Class 8 Mathematics · Chapter 6

6. We Distribute, Yet Things Multiply

Solutions to all Figure it Out and in-text questions. CBSE Class 8 Mathematics, NCERT Ganita Prakash Part 1.

6.1 Some Properties of Multiplication

1Consider 23 × 27. By how much does the product increase if (1) 23 is increased by 1? (2) 27 is increased by 1? (3) both are increased by 1?

(1) 24 × 27 = 23 × 27 + 27. The product increases by 27

(2) 23 × 28 = 23 × 27 + 23. The product increases by 23

(3) 24 × 28 = 23 × 27 + (23 + 27 + 1). The product increases by 51

2What would we get if we expanded (a + 1)(b + 1) by first taking (b + 1) as a single term?

(a + 1)(b + 1) = a(b + 1) + 1(b + 1) = ab + a + b + 1.

We get the same result: the increase is a + b + 1.

3(a + 1)(b − 1) = ab + b − a − 1. Will the product always increase? Find 3 examples where the product decreases.

No. The product increases only when b − a − 1 is positive, that is, when b is more than a + 1.

Examples where it decreases (b ≤ a):

a = 5, b = 3: 15 becomes 6 × 2 = 12.
a = 7, b = 7: 49 becomes 8 × 6 = 48.
a = 10, b = 2: 20 becomes 11 × 1 = 11.

4What happens when a and b are negative integers? Check with a = −5, b = 8 and a = −4, b = −5.

a = −5, b = 8: ab = −40. (a + 1)(b − 1) = −4 × 7 = −28. Change = b − a − 1 = 8 + 5 − 1 = 12 ✔.

a = −4, b = −5: ab = 20. (a + 1)(b − 1) = −3 × −6 = 18. Change = −5 + 4 − 1 = −2 ✔.

The identities hold for negative integers also.

5Use Identity 1 to find how the product changes when (i) one number is decreased by 2 and the other increased by 3; (ii) both are decreased, one by 3 and the other by 4.

(i) (a − 2)(b + 3) = ab + 3a − 2b − 6. The change is 3a − 2b − 6

(ii) (a − 3)(b − 4) = ab − 4a − 3b + 12. The change is −4a − 3b + 12

Check: a = 10, b = 8. (i) 8 × 11 = 88 = 80 + 30 − 16 − 6 ✔. (ii) 7 × 4 = 28 = 80 − 40 − 24 + 12 ✔.

6Expand (i) (a − u)(b + v) (ii) (a − u)(b − v).

(i) (a − u)(b + v) = ab + av − ub − uv

(ii) (a − u)(b − v) = ab − av − ub + uv

Figure it Out (Page 142)

1In the multiplication grid, the middle number of a 3 × 3 frame is pq. Write the expressions for the other numbers.

(p − 1)(q − 1)	(p − 1)q	(p − 1)(q + 1)
p(q − 1)	pq	p(q + 1)
(p + 1)(q − 1)	(p + 1)q	(p + 1)(q + 1)

2Expand the following products.

(i) (3 + u)(v − 3) = 3v − 9 + uv − 3u

(ii) 23(15 + 6a) = 10 + 4a

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

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

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

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

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

(a + 2)(b − 4) = ab − 4a + 2b − 8. For no change, −4a + 2b − 8 = 0, so b = 2a + 4.

Examples: 1 × 6 = 3 × 2 = 6; 2 × 8 = 4 × 4 = 16; 3 × 10 = 5 × 6 = 30.

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

(i) = 4a + 4ab − 12b² + ab + ab² − 3b³ = 4a + 5ab + ab² − 12b² − 3b³

(ii) = 4y² + 44yz − 12y + 7y + 77z − 21 = 4y² + 44yz − 5y + 77z − 21

5Expand (i) (a − b)(a + b) (ii) (a − b)(a² + ab + b²) (iii) (a − b)(a³ + a²b + ab² + b³). Do you see a pattern?

(i) a² + ab − ab − b² = a² − b²

(ii) a³ + a²b + ab² − a²b − ab² − b³ = a³ − b³

(iii) Similarly, all middle terms cancel. Result = a⁴ − b⁴

Next identity: (a − b)(a⁴ + a³b + a²b² + ab³ + b⁴) = a⁵ − b⁵. On expanding, the middle terms cancel in pairs.

Fast Multiplication

7Describe a general rule to multiply a number by 11 in one line. Evaluate (i) 94 × 11 (ii) 495 × 11 (iii) 3279 × 11 (iv) 4791256 × 11.

Rule: Write the units digit as it is. Then, moving left, write the sum of each pair of neighbouring digits (carry over if the sum is 10 or more). Finally write the first digit (plus any carry).

(i) 94 × 11 = 9 | 9 + 4 | 4 = 9 | 13 | 4 = 1034

(ii) 495 × 11 = 4 | 13 | 14 | 5 = 5445

(iii) 3279 × 11 = 36069

(iv) 4791256 × 11 = 5,27,03,816

Also, 5678 × 11 = 62458 and 3874 × 11 = 42614.

8Multiply 3874 by 101 in one line. What is a general rule for 101? Extend it to 1001, 10001, …

3874 × 101 = 387400 + 3874 = 391274

Rule for 101: Write the number, then add the same number shifted two places left (dcba00 + dcba). The last two digits stay as they are.

For 1001: add the number shifted three places (dcba000 + dcba). For 10001: shift four places, and so on.

9Find (i) 89 × 101 (ii) 949 × 101 (iii) 265831 × 1001 (iv) 1111 × 1001 (v) 9734 × 99 (vi) 23478 × 999.

(i) 8900 + 89 = 8989

(ii) 94900 + 949 = 95849

(iii) 265831000 + 265831 = 26,60,96,831

(iv) 1111000 + 1111 = 11,12,111

(v) 9734 × (100 − 1) = 973400 − 9734 = 9,63,666

(vi) 23478 × (1000 − 1) = 23478000 − 23478 = 2,34,54,522

6.2 Special Cases of the Distributive Property

10What if we write 65² as (30 + 35)² or (52 + 13)²? Check the area.

(30 + 35)² = 900 + 2 × 30 × 35 + 1225 = 900 + 2100 + 1225 = 4225.

(52 + 13)² = 2704 + 2 × 52 × 13 + 169 = 2704 + 1352 + 169 = 4225.

We get the same area, 4225 sq. units.

11Is (a + b)² always greater than a² + b²? If not, when is it greater?

(a + b)² = a² + b² + 2ab. So it is greater only when 2ab is positive.

That is, when a and b are both positive or both negative. If one of them is 0, both are equal. If one is negative and the other positive, (a + b)² is smaller.

12Use Identity 1A to find 104² and 37².

104² = (100 + 4)² = 10000 + 800 + 16 = 10816

37² = (30 + 7)² = 900 + 420 + 49 = 1369

13Write the expressions for (i) (m + 3)² (ii) (6 + p)². Expand (3j + 2k)².

(i) m² + 6m + 9

(ii) 36 + 12p + p²

(3j + 2k)² = (3j)² + 2 × 3j × 2k + (2k)² = 9j² + 12jk + 4k²

By distributive property: (3j + 2k)(3j + 2k) = 9j² + 6jk + 6jk + 4k² = 9j² + 12jk + 4k².

14Find the general expansion of (a − b)² using geometry.

Take a square of side a. Cut off two strips of size a × b along two sides. This removes 2ab, but the small b × b corner square is removed twice.

So we add it back once: (a − b)² = a² − 2ab + b².

15Use (a − b)² to find (a) 99² (b) 58².

(a) (100 − 1)² = 10000 − 200 + 1 = 9801

(b) (60 − 2)² = 3600 − 240 + 4 = 3364

16Expand using Identity 1B: (i) (b − 6)² (ii) (−2a + 3)² (iii) (7y − 34z)²

(i) b² − 12b + 36

(ii) (3 − 2a)² = 9 − 12a + 4a²

(iii) 49y² − 2 × 7y × 34z + 916z² = 49y² − 21y2z + 916z²

17Pattern 1: Take a pair of numbers. Can you write twice the sum of their squares as a sum of two squares?

Yes. 2(a² + b²) = (a + b)² + (a − b)².

Example: 2(7² + 3²) = 2 × 58 = 116 = 10² + 4².

18Use Identity 1C to calculate 98 × 102 and 45 × 55. Why is a² = (a + b)(a − b) + b² true?

98 × 102 = (100 − 2)(100 + 2) = 10000 − 4 = 9996

45 × 55 = (50 − 5)(50 + 5) = 2500 − 25 = 2475

(a + b)(a − b) + b² = a² − b² + b² = a². So it is true.

19Show that (a + b)(a − b) = a² − b² geometrically.

From a square of side a, cut away a small square of side b from a corner. The remaining area is a² − b².

Cut the remaining L-shape into two rectangles. Move one rectangle and join it to the other. We get one rectangle of length (a + b) and width (a − b). So a² − b² = (a + b)(a − b).

Figure it Out (Page 149)

1Which is greater: (a − b)² or (b − a)²? Justify.

Both are equal. b − a = −(a − b), and (−x)² = x². Also, both expand to a² − 2ab + b².

2Express 100 as the difference of two squares.

100 = (a + b)(a − b). Take a + b = 50 and a − b = 2. Then a = 26, b = 24.

So 100 = 26² − 24². (Also 100 = 10² − 0².)

3Find 406², 72², 145², 1097² and 124² using the identities.

406² = (400 + 6)² = 160000 + 4800 + 36 = 164836

72² = (70 + 2)² = 4900 + 280 + 4 = 5184

145² = (150 + 5)(150 − 5) + 25 → 150 × 140 + 25 = 21025

1097² = (1100 − 3)² = 1210000 − 6600 + 9 = 1203409

124² = (120 + 4)² = 14400 + 960 + 16 = 15376

4Do Patterns 1 and 2 hold only for counting numbers? Do they hold for negative integers and fractions?

They hold for all numbers. The identities come from the distributive property, which is true for negative integers and fractions too.

Example: a = −3, b = 12: (a + b)(a − b) = (−52)(−72) = 354 and a² − b² = 9 − 14 = 354.

6.3 Mind the Mistake, Mend the Mistake

20Check each simplification. Explain the mistake and write the correct expression.

1. −3p(−5p + 2q): −3p was added instead of multiplied. Correct: 15p² − 6pq
2. 2(x − 1) + 3(x + 4): 2 and 3 were not multiplied with −1 and 4. Correct: 2x − 2 + 3x + 12 = 5x + 10
3. y + 2(y + 2): this is not (y + 2)². Correct: y + 2y + 4 = 3y + 4
4. (5m + 6n)²: the middle term is missing. Correct: 25m² + 60mn + 36n²
5. (−q + 2)² = q² − 4q + 4: Correct
6. 3a(2b × 3c): multiplication does not distribute over multiplication. Correct: 3a × 6bc = 18abc
7. 12(10s − 6) + 3 = 5s − 3 + 3 = 5s: Correct
8. 5w² + 6w: these are unlike terms and cannot be added. Correct: 5w² + 6w (or w(5w + 6))
9. 2a³ + 3a³ + 6a²b + 6ab²: 6a²b and 6ab² are unlike terms. Correct: 5a³ + 6a²b + 6ab²
10. (x + 2)(x + 5) = x² + 7x + 10: Correct
11. (a + 2)(b + 4): only first terms and last terms were multiplied. Correct: ab + 4a + 2b + 8
12. ab² + a²b + a²b² = ab(a + b + ab): Correct

6.4 This Way or That Way, All Ways Lead to the Bay

21Circle pattern: Draw the next figure. How many circles in Step 10? Write an expression for Step k. Find Step 15.

Step 1 = 3, Step 2 = 8, Step 3 = 15, so Step 4 = 24 circles (a 5 × 5 square with one circle missing).

Step k = k² + 2k.

Step 10 = 100 + 20 = 120

Step 15 = 225 + 30 = 255

22Square tile pattern: How many tiles in each figure? In Step 4 and Step 10? Write an expression for Step n.

The tiles form a border around a square. Step 1 = 3² − 1² = 8, Step 2 = 4² − 2² = 12, Step 3 = 5² − 3² = 16.

Step 4 = 6² − 4² = 20. Step 10 = 12² − 10² = 44

Step n = (n + 2)² − n² = 4n + 4

Another method: 4 sides of n + 1 tiles each = 4(n + 1).

23Check that (m + n)² − 4mn = (n − m)².

(m + n)² − 4mn = m² + 2mn + n² − 4mn = m² − 2mn + n² = (n − m)² ✔

24Verify that x² − xy, x(x + 2y) − 3xy and x(x − y) are equivalent. If x = 8 and y = 3, find the area.

x(x + 2y) − 3xy = x² + 2xy − 3xy = x² − xy. Also x(x − y) = x² − xy. All three are the same.

Area = 8 × (8 − 3) = 40 sq. units

25Write an expression for the area of the dashed region. Substitute p = 6, r = 3.5 and s = 9.

Method 1: The dashed region is a rectangle of length (s − r) and width (p − r). Area = (s − r)(p − r).

Method 2: Whole rectangle minus the L-shaped strip = ps − (rs + rp − r²) = ps − rs − rp + r².

Area = (9 − 3.5)(6 − 3.5) = 5.5 × 2.5 = 13.75 sq. units

Figure it Out (Page 154)

1Compute these products using the suggested identity.

(i) 46² = (40 + 6)² = 1600 + 480 + 36 = 2116

(ii) 397 × 403 = (400 − 3)(400 + 3) = 160000 − 9 = 159991

(iii) 91² = (100 − 9)² = 10000 − 1800 + 81 = 8281

(iv) 43 × 45 = (44 − 1)(44 + 1) = 1936 − 1 = 1935

2Use a suitable identity or the distributive property to find each product.

(i) (p − 1)(p + 11) = p² + 11p − p − 11 = p² + 10p − 11

(ii) (3a − 9b)(3a + 9b) = 9a² − 81b²

(iii) −(2y + 5)(3y + 4) = −(6y² + 8y + 15y + 20) = −6y² − 23y − 20

(iv) (6x + 5y)² = 36x² + 60xy + 25y²

(v) (2x − 12)² = 4x² − 2x + 14

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

3Identify the appropriate algebraic expression(s).

(i) Two more than a square number: s² + 2

(ii) The sum of the squares of two consecutive numbers: m² + (m + 1)² and m² + (m − 1)²

4In a calendar, find the products of numbers along each diagonal of a 2 by 2 square. What do you observe? Explain.

Example: 3 × 11 = 33 and 4 × 10 = 40. The difference is 7.

Label the square a, a + 1, a + 7, a + 8.

(a + 1)(a + 7) − a(a + 8) = a² + 8a + 7 − a² − 8a = 7

So the two diagonal products always differ by 7.

5Verify which of the following statements are true.

(i) (k + 1)(k + 2) − (k + 3) is always 2.

False. It equals k² + 2k − 1. For k = 1, it is 2; for k = 2, it is 7.

(ii) (2q + 1)(2q − 3) is a multiple of 4.

False. It equals 4q² − 4q − 3, which is 3 less than 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.

True. (2n)² = 4n². (2n + 1)² = 4n(n + 1) + 1, and n(n + 1) is even, so 4n(n + 1) is a multiple of 8.

(iv) (6n + 2)² − (4n + 3)² is 5 less than a square number.

False. It equals 20n² − 5. But 20n² is not a square (for n ≠ 0). For n = 1, we get 15, and 20 is not a square.

6A number leaves remainder 3 and another leaves remainder 5 when divided by 7. Find the remainders of their sum, difference and product.

Let the numbers be 7a + 3 and 7b + 5.

Sum = 7(a + b) + 8 = 7(a + b + 1) + 1 → remainder 1

Difference (7b + 5) − (7a + 3) = 7(b − a) + 2 → remainder 2. (The other way, it is 2 short of a multiple of 7, so remainder 5.)

Product = 49ab + 35a + 21b + 15 = 7(7ab + 5a + 3b + 2) + 1 → remainder 1

7Choose three consecutive numbers, square the middle one and subtract the product of the other two. What pattern do you notice?

Example: 5² − 4 × 6 = 25 − 24 = 1. 10² − 9 × 11 = 1. The answer is always 1

Algebra: n² − (n − 1)(n + 1) = n² − (n² − 1) = 1.

8Add any two numbers. Multiply this by half of the sum of the two numbers. Prove that the result is half of the square of the sum.

Expression: (a + b) × 12(a + b) = 12(a + b)².

This is half of the square of the sum.

9Which is larger? Find out without fully computing the product.

(i) 14 × 26 = (20 − 6)(20 + 6) = 400 − 36. 16 × 24 = (20 − 4)(20 + 4) = 400 − 16. So 16 × 24 is larger.

(ii) 25 × 75 = (50 − 25)(50 + 25) = 2500 − 625. 26 × 74 = (50 − 24)(50 + 24) = 2500 − 576. So 26 × 74 is larger.

10A park has two square plots of area g² sq. ft. each. The remaining area is a path w ft. wide. Write an expression for the area to be tiled.

From the figure, the path is w ft. wide along the edges and 2w ft. wide between the two plots.

Length of park = w + g + 2w + g + w = 2g + 4w. Breadth = g + 2w.

Tiled area = (2g + 4w)(g + 2w) − 2g² = 2g² + 8gw + 8w² − 2g² = 8gw + 8w² sq. ft.

If the path between the plots is only w ft. wide, the length is 2g + 3w and the tiled area is 7gw + 6w² sq. ft.

11For each pattern: (i) draw the next figure (ii) find the basic units in Step 10 (iii) write an expression for Step y.

(a) Yellow pattern

It is a block of y rows and (y + 2) columns with two tails of (y + 2) squares each.

Step 1 = 9, Step 2 = 16, so Step 4 = 6² = 36.

Step y = y(y + 2) + 2(y + 2) = (y + 2)²

Step 10 = 12² = 144

(b) Blue pattern

Step 1 = 2² + 1 = 5, Step 2 = 3² + 2 = 11, Step 3 = 4² + 3 = 19. Step 4 = 5² + 4 = 29 (a 5 × 5 square with a row of 4 below).

Step y = (y + 1)² + y

Step 10 = 121 + 10 = 131

In the textbook figure, Step 3 of pattern (a) seems to show one extra square in each tail. The intended pattern 9, 16, 25, … gives (y + 2)².

Puzzle: Coin Conjoin

1Find the minimum moves needed to flip triangles of 10 and 15 coins. Is there a simple way to calculate the minimum for any triangle?

10 coins: move the 3 corner coins. 3 moves

15 coins: 5 moves

Rule: minimum moves = the number of coins ÷ 3, ignoring the remainder. For example, 3 → 1, 6 → 2, 10 → 3, 15 → 5, 21 → 7.

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