CBSE Class 8 Mathematics We Distribute Yet Things Multiply MCQs Set 11

Answer: (b) ab + ac

Explanation:
1. The multiplier a must multiply every term inside the bracket.
2. a × b = ab and a × c = ac.
3. So a(b + c) = ab + ac. Option (a) multiplies only the first term; (c) and (d) change the operations.

Teacher's Note:
1. Distributing means sharing the multiplier with each term.
2. A common slip is multiplying only the first term.

Answer: (c) a² + 2ab + b²

Explanation:
1. (a + b)² = (a + b)(a + b) = a² + ab + ba + b².
2. ab and ba are like terms, so together they make 2ab.
3. Hence (a + b)² = a² + 2ab + b².

Teacher's Note:
1. This is Identity 1A.
2. Never drop the middle term 2ab.

Answer: (a) a² − 2ab + b²

Explanation:
1. (a − b)(a − b) = a² − ab − ba + b².
2. −ab − ba = −2ab, while (−b) × (−b) = +b².
3. So (a − b)² = a² − 2ab + b².

Teacher's Note:
1. This is Identity 1B.
2. The middle term is negative but the last term stays positive.

Answer: (d) a² − b²

Explanation:
1. (a + b)(a − b) = a² − ab + ba − b².
2. −ab + ba = 0, so the middle terms disappear.
3. The product is a² − b².

Teacher's Note:
1. This is Identity 1C.
2. It turns a product into a difference of two squares.

Answer: (c) 7pq and −4qp

Explanation:
1. Like terms must have exactly the same letter-numbers with the same powers.
2. qp = pq, so 7pq and −4qp both contain pq.
3. So 7pq and −4qp are like terms (they add to 3pq). The other pairs differ in letters or powers.

Teacher's Note:
1. The order of letters in a product does not matter.
2. Powers must also match: x² and x are unlike.

Answer: (b) an identity

Explanation:
1. An equality that holds for all values of its letter-numbers is called an identity.
2. For example, a(b + 8) = ab + 8a is true whatever a and b are.
3. So the answer is an identity.

Teacher's Note:
1. An equation is true only for some values; an identity is true for all.
2. All the formulas of this chapter are identities.

Answer: (a) 528

Explanation:
1. 48 × 11 = 48 × (10 + 1) = 48 × 10 + 48 × 1.
2. = 480 + 48.
3. = 528.

Teacher's Note:
1. Multiplying by 10 just adds a zero.
2. Quick rule: write 4, then 4 + 8 = 12 (carry 1), then 8: 528.

Answer: (d) x² + 10x + 24

Explanation:
1. Multiply each term of the first bracket by each term of the second.
2. x² + 6x + 4x + 24.
3. Adding the like terms 6x + 4x = 10x gives x² + 10x + 24.

Teacher's Note:
1. The numbers add to give the middle coefficient and multiply to give the constant.
2. Check with x = 1: 5 × 7 = 35 and 1 + 10 + 24 = 35.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. 9² − 4² = 81 − 16 = 65 and 13 × 5 = 65, so A is true.
2. R is the identity a² − b² = (a + b)(a − b), which is true.
3. With a = 9, b = 4 it gives (9 + 4)(9 − 4) = 13 × 5, so R correctly explains A.

Teacher's Note:
1. A difference of two squares is a sum times a difference.
2. Here 13 = 9 + 4 and 5 = 9 − 4.

Answer: (d) A is false, but R is true.

Explanation:
1. (3 + 5)² = 8² = 64, but 3² + 5² = 9 + 25 = 34. So A is false.
2. R is Identity 1A, which is true.
3. R even shows the missing middle term 2 × 3 × 5 = 30, and 34 + 30 = 64. So A is false but R is true.

Teacher's Note:
1. The square of a sum is not the sum of the squares.
2. The difference is always the middle term 2ab.

Answer: (c) (i)-(q), (ii)-(s), (iii)-(p), (iv)-(r)

Explanation:
1. Brahmagupta gave the first explicit statement of the property: (i)-(q).
2. Sridharacharya (about 750 CE) gave the quick squaring rule: (ii)-(s). Bhaskaracharya wrote Lilavati in 1150 CE: (iii)-(p).
3. Āryabhaṭa used the law implicitly in algebra: (iv)-(r). So the answer is (c).

Teacher's Note:
1. Brahmagupta's statement is in the Brahmasphuṭasiddhānta, Verse 12.55.
2. Link each name with one key fact.

Answer: (b) 1600

Explanation:
1. Each side is 30 + 10 = 40 units long.
2. Area = 40² = 1600 square units.
3. Check with the parts: 30² + 2 × (30 × 10) + 10² = 900 + 600 + 100 = 1600. Answer: 1600.

Teacher's Note:
1. The picture shows (a + b)² = a² + 2ab + b².
2. There are two 30 × 10 rectangles, not one.

Answer: (b) associative property of multiplication

Explanation:
1. The three factors stay in the same order.
2. Only the brackets move, that is, the grouping changes. There is no addition at all.
3. Changing the grouping uses the associative property of multiplication.

Teacher's Note:
1. Commutative changes the order; associative changes the grouping.
2. Distributive needs both × and +.

Answer: (c) a³ + 3a²b + 3ab² + b³

Explanation:
1. a(a² + 2ab + b²) = a³ + 2a²b + ab² and b(a² + 2ab + b²) = a²b + 2ab² + b³.
2. Like terms: 2a²b + a²b = 3a²b and ab² + 2ab² = 3ab².
3. Result: a³ + 3a²b + 3ab² + b³, which is (a + b)³.

Teacher's Note:
1. (a + b)(a + b)² is the same as (a + b)³.
2. Check with a = b = 1: 2 × 4 = 8 and 1 + 3 + 3 + 1 = 8.

Answer: (b) a³ − b³

Explanation:
1. a(a² + ab + b²) = a³ + a²b + ab².
2. −b(a² + ab + b²) = −a²b − ab² − b³.
3. Adding, a²b and ab² cancel in pairs, leaving a³ − b³. Option (c) is (a − b)³, a different expression.

Teacher's Note:
1. Look for terms that cancel in pairs.
2. Check with a = 2, b = 1: 1 × 7 = 7 and 8 − 1 = 7.

Answer: (a) 2304

Explanation:
1. With a = 48, b = 2: 48² = (48 + 2)(48 − 2) + 2².
2. = 50 × 46 + 4 = 2300 + 4.
3. = 2304.

Teacher's Note:
1. Choose b so that one factor is a round number.
2. Do not forget to add b² at the end; 2300 is the trap.

Answer: (d) 11² + 3²

Explanation:
1. Adding Identities 1A and 1B gives 2(a² + b²) = (a + b)² + (a − b)².
2. With a = 7, b = 4: (7 + 4)² + (7 − 4)² = 11² + 3².
3. Check: 121 + 9 = 130 and 2(49 + 16) = 130. Answer: 11² + 3².

Teacher's Note:
1. The two new numbers are the sum and the difference of 7 and 4.
2. The other options give 170, 125 and 170.

Answer: (c) Both statements are true.

Explanation:
1. b − a = −(a − b), and a number and its opposite have the same square. So Statement I is true.
2. (a² + 2ab + b²) − (a² − 2ab + b²) = 4ab. So Statement II is true.
3. Hence both statements are true.

Teacher's Note:
1. Squaring removes a minus sign.
2. Subtracting 1B from 1A leaves only the middle terms.

Answer: (a) Only Statement I is true.

Explanation:
1. Integers also obey the distributive property, so Statement I is true.
2. The first term has a² and the last term has a; these are different letter-numbers.
3. Unlike terms cannot be combined, so Statement II is false. Only Statement I is true.

Teacher's Note:
1. a² and a are unlike terms.
2. Example for I: −2(−3 + 5) = −4 and 6 − 10 = −4.

Answer: (b) Only Statement II is true.

Explanation:
1. 5w² and 6w are unlike terms, so 5w² + 6w cannot be simplified. Statement I is false.
2. 2a³ and 3a³ are like terms, so they add to 5a³. Statement II is true.
3. So only Statement II is true.

Teacher's Note:
1. Only like terms can be added.
2. Test with w = 1: 5 + 6 = 11 works, but w = 2 gives 32 against 44.

Answer: (d) A is false, but R is true.

Explanation:
1. (2q + 1)(2q − 3) = 4q² − 6q + 2q − 3 = 4q² − 4q − 3, so R is true.
2. 4q² − 4q − 3 = 4(q² − q) − 3, which is 3 less than a multiple of 4.
3. For q = 1 the product is 3 × (−1) = −3, not a multiple of 4. So A is false but R is true.

Teacher's Note:
1. Write the expansion as 4 × (something) + remainder part.
2. Both factors are odd, so the product is odd and cannot be a multiple of 4.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Treat (a + 1) as a single term: (a + 1)(b + 1) = (a + 1)b + (a + 1).
2. = ab + b + a + 1, so R is true.
3. The new product is ab plus a + b + 1, which is exactly what A says. Both are true and R explains A.

Teacher's Note:
1. Example: 3 × 4 = 12 and 4 × 5 = 20; increase 8 = 3 + 4 + 1.
2. This is Identity 1 with m = n = 1.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. A is true: the order of factors does not change a product.
2. R is the distributive property, which is also true.
3. But A is an example of the commutative property, not of distributivity. So both are true but R does not explain A.

Teacher's Note:
1. Match each statement with the property it really uses.
2. No addition appears in A, so distributivity is not involved.

Answer: (c) (i)-(r), (ii)-(p), (iii)-(s), (iv)-(q)

Explanation:
1. 98 × 102 = 100² − 2² = 9996 and 45 × 55 = 50² − 5² = 2475.
2. 46² = (40 + 6)² = 1600 + 480 + 36 = 2116.
3. 91² = (100 − 9)² = 10000 − 1800 + 81 = 8281. So the answer is (c).

Teacher's Note:
1. Pick the identity that makes the numbers round.
2. Units digits help: 46² ends in 6, 91² ends in 1.

Answer: (b) 27

Explanation:
1. Write 37 × 63 = (36 + 1)(64 − 1).
2. = 36 × 64 − 36 + 64 − 1 = 36 × 64 + 27.
3. So the difference is 27. Check: 2331 − 2304 = 27.

Teacher's Note:
1. Express one product in terms of the other.
2. Identity 1 with m = 1 and n = −1 is used.

Answer: (c) A is true, but R is false.

Explanation:
1. (−q + 2)² = (2 − q)² = 4 − 4q + q², so A is true.
2. The correct identity is (a − b)² = a² − 2ab + b², with +b², not −b². So R is false.
3. Hence A is true but R is false.

Teacher's Note:
1. The last term of a square is always positive.
2. Check R with a = 3, b = 1: 4 ≠ 9 − 6 − 1 = 2.

Answer: (d) Neither statement is true.

Explanation:
1. (6n + 2)² − (4n + 3)² = 36n² + 24n + 4 − 16n² − 24n − 9 = 20n² − 5. For n = 1 this is 15, and 15 + 5 = 20 is not a square. Statement I is false.
2. (k + 1)(k + 2) − (k + 3) = k² + 3k + 2 − k − 3 = k² + 2k − 1, which is 2 only when k = 1 (for k = 2 it is 7). Statement II is false.
3. So neither statement is true.

Teacher's Note:
1. One counter-example is enough to show a 'for every' claim is false.
2. Always simplify before testing.

Answer: (a) (i)-(q), (ii)-(p), (iii)-(s), (iv)-(r)

Explanation:
1. Two more than a square is s² + 2: (i)-(q). Twice a square is 2s²: (ii)-(p).
2. For consecutive numbers m and m + 1, the sum of their squares is m² + (m + 1)²: (iii)-(s).
3. The square of their sum is (m + (m + 1))²: (iv)-(r). Answer: (a).

Teacher's Note:
1. Read the last operation first: 'sum of squares' ends in adding, 'square of sum' ends in squaring.
2. Twice means multiply by 2.

Answer: (d) 240 cm²

Explanation:
1. Shaded area = 23² − 17².
2. Using a² − b² = (a + b)(a − b): (23 + 17)(23 − 17) = 40 × 6.
3. = 240 cm², much quicker than 529 − 289.

Teacher's Note:
1. Area left = big square − small square.
2. 36 cm² is (23 − 17)², a common mistake.

Answer: (c) 13.75

Explanation:
1. The dashed rectangle measures (s − r) by (p − r).
2. (p − r)(s − r) = ps − pr − sr + r² = 54 − 21 − 31.5 + 12.25.
3. = 13.75 square units. Directly: 2.5 × 5.5 = 13.75.

Teacher's Note:
1. Subtract the strip width from each side.
2. The r² term is added back because it was removed twice.