Chapter 6 exercise answers: We Distribute, Yet Things Multiply

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1. Each cell in the frame equals its own row-number times its own column-number.
2. The 3 rows are p−1, p, p+1 and the 3 columns are q−1, q, q+1.
3. Multiply row × column and expand each bracket using the distributive property.
4. Top row: (p−1)(q−1) = pq−p−q+1, (p−1)q = pq−q, (p−1)(q+1) = pq+p−q−1.
5. Middle row: p(q−1) = pq−p and p(q+1) = pq+p (the centre is the given pq).
6. Bottom row: (p+1)(q−1) = pq+q−p−1, (p+1)q = pq+q, (p+1)(q+1) = pq+p+q+1.

Answertop-left = pq−p−q+1, top-middle = pq−q, top-right = pq+p−q−1, middle-left = pq−p, middle-right = pq+p, bottom-left = pq+q−p−1, bottom-middle = pq+q, bottom-right = pq+p+q+1.

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Open NCERT p. 143Matches NCERT’s answer

(i) (3 + u) (v − 3)

1. Multiply every term of (3+u) by every term of (v−3).
2. 3×v=3v, 3×(−3)=−9, u×v=uv, u×(−3)=−3u.
3. Add the four pieces: uv+3v−3u−9.

Answeruv + 3v − 3u − 9

(ii) 2/3 (15 + 6a)

1. Multiply 2/3 by each term inside the bracket.
2. 2/3×15=10, 2/3×6a=4a.

Answer10 + 4a

(iii) (10a + b) (10c + d)

1. Multiply every term of (10a+b) by every term of (10c+d).
2. 10a×10c=100ac, 10a×d=10ad, b×10c=10bc, b×d=bd.

Answer100ac + 10ad + 10bc + bd

(iv) (3 − x) (x − 6)

1. Multiply every term of (3−x) by every term of (x−6).
2. 3×x=3x, 3×(−6)=−18, (−x)×x=−x², (−x)×(−6)=6x.
3. Add: −x²+3x+6x−18 = −x²+9x−18.

Answer−x² + 9x − 18

(v) (−5a + b) (c + d)

1. Multiply every term of (−5a+b) by every term of (c+d).
2. −5a×c=−5ac, −5a×d=−5ad, b×c=bc, b×d=bd.

Answer−5ac − 5ad + bc + bd

(vi) (5 + z) (y + 9)

1. Multiply every term of (5+z) by every term of (y+9).
2. 5×y=5y, 5×9=45, z×y=yz, z×9=9z.
3. Add: yz+5y+9z+45.

Answeryz + 5y + 9z + 45

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Open NCERT p. 143Checked by computerAnswers can differ: one example

1. Let the two numbers be x and y, so xy is the product.
2. After the change the numbers are (x+2) and (y−4), and their product must still equal xy.
3. Expand: xy = xy − 4x + 2y − 8, which simplifies to y = 2x + 4.
4. Pick any x and compute y = 2x+4: x=1→y=6, x=2→y=8, x=3→y=10.

Answer(1, 6), (2, 8) and (3, 10) all work — e.g. 1×6=6 and (1+2)×(6−4)=3×2=6.

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(i) (a + ab − 3b²) (4 + b)

1. Multiply every term of (a+ab−3b²) by 4, then by b.
2. ×4: 4a+4ab−12b². ×b: ab+ab²−3b³.
3. Add and combine the like terms 4ab and ab: 4a+5ab+ab²−12b²−3b³.

Answer4a + 5ab + ab² − 12b² − 3b³

(ii) (4y + 7) (y + 11z − 3)

1. Multiply every term of (4y+7) by every term of (y+11z−3).
2. 4y×y=4y², 4y×11z=44yz, 4y×(−3)=−12y, 7×y=7y, 7×11z=77z, 7×(−3)=−21.
3. Combine the y terms (−12y+7y=−5y): 4y²+44yz−5y+77z−21.

Answer4y² + 44yz − 5y + 77z − 21

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(i) (a − b) (a + b)

1. Multiply (a−b) by (a+b): a×a=a², a×b=ab, −b×a=−ab, −b×b=−b².
2. The two ab terms cancel.

Answera² − b²

(ii) (a − b) (a² + ab + b²)

1. Multiply every term of (a−b) by every term of (a²+ab+b²).
2. a× gives a³+a²b+ab²; −b× gives −a²b−ab²−b³.
3. The a²b and ab² terms cancel in pairs.

Answera³ − b³

(iii) (a − b)(a³ + a²b + ab² + b³)

1. Multiply every term of (a−b) by every term of (a³+a²b+ab²+b³).
2. a× gives a⁴+a³b+a²b²+ab³; −b× gives −a³b−a²b²−ab³−b⁴.
3. All the middle terms cancel in pairs.

Answera⁴ − b⁴

Next identity

1. In each line, (a − b) multiplies a bracket whose terms run from the highest power of a down to the same power of b, and all the middle terms cancel: a² − b², then a³ − b³, then a⁴ − b⁴.
2. So the next line is (a − b)(a⁴ + a³b + a²b² + ab³ + b⁴).
3. Check by expanding: a × the bracket gives a⁵ + a⁴b + a³b² + a²b³ + ab⁴, and −b × the bracket gives −a⁴b − a³b² − a²b³ − ab⁴ − b⁵.
4. All the middle terms cancel in pairs, leaving a⁵ − b⁵.

Answer(a − b)(a⁴ + a³b + a²b² + ab³ + b⁴) = a⁵ − b⁵

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Open NCERT p. 149Matches NCERT’s answer

1. (b−a) is just −(a−b), so squaring removes the minus sign: (b−a)² = (−(a−b))² = (a−b)².
2. Expand both to check: (a−b)² = a²−2ab+b² and (b−a)² = b²−2ab+a² — the same expression.

AnswerNeither is greater — they are always equal.

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Open NCERT p. 149Checked by computerAnswers can differ: one example

1. Use a²−b² = (a+b)(a−b), so choose (a+b) and (a−b) that multiply to 100.
2. Pick a+b = 50 and a−b = 2, which gives a = 26 and b = 24.
3. Check: 26² − 24² = 676 − 576 = 100.

Answer26² − 24² = 676 − 576 = 100

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Open NCERT p. 149Matches NCERT’s answer

1. 406²: write 406 = 400+6, so 406² = 400²+2×400×6+6² = 160000+4800+36 = 164836.
2. 72²: write 72 = 70+2, so 72² = 70²+2×70×2+2² = 4900+280+4 = 5184.
3. 145²: write 145 = 150−5, so 145² = 150²−2×150×5+5² = 22500−1500+25 = 21025.
4. 1097²: write 1097 = 1100−3, so 1097² = 1100²−2×1100×3+3² = 1210000−6600+9 = 1203409.
5. 124²: write 124 = 120+4, so 124² = 120²+2×120×4+4² = 14400+960+16 = 15376.

Answer406²=164836, 72²=5184, 145²=21025, 1097²=1203409, 124²=15376

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1. Pattern 1, 2(a²+b²)=(a+b)²+(a−b)², and Pattern 2, (a+b)(a−b)=a²−b², were both proved just by expanding brackets and collecting like terms.
2. That proof never used the fact that a and b were counting numbers, so it works for any numbers at all.
3. Trying negative numbers or fractions for a and b still makes both sides come out equal.

AnswerNo — both patterns hold for negative integers and for fractions too, not only counting numbers, because the distributive-property proof never depended on the kind of number used.

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Open NCERT p. 154Matches NCERT’s answer

(i) 46² using Identity 1A for (a + b)²

1. Write 46 = 40+6.
2. 46² = 40²+2×40×6+6² = 1600+480+36.

Answer2116

(ii) 397 × 403 using Identity 1C

1. 397 and 403 are both 3 away from 400: 397=400−3, 403=400+3.
2. 397×403 = 400²−3² = 160000−9.

Answer159991

(iii) 91² using Identity 1B for (a − b)²

1. Write 91 = 100−9.
2. 91² = 100²−2×100×9+9² = 10000−1800+81.

Answer8281

(iv) 43 × 45 using Identity 1C

1. 43 and 45 are both 1 away from 44: 43=44−1, 45=44+1.
2. 43×45 = 44²−1² = 1936−1.

Answer1935

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(i) (p − 1) (p + 11)

1. Multiply every term: p×p=p², p×11=11p, −1×p=−p, −1×11=−11.
2. Combine 11p−p=10p.

Answerp² + 10p − 11

(ii) (3a − 9b) (3a + 9b)

1. Use (x + y)(x − y) = x² − y², taking x = 3a and y = 9b.
2. (3a − 9b)(3a + 9b) = (3a)² − (9b)² = 9a² − 81b².

Answer9a² − 81b²

(iii) −(2y + 5) (3y + 4)

1. First expand (2y + 5)(3y + 4): 6y² + 8y + 15y + 20 = 6y² + 23y + 20.
2. The minus sign in front changes the sign of every term: −(6y² + 23y + 20) = −6y² − 23y − 20.

Answer−6y² − 23y − 20

(iv) (6x + 5y)²

1. Use (a+b)²=a²+2ab+b² with a=6x, b=5y.
2. (6x)²+2(6x)(5y)+(5y)² = 36x²+60xy+25y².

Answer36x² + 60xy + 25y²

(v) (2x − 1/2)²

1. Use (a−b)²=a²−2ab+b² with a=2x, b=1/2.
2. (2x)²−2(2x)(1/2)+(1/2)² = 4x²−2x+1/4.

Answer4x² − 2x + 1/4

(vi) (7p) × (3r) × (p + 2)

1. Multiply the first two factors: 7p × 3r = 21pr.
2. Multiply 21pr by each term of (p + 2): 21pr × p = 21p²r and 21pr × 2 = 42pr.

Answer21p²r + 42pr

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(i) Two more than a square number.

1. A square number can be written as s². Two more than it is s² + 2.
2. 2 + s is two more than s, and s need not be a square; (s + 2)² is the square of a sum; s² + 4 is four more; 2s² is twice a square; 2²s is just 4s.
3. So only s² + 2 fits.

Answers² + 2

(ii) The sum of the squares of two consecutive numbers

1. Two consecutive numbers are a number m and the next number m + 1, so the sum of their squares is m² + (m + 1)².
2. (m + n)² and (m + (m + 1))² square a sum instead of adding two squares, m² + n² uses any two numbers, and m² + 1 is m² + 1², where m and 1 are next to each other only when m is 0 or 2.
3. (2m)² + (2m + 1)² also adds the squares of two consecutive numbers, but only when the smaller one is even (it can never give 1² + 2² = 5), so the book picks the general form m² + (m + 1)².
4. m² + (m − 1)² gives the same kind of sum with m as the larger number; the book's answer is written as m² + (m + 1)².

Answerm² + (m + 1)²

Open NCERT p. 155Matches NCERT’s answer

1. Label a 2×2 square in the calendar as a, a+1 (top row) and a+7, a+8 (bottom row, one week later).
2. One diagonal product is a×(a+8); the other is (a+1)×(a+7).
3. Expand both: a(a+8)=a²+8a, and (a+1)(a+7)=a²+8a+7.
4. Subtracting shows the second diagonal is always exactly 7 more than the first.

AnswerThe two diagonal products always differ by 7 — (a+1)(a+7) is always 7 more than a(a+8), for every choice of a.

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

1. Expand: (k+1)(k+2) = k²+3k+2, so the expression is k²+3k+2−(k+3) = k²+2k−1.
2. This depends on k (e.g. k=0 gives −1, not 2), so it is not always 2.

AnswerFalse — it is k²+2k−1, not the constant 2.

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

1. Expand: (2q+1)(2q−3) = 4q²−4q−3 = 4(q²−q)−3.
2. This is always 3 less than a multiple of 4, so it always leaves remainder 1 on division by 4 — never 0.

AnswerFalse — it is never a multiple of 4.

(iii) Squares of even numbers are multiples of 4, and squares of odd …

1. An even number is 2k; its square is 4k², clearly a multiple of 4.
2. An odd number is 2k+1; its square is 4k²+4k+1 = 4k(k+1)+1.
3. k(k+1) is a product of consecutive integers, so it is always even, making 4k(k+1) a multiple of 8.

AnswerTrue — both parts of the statement hold.

(iv) (6n + 2)² − (4n + 3)² is 5 less …

1. Expand: (6n+2)²−(4n+3)² = (36n²+24n+4)−(16n²+24n+9) = 20n²−5.
2. For this to be '5 less than a square', 20n² would have to be a perfect square for every n, but at n=1 it is 20, which is not a perfect square.

AnswerFalse — e.g. at n=1 the value is 15, and 15+5=20 is not a perfect square.

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Open NCERT p. 155Matches NCERT’s answer

1. Write the first number as 7a + 3 and the second as 7b + 5, for whole numbers a and b.
2. Sum: (7a + 3) + (7b + 5) = 7(a + b) + 8 = 7(a + b + 1) + 1, so the remainder is 1.
3. Difference: take the number that leaves 5 minus the number that leaves 3 (as the book does): (7b + 5) − (7a + 3) = 7(b − a) + 2, so the remainder is 2.
4. Product: (7a + 3)(7b + 5) = 49ab + 35a + 21b + 15. The first three terms are multiples of 7, and 15 = 14 + 1, so the remainder is 1.

AnswerSum: remainder 1; difference (second number minus first): remainder 2; product: remainder 1.

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1. Let the three consecutive numbers be n−1, n, n+1.
2. Square the middle number: n². The product of the outer two is (n−1)(n+1) = n²−1.
3. Subtract: n² − (n²−1) = 1, for every n.

AnswerThe result is always 1: n² − (n−1)(n+1) = 1, and expanding (n−1)(n+1) to n²−1 confirms it.

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Open NCERT p. 156Matches NCERT’s answer

1. Let the two numbers be x and y.
2. Add them: x+y. Multiply by half the sum: (x+y) × (1/2)(x+y).
3. This equals (1/2)(x+y)², which is exactly half the square of the sum — proving the claim.

Answer(x + y)² / 2

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(i) 14 × 26 or 16 × 24

1. Both pairs add up to 40, so write each around 20: 14 × 26 = (20 − 6)(20 + 6) and 16 × 24 = (20 − 4)(20 + 4).
2. By (a + b)(a − b) = a² − b²: 14 × 26 = 20² − 6² and 16 × 24 = 20² − 4².
3. Both start from the same 20², and 4² is less than 6², so less is taken away in 16 × 24 (6² − 4² = 20 less).

Answer16 × 24 is larger

(ii) 25 × 75 or 26 × 74

1. Both pairs add up to 100, so write each around 50: 25 × 75 = (50 − 25)(50 + 25) and 26 × 74 = (50 − 24)(50 + 24).
2. So 25 × 75 = 50² − 25² and 26 × 74 = 50² − 24².
3. 24² is less than 25², so 26 × 74 is larger (by 25² − 24² = 49).

Answer26 × 74 is larger

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Open NCERT p. 156Checked by computerReads two ways: both answers shown

1. The words say the path is w ft wide, but the strip between the two plots has no label. So the question can be read two ways.
2. Reading 1 (NCERT's answer key uses this reading, and the drawing agrees: the strip between the plots is drawn about twice as wide as the strips along the edges): the middle strip is 2w wide, a w-wide path round each plot.
3. Length of the park = w + g + 2w + g + w = 2g + 4w ft. Breadth = w + g + w = g + 2w ft.
4. Area of the park = (2g + 4w)(g + 2w) = 2g² + 8gw + 8w² sq. ft. The two green plots cover 2g² sq. ft.
5. Area to be tiled = 2g² + 8gw + 8w² − 2g² = 8gw + 8w² = 8w(w + g) sq. ft.
6. Reading 2 (every strip, the middle one too, is w wide, as the words alone suggest): length = w + g + w + g + w = 2g + 3w ft, breadth = g + 2w ft.
7. Area of the park = (2g + 3w)(g + 2w) = 2g² + 7gw + 6w². Area to be tiled = 7gw + 6w² sq. ft.

AnswerRead with the middle strip 2w wide, as drawn (NCERT's key): 8w(w + g) = 8w² + 8gw sq. ft. Read with every strip w wide: 7gw + 6w² sq. ft.

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(ii) staircase pattern How many basic units are there in Step 10?

1. Each step is a block of y rows and (y + 2) columns, with an arm of (y + 2) squares going up and another going down: Step 1 = 3 + 3 + 3 = 9, Step 2 = 8 + 4 + 4 = 16.
2. The Step 3 drawing on the page shows 27 squares, with arms 6 long instead of 5. That looks like a drawing slip: the pattern the book means is 9, 16, 25, 36, …, the squares (y + 2)².
3. Step 10 = (10 + 2)² = 12² = 144.

Answer144

(iii) staircase pattern Write an expression to describe the number of basic units in Step …

1. Block plus two arms: y(y + 2) + 2(y + 2) = (y + 2)(y + 2) = (y + 2)².
2. Check: Step 1 = 3² = 9, Step 2 = 4² = 16, Step 4 = 6² = 36 (a 4 by 6 block with arms of 6).

Answer(y + 2)²

(ii) square-dot pattern How many basic units are there in Step 10?

1. Each step is a square of (y + 1) by (y + 1) dots with one more row of y dots below it: Step 1 = 2² + 1 = 5, Step 2 = 3² + 2 = 11, Step 3 = 4² + 3 = 19.
2. Step 10 = 11² + 10 = 121 + 10 = 131.

Answer131

(iii) square-dot pattern Write an expression to describe the number of basic units in Step …

1. Step y = (y + 1)² + y = y² + 2y + 1 + y = y² + 3y + 1.

Answery² + 3y + 1

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