CBSE Class 8 Mathematics Part I Chapter 6 Questions and Answers

Class 8 Maths Ganita Prakash

Class 8 Maths Ganita Prakash Chapter 6 We Distribute, Yet Things Multiply NCERT Solutions

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Textbook Page 142

Figure it Out

1. Observe the multiplication grid below. Each number inside the grid is formed by multiplying two numbers. If the middle number of a 3 × 3 frame is given by the expression pq, as shown in the figure, write the expressions for the other numbers in the grid.

Solution:

2. Expand the following products.
(i) (3 + u) (v – 3)
(ii) 2/3 (15 + 6a)
(iii) (10a + b) (10c + d)
(iv) (3 – x) (x – 6)
(v) (–5a + b) (c + d)
(vi) (5 + z) (y + 9)

Solution:

(i) (3 + u) (v – 3)
= (3 + u)v – (3 + u)3
= 3v + uv – (9 + 3u)
= 3v + uv – 9 – 3u
= uv + 3v – 3u – 9

(iii) (10a + b) (10c + d)
= (10a + b)10c + (10a + b)d
= 100ac + 10bc + 10ad + bd

(iv) (3 – x) (x – 6)
= (3 – x)x – (3 – x)6
= 3x – x² – (18 – 6x)
= 3x – x² – 18 + 6x
= – x² + 6x + 3x – 18
= – x² + 9x – 18

(v) (–5a + b) (c + d)
= (–5a + b)c + (–5a + b)d
= – 5ac + bc – 5ad + bd
= – 5ac – 5ad + bc + bd

(vi) (5 + z) (y + 9)
= (5 + z)y + (5 + z)9
= 5y + zy + 45 + 9z
= 5y + 9z + zy + 45

3. Find 3 examples where the product of two numbers remains unchanged when one of them is increased by 2 and the other is decreased by 4.

Solution:
If the two numbers are x and y, then:
x × y = (x + 2) × (y − 4)
xy = (x + 2)y – (x + 2)4
xy = xy + 2y – (4x + 8)
xy = xy + 2y – 4x – 8
xy – xy = 2y – 4x – 8
0 = 2y – 4x – 8
4x + 8 = 2y
2(2x + 4) = 2y
y = 2x + 4.

Examples:
(i) x = 1, y = 6 → Product = 1 × 6 = 6
Check: (1 + 2) × (6 − 4) = 3 × 2 = 6.

(ii) x = 2, y = 8 → Product = 16
Check: (2 + 2) × (8 − 4) = 4 × 4 = 16.

(iii) x = 5, y = 14 → Product = 5 × 14 = 70
Check: (5 + 2) × (14 − 4) = 7 × 10 = 70.

Therefore, (1, 6), (2, 8), and (5, 14) are three valid examples.

4. Expand (i) (a + ab – 3b²) (4 + b), and (ii) (4y + 7) (y + 11z – 3).

Solution:
(i) (a + ab – 3b²) (4 + b)
= (a + ab – 3b²)4 + (a + ab – 3b²)b
= 4a + 4ab – 12b² + ab + ab² – 3b³
= – 3b³ – 12b² + ab² + 4ab + ab + 4a
= – 3b³ – 12b² + ab² + 5ab + 4a.

(ii) (4y + 7) (y + 11z – 3)
= (4y + 7)y + (4y + 7)11z – (4y + 7)3
= 4y² + 7y + 44yz + 77z – (12y + 21)
= 4y² + 7y + 44yz + 77z – 12y – 21
= 4y² + 7y – 12y + 44yz + 77z – 21
= 4y² – 5y + 44yz + 77z – 21.

5. Expand (i) (a – b) (a + b), (ii) (a – b) (a² + ab + b²), and (iii) (a – b)(a³ + a²b + ab² + b³). Do you see a pattern? What would be the next identity in the pattern that you see? Can you check it by expanding?

Solution:
(i) (a − b)(a + b)
= (a − b)a + (a − b)b
= a² – ab + ab – b²
= a² – b².

(ii) (a – b) (a² + ab + b²)
= (a – b)a² + (a – b)ab + (a – b)b²
= a³ – a²b + a²b – ab² + ab² – b³
= a³ – b³.

(iii) (a – b)(a³ + a²b + ab² + b³)
= (a – b)a³ + (a – b)a²b + (a – b)ab² + (a – b)b³
= a⁴ – a³b + a³b – a²b² + a²b² – ab³ + ab³ – b⁴
= a⁴ – b⁴.

The next identity would be: (a − b)(a⁴ + a³b + a²b² + ab³ + b⁴) = a⁵ − b⁵.

Textbook Page 149

Figure it Out

1. Which is greater: (a – b)² or (b – a)²? Justify your answer.

Solution:
(a – b)² = a² + b² – 2ab
(b – a)² = b² + a² – 2ba = a² + b² – 2ab
Therefore, (a – b)² = (b – a)².

2. Express 100 as the difference of two squares.

Solution:
Let a² − b² = 100
We know that a² − b² = (a + b)(a − b)
∴ (a + b)(a − b) = 100
(a + b)(a – b) = 50 × 2
So,
a + b = 50 ………. (i)
a – b = 2 ………… (ii)
Adding (i) and (ii): a + b + a – b = 50 + 2
2a = 52
a = 52/2 = 26 ………. (iii)
Substituting a in (i): 26 + b = 50
b = 50 – 26 = 24 ……… (iv)
Therefore, 26² − 24² = 100.

3. Find 406², 72², 145², 1097², and 124² using the identities you have learnt so far.

Solution:

(i) 406²
= (400 + 6)²
= 400² + 6² + 2 × 400 × 6
= 160000 + 36 + 4800
= 164836.

(ii) 72²
= (70 + 2)²
= 70² + 2² + 2 × 70 × 2
= 4900 + 4 + 280
= 5184.

(iii) 145²
= (150 – 5)²
= 150² + 5² – 2 × 150 × 5
= 22500 + 25 – 1500
= 21025.

(iv) 1097²
= (1100 – 3)²
= 1100² + 3² – 2 × 1100 × 3
= 1210000 + 9 – 6600
= 1203409.

(v) 124²
= (130 – 6)²
= 130² + 6² – 2 × 130 × 6
= 16900 + 36 – 1560
= 15376.

4. Do Patterns 1 and 2 hold only for counting numbers? Do they hold for negative integers as well? What about fractions? Justify your answer.

Solution:
Pattern 1: 2(a² + b²) = (a + b)² + (a – b)².
(i) Let a = 6, b = 4
L.H.S = 2(a² + b²) = 2(6² + 4²) = 2(36 + 16) = 2(52) = 104.
R.H.S = (a + b)² + (a – b)² = (6 + 4)² + (6 – 4)² = 10² + 2² = 100 + 4 = 104.
Therefore, Pattern 1 holds for counting numbers.

(ii) Let a = -10, b = -5
L.H.S = 2(a² + b²) = 2[(-10)² + (-5)²] = 2[100 + 25] = 2(125) = 250.
R.H.S = (a + b)² + (a – b)² = [(-10) + (-5)]² + [(-10) – (-5)]² = (-15)² + (-5)² = 225 + 25 = 250.
Therefore, Pattern 1 holds for negative integers.

(iii)

Therefore, Pattern 1 holds for fractions.

Pattern 2: (a + b) × (a – b) = a² – b².
(i) Let a = 6, b = 4
L.H.S = (a + b) × (a – b) = (6 + 4)(6 – 4) = 10 × 2 = 20.
R.H.S = a² – b² = 6² – 4² = 36 – 16 = 20.
Therefore, Pattern 2 holds for counting numbers.

(ii) Let a = -10, b = -5
L.H.S = (a + b) × (a – b) = [(-10) + (-5)] × [(-10) – (-5)] = [-15] × [-5] = 75.
R.H.S = a² – b² = [(-10)² – (-5)²] = 100 – 25 = 75.
Therefore, Pattern 2 holds for negative numbers.

(iii)

Therefore, Pattern 2 holds for fractions.

Textbook Page 154

Figure it Out

1. Compute these products using the suggested identity.
(i) 46² using Identity 1A for (a + b)²
(ii) 397 × 403 using Identity 1C for (a + b) (a – b)
(iii) 91² using Identity 1B for (a – b)²
(iv) 43 × 45 using Identity 1C for (a + b) (a – b)

Solution:
(i) Identity (a + b)² = a² + 2ab + b²
46² = (40 + 6)² = (40)² + (6)² + 2 × 40 × 6 = 1600 + 36 + 480 = 2116.

(ii) Identity (a + b) (a – b) = a² – b²
397 × 403 = (400 – 3) (400 + 3) = 400² – 3² = 160000 – 9 = 159991.

(iii) Identity (a – b)² = a² – 2ab + b²
91² = (100 – 9)² = (100)² + (9)² – 2 × 100 × 9 = 10000 + 81 – 1800 = 8281.

(iv) Identity (a + b) (a – b) = a² – b²
43 × 45 = (44 – 1) (44 + 1) = 44² – 1² = 1936 – 1 = 1935.

2. Use either a suitable identity or the distributive property to find each of the following products.
(i) (p – 1) (p + 11)
(ii) (3a – 9b) (3a + 9b)
(iii) –(2y + 5) (3y + 4)
(iv) (6x + 5y)²
(v) (2x – 1/2)²
(vi) (7p) × (3r) × (p + 2)

Solution:
(i) (p – 1) (p + 11) = p(p + 11) – 1(p + 11) = p² + 11p – p – 11 = p² + 10p – 11.

(ii) (3a – 9b) (3a + 9b) = (3a)² – (9b)² = 9a² – 81b².

(iii) –(2y + 5) (3y + 4) = – [ 2y(3y + 4) + 5(3y + 4) ] = – [ 6y² + 8y + 15y + 20 ] = – [ 6y² + 23y + 20 ] = -6y² – 23y – 20.

(iv) (6x + 5y)² = (6x)² + (5y)² + 2 × 6x × 5y = 36x² + 25y² + 60xy.

(v) (2x – 1/2)²:

(vi) (7p) × (3r) × (p + 2) = 21pr (p + 2) = 21p²r + 42pr.

3. For each statement identify the appropriate algebraic expression(s).
(i) Two more than a square number.
Options: 2 + s ; (s + 2)² ; s² + 2 ; s² + 4 ; 2s² ; 2²s
Solution: Let the number be s. Square number = s². Therefore: s² + 2.

(ii) The sum of the squares of two consecutive numbers.
Options: m² + n² ; (m + n)² ; m² + 1 ; m² + (m + 1)² ; m² + (m – 1)² ; (m + (m + 1))² ; (2m)² + (2m + 1)²
Solution: Let the two consecutive numbers be m and (m + 1). Therefore: m² + (m + 1)² (or m² + (m – 1)²).

4. Consider any 2 by 2 square of numbers in a calendar, as shown in the figure.

Find products of numbers lying along each diagonal — 4 × 12 = 48, 5 × 11 = 55. Do this for the other 2 by 2 squares. What do you observe about the diagonal products? Explain why this happens.
Hint: Label the numbers in each 2 by 2 square as:

Solution:

(i)

First diagonal: 16 × 24 = 384.
Second diagonal: 23 × 17 = 391.
Difference: 391 – 384 = 7.

(ii)

First diagonal: 7 × 15 = 105.
Second diagonal: 14 × 8 = 112.
Difference: 112 – 105 = 7.
We observe that the difference of the diagonal products of any 2 by 2 square of numbers from the calendar is always 7.

5. Verify which of the following statements are true.

(i) (k + 1) (k + 2) – (k + 3) is always 2.
Solution: False.
Explanation: (k + 1) (k + 2) – (k + 3) = k² + 3k + 2 – k – 3 = k² + 2k – 1.
If k = 2, value = 4 + 4 – 1 = 7 ≠ 2.

(ii) (2q + 1) (2q – 3) is a multiple of 4.
Solution: False.
Explanation: (2q + 1) (2q – 3) = 4q² – 4q – 3 = 4(q² – q) – 3.
Leaves a remainder of 1 (or -3), so it is not a multiple of 4.

(iii) Squares of even numbers are multiples of 4, and squares of odd numbers are 1 more than multiples of 8.
Solution: True.
Explanation:
1. Even numbers: (2n)² = 4n² (multiple of 4).
2. Odd numbers: (2n + 1)² = 4n² + 4n + 1 = 4n(n + 1) + 1. Since n(n + 1) is always even (say 2k), this equals 4(2k) + 1 = 8k + 1 (1 more than a multiple of 8).

(iv) (6n + 2)² – (4n + 3)² is 5 less than a square number.
Solution: False.
Explanation: (36n² + 24n + 4) – (16n² + 24n + 9) = 20n² – 5. Since 20 is not a square number, 20n² is not necessarily a square number.

6. A number leaves a remainder of 3 when divided by 7, and another number leaves a remainder of 5 when divided by 7. What is the remainder when their sum, difference, and product are divided by 7?

Solution:
Let the numbers be 7a + 3 and 7b + 5.
(i) Sum: (7a + 3) + (7b + 5) = 7(a + b) + 8 = 7(a + b + 1) + 1 → Remainder is 1.
(ii) Difference: (7a + 3) − (7b + 5) = 7(a − b) − 2 = 7(a – b – 1) + 5 → Remainder is 5.
(iii) Product: (7a + 3)(7b + 5) = 49ab + 35a + 21b + 15 = 7(7ab + 5a + 3b + 2) + 1 → Remainder is 1.

7. Choose three consecutive numbers, square the middle one, and subtract the product of the other two. Repeat the same with other sets of numbers. What pattern do you notice? How do we write this as an algebraic equation? Expand both sides of the equation to check that it is a true identity.

Solution:
Let three consecutive numbers be (n − 1), n, (n + 1).
n² − (n − 1)(n + 1) = n² − (n² − 1) = 1.
Therefore, the result is always 1.
Algebraic identity: n² − (n − 1)(n + 1) = 1.
Check: With 2, 3, 4: 3² − (2 × 4) = 9 − 8 = 1.

8. What is the algebraic expression describing the following steps—add any two numbers. Multiply this by half of the sum of the two numbers? Prove that this result will be half of the square of the sum of the two numbers.

Solution:
Let the two numbers be a and b.
Sum = (a + b).
Multiplying by half of the sum = (a + b) × 1/2(a + b) = 1/2(a + b)².
Therefore, the result is indeed half of the square of the sum of the two numbers.

9. Which is larger? Find out without fully computing the product.
(i) 14 × 26 or 16 × 24
(ii) 25 × 75 or 26 × 74

Solution:
(i) Let a = 14 × 26 and b = 16 × 24 = (14 + 2)(26 – 2) = 14 × 26 + 2 × 26 – 14 × 2 – 4 = a + 20.
∴ b > a → 16 × 24 > 14 × 26.

(ii) Let a = 25 × 75 and b = 26 × 74 = (25 + 1)(75 – 1) = 25 × 75 + 75 – 25 – 1 = a + 49.
∴ b > a → 26 × 74 > 25 × 75.

10. A tiny park is coming up in Dhauli. The plan is shown in the figure. The two square plots, each of area g² sq. ft., will have a green cover. All the remaining area is a walking path w ft. wide that needs to be tiled. Write an expression for the area that needs to be tiled.

Solution:
Side of each green square plot = g ft.
Length of the park = w + g + w + w + g + w = (4w + 2g) ft.
Breadth of the park = w + g + w = (2w + g) ft.
Total area of the park = (4w + 2g)(2w + g) = 8w² + 8gw + 2g² sq. ft.
Area of two green plots = 2g² sq. ft.
Area to be tiled = (8w² + 8gw + 2g²) – 2g² = 8w² + 8gw sq. ft. (or 8w(w + g) sq. ft.).

11. For each pattern shown below,
(i) Draw the next figure in the sequence.
(ii) How many basic units are there in Step 10?
(iii) Write an expression to describe the number of basic units in Step y.

Solution: