Official NCERT 2026-27 Solutions
NCERT Solutions for Class 8 Maths Chapter 6: We Distribute, Yet Things Multiply (Ganita Prakash)
Authentic handwritten solutions in Topper Notebook style. Covers algebraic identities, distributive property proofs, calendar matrix invariants, geometric area tiling expressions, and sequence modeling for all exercises from Pages 31 to 35 of Ganita Prakash Part I.
Quick Chapter Navigation & Blueprint
- Part A: Algebraic Translations & Calendar Matrices (Questions 3 & 4) — Translating verbal statements into algebraic forms and proving the 7-difference calendar diagonal invariant.
- Part B: Rigorous Identity Verification & Modulo Arithmetic (Questions 5 & 6) — True/false algebraic analysis, odd/even square divisibility, and remainder arithmetic modulo 7.
- Part C: Difference of Squares & Mental Multiplication (Questions 7 to 9) — Proving n2 − (n − 1)(n + 1) = 1, expanding half-sums, and comparing product magnitudes via identities.
- Part D: Practical Geometry & Sequence Modeling (Questions 10 & 11) — Park tiling area expression 8w(g + w) and visual sequence formulas (y + 2)2 and (y + 1)2 + y.
Official Textbook PDF Reference: Download NCERT Class 8 Maths Chapter 6 (hegp106.pdf)
Part A: Algebraic Translations & Calendar Matrices (Pages 31–32)
Question 3 (Page 31):
For each statement, identify the appropriate algebraic expression(s):
(i) Two more than a square number.
Options: 2 + s, (s + 2)2, s2 + 2, s2 + 4, 2s2, 22s.
(ii) The sum of the squares of two consecutive numbers.
Options: m2 + n2, (m + n)2, m2 + 1, m2 + (m + 1)2, m2 + (m − 1)2.
For each statement, identify the appropriate algebraic expression(s):
(i) Two more than a square number.
Options: 2 + s, (s + 2)2, s2 + 2, s2 + 4, 2s2, 22s.
(ii) The sum of the squares of two consecutive numbers.
Options: m2 + n2, (m + n)2, m2 + 1, m2 + (m + 1)2, m2 + (m − 1)2.
Topper’s Translation Breakdown:
- (i) Two more than a square number:
• Let the base number be s. The square of this number is s2.
• “Two more than” indicates adding 2: s2 + 2. - (ii) The sum of the squares of two consecutive numbers:
• Let the first number be m. The next consecutive integer is (m + 1) [or the preceding integer is (m − 1)].
• Squares of the two numbers are m2 and (m + 1)2.
• Their sum is: m2 + (m + 1)2 [or m2 + (m − 1)2].
Correct Algebraic Expressions:
(i) s2 + 2
(ii) m2 + (m + 1)2 and m2 + (m − 1)2
(i) s2 + 2
(ii) m2 + (m + 1)2 and m2 + (m − 1)2
Question 4 (Page 31–32 – Calendar Matrix Invariant):
Consider any 2 by 2 square of numbers in a calendar (e.g., [3, 4] and [10, 11]).
Find the products of the numbers along each diagonal (e.g., 3 × 11 = 33, 4 × 10 = 40). What do you observe about the difference between diagonal products? Explain algebraically why this happens.
Consider any 2 by 2 square of numbers in a calendar (e.g., [3, 4] and [10, 11]).
Find the products of the numbers along each diagonal (e.g., 3 × 11 = 33, 4 × 10 = 40). What do you observe about the difference between diagonal products? Explain algebraically why this happens.
Algebraic Proof of the Invariant Difference:
- In any monthly calendar, the days in each week increase by 1 horizontally, and by 7 vertically (since there are 7 days in a week).
- Let the top-left number of any 2 × 2 square be a:
• Top-right number = a + 1
• Bottom-left number = a + 7
• Bottom-right number = (a + 7) + 1 = a + 8 - Now, compute the product along each diagonal:
• Main diagonal (Top-Left × Bottom-Right):
P1 = a(a + 8) = a2 + 8a.
• Off diagonal (Bottom-Left × Top-Right):
P2 = (a + 7)(a + 1) = a2 + 8a + 7. - Subtracting P1 from P2:
Difference = P2 − P1 = (a2 + 8a + 7) − (a2 + 8a) = 7! - Notice that the variable a cancels out completely! This proves the difference is an absolute invariant.
Conclusion: The difference between the diagonal products in ANY 2 × 2 calendar block is always exactly 7, regardless of the month, year, or starting number!
Part B: Identity Proofs & Remainder Modulo Arithmetic (Pages 32–33)
Question 5 (Page 32):
Verify which of the following statements are True or False:
(i) (k + 1)(k + 2) − (k + 3) is always 2.
(ii) (2q + 1)(2q − 3) is a multiple of 4.
(iii) Squares of even numbers are multiples of 4, and squares of odd numbers are 1 more than multiples of 8.
(iv) (6n + 2)2 − (4n + 3)2 is 5 less than a square number.
Verify which of the following statements are True or False:
(i) (k + 1)(k + 2) − (k + 3) is always 2.
(ii) (2q + 1)(2q − 3) is a multiple of 4.
(iii) Squares of even numbers are multiples of 4, and squares of odd numbers are 1 more than multiples of 8.
(iv) (6n + 2)2 − (4n + 3)2 is 5 less than a square number.
Topper’s Rigorous Verification:
- (i) False: Expanding: (k + 1)(k + 2) − (k + 3) = k2 + 3k + 2 − k − 3 = k2 + 2k − 1. This depends directly on k and is only 2 if k = 1.
- (ii) False: Expanding: (2q + 1)(2q − 3) = 4q2 − 6q + 2q − 3 = 4q2 − 4q − 3 = 4(q2 − q − 1) + 1. This is always an odd number (leaves remainder 1 mod 4), so it is NEVER a multiple of 4.
- (iii) True:
• Square of even number: (2m)2 = 4m2 → always a multiple of 4.
• Square of odd number: (2m + 1)2 = 4m2 + 4m + 1 = 4m(m + 1) + 1. Since m(m + 1) is the product of two consecutive integers, it is always even, so m(m + 1) = 2k.
Therefore, 4(2k) + 1 = 8k + 1 → exactly 1 more than a multiple of 8! - (iv) False: Expanding: (36n2 + 24n + 4) − (16n2 + 24n + 9) = 20n2 − 5 = 5(4n2 − 1). This is not generally of the form S2 − 5.
Answers: (i) False | (ii) False | (iii) True | (iv) False.
Question 6 (Page 32 – Remainder Modulo 7):
A number leaves a remainder of 3 when divided by 7, and another leaves a remainder of 5 when divided by 7. What is the remainder when their (i) sum, (ii) difference, and (iii) product are divided by 7?
A number leaves a remainder of 3 when divided by 7, and another leaves a remainder of 5 when divided by 7. What is the remainder when their (i) sum, (ii) difference, and (iii) product are divided by 7?
Topper’s Modulo Arithmetic Steps:
- Let the two numbers be:
n1 = 7a + 3
n2 = 7b + 5 - (i) Sum:
n1 + n2 = (7a + 3) + (7b + 5) = 7(a + b) + 8 = 7(a + b + 1) + 1.
Remainder = 1. - (ii) Difference:
n2 − n1 = (7b + 5) − (7a + 3) = 7(b − a) + 2.
Remainder = 2. - (iii) Product:
n1 × n2 = (7a + 3)(7b + 5) = 49ab + 35a + 21b + 15
= 7(7ab + 5a + 3b + 2) + 1.
Remainder = 1.
Remainders when divided by 7:
(i) Sum remainder = 1
(ii) Difference remainder = 2
(iii) Product remainder = 1
(i) Sum remainder = 1
(ii) Difference remainder = 2
(iii) Product remainder = 1
Part C: Mental Math & Practical Geometry (Pages 32–34)
Question 7 (Page 32):
Choose three consecutive numbers, square the middle one, and subtract the product of the other two. Repeat with other sets. What pattern do you notice? Write this as an algebraic identity and prove it.
Choose three consecutive numbers, square the middle one, and subtract the product of the other two. Repeat with other sets. What pattern do you notice? Write this as an algebraic identity and prove it.
Algebraic Proof of the Consecutive Square Difference:
- Let the three consecutive integers be (n − 1), n, and (n + 1).
- The middle number is n; its square is n2.
- The product of the outer two numbers is (n − 1)(n + 1).
- Using the Difference of Squares identity (a − b)(a + b) = a2 − b2:
(n − 1)(n + 1) = n2 − 12 = n2 − 1. - Subtracting this product from the square of the middle number:
Result = n2 − (n2 − 1) = n2 − n2 + 1 = 1! - Examples:
• For 4, 5, 6: 52 − (4 × 6) = 25 − 24 = 1.
• For 9, 10, 11: 102 − (9 × 11) = 100 − 99 = 1.
• For 99, 100, 101: 1002 − (99 × 101) = 10000 − 9999 = 1.
Identity: n2 − (n − 1)(n + 1) = 1. The result is always 1.
Question 8 (Page 32):
What is the algebraic expression describing: “Add any two numbers. Multiply this by half of the sum of the two numbers”? Prove that this result will be half of the square of the sum.
What is the algebraic expression describing: “Add any two numbers. Multiply this by half of the sum of the two numbers”? Prove that this result will be half of the square of the sum.
Topper’s Algebraic Derivation:
- Let the two arbitrary numbers be a and b.
- Step 1 (Sum): a + b.
- Step 2 (Half of sum): (1/2)(a + b).
- Step 3 (Multiply): (a + b) × [(1/2)(a + b)].
- Using associativity and exponent laws:
= (1/2) × (a + b)(a + b)
= (1/2)(a + b)2.
Algebraic Expression: ½(a + b)2, which by definition is exactly half the square of their sum!
Question 9 (Page 33):
Which is larger? Find out without fully computing the product:
(i) 14 × 26 or 16 × 24
(ii) 25 × 75 or 26 × 74
Which is larger? Find out without fully computing the product:
(i) 14 × 26 or 16 × 24
(ii) 25 × 75 or 26 × 74
Comparison via Algebraic Distance from Mean:
- (i) Comparing 14 × 26 and 16 × 24:
• Notice that the sum in both pairs is 40: 14 + 26 = 40, and 16 + 24 = 40.
• Average = 20.
• 14 × 26 = (20 − 6)(20 + 6) = 202 − 62 = 400 − 36 = 364.
• 16 × 24 = (20 − 4)(20 + 4) = 202 − 42 = 400 − 16 = 384.
• Since 400 − 16 > 400 − 36: 16 × 24 > 14 × 26. - (ii) Comparing 25 × 75 and 26 × 74:
• Both pairs sum to 100. Average = 50.
• 25 × 75 = (50 − 25)(50 + 25) = 502 − 252 = 2500 − 625.
• 26 × 74 = (50 − 24)(50 + 24) = 502 − 242 = 2500 − 576.
• Since 242 < 252, less is subtracted in the second product: 26 × 74 > 25 × 75.
Answers:
(i) 16 × 24 is larger (by 20).
(ii) 26 × 74 is larger (by 49).
(i) 16 × 24 is larger (by 20).
(ii) 26 × 74 is larger (by 49).
Question 10 (Page 33 – Dhauli Park Tiling Area):
A rectangular park has two square plots of area g2 sq. ft. each for green cover. The rest of the area is a walking path of uniform width w ft. Write an algebraic expression for the area that needs to be tiled.
A rectangular park has two square plots of area g2 sq. ft. each for green cover. The rest of the area is a walking path of uniform width w ft. Write an algebraic expression for the area that needs to be tiled.
Geometric Dimension Expansion:
- The side length of each green square plot is g ft.
- Total Length of Park:
Length = w (outer border) + g (first plot) + 2w (middle path) + g (second plot) + w (outer border)
= 2g + 4w ft. - Total Breadth of Park:
Breadth = w + g + w = g + 2w ft. - Total Park Area:
Area = (2g + 4w)(g + 2w)
= 2(g + 2w)(g + 2w) = 2(g + 2w)2
= 2(g2 + 4gw + 4w2) = 2g2 + 8gw + 8w2 sq. ft. - Area to be Tiled:
Tiled Area = Total Area − Area of the 2 Green Squares
= (2g2 + 8gw + 8w2) − 2g2
= 8gw + 8w2 = 8w(g + w) sq. ft.
Expression for Tiled Area: 8w(g + w) sq. ft. (or 8gw + 8w2 sq. ft.).
Question 11 (Pages 33–35 – Pattern Sequences):
For each pattern shown below, find: (i) the next figure, (ii) basic units in Step 10, (iii) expression for Step y.
(a) Sequence of squares growing around a core.
(b) Sequence of square grids with tail units.
For each pattern shown below, find: (i) the next figure, (ii) basic units in Step 10, (iii) expression for Step y.
(a) Sequence of squares growing around a core.
(b) Sequence of square grids with tail units.
Mathematical Sequence Modeling:
- For Pattern (a):
• Step 1: Side = 3 → Units = 32 = (1 + 2)2 = 9
• Step 2: Side = 4 → Units = 42 = (2 + 2)2 = 16
• Step 3: Side = 5 → Units = 52 = (3 + 2)2 = 25
• Next figure (Step 4): Side = 6 → Units = 62 = 36
• General Formula for Step y: (y + 2)2
• Step 10 Units: (10 + 2)2 = 122 = 144 - For Pattern (b):
• Step 1: 22 + 1 = 5
• Step 2: 32 + 2 = 11
• Step 3: 42 + 3 = 19
• Next figure (Step 4): 52 + 4 = 29
• General Formula for Step y: (y + 1)2 + y
• Step 10 Units: (10 + 1)2 + 10 = 112 + 10 = 121 + 10 = 131
Summary:
Pattern (a): Next = 36 | Step 10 = 144 | Formula = (y + 2)2
Pattern (b): Next = 29 | Step 10 = 131 | Formula = (y + 1)2 + y
Pattern (a): Next = 36 | Step 10 = 144 | Formula = (y + 2)2
Pattern (b): Next = 29 | Step 10 = 131 | Formula = (y + 1)2 + y