Chapter 6: We Distribute, Yet Things Multiply
Class 8 Mathematics – NCERT Ganita Prakash
Complete Interactive Question Bank
Section A: Multiple Choice Questions (20 Questions)
1.
What is the expanded form of \((x + 4)(x + 3)\)?
a) \(x^2 + 7x + 12\)
Using distributive property: \((x+4)(x+3) = x(x+3) + 4(x+3) = x^2 + 3x + 4x + 12 = x^2 + 7x + 12\)
2.
The expression \((a - 7)^2\) equals:
b) \(a^2 - 14a + 49\)
Using identity: \((a-b)^2 = a^2 - 2ab + b^2\). Here \(b=7\), so \(a^2 - 2×a×7 + 49 = a^2 - 14a + 49\)
3.
Which identity is used to find \(98 \times 102\) quickly?
c) \((a + b)(a - b)\)
\(98 \times 102 = (100-2)(100+2) = 100^2 - 2^2\). This uses difference of squares identity.
4.
The product \(45 \times 55\) can be written as:
a) \((50 - 5)(50 + 5)\)
\(45 = 50-5\) and \(55 = 50+5\), so \(45×55 = (50-5)(50+5)\)
5.
If \(a\) and \(b\) are integers, \((a - b)^2\) is always equal to:
a) \((b - a)^2\)
Squaring removes the sign: \((a-b)^2 = [-(b-a)]^2 = (b-a)^2\)
6.
The value of \(101^2\) using the identity is:
a) 10201
\(101^2 = (100+1)^2 = 100^2 + 2×100×1 + 1^2 = 10000 + 200 + 1 = 10201\)
7.
The expression \(3p(2q - 5)\) expands to:
a) \(6pq - 15p\)
\(3p(2q-5) = 3p×2q - 3p×5 = 6pq - 15p\)
8.
If the product of two numbers is \(ab\), and both are increased by 1, the new product is:
b) \(ab + a + b + 1\)
\((a+1)(b+1) = ab + a + b + 1\)
9.
The product \((x + 2)(x - 2)\) simplifies to:
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?
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
If second number increases by 1, product increases by first number
12.
The expression \((2x + 5)^2\) equals:
c) \(4x^2 + 20x + 25\)
\((2x+5)^2 = (2x)^2 + 2×2x×5 + 5^2 = 4x^2 + 20x + 25\)
[MCQs 13-20 continue in similar pattern...]
Section B: Assertion & Reasoning Questions (20 Questions)
1.
Assertion (A): \((a + b)^2 = a^2 + 2ab + b^2\) for all integers \(a, b\).
Reason (R): The distributive property holds for integers.
Reason (R): The distributive property holds for integers.
a) Both A and R are true and R explains A.
The distributive property is used to derive \((a+b)^2 = a^2 + 2ab + b^2\)
2.
A: \((a - b)^2\) and \((b - a)^2\) are equal.
R: Squaring a negative gives a positive.
R: Squaring a negative gives a positive.
a) Both true, R explains A.
\((a-b)^2 = [-(b-a)]^2 = (b-a)^2\) because square of negative is positive
3.
A: \(99 \times 101 = 9999\).
R: \(99 \times 101 = (100-1)(100+1) = 100^2 - 1^2\).
R: \(99 \times 101 = (100-1)(100+1) = 100^2 - 1^2\).
a) Both true, R explains A.
R shows the method using difference of squares identity
[Assertion-Reasoning questions 4-20 continue...]
Section C: True/False Questions (10 Questions)
1.
\((p + q)^2 = p^2 + q^2\)
False
Correct identity: \((p+q)^2 = p^2 + 2pq + q^2\)
2.
\((a - b)^2 = (b - a)^2\)
True
Squaring removes the sign difference
3.
\(a(b + c) = ab + ac\) for all real numbers
True
This is the distributive property
4.
\(11 \times 12 = 132\) can be found using \(10 \times 12 + 1 \times 12\)
True
\(11 \times 12 = (10+1) \times 12 = 10 \times 12 + 1 \times 12 = 120 + 12 = 132\)
5.
\((x + 3)(x - 3) = x^2 - 9\)
True
Difference of squares: \((x+3)(x-3) = x^2 - 3^2 = x^2 - 9\)
6.
\(2(a^2 + b^2) = (a+b)^2 + (a-b)^2\)
True
Adding the two square identities gives this result
7.
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
[True/False questions 8-10 continue...]
Section D: Short Answer I (2 Marks × 15 Questions)
1.
Expand: \((3x + 2)(x + 5)\)
\(3x^2 + 17x + 10\)
\((3x+2)(x+5) = 3x(x+5) + 2(x+5) = 3x^2 + 15x + 2x + 10 = 3x^2 + 17x + 10\)
2.
Find \(96^2\) using \((a - b)^2\) identity.
9216
\(96^2 = (100-4)^2 = 100^2 - 2×100×4 + 4^2 = 10000 - 800 + 16 = 9216\)
3.
Simplify: \(4a(3b - 2) + 5a\)
\(12ab - 3a\)
\(4a(3b-2) + 5a = 12ab - 8a + 5a = 12ab - 3a\)
4.
How much does \(23 \times 27\) increase if 23 is increased by 1?
27
When first number increases by 1, product increases by second number
5.
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
[Short Answer I questions 6-15 continue...]
Section E: Short Answer II (3 Marks × 10 Questions)
1.
Expand \((a + b)(a^2 + 2ab + b^2)\) and simplify.
\(a^3 + 3a^2b + 3ab^2 + b^3\)
\((a+b)(a^2+2ab+b^2) = a(a^2+2ab+b^2) + b(a^2+2ab+b^2) = a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3 = a^3 + 3a^2b + 3ab^2 + b^3\)
2.
Multiply \(3874 \times 101\) using distributive property in one line.
391374
\(3874 \times 101 = 3874 \times (100+1) = 387400 + 3874 = 391374\)
3.
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
4.
If \(x = 8, y = 3\), find the area of the shaded region from page 95.
25
Area = \((n-m)^2\) where n=8, m=3 gives \((8-3)^2 = 5^2 = 25\)
[Short Answer II questions 5-10 continue...]
Section F: Long Answer Questions (5 Marks × 10 Questions)
1.
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. \((a+b)(a-b) = a(a-b) + b(a-b) = a^2 - ab + ba - b^2 = a^2 - b^2\)
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 \(a(b+c) = ab + ac\)
2.
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
3.
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. \(k(k+2) = k^2 + 2k\)
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
[Long Answer questions 4-10 continue...]
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 \( dcba \): \( dcba \times 101 = dcba \times (100 + 1) = dcba00 + dcba \).
1.
Which property is Rahul using here?
b) Distributive property
Using \(a(b+c) = ab + ac\)
2.
What is \(495 \times 11\) using this method?
a) 5445
\(495 \times 11 = 495 \times (10+1) = 4950 + 495 = 5445\)
3.
Using the rule for multiplying by 101, what is \(3874 \times 101\)?
a) 391374
\(3874 \times 101 = 3874 \times (100+1) = 387400 + 3874 = 391374\)
4.
Which is NOT true about multiplying by 11?
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:
| \( a \) | \( a+1 \) |
| \( a+7 \) | \( a+8 \) |
1.
What is the difference between the diagonal products?
b) 7
\(a(a+8) - (a+1)(a+7) = a^2+8a - (a^2+8a+7) = -7\), absolute difference = 7
[Case 2 sub-questions 2-4 continue...]
[Cases 3-5 with their sub-questions continue...]
