Class 8 Mathematics Chapter 6 Question Bank CBSE Board Pattern

Section A — MCQs (10 questions, 1 mark each)

  1. The expansion of \((a + 1)(b + 1)\) using the distributive property is
    (a) \(ab + a + b\)
    (b) \(ab + a + b + 1\)
    (c) \(ab + 1\)
    (d) \(a + b + 1\)

  2. If one number is increased by \(m\) and the other by \(n\), the increase in the product \(ab\) is
    (a) \(an + bm\)
    (b) \(an + bm + mn\)
    (c) \(mn\)
    (d) \(an + bm - mn\)

  3. Using the distributive property, \(3a^2(a - b + \frac{1}{5})\) expands to
    (a) \(3a^3 - 3a^2b + \frac{3}{5}a^2\)
    (b) \(3a^3 - 3a^2b + \frac{3}{5}a\)
    (c) \(3a^2 - 3ab + \frac{3}{5}a\)
    (d) \(3a^3 + 3a^2b + \frac{3}{5}a\)

  4. The identity \((a + b)^2 = a^2 + 2ab + b^2\) is obtained by
    (a) only commutativity
    (b) distributivity applied to \((a + b)(a + b)\)
    (c) only addition
    (d) subtraction of like terms

  5. The product \((a - b)(a + b)\) equals
    (a) \(a^2 + b^2\)
    (b) \(a^2 - b^2\)
    (c) \(a^2 - 2ab + b^2\)
    (d) \(a^2 + 2ab + b^2\)

Assertion-Reason Questions

  1. Assertion (A): \((a + 1)(b - 1) = ab + b - a - 1\).
    Reason (R): This follows from Identity 1 by taking \(m = 1\), \(n = -1\).
    (a) Both A and R are true and R is the correct explanation of A.
    (b) Both A and R are true but R is not the correct explanation of A.
    (c) A is true but R is false.
    (d) A is false but R is true.

  2. Assertion (A): The distributive property holds for all integers, including negative integers.
    Reason (R): Integers satisfy \(x(y + z) = xy + xz\) for any integers \(x, y, z\).
    (a) Both A and R are true and R is the correct explanation of A.
    (b) Both A and R are true but R is not the correct explanation of A.
    (c) A is true but R is false.
    (d) A is false but R is true.

  3. The square of the difference of two numbers is given by
    (a) \((a - b)^2 = a^2 + 2ab + b^2\)
    (b) \((a - b)^2 = a^2 - 2ab + b^2\)
    (c) \((a - b)^2 = a^2 - b^2\)
    (d) \((a - b)^2 = a^2 + b^2\)

  4. When multiplying a number by 11 using the distributive property, a 4-digit number \(dcba\) gives the product with digits formed by
    (a) simple addition without carry
    (b) successive addition with possible carry-over
    (c) only subtraction
    (d) multiplication of each digit by 11

  5. The pattern \(2(a^2 + b^2) = (a + b)^2 + (a - b)^2\) is derived from
    (a) adding the expansions of \((a + b)^2\) and \((a - b)^2\)
    (b) subtracting the expansions
    (c) only the distributive property on one term
    (d) commutativity alone

Section B — Very Short Answer (6 questions, 2 marks each)

  1. Expand \((3 + u)(v - 3)\) using the distributive property.

  2. Using Identity 1, find the change in the product when one number is decreased by 2 and the other is increased by 3.

  3. Expand \((a - b)(a + b)\) and state the identity obtained.

  4. Write the general rule (in one line) for multiplying any number by 11 using distributivity.

  5. Expand \((b - 6)^2\) using both the identity and the distributive property.

  6. Verify that \((a + 1)(b - 1) = ab + b - a - 1\) by substituting \(a = 23\), \(b = 27\).

Section C — Short Answer (5 questions, 3 marks each)

  1. Expand \((10a + b)(10c + d)\) and simplify.

  2. Find the value of \(104^2\) using Identity 1A after expressing 104 as a sum of convenient numbers.

  3. Expand \((a + ab - 3b^2)(4 + b)\) and collect like terms.

  4. Using Identity 1C, evaluate \(98 \times 102\) and \(45 \times 55\).

  5. Expand \((6x + 5)^2\) by both the identity and direct application of the distributive property; show they give the same result.

Section D — Long Answer (3 questions, 5 marks each)

  1. Prove the identity \((a + b)^2 = a^2 + 2ab + b^2\) using the distributive property. Also prove the related identity \((a - b)^2 = a^2 - 2ab + b^2\) by replacing \(b\) with \(-b\).

  2. A 4-digit number \(dcba\) is to be multiplied by 101. Using the distributive property, derive the one-line multiplication rule and apply it to compute \(3874 \times 101\). Extend the rule briefly for multiplication by 1001.

  3. Using the identities learnt, compute \(406^2\), \(72^2\) and \(145^2\). Show each step with the chosen identity and verify one of them by direct expansion.

Section E — Case/Source-Based (2 questions, 4 marks each)

Case 1: In a park layout, two square green plots each of area \(g^2\) sq. ft. are separated by a walking path of width \(w\) ft. The overall rectangular plot has length \(2g + 4w\) ft. and breadth \(g + 2w\) ft.

(i) Write the total area of the rectangular plot.
(ii) Write the area of the two green squares.
(iii) Find an expression for the area that needs to be tiled (the walking path).
(iv) Substitute \(g = 5\), \(w = 2\) and compute the tiled area.

Case 2: A shopkeeper uses the distributive property to multiply a 3-digit number by 11 quickly. He writes the number as \(cba\) and applies the rule \(cba \times 11 = c(c + b)(b + a)a\) (with carry if needed).

(i) Multiply 495 by 11 using this method and show the steps.
(ii) Multiply 3279 by 11 using the same method.
(iii) State the general one-line rule for any number of digits.
(iv) Verify one of the results by the distributive property: number \(\times (10 + 1)\).

Answer Key Attempt all questions first,
then tap to reveal

Section A

  1. (b) — correct expansion by distributivity gives the extra +1 term.
  2. (b) — Identity 1 directly yields \(an + bm + mn\).
  3. (b) — each term multiplied and like terms collected.
  4. (b) — distributivity on \((a + b)(a + b)\).
  5. (b) — standard Identity 1C.
  6. (a) — both true; R explains how the expression is obtained.
  7. (a) — R justifies why the identities hold for negative integers.
  8. (b) — Identity 1B.
  9. (b) — successive addition with carry-over.
  10. (a) — adding the two expansions cancels the \(\pm 2ab\) terms.

Section B

  1. \(3v - 9 + uv - 3u\) (1 mark for each pair of terms).
  2. Increase = \(3a - 2b - 6\) (Identity 1 with \(m = -2\), \(n = 3\)).
  3. \(a^2 - b^2\) (Identity 1C).
  4. Successive addition of adjacent digits from right to left with carry.
  5. Identity: \(b^2 - 12b + 36\); distributive yields the same.
  6. Both sides equal 598 (substitution verifies the identity).

Section C

  1. \(100ac + 10ad + 10bc + bd\) (correct distribution of each term).
  2. \(104^2 = (100 + 4)^2 = 10816\) (1 mark each for decomposition, identity, result).
  3. \(4a + 5ab + ab^2 - 12b^2 - 3b^3\) (all like terms collected).
  4. \(98 \times 102 = 9996\), \(45 \times 55 = 2475\) (each using \(a^2 - b^2\)).
  5. Both methods give \(36x^2 + 60x + 25\) (identity vs. four-term distribution).

Section D

  1. \((a + b)(a + b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2\) (full distribution + commutativity). Replacement of \(b\) by \(-b\) gives the second identity.
  2. Rule derived: \(dcba00 + dcba\) with overlapping addition; \(3874 \times 101 = 391274\). Extension to 1001 shown.
  3. \(406^2 = 164836\), \(72^2 = 5184\), \(145^2 = 21025\) (each identity applied correctly).

Section E

Case 1

(i) \((2g + 4w)(g + 2w)\)
(ii) \(2g^2\)
(iii) \((2g + 4w)(g + 2w) - 2g^2 = 8w(g + w)\)
(iv) 96 sq. ft.

Case 2

(i) 5445 (with carry steps)
(ii) 36069
(iii) Successive adjacent-digit addition with carry.
(iv) Verification matches.

All questions are answerable from the NCERT chapter text. Reviewed by GFIS faculty.