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NCERT Solutions · Class 8 Maths · Ganita Prakash Part 1 · Chapter 6

Chapter 6: We Distribute, Yet Things Multiply (Algebra)

Step-by-step answers to every "Figure it Out" and in-text question of Chapter 6, We Distribute, Yet Things Multiply (NCERT Class 8 Maths, Ganita Prakash Part 1, 2026-27): increments in products, expanding products, quick multiplication by 11, 101 and 99, the identities (a ± b)² and (a + b)(a − b), mending algebra mistakes, and pattern and area problems. All 43 questions are answered, with the key answer highlighted.

6.1 Some Properties of Multiplication

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? Is there a general pattern?

Solution

$23 \times 27 = 621$ .

$24 \times 27 = 648$ , so the product increases by 27 (the other number).
$23 \times 28 = 644$ , so it increases by 23.
$24 \times 28 = 672$ , so it increases by 51 $= 23 + 27 + 1$ .

In general, $a (b + 1) = ab + a$ and $(a + 1) (b + 1) = ab + (a + b + 1)$ .

The increases are 27, 23 and 51. Increasing one number by 1 adds the other number; increasing both by 1 adds a + b + 1.

What do we get if we expand (a + 1)(b + 1) by first taking (b + 1) as a single term?

Solution

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

This is the same result as before, so the product again increases by $a + b + 1$ .

(a + 1)(b + 1) = ab + a + b + 1, the same as before.

Will the product always increase when one number is increased by 1 and the other decreased by 1? Find 3 examples where it decreases. What happens for negative integers, e.g. a = −5, b = 8 and a = −4, b = −5?

Solution

$(a + 1) (b - 1) = ab + b - a - 1$ , so the change is $b - a - 1$ . The product decreases whenever $b - a - 1 < 0$ , that is, when $b \leq a$ .

$a = 5, b = 3$ : $6 \times 2 = 12 < 15$
$a = 10, b = 10$ : $11 \times 9 = 99 < 100$
$a = 7, b = 4$ : $8 \times 3 = 24 < 28$

Negative integers. The identities still hold:

$a = - 5, b = 8$ : $ab = - 40$ and $(a + 1) (b + 1) = (- 4) (9) = - 36$ , an increase of $4 = a + b + 1$ . Also $(a + 1) (b - 1) = (- 4) (7) = - 28$ , a change of $12 = b - a - 1$ .
$a = - 4, b = - 5$ : $ab = 20$ and $(a + 1) (b + 1) = (- 3) (- 4) = 12$ , a change of $- 8 = a + b + 1$ . Also $(a + 1) (b - 1) = (- 3) (- 6) = 18$ , a change of $- 2 = b - a - 1$ .

No. The change is b − a − 1, so the product falls whenever b ≤ a (e.g. 6 × 2 < 5 × 3, 11 × 9 < 10 × 10, 8 × 3 < 7 × 4). The same expressions work for negative integers.

Use Identity 1 to find how the product changes when (i) one number is decreased by 2 and the other increased by 3; (ii) both numbers are decreased, one by 3 and the other by 4. Verify without converting subtractions to additions.

Solution

(i) Take $m = - 2$ , $n = 3$ in $(a + m) (b + n) = ab + mb + an + mn$ :

$(a - 2) (b + 3) = ab - 2 b + 3 a - 6$

The product changes by $3 a - 2 b - 6$ .

Check with 23 × 27: $21 \times 30 = 630$ and $621 + 69 - 54 - 6 = 630$ ✓

(ii) Take $m = - 3$ , $n = - 4$ :

$(a - 3) (b - 4) = ab - 4 a - 3 b + 12$

The product changes by $- 4 a - 3 b + 12$ .

Check: $20 \times 23 = 460$ and $621 - 92 - 81 + 12 = 460$ ✓

Directly: $(a - 3) (b - 4) = (a - 3) b - (a - 3) 4 = ab - 3 b - 4 a + 12$ , the same.

(i) Change = 3a − 2b − 6; (ii) change = −4a − 3b + 12.

Figure it Out (page 142)

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

Solution

The rows go up by 1 (row number $p - 1, p, p + 1$ ) and the columns go up by 1 ( $q - 1, q, q + 1$ ). For example, 4 × 6 has the frame 3 × 5 to 5 × 7.

(p − 1)(q − 1)

(p − 1)q

(p − 1)(q + 1)

p(q − 1)

p(q + 1)

(p + 1)(q − 1)

(p + 1)q

(p + 1)(q + 1)

Expanded: $pq - p - q + 1$ , $pq - q$ , $pq + p - q - 1$ ; $pq - p$ , $pq$ , $pq + p$ ; $pq - p + q - 1$ , $pq + q$ , $pq + p + q + 1$ .

Top row: (p − 1)(q − 1), (p − 1)q, (p − 1)(q + 1); middle row: p(q − 1), pq, p(q + 1); bottom row: (p + 1)(q − 1), (p + 1)q, (p + 1)(q + 1).

Expand: (i) (3 + u)(v − 3) (ii) ⅔(15 + 6a) (iii) (10a + b)(10c + d) (iv) (3 − x)(x − 6) (v) (−5a + b)(c + d) (vi) (5 + z)(y + 9)

Solution

$(3 + u) (v - 3) = 3 v - 9 + uv - 3 u$
$\frac{2}{3} (15 + 6 a) = 10 + 4 a$
$(10 a + b) (10 c + d) = 100 a c + 10 a d + 10 b c + b d$
$(3 - x) (x - 6) = 3 x - 18 - x^{2} + 6 x = - x^{2} + 9 x - 18$
$(- 5 a + b) (c + d) = - 5 a c - 5 a d + b c + b d$
$(5 + z) (y + 9) = 5 y + 45 + y z + 9 z$

(i) uv + 3v − 3u − 9 (ii) 10 + 4a (iii) 100ac + 10ad + 10bc + bd (iv) −x² + 9x − 18 (v) −5ac − 5ad + bc + bd (vi) yz + 5y + 9z + 45

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

$(a + 2) (b - 4) = ab - 4 a + 2 b - 8$ . This equals $ab$ when $- 4 a + 2 b - 8 = 0$ , that is, when $b = 2 a + 4$ .

$a = 1, b = 6$ : $1 \times 6 = 6$ and $3 \times 2 = 6$
$a = 2, b = 8$ : $2 \times 8 = 16$ and $4 \times 4 = 16$
$a = 3, b = 10$ : $3 \times 10 = 30$ and $5 \times 6 = 30$

Any pair with b = 2a + 4 works: 1 × 6 = 3 × 2, 2 × 8 = 4 × 4, 3 × 10 = 5 × 6.

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

Solution

(i) Multiply each term of the first bracket by 4 and by $b$ :

$4 a + 4 ab - 12 b^{2} + ab + a b^{2} - 3 b^{3} = 4 a + 5 ab - 12 b^{2} + a b^{2} - 3 b^{3}$

(ii)

$4 y^{2} + 44 y z - 12 y + 7 y + 77 z - 21 = 4 y^{2} + 44 y z - 5 y + 77 z - 21$

(i) 4a + 5ab − 12b² + ab² − 3b³ (ii) 4y² + 44yz − 5y + 77z − 21

Expand (i) (a − b)(a + b) (ii) (a − b)(a² + ab + b²) (iii) (a − b)(a³ + a²b + ab² + b³). Do you see a pattern? What is the next identity?

Solution

$(a - b) (a + b) = a^{2} + ab - ab - b^{2} = a^{2} - b^{2}$
$(a - b) (a^{2} + ab + b^{2}) = a^{3} + a^{2} b + a b^{2} - a^{2} b - a b^{2} - b^{3} = a^{3} - b^{3}$
$(a - b) (a^{3} + a^{2} b + a b^{2} + b^{3}) = a^{4} - b^{4}$ (all the middle terms cancel in pairs)

Pattern: $(a - b) \times$ (all terms of degree $n - 1$ in $a$ and $b$ ) $= a^{n} - b^{n}$ .

Next identity: $(a - b) (a^{4} + a^{3} b + a^{2} b^{2} + a b^{3} + b^{4}) = a^{5} - b^{5}$ .

To check it, multiply by $a$ to get $a^{5} + a^{4} b + a^{3} b^{2} + a^{2} b^{3} + a b^{4}$ . Multiply by $- b$ to get $- a^{4} b - a^{3} b^{2} - a^{2} b^{3} - a b^{4} - b^{5}$ . Adding the two leaves $a^{5} - b^{5}$ ✓

a² − b², a³ − b³, a⁴ − b⁴; next: (a − b)(a⁴ + a³b + a²b² + ab³ + b⁴) = a⁵ − b⁵.

Fast Multiplications Using the Distributive Property

Multiply by 11 in one line: (a) 3874 × 11 (b) 5678 × 11. Describe a general rule for multiplying any number by 11.

Solution

$number \times 11 = number \times 10 + number$ , so each digit of the answer is the sum of two neighbouring digits.

Rule:

Write the units digit.
Moving left, write the sum of each digit and its right-hand neighbour, carrying 1 whenever a sum is 10 or more.
Finally write the first digit, plus any carry.

(a) 3874 → 3 | 3+8 | 8+7 | 7+4 | 4 = 3 | 11 | 15 | 11 | 4. With carries this gives 42614.

(b) 5678 → 5 | 11 | 13 | 15 | 8. With carries this gives 62458.

3874 × 11 = 42614, 5678 × 11 = 62458. Rule: keep the end digits and write the sums of neighbouring digits in between, carrying where needed.

Evaluate (i) 94 × 11 (ii) 495 × 11 (iii) 3279 × 11 (iv) 4791256 × 11

Solution

9 | 13 | 4 → 1034
4 | 13 | 14 | 5 → 5445
3 | 5 | 9 | 16 | 9 → 36069
4 | 11 | 16 | 10 | 3 | 7 | 11 | 6 → 52703816

(i) 1034 (ii) 5445 (iii) 36069 (iv) 52703816

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

Solution

$d c ba \times 101 = d c ba 00 + d c ba$ . In each column, a digit is added to the digit two places to its left:

$3874 \times 101$ : 3 | 8 | 7+3 | 4+8 | 7 | 4 = 3 | 8 | 10 | 12 | 7 | 4. With carries this gives 391274.

Rule for 101: write the number twice, the second copy shifted 2 places to the right, and add.

Rule for 1001: shift 3 places. For 10001: shift 4 places, and so on. A number with fewer digits than the shift is simply written twice, as in 89 × 101 = 8989.

3874 × 101 = 391274. For 101, 1001, 10001, … add the number to itself shifted 2, 3, 4, … places.

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

Solution

$8900 + 89 =$ 8989
$94900 + 949 =$ 95849
$265831000 + 265831 =$ 266096831
$1111000 + 1111 =$ 1112111
$9734 \times (100 - 1) = 973400 - 9734 =$ 963666
$23478 \times (1000 - 1) = 23478000 - 23478 =$ 23454522

(i) 8989 (ii) 95849 (iii) 266096831 (iv) 1112111 (v) 963666 (vi) 23454522

6.2 Special Cases of the Distributive Property

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

Solution

In each case, split the square into two squares and two equal rectangles:

$(30 + 35)^{2} = 3 0^{2} + 2 (30 \times 35) + 3 5^{2} = 900 + 2100 + 1225 = 4225$
$(52 + 13)^{2} = 5 2^{2} + 2 (52 \times 13) + 1 3^{2} = 2704 + 1352 + 169 = 4225$

Both give 4225 sq. units, the same as (60 + 5)².

If a and b are any two integers, is (a + b)² always greater than a² + b²? If not, when is it greater?

Solution

$(a + b)^{2} = a^{2} + b^{2} + 2 ab$ , so the answer depends on the sign of $ab$ :

It is greater when $ab > 0$ , that is, when $a$ and $b$ are both positive or both negative. Example: $(2 + 3)^{2} = 25 > 13$ .
It is equal when $a = 0$ or $b = 0$ .
It is smaller when $a$ and $b$ have opposite signs. Example: $(5 + (- 2))^{2} = 9 < 29$ .

Not always. (a + b)² > a² + b² only when ab > 0 (a and b non-zero and of the same sign).

Use Identity 1A to find 104² and 37².

Solution

$10 4^{2} = (100 + 4)^{2} = 10000 + 800 + 16 =$ 10816
$3 7^{2} = (30 + 7)^{2} = 900 + 420 + 49 =$ 1369

104² = 10816, 37² = 1369

Write the expressions for (i) (m + 3)² (ii) (6 + p)². Expand (3j + 2k)² using the identity and using the distributive property.

Solution

$(m + 3)^{2} = m^{2} + 6 m + 9$
$(6 + p)^{2} = 36 + 12 p + p^{2}$

By the identity: $(3 j + 2 k)^{2} = (3 j)^{2} + 2 (3 j) (2 k) + (2 k)^{2} = 9 j^{2} + 12 j k + 4 k^{2}$ .

By distributing: $(3 j + 2 k) (3 j + 2 k) = 9 j^{2} + 6 j k + 6 j k + 4 k^{2} = 9 j^{2} + 12 j k + 4 k^{2}$ .

m² + 6m + 9; p² + 12p + 36; (3j + 2k)² = 9j² + 12jk + 4k² both ways.

Find the expansion of (a − b)² using geometry, and use it to find 99² and 58².

Solution

Start with a square of side $a$ and cut off a strip of width $b$ along two adjacent sides. What is left is a square of side $a - b$ .

Removing two $a \times b$ rectangles takes away the $b \times b$ corner square twice, so we add it back once:

$(a - b)^{2} = a^{2} - ab - ab + b^{2} = a^{2} - 2 ab + b^{2}$

$9 9^{2} = (100 - 1)^{2} = 10000 - 200 + 1 =$ 9801
$5 8^{2} = (60 - 2)^{2} = 3600 - 240 + 4 =$ 3364

(a − b)² = a² − 2ab + b²; 99² = 9801, 58² = 3364.

Expand using Identity 1B and the distributive property: (i) (b − 6)² (ii) (−2a + 3)² (iii) (7y − ¾z)²

Solution

$(b - 6)^{2} = b^{2} - 12 b + 36$
$(- 2 a + 3)^{2} = (3 - 2 a)^{2} = 9 - 12 a + 4 a^{2}$
$(7 y - \frac{3}{4} z)^{2} = 49 y^{2} - 2 \cdot 7 y \cdot \frac{3}{4} z + \frac{9}{16} z^{2} = 49 y^{2} - \frac{21}{2} y z + \frac{9}{16} z^{2}$

Distributing gives the same results. For example, $(b - 6) (b - 6) = b^{2} - 6 b - 6 b + 36$ .

(i) b² − 12b + 36 (ii) 4a² − 12a + 9 (iii) 49y² − (21/2)yz + (9/16)z²

Pattern 1 and Pattern 2: twice a sum of two squares is a sum of two squares; and 9 × 9 − 1 × 1 = 10 × 8, etc. Explain them.

Solution

Pattern 1: $(a + b)^{2} + (a - b)^{2} = 2 (a^{2} + b^{2})$ . For example, $2 (4^{2} + 1^{2}) = 34 = 5^{2} + 3^{2}$ , and $2 (7^{2} + 2^{2}) = 106 = 9^{2} + 5^{2}$ .

Pattern 2: $a^{2} - b^{2} = (a + b) (a - b)$ . For example, $9^{2} - 1^{2} = 80 = 10 \times 8$ and $7^{2} - 2^{2} = 45 = 9 \times 5$ .

Pattern 1 is 2(a² + b²) = (a + b)² + (a − b)²; Pattern 2 is a² − b² = (a + b)(a − b).

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

Solution

$98 \times 102 = (100 - 2) (100 + 2) = 10000 - 4 =$ 9996
$45 \times 55 = (50 - 5) (50 + 5) = 2500 - 25 =$ 2475

Geometrically:

Remove a $b \times b$ square from the corner of an $a \times a$ square. The L-shape left has area $a^{2} - b^{2}$ .
Cut the L into two rectangles: $a \times (a - b)$ and $b \times (a - b)$ .
Move the small rectangle to the end of the big one. Together they form one rectangle of length $a + b$ and width $a - b$ .

So $a^{2} - b^{2} = (a + b) (a - b)$ .

Sridharacharya's form: $(a + b) (a - b) + b^{2} = a^{2} - b^{2} + b^{2} = a^{2}$ .

98 × 102 = 9996, 45 × 55 = 2475; a² = (a + b)(a − b) + b² because (a + b)(a − b) = a² − b².

Figure it Out (page 149)

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

Solution

$(b - a) = - (a - b)$ , and a number and its negative have the same square. Also, both expand to $a^{2} - 2 ab + b^{2}$ .

Neither: they are always equal.

Express 100 as the difference of two squares.

Solution

We need $(a + b) (a - b) = 100$ , where $a + b$ and $a - b$ are both even or both odd. The factor pairs that work are $50 \times 2$ and $10 \times 10$ :

$a + b = 50$ , $a - b = 2$ gives $a = 26$ , $b = 24$
$a + b = 10$ , $a - b = 10$ gives $a = 10$ , $b = 0$

100 = 26² − 24² (= 676 − 576), and also 100 = 10² − 0².

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

Solution

$(400 + 6)^{2} = 160000 + 4800 + 36 =$ 164836
$(70 + 2)^{2} = 4900 + 280 + 4 =$ 5184
$(150 - 5)^{2} = 22500 - 1500 + 25 =$ 21025
$(1100 - 3)^{2} = 1210000 - 6600 + 9 =$ 1203409
$(125 - 1)^{2} = 15625 - 250 + 1 =$ 15376

164836, 5184, 21025, 1203409, 15376

Do Patterns 1 and 2 hold only for counting numbers? What about negative integers and fractions?

Solution

Both patterns come from the distributive property, which holds for all numbers: integers, fractions and decimals. So both patterns hold for every $a$ and $b$ .

Example with $a = - 3$ , $b = \frac{1}{2}$ :

Pattern 1: $2 (9 + \frac{1}{4}) = \frac{37}{2}$ and $(- \frac{5}{2})^{2} + (- \frac{7}{2})^{2} = \frac{25 + 49}{4} = \frac{37}{2}$ ✓
Pattern 2: $9 - \frac{1}{4} = \frac{35}{4}$ and $(- \frac{5}{2}) (- \frac{7}{2}) = \frac{35}{4}$ ✓

They hold for all numbers: negative integers and fractions too, because they are identities.

6.3 Mind the Mistake, Mend the Mistake

Check each of the 12 simplifications, explain any mistake, and write the correct expression.

Solution

#	Given	Verdict and what went wrong	Correct result
1	−3p(−5p + 2q) = p − 2q	Wrong: −3p was added to the first term instead of multiplying it	15p² − 6pq
2	2(x − 1) + 3(x + 4) = 5x + 3	Wrong: 2 and 3 were multiplied only with x, not with −1 and 4	5x + 10
3	y + 2(y + 2) = (y + 2)²	Wrong: y + 2(…) is not (y + 2) × (…)	3y + 4
4	(5m + 6n)² = 25m² + 36n²	Wrong: the middle term 2 × 5m × 6n is missing	25m² + 60mn + 36n²
5	(−q + 2)² = q² − 4q + 4	Correct	q² − 4q + 4
6	3a(2b × 3c) = 54a²bc	Wrong: 3a was "distributed" over a product; it multiplies only once	18abc
7	½(10s − 6) + 3 = 5s	Correct	5s
8	5w² + 6w = 11w²	Wrong: w² and w are unlike terms	5w² + 6w (cannot be combined)
9	2a³ + 3a³ + 6a²b + 6ab² = 5a³ + 12a²b²	Wrong: 6a²b and 6ab² are unlike terms	5a³ + 6a²b + 6ab²
10	(x + 2)(x + 5) = x² + 7x + 10	Correct	x² + 7x + 10
11	(a + 2)(b + 4) = ab + 8	Wrong: the cross terms 4a and 2b are missing	ab + 4a + 2b + 8
12	ab² + a²b + a²b² = ab(a + b + ab)	Correct (ab is a common factor)	ab(a + b + ab)

Correct as given: 5, 7, 10, 12. Corrected: (1) 15p² − 6pq (2) 5x + 10 (3) 3y + 4 (4) 25m² + 60mn + 36n² (6) 18abc (8) 5w² + 6w (9) 5a³ + 6a²b + 6ab² (11) ab + 4a + 2b + 8.

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

Circle pattern (3, 8, 15, …): draw the next figure. How many circles in Step 4, in Step 10, and in Step k? Use the formula to find Step 15.

Solution

Step $k$ is a $(k + 1) \times (k + 1)$ square of circles with one missing, or equally $k$ rows of $k + 2$ circles:

Step 4: a 5 × 5 square of circles with one missing, 5² − 1 = 24 circles

Step	1	2	3	4	10	15	k
Circles	3	8	15	24	120	255	k² + 2k

Step 15: $1 5^{2} + 2 \times 15 = 225 + 30 = 255$ .

Step 4: 24 circles, Step 10: 120, Step k: k² + 2k, Step 15: 255.

Square-tile pattern: how many tiles in each figure? In Step 4? Step 10? Write an expression for Step n using more than one method.

Solution

Step $n$ is an $(n + 2) \times (n + 2)$ square with an $n \times n$ hole:

Step 4: a 6 × 6 square with a 4 × 4 hole, 6² − 4² = 20 tiles

Counts: Step 1 = 8, Step 2 = 12, Step 3 = 16, Step 4 = 20, Step 10 = 44.

Method 1 (square minus hole): $(n + 2)^{2} - n^{2} = n^{2} + 4 n + 4 - n^{2} = 4 n + 4$

Method 2 (four sides of $n + 1$ each, going round): $4 (n + 1) = 4 n + 4$

Method 3 (two full rows of $n + 2$ and two columns of $n$ ): $2 (n + 2) + 2 n = 4 n + 4$

8, 12, 16, …; Step 4 = 20, Step 10 = 44; Step n = 4n + 4 = (n + 2)² − n².

Check that Tadang's and Yusuf's answers for the shaded square agree: (m + n)² − 4mn = (n − m)².

Solution

$(m + n)^{2} - 4 mn = m^{2} + 2 mn + n^{2} - 4 mn = m^{2} - 2 mn + n^{2} = (n - m)^{2}$

Both equal m² − 2mn + n².

H-shaped figure (Fig. 1): verify that Anusha's, Vaishnavi's and Aditya's expressions are equivalent, and find the area for x = 8, y = 3.

Solution

Anusha: $x^{2} - x y$
Vaishnavi: $x (x + 2 y) - 3 x y = x^{2} + 2 x y - 3 x y = x^{2} - x y$
Aditya: $2 x \times \frac{x - y}{2} = x (x - y) = x^{2} - x y$

All three equal $x^{2} - x y$ . For $x = 8, y = 3$ : $64 - 24 = 40$ .

All three equal x² − xy; the area is 40 sq. units.

Write an expression for the area of the dashed region (rectangle s × p with an L-shaped strip of width r along one side and the bottom). Substitute p = 6, r = 3.5, s = 9.

Solution

The dashed part is a rectangle of width $s - r$ and height $p - r$ .

Method 1: Area $= (s - r) (p - r)$ .

Method 2: Subtract the L-strip from the whole rectangle. The strip is $r p + r s - r^{2}$ (its corner square is counted twice in $r p + r s$ ), so:

Area $= s p - (r p + r s - r^{2}) = p s - p r - r s + r^{2}$ .

Expanding Method 1 gives the same expression.

With the values: $(9 - 3.5) (6 - 3.5) = 5.5 \times 2.5 = 13.75$ .

Check with Method 2: $54 - 21 - 31.5 + 12.25 = 13.75$ ✓

Area = (s − r)(p − r) = ps − pr − rs + r² = 13.75 sq. units.

Figure it Out (page 154)

Compute using the suggested identity: (i) 46² (ii) 397 × 403 (iii) 91² (iv) 43 × 45

Solution

$(40 + 6)^{2} = 1600 + 480 + 36 =$ 2116
$(400 - 3) (400 + 3) = 160000 - 9 =$ 159991
$(100 - 9)^{2} = 10000 - 1800 + 81 =$ 8281
$(44 - 1) (44 + 1) = 1936 - 1 =$ 1935

2116, 159991, 8281, 1935

Find: (i) (p − 1)(p + 11) (ii) (3a − 9b)(3a + 9b) (iii) −(2y + 5)(3y + 4) (iv) (6x + 5y)² (v) (2x − ½)² (vi) (7p) × (3r) × (p + 2)

Solution

$p^{2} + 11 p - p - 11 = p^{2} + 10 p - 11$
$(3 a)^{2} - (9 b)^{2} = 9 a^{2} - 81 b^{2}$
$- (6 y^{2} + 8 y + 15 y + 20) = - 6 y^{2} - 23 y - 20$
$36 x^{2} + 60 x y + 25 y^{2}$
$4 x^{2} - 2 x + \frac{1}{4}$
$21 p r (p + 2) = 21 p^{2} r + 42 p r$

(i) p² + 10p − 11 (ii) 9a² − 81b² (iii) −6y² − 23y − 20 (iv) 36x² + 60xy + 25y² (v) 4x² − 2x + ¼ (vi) 21p²r + 42pr

Identify the expression(s): (i) two more than a square number; (ii) the sum of the squares of two consecutive numbers.

Solution

(i) $s^{2} + 2$ . ( $(s + 2)^{2}$ is the square of 2 more, and $2 + s$ adds 2 to $s$ , not to its square.)

(ii) $m^{2} + (m + 1)^{2}$ , and also $m^{2} + (m - 1)^{2}$ , since $m - 1$ and $m$ are consecutive too.

$(2 m)^{2} + (2 m + 1)^{2}$ is also such a sum, but only for pairs that start with an even number, so it does not cover all cases.

(i) s² + 2 (ii) m² + (m + 1)² (equivalently m² + (m − 1)²).

In a calendar, take any 2 by 2 square and multiply the numbers along each diagonal. What do you observe? Explain.

Solution

Examples:

4 × 12 = 48 and 5 × 11 = 55
10 × 18 = 180 and 11 × 17 = 187
20 × 28 = 560 and 21 × 27 = 567

The other diagonal is always 7 more. Label the square as $a$ , $a + 1$ (top) and $a + 7$ , $a + 8$ (bottom):

$(a + 1) (a + 7) - a (a + 8) = a^{2} + 8 a + 7 - a^{2} - 8 a = 7$

The diagonal products always differ by 7, because (a + 1)(a + 7) − a(a + 8) = 7 (a row of a week has 7 days).

True or false? (i) (k + 1)(k + 2) − (k + 3) is always 2. (ii) (2q + 1)(2q − 3) is 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. (iv) (6n + 2)² − (4n + 3)² is 5 less than a square number.

Solution

False. The expression is $k^{2} + 3 k + 2 - k - 3 = k^{2} + 2 k - 1$ . It is 2 only for $k = 1$ (or $k = - 3$ ). For $k = 2$ it is 7.
False. The expression is $4 q^{2} - 4 q - 3$ , which is 3 less than a multiple of 4. It is always odd, so it is never a multiple of 4.
True. $(2 m)^{2} = 4 m^{2}$ , and $(2 m + 1)^{2} = 4 m (m + 1) + 1$ . Since $m (m + 1)$ is even, $4 m (m + 1)$ is a multiple of 8.
False. The expression is $36 n^{2} + 24 n + 4 - (16 n^{2} + 24 n + 9) = 20 n^{2} - 5$ , which is 5 less than $20 n^{2}$ . But $20 n^{2}$ is not a perfect square unless $n = 0$ (for example, $n = 1$ gives 20).

(i) False (ii) False (iii) True (iv) False: it equals 20n² − 5, and 20n² is a square only for n = 0.

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

Solution

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

Sum: $7 (a + b) + 8 = 7 (a + b + 1) + 1$ , so the remainder is 1.
Difference (second − first): $7 (b - a) + 2$ , so the remainder is 2. (First − second leaves $- 2$ , which is a remainder of 5.)
Product: $49 ab + 35 a + 21 b + 15 = 7 (7 ab + 5 a + 3 b + 2) + 1$ , so the remainder is 1.

Check with 10 and 12: $22 = 21 + 1$ , $12 - 10 = 2$ , $120 = 119 + 1$ ✓

Sum → 1, difference → 2, product → 1.

Square the middle of three consecutive numbers and subtract the product of the other two. What pattern do you notice? Write it algebraically and verify.

Solution

Examples: $5^{2} - 4 \times 6 = 1$ and $1 0^{2} - 9 \times 11 = 1$ . The answer is always 1.

Algebraically, $n^{2} - (n - 1) (n + 1) = 1$ . Expanding: $n^{2} - (n^{2} - n + n - 1) = n^{2} - n^{2} + 1 = 1$ ✓

Always 1: n² − (n − 1)(n + 1) = 1.

Add two numbers and multiply by half their sum. Write the expression and prove it is half the square of the sum.

Solution

$(a + b) \times \frac{1}{2} (a + b) = \frac{1}{2} (a + b) (a + b) = \frac{1}{2} (a + b)^{2}$

(a + b) × ½(a + b) = ½(a + b)².

Which is larger (without full multiplication)? (i) 14 × 26 or 16 × 24 (ii) 25 × 75 or 26 × 74

Solution

(i) $14 \times 26 = (16 - 2) (24 + 2) = 16 \times 24 + 32 - 48 - 4 = 16 \times 24 - 20$ . So $16 \times 24$ is larger.

Alternatively, $(20 - 6) (20 + 6) = 400 - 36$ and $(20 - 4) (20 + 4) = 400 - 16$ .

(ii) $25 \times 75 = 5 0^{2} - 2 5^{2}$ and $26 \times 74 = 5 0^{2} - 2 4^{2}$ . So $26 \times 74$ is larger, by $2 5^{2} - 2 4^{2} = 49$ .

(i) 16 × 24 (larger by 20) (ii) 26 × 74 (larger by 49).

A tiny park has two square plots of area g² each, surrounded by a walking path w ft wide. Write an expression for the area to be tiled.

Solution

Plan of the park: the hatched path is w ft wide all round and 2w ft between the plots

Length of the park $= w + g + 2 w + g + w = 2 g + 4 w$
Breadth $= w + g + w = g + 2 w$
Whole park $= (2 g + 4 w) (g + 2 w) = 2 g^{2} + 8 g w + 8 w^{2}$
Tiled area $= 2 g^{2} + 8 g w + 8 w^{2} - 2 g^{2} = 8 g w + 8 w^{2}$

Tiled area = 8w² + 8gw = 8w(w + g) sq. ft.

For each pattern (a) yellow and (b) blue: (i) draw the next figure (ii) find the number of units in Step 10 (iii) write an expression for Step y.

Solution

(a) Step $y$ is a $y \times (y + 2)$ block with an arm of $y + 2$ squares rising from its right end and another hanging from its left end:

Step 1: $3 + 3 + 3 = 9$
Step 2: $8 + 4 + 4 = 16$
Step 3: $15 + 5 + 5 = 25$

Total: $y (y + 2) + 2 (y + 2) = (y + 2)^{2}$ .

(In the printed Step 3 the arms are drawn 6 squares long. Following the rule of Steps 1 and 2, they should have 5.)

(a) Step 4: a 4 × 6 block with an arm of 6 at each end, 24 + 12 = 36 = 6² units

(b) Step $y$ is a $(y + 1) \times (y + 1)$ square with a row of $y$ squares below it: $5, 11, 19, \dots$

(b) Step 4: a 5 × 5 square and a row of 4 below it, 5² + 4 = 29 units

Pattern	Step 4	Step 10	Step y
(a)	36	144	(y + 2)²
(b)	29	131	(y + 1)² + y = y² + 3y + 1

(a) Step 10 = 144, Step y = (y + 2)²; (b) Step 10 = 131, Step y = (y + 1)² + y.

Puzzle: Coin Conjoin

Turn a 10-coin triangle upside down by moving one coin at a time. What is the minimum? What about 15 coins and bigger triangles?

Solution

The coins that can stay where they are are the ones shared by the triangle and its upside-down copy. Only the coins outside that shared part need to move.

3 coins: 1 move.
6 coins: 2 moves.
10 coins: 3 moves. Move the two bottom corner coins up to the ends of the second row, and move the top coin to below the middle of the bottom row.
15 coins: 5 moves. The upside-down copy can overlap the original in at most 10 coins, so 5 coins must move.

In general, the minimum is the number of coins divided by 3, ignoring any remainder:

Coins	3	6	10	15	21	28	36
Minimum moves	1	2	3	5	7	9	12

10 coins: 3 moves; 15 coins: 5 moves; in general, the number of coins ÷ 3 (whole-number part).

← Chapter 5: Number Play Chapter 7: Proportional Reasoning-1 →

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