CBSE Class 8 Mathematics We Distribute Yet Things Multiply Very Short Answer Questions with Answers

Very Short Answer Questions

Question 1. State the distributive property of multiplication over addition using the letter-numbers a, b and c.
Answer:
For any three numbers, a(b + c) = ab + ac. The multiplier a multiplies each term inside the bracket.

Teacher's Note:
1. It links multiplication with addition.
2. Example: 7 × 13 = 7 × 10 + 7 × 3 = 91.

Question 2. Write the expansion of (a + b)².
Answer:
(a + b)² = a² + 2ab + b² (Identity 1A).

Teacher's Note:
1. The middle term 2ab comes from ab + ba.
2. It is the most-used identity of the chapter.

Question 3. Write the expansion of (a − b)².
Answer:
(a − b)² = a² − 2ab + b² (Identity 1B).

Teacher's Note:
1. Only the middle term is negative.
2. b² stays positive because (−b)(−b) = b².

Question 4. Write the product (a + b)(a − b) in its simplest form.
Answer:
(a + b)(a − b) = a² − ab + ba − b² = a² − b² (Identity 1C), since −ab and +ba cancel.

Teacher's Note:
1. The two middle terms always cancel.
2. It is called the difference of two squares.

Question 5. What is meant by an identity? Give one example.
Answer:
An identity is an equality of two algebraic expressions that is true for every value of the letter-numbers in it.
Example: a(b + 8) = ab + 8a.

Teacher's Note:
1. An identity differs from an equation, which is true only for some values.
2. Any expansion in this chapter is an example.

Question 6. What are like terms? Write one pair of like terms.
Answer:
Terms that have exactly the same letter-numbers (with the same powers) are like terms; only these can be added into one term.
Example: 3ab and −7ba.

Teacher's Note:
1. ab and ba are the same product.
2. The number in front may differ.

Question 7. Expand 2p(3p − q + 4).
Answer:
2p × 3p − 2p × q + 2p × 4 = 6p² − 2pq + 8p.

Teacher's Note:
1. Multiply 2p with all three terms.
2. Keep the sign of each term.

Question 8. Expand (x + 2)(x + 9).
Answer:
(x + 2)(x + 9) = x² + 9x + 2x + 18 = x² + 11x + 18.

Teacher's Note:
1. 2 + 9 gives the middle coefficient, 2 × 9 the constant.
2. Collect like terms at the end.

Question 9. Find 52 × 11 by splitting 11 as 10 + 1.
Answer:
52 × 11 = 52 × (10 + 1) = 520 + 52 = 572.

Teacher's Note:
1. Times 10, then add the number once.
2. One-line rule: 5, 5 + 2, 2 gives 572.

Question 10. Use the identity (a + b)(a − b) = a² − b² to find 31 × 29.
Answer:
31 × 29 = (30 + 1)(30 − 1) = 30² − 1² = 900 − 1 = 899.

Teacher's Note:
1. Take a as the number midway between the factors.
2. Here a = 30 and b = 1.

Question 11. Name the property that allows (a + b)c to be written as c(a + b).
Answer:
The commutative property of multiplication: the two factors (a + b) and c can change places.

Teacher's Note:
1. The bracket acts as one factor here.
2. Nothing is distributed in this step.

Question 12. Who gave the first explicit statement of the distributive property, in which work, and by what name did he call multiplication by parts?
Answer:
Brahmagupta, in the Brahmasphuṭasiddhānta (Verse 12.55). He called multiplication by parts khaṇḍa-guṇanam.

Teacher's Note:
1. Khaṇḍa means part; guṇana means multiplication.
2. His fast methods were called iṣṭa-guṇana.

Question 13. Write 65² as (60 + 5)² and use Identity 1A to find its value.
Answer:
65² = 60² + 2(60 × 5) + 5² = 3600 + 600 + 25 = 4225.

Teacher's Note:
1. Split into a round number plus a small number.
2. Twice 60 × 5 gives 600.

Question 14. Which is greater, (a − b)² or (b − a)²? Justify your answer in one line.
Answer:
Neither; they are equal, because b − a = −(a − b) and a number and its opposite have the same square.

Teacher's Note:
1. Example: (5 − 2)² = 9 and (2 − 5)² = 9.
2. Squaring removes the minus sign.

Question 15. By how much does the product ab change when a is increased by 1 and b is decreased by 1?
Answer:
(a + 1)(b − 1) = ab − a + b − 1, so the product changes by b − a − 1. It increases only when b is more than a + 1.

Teacher's Note:
1. The change can be positive, negative or zero.
2. Example: a = 3, b = 7: 4 × 6 − 21 = 3 = 7 − 3 − 1.

Question 16. Use Identity 1 to expand (a − 2)(b + 3).
Answer:
Put m = −2 and n = 3 in (a + m)(b + n) = ab + mb + an + mn:
(a − 2)(b + 3) = ab + 3a − 2b − 6.

Teacher's Note:
1. A decrease is a negative value of m.
2. mn = (−2)(3) = −6.

Question 17. Expand \( \left(2x - \frac{1}{2}\right)^2 \).
Answer:
\( (2x)^2 - 2(2x)\left(\frac{1}{2}\right) + \left(\frac{1}{2}\right)^2 \) = \( 4x^2 - 2x + \frac{1}{4} \).

Teacher's Note:
1. Use Identity 1B with a = 2x and b = 1/2.
2. Square the whole term 2x, not just x.

Question 18. Find 99² using Identity 1B.
Answer:
99² = (100 − 1)² = 10000 − 200 + 1 = 9801.

Teacher's Note:
1. Numbers just below a round number suit Identity 1B.
2. Middle term: 2 × 100 × 1 = 200.

Question 19. Find 397 × 403 using Identity 1C.
Answer:
397 × 403 = (400 − 3)(400 + 3) = 400² − 3² = 160000 − 9 = 159991.

Teacher's Note:
1. 400 is midway between 397 and 403.
2. Only two squares are needed.

Question 20. Find 41² using Sridharacharya’s rule a² = (a + b)(a − b) + b² with b = 1.
Answer:
41² = (41 + 1)(41 − 1) + 1² = 42 × 40 + 1 = 1680 + 1 = 1681.

Teacher's Note:
1. b = 1 makes one factor 40, a round number.
2. Remember to add b².

Question 21. Write 2(7² + 2²) as the sum of two squares.
Answer:
2(7² + 2²) = (7 + 2)² + (7 − 2)² = 9² + 5² = 81 + 25 = 106.

Teacher's Note:
1. Use 2(a² + b²) = (a + b)² + (a − b)².
2. Check: 2(49 + 4) = 106.

Question 22. A student wrote 2(x − 1) + 3(x + 4) = 2x − 1 + 3x + 4 = 5x + 3. Point out the mistake and write the correct answer.
Answer:
The 2 and the 3 were multiplied with only the first term of each bracket.
Correct: 2x − 2 + 3x + 12 = 5x + 10.

Teacher's Note:
1. Every term inside the bracket must be multiplied.
2. Check with x = 1: 0 + 15 = 15 = 5 + 10.

Question 23. A student wrote (5m + 6n)² = 25m² + 36n². Point out the mistake and write the correct expansion.
Answer:
The middle term 2(5m)(6n) was left out.
Correct: (5m + 6n)² = 25m² + 60mn + 36n².

Teacher's Note:
1. The square of a sum has three terms.
2. Middle term = 2 × first × second.

Question 24. The number of circles at Step k of a pattern is k² + 2k. How many circles are there at Step 15?
Answer:
Put k = 15: 15² + 2 × 15 = 225 + 30 = 255 circles.

Teacher's Note:
1. Substitute, then square before adding.
2. Also 15 × 17 = 255, since k² + 2k = k(k + 2).

Question 25. Express 100 as the difference of the squares of two whole numbers.
Answer:
Using a² − b² = (a + b)(a − b) with a + b = 50 and a − b = 2, we get a = 26, b = 24.
So 100 = 26² − 24² = 676 − 576.

Teacher's Note:
1. Write 100 as a product of two even numbers.
2. Another answer: 10² − 0² (using 10 × 10).

Question 26. The number of square tiles at Step n of a pattern is (n + 2)² − n². Simplify this expression and find the number of tiles at Step 10.
Answer:
(n + 2)² − n² = n² + 4n + 4 − n² = 4n + 4.
At Step 10: 4 × 10 + 4 = 44 tiles (check: 12² − 10² = 44).

Teacher's Note:
1. The n² terms cancel.
2. Simplified forms make counting quick.

Question 27. Four identical rectangles of sides m and n are placed inside a square of side (m + n), leaving a square hole in the middle. Show that (m + n)² − 4mn = (n − m)².
Answer:
(m + n)² − 4mn = m² + 2mn + n² − 4mn = m² − 2mn + n² = (n − m)², so both ways give the same area.

Teacher's Note:
1. Two ways of finding one area must agree.
2. −2mn appears in both (n − m)² and (m − n)².

Question 28. One number leaves a remainder of 3 when divided by 7 and another leaves a remainder of 5. What remainder does their product leave when divided by 7?
Answer:
Let the numbers be 7a + 3 and 7b + 5. Their product is 49ab + 35a + 21b + 15 = 7(7ab + 5a + 3b) + 15.
15 leaves remainder 1 on division by 7, so the remainder is 1.

Teacher's Note:
1. Only the product of the remainders matters: 3 × 5 = 15.
2. Check: 10 × 12 = 120 = 7 × 17 + 1.

Question 29. Is (k + 1)(k + 2) − (k + 3) always equal to 2? Justify your answer using algebra.
Answer:
No. (k + 1)(k + 2) − (k + 3) = k² + 3k + 2 − k − 3 = k² + 2k − 1.
This equals 2 only for k = 1; for k = 2 it is 7.

Teacher's Note:
1. Simplify fully before deciding.
2. One counter-example disproves 'always'.

Question 30. Which is larger, 16 × 24 or 14 × 26, and by how much? Answer without multiplying out completely.
Answer:
14 × 26 = (16 − 2)(24 + 2) = 16 × 24 + 32 − 48 − 4 = 16 × 24 − 20.
So 16 × 24 is larger, by 20.

Teacher's Note:
1. Both pairs add to 40; closer factors give the bigger product.
2. Check: 384 − 364 = 20.