Class 8 Maths Ganita Prakash Part 1 Chapter 6: We Distribute, Yet Things Multiply MCQs with Answers

(23 + 1)(27 + 1) = 23 × 27 + 23 + 27 + 1.
Increase in the product = 23 + 27 + 1 = 51.
The extra 1 comes from multiplying the two increases together. Therefore, 24 × 28 is 51 more than 23 × 27.

(a + 1)(b − 1) = ab − a + b − 1.
New product − original product = (ab − a + b − 1) − ab
= b − a − 1.
The change depends on a and b, so it need not always be positive.

(a + m)(b + n) = a(b + n) + m(b + n)
= ab + an + mb + mn.
Each term in the first bracket is multiplied by each term in the second bracket. Since mb = bm, the result is ab + an + bm + mn.

(3a/2)(a − b + 1/5)
= (3a/2) × a − (3a/2) × b + (3a/2) × (1/5)
= 3a²/2 − 3ab/2 + 3a/10.
The factor 3a/2 multiplies all three terms, and the denominator of the last term is 2 × 5 = 10.

(a + 2)(b − 4) = ab − 4a + 2b − 8.
Since this equals ab, −4a + 2b − 8 = 0.
Therefore, 2b = 4a + 8 and b = 2a + 4.
This relation makes the extra terms cancel.

(3 − x)(x − 6) = 3x − 18 − x² + 6x
= −x² + 9x − 18.
In (−x)(−6), the coefficients multiply as (−1) × (−6) = 6, giving 6x. The like terms 3x and 6x add to 9x.

(10a + b)(10c + d)
= (10a)(10c) + (10a)d + b(10c) + bd
= 100ac + 10ad + 10bc + bd.
There are four partial products. In particular, multiplying the two terms with a factor of 10 gives a factor of 100.

(a + b)(a² + 2ab + b²)
= a³ + 2a²b + ab² + a²b + 2ab² + b³
= a³ + 3a²b + 3ab² + b³.
The two terms containing a²b add together, and so do the two terms containing ab².

(a − b)(a² + ab + b²)
= a³ + a²b + ab² − a²b − ab² − b³.
The terms +a²b and −a²b cancel. The terms +ab² and −ab² also cancel.
Therefore, the expression simplifies to a³ − b³.

11 = 10 + 1.
3279 × 11 = 3279 × (10 + 1)
= 32790 + 3279 = 36069.
The distributive property gives two partial products: ten times 3279 and one more 3279. Their sum is the required product.

101 = 100 + 1.
949 × 101 = 949 × (100 + 1)
= 94900 + 949 = 95849.
Multiplication by 101 adds the original number to one hundred times the number, so both partial products are included.

99 = 100 − 1.
9734 × 99 = 9734 × (100 − 1)
= 973400 − 9734 = 963666.
One hundred groups of 9734 are one group too many, so subtracting one group gives the product for 99 groups.

Area of each rectangle = a × b = ab.
Total area of the two rectangles = ab + ab = 2ab.
The whole square therefore has area a² + 2ab + b². The term 2ab represents the two rectangles, one for each cross-product.

(3j + 2k)² = (3j)² + 2(3j)(2k) + (2k)²
= 9j² + 12jk + 4k².
The squares give 9j² and 4k². The two cross-products are 6jk each, so together they give the middle term 12jk.

99² = (100 − 1)²
= 100² − 2 × 100 × 1 + 1²
= 10000 − 200 + 1 = 9801.
The final term is positive because the square of −1 is +1.

(−2a + 3)² = (−2a)² + 2(−2a)(3) + 3²
= 4a² − 12a + 9.
The coefficient of a² is (−2)² = 4. The coefficient of a is 2 × (−2) × 3 = −12. The constant term is 3² = 9.

(a + b)² + (a − b)²
= (a² + 2ab + b²) + (a² − 2ab + b²).
The terms +2ab and −2ab cancel.
The remaining terms give a² + a² + b² + b² = 2a² + 2b².

98 × 102 = (100 − 2)(100 + 2).
For a product of a sum and a difference, (a − b)(a + b) = a² − b².
Therefore, 98 × 102 = 100² − 2²
= 10000 − 4 = 9996.

197² = (197 + 3)(197 − 3) + 3²
= 200 × 194 + 9
= 38800 + 9 = 38809.
The product of the sum and difference is 3² less than 197², so adding 9 gives the required square.

b − a = −(a − b).
Therefore, (b − a)² = [−(a − b)]² = (a − b)².
A number and its opposite have the same square because the product of two negative factors is positive. Thus, the two expressions are always equal.

−3p(−5p + 2q) = (−3p)(−5p) + (−3p)(2q)
= 15p² − 6pq.
The first coefficient is (−3) × (−5) = 15, and p × p = p². The second coefficient is (−3) × 2 = −6, with variable part pq.

(5m + 6n)² = (5m)² + 2(5m)(6n) + (6n)²
= 25m² + 60mn + 36n².
The omitted term is twice the product of 5m and 6n. Therefore, the missing term is 60mn.

2(x − 1) + 3(x + 4)
= 2x − 2 + 3x + 12
= (2x + 3x) + (−2 + 12) = 5x + 10.
Each multiplier applies to both terms inside its bracket before like terms are combined.

A complete square grid at Step k has (k + 1)² dots. One corner dot is missing.
Number of dots = (k + 1)² − 1.
At Step 15, this is 16² − 1 = 256 − 1 = 255.
Equivalently, 15(15 + 2) = 255.

The empty square has side n units. A border one tile thick adds 1 unit on each side, making the outer side n + 2 units.
Number of tiles = (n + 2)² − n² = 4n + 4.
At Step 10, the number is 4 × 10 + 4 = 44.

Area of the outer square = (m + n)².
Total area of the four rectangles = 4mn.
Area of the central square = (m + n)² − 4mn
= m² + 2mn + n² − 4mn
= n² − 2mn + m² = (n − m)².

(a + 1)(a + 7) = a² + 8a + 7.
a(a + 8) = a² + 8a.
Their difference = (a² + 8a + 7) − (a² + 8a) = 7.
Therefore, the top-right × bottom-left product is always 7 more than the other diagonal product.

Let the odd integer be 2k + 1.
(2k + 1)² = 4k² + 4k + 1 = 4k(k + 1) + 1.
One of the consecutive integers k and k + 1 is even, so 4k(k + 1) is divisible by 8.
Therefore, the remainder is 1.

The square of the middle integer is n².
The product of the other two is (n − 1)(n + 1) = n² − 1.
Required difference = n² − (n² − 1) = 1.
Therefore, the result is always 1, whichever three consecutive integers are chosen.

The chapter introduces distributivity as “a property relating multiplication and addition that is captured concisely using algebra”. It is the central theme for exploring multiplication patterns and their algebraic descriptions in this chapter. 

When ‘b’ is increased by 1, the new product is a(b+1) = ab + a. This shows an increase of ‘a’ compared to the original product ‘ab’.

“a(b+c)=ab+ac”. This formula is then visualized with a diagram illustrating areas of rectangles to represent the multiplication.

When expanding (a+1)(b+1), the result is ab + b + a + 1. Comparing this to the original product ‘ab’, the increase is found to be a+b+1.

“Mathematical statements that express the equality of two algebraic expressions”. Examples include a(b+8)=ab+8a and (a + 1)(b – 1) = ab + b – a – 1.

Identity 1 is given as “(a + m)(b + n) = ab + mb + an + mn”. This identity is also visually represented by dividing a larger rectangle into four smaller rectangular areas.

The expansion is derived as (a + u)(b – v) = (a + u)b – (a + u)v = ab + ub – (av + uv) = ab + ub – av – uv. This demonstrates how integer multiplication rules apply.

The expansion is shown step-by-step: (a + b)(a + b) = (a + b)a + (a + b)b = a²+ba + ab + b². Since ba = ab, combining like terms yields a² + 2ab + b².

The “Pinch of History” section states that “The first explicit statement of the distributive property was given by Brahmagupta in his work Brahmasphuṭasiddhānta (Verse 12.55), who referred to the use of the property for multiplication as khanda-gunanam (multiplication by parts)”.

“The multiplier is broken up into two or more parts whose sum is equal to it; the multiplicand is then multiplied by each of these and the results added”. This is equivalent to (a + b)c = ac + bc.

The chapter uses an example of a square with side 65 units, splitting it into smaller regions to show its area is (60 + 5)². This visually represents the sum’s square.

Identity 1A is explicitly stated as “(a + b)² = a²+2ab + b²”. This formula is derived using the distributive property, multiplying (a + b) by (a + b).

Identity 1B is given as “(a – b)² = a² + b² – 2ab”. This identity can be derived by applying the distributive property to (a – b) × (a – b) or by substituting -b into the (a+b)² identity.

Pattern 2 introduces the relationship a²-b²=(a+b)×(a-b). Identity 1C confirms this: “(a + b) × (a – b)=a² – b²”.

Adding (a+b)² = a² + 2ab + b² and (a – b)² = a² – 2ab + b² results in 2a²+2b². This simplifies to 2(a²+b²).

The expression a² = (a + b)(a – b) + b² is a rearrangement of the identity a² – b²=(a + b)(a – b). It highlights how squaring a number can be computed using products of sums and differences.

The chapter shows that dcba × 11 = dcba × (10 + 1) = dcba × 10 + dcba. This is represented visually as adding the number to itself shifted one place to the left.

The “Math Talk” section states that “Such methods… were discussed extensively in the ancient mathematical works of Brahmagupta (628 CE), Sridharacharya (750 CE) and Bhaskaracharya (Lilavati, 1150 CE)”. Brahmagupta specifically used the term ‘ista-gunana’.

The formula derived for the number of circles at Step k is k² + 2k. For Step 15, this would be 15² + 2 × 15 = 225 + 30 = 255.

Tadang’s method calculates the area as (m + n)² – 4mn. Yusuf’s method states the area is (n – m)². The text then asks to verify that these two expressions are equivalent: (m + n)² – 4mn = (n – m)².