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

Short Answer Questions

Question 1. (i) Write the distributive property of multiplication over addition for three numbers a, b and c.
(ii) Using this property, evaluate 18 × 106 in two steps.
Answer:
(i) For any three numbers, a(b + c) = ab + ac. This is the distributive property of multiplication over addition.
(ii) 18 × 106 = 18 × (100 + 6) = 18 × 100 + 18 × 6
= 1800 + 108 = 1908.
Breaking 106 into 100 + 6 turns one hard multiplication into two easy ones.
In simple words: Split 106 into 100 and 6, multiply each by 18 and add.

Teacher's Note:
1. Choose a split with a round number.
2. Both parts are multiplied by 18.

Question 2. Expand and simplify, naming the property you use:
(i) 5p(2p − 3q + 4)
(ii) (y + 7)(y + 2)
Answer:
The distributive property is used in both parts.
(i) 5p × 2p − 5p × 3q + 5p × 4 = 10p² − 15pq + 20p.
(ii) y(y + 2) + 7(y + 2) = y² + 2y + 7y + 14 = y² + 9y + 14.
In (ii), 2y and 7y are like terms, so they add to 9y.
In simple words: Multiply every term inside by the outside term, then add the like terms.

Teacher's Note:
1. In (ii) the property is used twice.
2. Check (ii) with y = 1: 8 × 3 = 24 = 1 + 9 + 14.

Question 3. Use the identity (a + b)² = a² + 2ab + b² to write:
(i) the value of 43², by taking 43 = 40 + 3;
(ii) the expansion of (2x + 5)².
Answer:
(i) With a = 40, b = 3: 43² = 40² + 2 × 40 × 3 + 3² = 1600 + 240 + 9 = 1849.
(ii) With a = 2x, b = 5: (2x + 5)² = (2x)² + 2 × 2x × 5 + 5² = 4x² + 20x + 25.
In simple words: Square the first part, add twice the product of the parts, and add the square of the second part.

Teacher's Note:
1. a and b can be numbers or whole terms like 2x.
2. (2x)² is 4x², not 2x².

Question 4. Use the identity (a − b)² = a² − 2ab + b² to write:
(i) the value of 97², by taking 97 = 100 − 3;
(ii) the expansion of (3m − 4)².
Answer:
(i) With a = 100, b = 3: 97² = 10000 − 600 + 9 = 9409.
(ii) With a = 3m, b = 4: (3m − 4)² = 9m² − 2 × 3m × 4 + 16 = 9m² − 24m + 16.
In simple words: Same as the square of a sum, but the middle term is subtracted.

Teacher's Note:
1. Only the middle term carries the minus sign.
2. Numbers close to 100 are easy with Identity 1B.

Question 5. Using the identity (a + b)(a − b) = a² − b², evaluate 52 × 48 and 105 × 95. Show the choice of a and b in each case.
Answer:
Take a as the number midway between the two factors.
52 × 48: a = 50, b = 2, so (50 + 2)(50 − 2) = 2500 − 4 = 2496.
105 × 95: a = 100, b = 5, so (100 + 5)(100 − 5) = 10000 − 25 = 9975.
The middle terms cancel, which is why only two squares remain.
In simple words: Write each product as (middle + gap)(middle − gap) and subtract the squares.

Teacher's Note:
1. a is the average of the two factors.
2. b is half the difference of the factors.

Question 6. Find 4352 × 11 by writing 11 as 10 + 1. Show both steps and state why the distributive property allows this.
Answer:
4352 × 11 = 4352 × (10 + 1) = 4352 × 10 + 4352 × 1
= 43520 + 4352 = 47872.
The split is allowed because a(b + c) = ab + ac: multiplying by a sum is the same as multiplying by each part and adding.
In simple words: Times 10 means adding a zero; then add the number once more.

Teacher's Note:
1. One-line check: 4, 4+3, 3+5, 5+2, 2 gives 47872.
2. No long multiplication is needed.

Question 7. A student wrote (4p + 7q)² = 16p² + 49q². Is this correct? Explain what the student forgot and write the correct expansion.
Answer:
No. The student squared each term separately and forgot the middle term.
(4p + 7q)(4p + 7q) = 16p² + 28pq + 28pq + 49q².
Correct expansion: 16p² + 56pq + 49q². The missing term is 2 × 4p × 7q = 56pq.
In simple words: The square of a sum always has the extra middle term 2ab.

Teacher's Note:
1. Check with p = q = 1: 11² = 121 but 16 + 49 = 65.
2. 16 + 56 + 49 = 121 confirms the correct form.

Question 8. Rupa found 25 × 17 × 4 by first multiplying 25 and 4. Write her steps and name the property of multiplication used at each step.
Answer:
25 × 17 × 4 = 25 × 4 × 17: the order of the factors is changed, using the commutative property.
= (25 × 4) × 17: the grouping is changed, using the associative property.
= 100 × 17 = 1700.
The two properties together let her make the round number 100 first.
In simple words: Rearrange and regroup so that 25 × 4 = 100 comes first.

Teacher's Note:
1. Look for pairs that make 10, 100 or 1000.
2. Order and grouping do not change a product.

Question 9. Two numbers a and b are multiplied. The first is then increased by 4 and the second decreased by 3.
(i) Using the identity (a + m)(b + n) = ab + mb + an + mn, write the change in the product.
(ii) Verify your answer for a = 12 and b = 20.
Answer:
(i) Take m = 4 and n = −3: (a + 4)(b − 3) = ab + 4b − 3a − 12.
So the product changes by 4b − 3a − 12.
(ii) For a = 12, b = 20: change = 80 − 36 − 12 = 32.
Directly: 16 × 17 = 272 and 12 × 20 = 240; 272 − 240 = 32. Verified.
In simple words: New product minus old product is 4b − 3a − 12, which gives 32 for 12 and 20.

Teacher's Note:
1. A decrease is written as a negative n.
2. mn = 4 × (−3) = −12.

Question 10. The product of two numbers stays the same when one of them is increased by 3 and the other is decreased by 6. Using algebra, find the relation between the two numbers and give two examples.
Answer:
Let the numbers be a and b. Then (a + 3)(b − 6) = ab − 6a + 3b − 18.
For the product to stay the same, −6a + 3b − 18 = 0, that is b = 2a + 6.
Example 1: a = 1, b = 8: 4 × 2 = 8 = 1 × 8.
Example 2: a = 2, b = 10: 5 × 4 = 20 = 2 × 10.
In simple words: The change must be zero, which happens when the second number is 6 more than twice the first.

Teacher's Note:
1. Set the extra terms equal to zero.
2. Divide −6a + 3b − 18 = 0 by 3 to simplify.

Question 11. Here is a student’s working:
3(y − 2) + 4(y + 5) = 3y − 2 + 4y + 5 = 7y + 3.
Find the mistake, explain what went wrong and write the correct simplified expression.
Answer:
The mistake is in the first step: 3 and 4 were multiplied only with the first term inside each bracket.
Correct working: 3(y − 2) + 4(y + 5) = 3y − 6 + 4y + 20 = 7y + 14.
Distributivity needs every term in the bracket to be multiplied.
In simple words: Multiply the −2 by 3 and the 5 by 4 as well.

Teacher's Note:
1. Check with y = 0: −6 + 20 = 14, not 3.
2. This is the most common slip with brackets.

Question 12. Twice the sum of the squares of 7 and 4 can itself be written as a sum of two squares. Find the two squares and state the identity that explains this pattern.
Answer:
2(7² + 4²) = 2(49 + 16) = 130, and 130 = 121 + 9 = 11² + 3².
Here 11 = 7 + 4 and 3 = 7 − 4. The identity is 2(a² + b²) = (a + b)² + (a − b)².
It follows by adding (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b², since +2ab and −2ab cancel.
In simple words: Use the sum and the difference of 7 and 4: 11² + 3².

Teacher's Note:
1. Adding Identities 1A and 1B removes the 2ab terms.
2. Try it with other pairs to see the pattern.

Question 13. Sridharacharya’s rule for squaring is a² = (a + b)(a − b) + b². Use it to find:
(i) 41², taking b = 1;
(ii) 68², taking b = 2.
Answer:
(i) 41² = (41 + 1)(41 − 1) + 1² = 42 × 40 + 1 = 1680 + 1 = 1681.
(ii) 68² = (68 + 2)(68 − 2) + 2² = 70 × 66 + 4 = 4620 + 4 = 4624.
Each time b is chosen so that one factor becomes a round number.
In simple words: Move the number to a round number, multiply, and add b².

Teacher's Note:
1. The rule is Identity 1C with b² moved across.
2. Pick b to reach the nearest multiple of 10.

Question 14. Expand (x − y)(x² + xy + y²) by using the distributive property and collecting like terms. Verify your result for x = 3 and y = 2.
Answer:
(x − y)(x² + xy + y²) = x(x² + xy + y²) − y(x² + xy + y²)
= x³ + x²y + xy² − x²y − xy² − y³.
x²y cancels −x²y and xy² cancels −xy², so the product is x³ − y³.
Check for x = 3, y = 2: (1)(9 + 6 + 4) = 19 and 27 − 8 = 19. Verified.
In simple words: All the middle terms cancel, leaving x³ − y³.

Teacher's Note:
1. Multiply x and then −y with all three terms.
2. A numerical check catches sign errors.

Question 15. Using the distributive property, find:
(i) 7236 × 101
(ii) 2548 × 99
Write each step.
Answer:
(i) 7236 × 101 = 7236 × (100 + 1) = 723600 + 7236 = 730836.
(ii) 2548 × 99 = 2548 × (100 − 1) = 254800 − 2548 = 252252.
Splitting the multiplier as 100 + 1 or 100 − 1 turns a long multiplication into a shift and one addition or subtraction.
In simple words: Multiply by 100, then add or subtract the number once.

Teacher's Note:
1. Distributivity works over subtraction too.
2. Times 100 means adding two zeros.

Question 16. For a growing design, Meera writes the number of triangles at Step k as k(k + 3), while Arun writes it as (k + 1)² + k − 1. Are the two expressions the same? Justify by expanding.
Answer:
Yes, they are the same.
Meera: k(k + 3) = k² + 3k.
Arun: (k + 1)² + k − 1 = k² + 2k + 1 + k − 1 = k² + 3k.
Both simplify to k² + 3k, so they are equal for every k. For example, at k = 4 both give 28.
In simple words: After simplifying, both become k² + 3k.

Teacher's Note:
1. Different ways of seeing a pattern can give the same rule.
2. Expand fully before comparing.

Question 17. Is the product (2n + 1)(2n + 5) a multiple of 4 for every whole number n? Expand it and justify your answer with algebra.
Answer:
(2n + 1)(2n + 5) = 4n² + 10n + 2n + 5 = 4n² + 12n + 5 = 4(n² + 3n) + 5.
4(n² + 3n) is a multiple of 4, so the product is 5 more than a multiple of 4, that is, 1 more than a multiple of 4.
So it is never a multiple of 4. In fact both factors are odd, so the product is odd.
In simple words: The product always leaves remainder 1 when divided by 4.

Teacher's Note:
1. Write the result as 4 × (something) + remainder.
2. Check n = 0: 1 × 5 = 5, not a multiple of 4.

Question 18. Without multiplying out completely, decide which is larger: 34 × 46 or 32 × 48. Use an identity and state the exact difference.
Answer:
32 × 48 = (34 − 2)(46 + 2) = 34 × 46 + 2 × 34 − 2 × 46 − 4
= 34 × 46 + 68 − 92 − 4 = 34 × 46 − 28.
So 34 × 46 is larger, by 28. (Check: 1564 and 1536.)
In simple words: Changing 34 × 46 into 32 × 48 takes away 28.

Teacher's Note:
1. Both pairs add to 80; the closer pair gives the bigger product.
2. Identity 1 with m = −2 and n = 2 is used.

Question 19. (i) Choose three consecutive numbers. Show with algebra that the square of the middle number minus the product of the other two is always 1.
(ii) What happens if the three numbers differ by 2 instead, as in n − 2, n, n + 2?
Answer:
(i) Let the numbers be n − 1, n, n + 1. Then n² − (n − 1)(n + 1) = n² − (n² − 1) = 1, using (a + b)(a − b) = a² − b². It is 1 for every n. (Example: 5, 6, 7: 36 − 35 = 1.)
(ii) n² − (n − 2)(n + 2) = n² − (n² − 4) = 4.
The answer is always the square of the common gap.
In simple words: The middle square is bigger than the outer product by the square of the gap.

Teacher's Note:
1. Identity 1C makes this a one-line proof.
2. A gap of 3 would give 9.

Question 20. One number leaves remainder 2 when divided by 5 and another leaves remainder 4 when divided by 5. Using algebra, find the remainder when (i) their sum and (ii) their product are divided by 5.
Answer:
Let the numbers be 5a + 2 and 5b + 4.
(i) Sum = 5a + 5b + 6 = 5(a + b + 1) + 1, so the remainder is 1.
(ii) Product = 25ab + 20a + 10b + 8 = 5(5ab + 4a + 2b + 1) + 3, so the remainder is 3.
Check with 12 and 14: 26 = 5 × 5 + 1 and 168 = 5 × 33 + 3.
In simple words: Only the remainders matter: 2 + 4 = 6 leaves 1, and 2 × 4 = 8 leaves 3.

Teacher's Note:
1. Write each number as 5 × (something) + remainder.
2. Pull out every multiple of 5.