded square?

a) 16

b) 25

c) 36

d) 49

a) 16

\((7-3)^2 = 4^2 = 16\)

Case 4: Algebraic Pattern Recognition

Anika observes this pattern in her notebook:
\( 2(2^2 + 1^2) = 3^2 + 1^2 \)
\( 2(3^2 + 1^2) = 4^2 + 2^2 \)
\( 2(5^2 + 3^2) = 8^2 + 2^2 \)
She realizes this follows the identity: \( 2(a^2 + b^2) = (a+b)^2 + (a-b)^2 \)

For \( a = 6, b = 4 \), what is \( 2(a^2 + b^2) \)?

a) 100

b) 104

c) 124

d) 144

b) 104

\(2(36+16) = 2×52 = 104\)

Using the identity, \( (a+b)^2 + (a-b)^2 \) for \( a=6, b=4 \) is:

a) 100 + 4

b) 104 + 4

c) 100 + 4

d) 104

\((10)^2 + (2)^2 = 100 + 4 = 104\)

Does this identity work for negative numbers? For \( a = -3, b = 5 \):

a) Yes, both sides equal 68

b) Yes, both sides equal 34

c) No, it only works for positive numbers

d) No, squares of negatives are positive

a) Yes, both sides equal 68

LHS: \(2(9+25)=68\), RHS: \((2)^2+(-8)^2=4+64=68\)

Another identity in the same chapter is:

a) \( (a+b)^3 = a^3 + b^3 \)

b) \( a^2 - b^2 = (a+b)(a-b) \)

c) \( a(b+c) = ab + c \)

d) \( (a-b)^2 = a^2 + b^2 \)

b) \( a^2 - b^2 = (a+b)(a-b) \)

This is Identity 1C (difference of squares)

Case 5: Error Analysis in Algebra

Mr. Sharma gives his class these expansions to check:
1. \( (5m + 6n)^2 = 25m^2 + 36n^2 \)
2. \( ab^2 + a^2b + a^2b^2 = ab(a + b + ab) \)
3. \( -3p(-5p + 2q) = -3p + 5p - 2q \)

What is the correct expansion of \( (5m + 6n)^2 \)?

a) \( 25m^2 + 36n^2 \)

b) \( 25m^2 + 60mn + 36n^2 \)

c) \( 5m^2 + 30mn + 6n^2 \)

d) \( 25m^2 + 30mn + 36n^2 \)

b) \( 25m^2 + 60mn + 36n^2 \)

Missing middle term \(2×5m×6n = 60mn\)

The correct simplification of \( ab^2 + a^2b + a^2b^2 \) is:

a) \( ab(a + b + ab) \)

b) \( ab(b + a + ab) \)

c) \( ab(ab + a + b) \)

d) Cannot be factored further

b) \( ab(b + a + ab) \)

Factor \(ab\) from each term: \(ab(b + a + ab)\)

The error in \( -3p(-5p + 2q) \) is:

a) Sign error

b) Forgot to multiply

c) Added instead of multiplied

d) All of these

Multiple errors: sign, multiplication, addition

Which mistake is most common when expanding \( (a+b)^2 \)?

a) Writing \( a^2 + b^2 \)

b) Writing \( a^2 + 2ab - b^2 \)

c) Writing \( a^2 + ab + b^2 \)

d) Writing \( 2a + 2b \)

a) Writing \( a^2 + b^2 \)

Most common error is forgetting the middle term \(2ab\)

🌈⚡Key To Enjoy Learning Maths☘🌈

Sunday, December 7, 2025