1. What does the phrase 'We Distribute, Yet Things Multiply' mean in a mathematical context?

Ans. The phrase suggests that while we may be distributing or sharing quantities, the result can lead to an increase or multiplication of those quantities. For example, in mathematics, when distributing a number across an addition (like in the expression a(b + c)), the result can often yield a greater total than the individual components alone.

2. How can distribution be used to simplify algebraic expressions?

Ans. Distribution is a technique used in algebra to simplify expressions by multiplying a single term by each term within parentheses. For instance, in the expression 3(x + 4), you would distribute the 3 to both x and 4, resulting in 3x + 12. This makes it easier to work with the expression in further calculations.

3. Can you give an example of real-life situations where distribution is applied?

Ans. Yes, distribution is commonly used in budgeting and finance. For instance, if a family has a monthly budget of $600 and wants to distribute it among groceries, utilities, and entertainment, they might allocate $300 for groceries, $200 for utilities, and $100 for entertainment. Here, the total is distributed but represents a significant amount of spending.

4. What is the importance of understanding multiplication in relation to distribution?

Ans. Understanding multiplication is crucial because distribution often combines both operations. When we distribute, we are essentially multiplying a factor by each term in a sum. This understanding helps in solving complex equations and simplifies calculations, making it easier to manage mathematical problems.

5. How does the concept of multiplication relate to growth in real-world scenarios?

Ans. In many real-world situations, multiplication indicates growth. For instance, in population studies, if a population grows by a certain percentage each year, the total population can be calculated by multiplying the current population by a growth factor. This illustrates how distribution of resources or factors can lead to significant increases over time.

Important Questions: We Distribute, Yet Things Multiply - Mathematics Class 8- New NCERT (Ganita Prakash) PDF Download

Solution:
3(2x + 5y) = 6x + 15y

Q2: Expand:
(7x - 4) × 2

Solution:
(7x - 4) × 2 = 14x - 8

Q3: Simplify:
(10a + 15b + 20c) ÷ 5

Solution:
(10a ÷ 5) + (15b ÷ 5) + (20c ÷ 5) = 2a + 3b + 4c

Q4: Simplify:
6(2x + 3) + 4(x - 5)

Solution:
6(2x + 3) = 12x + 18
4(x - 5) = 4x - 20
Adding them → 12x + 18 + 4x - 20 = 16x - 2

Q5: Simplify:
8(y - 2) - 3(2y + 1)

Solution:
8(y - 2) = 8y - 16
-3(2y + 1) = -6y - 3
So, 8y - 16 - 6y - 3 = 2y - 19

Q6: A fruit seller packs 8 baskets, each basket containing (12 apples + 15 oranges). Using distributive property, find the total number of fruits.

Solution:
8(12 + 15) = 8 × 12 + 8 × 15 = 96 + 120 = 216
Total fruits = 216

Q7: A builder uses 25 tiles for each row. If he lays (40 + 30) rows, how many tiles does he need in total?

Solution:
25(40 + 30) = 25 × 70 = 1750
Total tiles = 1750

Q8: The cost of one pen is Rs. (x + 5). Find the cost of 12 pens using distributive property.

Solution:
12(x + 5) = 12x + 60

Q9: A rectangular park has length (50 + 20) m and breadth 12 m. Find its area using distributive property.

Solution:
(50 + 20) × 12 = 50 × 12 + 20 × 12 = 600 + 240 = 840 m²

Q10: A shopkeeper sells (15 + 10) chocolates in a pack. If he sells 24 such packs, find the total number of chocolates.

Solution:
24(15 + 10) = 24 × 25 = 600
Total chocolates = 600

Q11: Simplify:
(x + 5)(x + 3)

Solution:
(x + 5)(x + 3) = x(x + 3) + 5(x + 3)
= x² + 3x + 5x + 15
= x² + 8x + 15

Q12: Expand and simplify:
(2x + 7)(3x - 4)

Solution:
(2x + 7)(3x - 4) = 2x(3x - 4) + 7(3x - 4)
= 6x² - 8x + 21x - 28
= 6x² + 13x - 28

Q13: If a = 15, b = 12, evaluate:
5(a + b) - 3(a - b)

Solution:
5(a + b) - 3(a - b) = 5a + 5b - 3a + 3b
= 2a + 8b
Substitute a = 15, b = 12 → 2(15) + 8(12) = 30 + 96 = 126

Q14: Factorize:
12x + 18y

Solution:
12x + 18y = 6(2x) + 6(3y) = 6(2x + 3y)

Q15: Verify distributive property:
9 × (14 + 6) = (9 × 14) + (9 × 6)

Solution:
LHS = 9 × (14 + 6) = 9 × 20 = 180
RHS = (9 × 14) + (9 × 6) = 126 + 54 = 180
Since LHS = RHS

Q16: Which term is the like term similar to 24a²bc?
(a) 13 × 8a × 2b × c × a
(b) 8 × 3 × a × b × c
(c) 3 × 8 × a × b × c × c
(d) 3 × 8 × a × b × b × c
Ans: (a)
Sol: To find out the similar term as 24a²bc, let us find the product of each of the equations,

Hence, we can get that option (a) is correct.
Q17: Fill in the blanks.
(a) (x + a) (x + b) = x² + (a + b)x + ________.
(b) The product of two terms with like signs is a ________ term.
(c) The product of two terms with unlike signs is a ________ term.
(d) (a – b) _________ = a² – 2ab + b²
(e) a² – b² = (a + b) __________.
(f) (a – b)² + ____________ = a² – b²
(g) (a + b)² – 2ab = ___________ + ____________
(h) The product of two polynomials is a ________
(i) The coefficient in – 37abc is __________.
(j) Number of terms in the expression a2 + bc × d is ________
Ans:
(a) ab
As per the standard identity 4, (x + a) (x + b) =  x² + (a + b)x + ab
(b) Positive
(c) Negative
(d) (a – b)²
As per standard identity 2, (a – b)² = a² – 2ab + b²
(e) (a – b)
As per standard identity 3, (a + b) (a – b) = a² – b²
(f) 2ab – 2b²
Let us solve the equation with x in the blank space. As per identity 2, (a – b)² = a² – 2ab + b².
Hence, a² – 2ab + b² + x = a² – b²
x = a² – b² – a² + 2ab – b²
x = 2ab – 2b²
(g) a² + b²
Using Identity 1 (a + b)² = a² + 2ab + b²,
(a + b)² – 2ab = a² + 2ab + b² – 2ab = a² + b²
(h) Polynomial
(i) -37
(j) 2
Q18: The length of a rectangular box is (x + 9y) and the area is x² + 12xy + 27y². Find the breadth.
Sol. Area of a rectangle = length x breadth, hence breadth = area / length.
breadth = x² + 12xy + 27y²
(x + 9y)
=  x² + 9xy + 3xy + 27y²
(x + 9y)
= x (x + 9y) + 3y (x + 9y)
(x + 9y)
breadth  = x + 3y
Q19: Using identities, find products for the below.
(a) 71²
(b) 99²
(c) 102²
(d) 998²
(e) 5.2²
(f) 297 × 303
(g) 78 × 82
(h) 8.9²
(i) 10.5 × 9.5
Sol:

(a) 71² = (70 + 1)² Identity applied (a + b)² = a² + 2ab + b²
= 70² + 2 (70 x 1) + 1²
= 4900 + 140 + 1
= 5041

(b) 99² = (100 – 1)² Identity applied (a – b)² = a² – 2ab + b²
= 100² – 2 (100 x 1) + 1²
= 10000 – 200 + 1
= 9801
(c) 102² = (100 + 2)² Identity applied (a + b)² = a² + 2ab + b²
= 100² + 2 (100 x 2) + 2²
= 10000 + 400 + 4
= 10404
(d) 998² = (1000 – 2)² Identity applied (a – b)² = a² – 2ab + b²
= 1000² – 2 (1000 x 2) + 2²
= 1000000 – 4000 + 4
= 996004
(e) 5.2² = (5 + 0.2)² Identity applied (a + b)² = a² + 2ab + b²
= 5² + 2 (5 x 0.2) + 0.2²
= 25 + 2 + 0.04
= 27.04
(f) 297 × 303 = (300 – 3) (300 + 3) Identity applied (a + b) (a – b) = a² – b²
= 300² – 3²
= 90000 – 9
= 89991
(g) 78 × 82 = (80 – 2) (80 + 2) Identity applied (a + b) (a – b) = a² – b²
= 80² – 2²
= 6400 – 4
= 6396
(h) 8.9² = (9.0 – 0.1)² Identity applied (a – b)² = a² – 2ab + b²
= 9.0² – 2 (9.0 x 0.1) + 0.1²
= 81 – 1.8 + 0.01
= 79.21
(i) 10.5 x 9.5 = (10 + 0.5) (10 – 0.5) Identity applied (a + b) (a – b) = a² – b²
= 10² – 0.5²
= 100 – 0.25
= 99.75

Q20: Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
(a) 5a, 3a², 7a4
(b) 2p, 4q, 8r
(c) xy, 2x²y, 2xy²
(d) a, 2b, 3c
Sol: The volume of a rectangular box is the product of its length, breadth and height, i.e Volume = length x breadth x height.
Volumes are calculated as below,
(a) length = 5a, breadth = 3a², height = 7a4
Volume = 5a x 3a² x 7a4
= 105a7
(b) length = 2p, breadth = 4q, height = 8r
Volume = 2p x 4q x 8
= 64pqr
(c) length = xy, breadth = 2x²y, height = 2xy²
Volume = xy x 2x²y x 2xy²
= 4x4y4
(d) length = a, breadth = 2b, height = 3c
Volume = a x 2b x 3c
= 6abc

Q21: Show that LHS = RHS for the below equations.
(a) (3x + 7)² – 84x = (3x – 7)²
(b) (9p – 5q)² + 180pq = (9p + 5q)²
(c) (4pq + 3q)² – (4pq – 3q)² = 48pq²
(d) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
Answer 18.
(a) LHS = (3x + 7)² – 84x
= (3x)² + 2 (3x x 7) + 7² – 84x
= 9x² + 42x + 49 – 84x
= 9x² – 42x + 49
RHS = (3x – 7)²
= (3x)² – 2 (3x x 7) + 7²
= 9x² – 42x + 49
Hence LHS = RHS
(b) LHS = (9p – 5q)² + 180pq
= (9p)² – 2 (9p x 5q) + (5q)² + 180pq
= 81p² + 90pq + 25q²
RHS = (9p + 5q)²
= (9p)² + 2 (9p x 5q) + (5q)²
= 81p² + 90pq + 25q²
Hence LHS = RHS

(c) LHS = (4pq + 3q)² – (4pq – 3q)²
= (4pq)² + 2 (4pq x 3q) + (3q)² – ((4pq)² – 2 (4pq x 3q) + (3q)²)
= 16p²q² + 24pq² + 9q² – 16p²q² + 24pq² – 9q²
= 48pq²
RHS = 48pq²
Hence LHS = RHS
(d) LHS = (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)
= a² + ab – ba – b² + b² + bc – cb – c² + c² + ca – ac – a²
= 0
RHS = 0
Hence LHS = RHS

Q22: Solve the below using correct identities.
(a) (48)²
(b) 181² – 19²
(c) 497 × 505
(d) 2.07 × 1.93
Sol:
(a) (48)²
= (50 – 2)²
As (a – b)² = a² – 2ab + b² , hence
(50 – 2)² = (50)² – 2 × 50 × 2 + (2)²
= 2500 – 200 + 4
= 2300 + 4
= 2304
(b) As a² – b² = (a – b) (a + b)
181² – 19² = (181 – 19) (181 + 19)
= 162 × 200
= 32400

(c) By using the identity (x + a) (x + b) = x² + (a + b) x + ab
497 x 505 = (500 – 3) (500 + 5)
= 500² + (–3 + 5) × 500 + (–3) (5)
= 250000 + 1000 – 15
= 250985
(d) As (a + b) (a – b) = a² – b²
2.07 × 1.93 = (2 + 0.07) (2 – 0.07)
= 2² – 0.07²
= 3.9951

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