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CLASS 8 · GANITA PRAKASH PART I · CHAPTER 6

We Distribute, Yet Things Multiply

Use distributivity and algebraic identities to explain product changes, expand expressions, and solve number and area patterns.

105 worked answersAll 11 end-of-chapter questions

What you will learn

  • Use distributivity to expand products and explain how products change.
  • Apply square and difference-of-squares identities to calculate efficiently.
  • Model tile, calendar and area patterns with algebra.

105 worked answers · short steps first, explanations when you need them

6.1 Some Properties of Multiplication

Textbook pages 137–141

Question increase-first-factor

By how much does 23×27 increase if 23 is increased by 1?

Steps only

(23+1)×27−23×27=27.

Answer

It increases by 27, the unchanged second factor.

Question increase-second-factor

By how much does 23×27 increase if 27 is increased by 1?

Steps only

23×(27+1)−23×27=23.

Answer

It increases by 23, the unchanged first factor.

Question increase-both-factors

By how much does 23×27 increase if both factors increase by 1?

Steps only

(a+1)(b+1)−ab=a+b+1.

For a=23,b=27: 23+27+1=51.

Answer

The product increases by 51.

Question increase-both-general

How much does ab increase when both a and b are increased by 1?

Steps only

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

Subtract the original product ab.

Answer

The increase is a+b+1.

Question increase-and-decrease

Expand (a+1)(b−1).

Steps only

(a+1)(b−1)=ab−a+b−1.

Answer

The change from ab is b−a−1.

Question decrease-example

Give an example where decreasing both positive factors by 1 decreases their product.

Steps only

Choose a=10,b=3.

10×3=30, while (10−1)(3−1)=9×2=18.

Answer

The product decreases from 30 to 18.

Question change-by-m-n

If ab changes to (a+m)(b+n), how much does the product change?

Steps only

(a+m)(b+n)=ab+an+mb+mn.

Subtract ab.

Answer

The change is an+bm+mn.

Question change-minus2-plus3

Find the change in ab when the factors become (a−2) and (b+3).

Steps only

(a−2)(b+3)=ab+3a−2b−6.

Subtract ab.

Answer

The change is 3a−2b−6.

Question change-minus3-minus4

Find the change in ab when the factors become (a−3)(b−4).

Steps only

(a−3)(b−4)=ab−4a−3b+12.

Subtract ab.

Answer

The change is −4a−3b+12.

Question expand-increment-both-ways

Expand (a+1)(b+1) by first treating (b+1) as one term.

Steps only

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

Answer

(a+1)(b+1)=ab+a+b+1. The increase from ab is a+b+1.

Question product-decreases-examples

Give three examples where increasing the first positive number by 1 and decreasing the second by 1 makes the product smaller.

Steps only

10×2=20; (10+1)(2−1)=11×1=11.

9×2=18; (9+1)(2−1)=10×1=10.

20×1=20; (20+1)(1−1)=21×0=0.

Answer

Examples: 20 becomes 11, 18 becomes 10, and 20 becomes 0; each product decreases.

Question product-change-negative-integers

For a=−5, b=8 and for a=−4, b=−5, compare ab with (a+1)(b−1).

Steps only

(−5)(8)=−40; (−5+1)(8−1)=−4×7=−28; change=12.

(−4)(−5)=20; (−4+1)(−5−1)=(−3)(−6)=18; change=−2.

Answer

The changes are +12 and −2, respectively, agreeing with b−a−1.

Question expand-two-negative-binomial-products

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

Steps only

(i) ab+av−ub−uv.

(ii) ab−av−ub+uv.

Answer

  • (i) ab−ub+av−uv;
  • (ii) ab−ub−av+uv.

Question verify-change-examples

Check the product changes when (i) a is reduced by 2 and b increased by 3, using a=8,b=6; (ii) a is reduced by 3 and b reduced by 4, using a=9,b=8.

Steps only

(i) 8×6=48; (8−2)(6+3)=6×9=54; change=+6.||Formula: 3a−2b−6=24−12−6=6.

(ii) 9×8=72; (9−3)(8−4)=6×4=24; change=−48.||Formula: −4a−3b+12=−36−24+12=−48.

Answer

The changes are +6 and −48, matching direct multiplication.

Expand and factor using distributivity

Textbook pages 142–144

Question pq-grid

If the centre of a 3×3 multiplication grid is pq, write all nine entries around it.

Steps only

The row factors are p−1,p,p+1.

The column factors are q−1,q,q+1.

Multiply the row factor by the column factor in each cell.

Answer

Rows: (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).

Question expand-3u-v3

Expand (3+u)(v−3).

Steps only

Multiply each term in the first bracket by every term in the second.

Collect like terms.

Answer

(3+u)(v−3)=uv+3v−3u−9.

Question expand-fraction

Expand (2/3)(15+6a).

Steps only

Multiply each term in the first bracket by every term in the second.

Collect like terms.

Answer

(2/3)(15+6a)=10+4a.

Question expand-10a-b

Expand (10a+b)(10c+d).

Steps only

Multiply each term in the first bracket by every term in the second.

Collect like terms.

Answer

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

Question expand-3x-x6

Expand (3−x)(x−6).

Steps only

Multiply each term in the first bracket by every term in the second.

Collect like terms.

Answer

(3−x)(x−6)=−x²+9x−18.

Question expand-minus5

Expand (−5a+b)(c+d).

Steps only

Multiply each term in the first bracket by every term in the second.

Collect like terms.

Answer

(−5a+b)(c+d)=−5ac−5ad+bc+bd.

Question expand-5z-y9

Expand (5+z)(y+9).

Steps only

Multiply each term in the first bracket by every term in the second.

Collect like terms.

Answer

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

Question unchanged-product-examples

Give three pairs (a,b) for which (a+2)(b−4)=ab.

Steps only

Expand: ab−4a+2b−8=ab.

So 2b=4a+8, hence b=2a+4.

Choose a=1,2,3.

Answer

Examples: (1,6), (2,8) and (3,10).

Question expand-polynomial-i

Expand (a+ab−3b²)(4+b).

Steps only

Multiply the polynomial by 4, then by b.

Combine the two ab terms.

Answer

4a+5ab+ab²−12b²−3b³.

Question expand-polynomial-ii

Expand (4y+7)(y+11z−3).

Steps only

4y(y+11z−3)=4y²+44yz−12y.

7(y+11z−3)=7y+77z−21.

Combine −12y+7y.

Answer

4y²+44yz+77z−5y−21.

Question difference-cubes

Expand (a−b)(a+b), (a−b)(a²+ab+b²) and (a−b)(a³+a²b+ab²+b³). What comes next?

Steps only

Expand and cancel opposite middle terms.

The results are a²−b², a³−b³ and a⁴−b⁴.

Answer

Next: (a−b)(a⁴+a³b+a²b²+ab³+b⁴)=a⁵−b⁵.

Question multiply-94-11

Calculate 94×11 using the place-value pattern.

Steps only

11=10+1, so 94×11=94×10+94.

Add the shifted copies with carries.

Answer

94×11=1034.

Question multiply-495-11

Calculate 495×11 using the place-value pattern.

Steps only

11=10+1, so 495×11=495×10+495.

Add the shifted copies with carries.

Answer

495×11=5445.

Question multiply-3279-11

Calculate 3279×11 using the place-value pattern.

Steps only

11=10+1, so 3279×11=3279×10+3279.

Add the shifted copies with carries.

Answer

3279×11=36069.

Question multiply-4791256-11

Calculate 4791256×11 using the place-value pattern.

Steps only

11=10+1, so 4791256×11=4791256×10+4791256.

Add the shifted copies with carries.

Answer

4791256×11=52703816.

Question multiply-by-101-rule

Describe a quick rule for dcba×101.

Steps only

101=100+1.

dcba×101=dcba00+dcba.

Answer

Add the number to itself shifted two places; align digits and carry as needed.

Question multiply-by-1001-rule

How can you multiply a number by 1001 or 10001 using place value?

Steps only

1001=1000+1, so append three zeroes and add the original number.

10001=10000+1, so append four zeroes and add the original number.

Answer

For 1001, add a three-place shift; for 10001, add a four-place shift.

Question by101-89

Evaluate 89×101 efficiently.

Steps only

101 is a power of ten plus 1; add one shifted copy.

Answer

89×101=8989.

Question by101-949

Evaluate 949×101 efficiently.

Steps only

101 is a power of ten plus 1; add one shifted copy.

Answer

949×101=95849.

Question by1001-265831

Evaluate 265831×1001 efficiently.

Steps only

1001 is a power of ten plus 1; add one shifted copy.

Answer

265831×1001=266096831.

Question by1001-1111

Evaluate 1111×1001 efficiently.

Steps only

1001 is a power of ten plus 1; add one shifted copy.

Answer

1111×1001=1112111.

Question by99-9734

Evaluate 9734×99 efficiently.

Steps only

99 is a power of ten minus 1; subtract one copy.

Answer

9734×99=963666.

Question by999-23478

Evaluate 23478×999 efficiently.

Steps only

999 is a power of ten minus 1; subtract one copy.

Answer

23478×999=23454522.

Question expand-fractional-binomial

Expand (3a/2)(a−b+1/5).

Steps only

(3a/2)(a−b+1/5)=(3a/2)a−(3a/2)b+(3a/2)(1/5).

=3a²/2−3ab/2+3a/10.

Answer

3a²/2−3ab/2+3a/10.

Question expand-sum-times-square-sum

Expand (a+b)(a²+2ab+b²) and combine like terms.

Steps only

a(a²+2ab+b²)+b(a²+2ab+b²).

=a³+2a²b+ab²+a²b+2ab²+b³.

=a³+3a²b+3ab²+b³.

Answer

a³+3a²b+3ab²+b³. The unlike terms in the expansion cannot be combined further.

Question multiply-3874-and-5678-by-11

Evaluate 3874×11 and 5678×11.

Steps only

3874×11=3874×(10+1)=38,740+3,874=42,614.

5678×11=5678×(10+1)=56,780+5,678=62,458.

Answer

3874×11=42,614 and 5678×11=62,458.

Question multiply-3874-by-101

Evaluate 3874×101 using place value.

Steps only

3874×101=3874×(100+1).

=387,400+3,874=391,274.

Answer

3874×101=391,274.

Question multiply-any-number-by-11

Describe a place-value method for multiplying a whole number of any number of digits by 11.

Steps only

N×11=N×10+N.

For 4,275: 4,275×11=47,025.

Answer

For every whole number N, N×11=N×10+N; add the one-place-shifted number to the original, carrying as needed.

Square identities and difference of squares

Textbook pages 145–150

Question square-binomial-a

What is the expansion of (a+b)²?

Steps only

(a+b)(a+b)=a²+ab+ab+b².

Answer

a²+2ab+b².

Question square-binomial-b

What is the expansion of (a−b)²?

Steps only

(a−b)(a−b)=a²−ab−ab+b².

Answer

a²−2ab+b².

Question difference-squares

Find (a+b)(a−b), and explain the difference-of-squares identity geometrically.

Steps only

Distribute: a²−ab+ab−b².

The middle terms cancel.

Answer

a²−b².

Question compare-opposite-squares

Which is greater: (a−b)² or (b−a)²?

Steps only

b−a=−(a−b).

Squaring removes the sign: (−x)²=x².

Answer

They are equal for every a and b.

Question 100-difference-squares

Give two square numbers whose difference is 100.

Steps only

676−576=100.

These are 26²−24².

Answer

100=26²−24².

Question square-406

Find 406² efficiently.

Steps only

Write 406 as a nearby round number plus or minus a small amount.

Apply (a±b)²=a²±2ab+b².

Answer

406²=164836.

Question square-72

Find 72² efficiently.

Steps only

Write 72 as a nearby round number plus or minus a small amount.

Apply (a±b)²=a²±2ab+b².

Answer

72²=5184.

Question square-145

Find 145² efficiently.

Steps only

Write 145 as a nearby round number plus or minus a small amount.

Apply (a±b)²=a²±2ab+b².

Answer

145²=21025.

Question square-1097

Find 1097² efficiently.

Steps only

Write 1097 as a nearby round number plus or minus a small amount.

Apply (a±b)²=a²±2ab+b².

Answer

1097²=1203409.

Question square-124

Find 124² efficiently.

Steps only

Write 124 as a nearby round number plus or minus a small amount.

Apply (a±b)²=a²±2ab+b².

Answer

124²=15376.

Question identities-integers-fractions

Do the distributive and square identities work for negative integers and fractions?

Steps only

The identities follow from multiplication and addition rules, not from a restriction to counting numbers.

Substitution of negative values or fractions preserves both sides.

Answer

Yes. They hold for negative integers and fractions as well.

Question square-area-65

A square has side 65 units. Split its side as 60+5 and find the four areas and the total area. Also check the splits 30+35 and 52+13.

Steps only

60²=3600; 60×5=300 (twice); 5²=25.||3600+300+300+25=4225.||30²+2(30×35)+35²=900+2100+1225=4225.||52²+2(52×13)+13²=2704+1352+169=4225.

Answer

The area is 4,225 square units in each split.

Question sum-square-compared-squares

For integers a and b, when is (a+b)² greater than a²+b²?

Steps only

(a+b)²−(a²+b²)=2ab.||Greater if ab>0; equal if ab=0; smaller if ab<0.

Answer

(a+b)² is greater exactly when ab>0.

Question square-sum-examples

Expand (m+3)² and (6+p)² using the identity; expand (6x+5)²; then expand (3j+2k)² using both the identity and the distributive property.

Steps only

(m+3)²=m²+6m+9; (6+p)²=p²+12p+36.

(6x+5)²=36x²+60x+25.

(3j+2k)²=9j²+12jk+4k².

Distribute: (3j+2k)(3j+2k)=9j²+6jk+6jk+4k².

Answer

m²+6m+9; p²+12p+36; 36x²+60x+25; 9j²+12jk+4k².

Question square-difference-examples

Expand (b−6)², (−2a+3)² and (7y−3z/4)² using both the square identity and distribution. Then find 99² and 58².

Steps only

(b−6)²=b²−12b+36; (−2a+3)²=4a²−12a+9.

(7y−3z/4)²=49y²−21yz/2+9z²/16.

99²=9801; 58²=3364.

Answer

The three expansions are b²−12b+36, 4a²−12a+9, and 49y²−21yz/2+9z²/16; 99²=9,801 and 58²=3,364.

Question square-pattern-sum-difference-identity

Show that twice the sum of two squares equals the square of their sum plus the square of their difference.

Steps only

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

Answer

2(a²+b²)=(a+b)²+(a−b)².

Question square-55-area-check

Use (60−5)² to find 55².

Steps only

55²=(60−5)²=60²−2×60×5+5².

=3600−600+25=3025.

Answer

55²=3,025.

Question square-104-and-37

Use a square identity to find 104² and 37².

Steps only

104²=(100+4)²=10000+800+16=10816.

37²=(40−3)²=1600−240+9=1369.

Answer

104²=10,816 and 37²=1,369.

Question products-98x102-45x55

Use a difference of squares to find 98×102 and 45×55.

Steps only

98×102=(100−2)(100+2)=10000−4=9996.

45×55=(50−5)(50+5)=2500−25=2475.

Answer

98×102=9,996 and 45×55=2,475.

Question square-nearby-31-and-197

Use a²=(a+b)(a−b)+b² to find 31² and 197².

Steps only

31²=32×30+1=961.

197²=200×194+9=38809.

Answer

31²=961 and 197²=38,809.

6.3 Mind the Mistake, Mend the Mistake

Textbook pages 150

Question 1

Check the simplification of −3p(−5p+2q). Decide whether it is correct; if not, write the correct result.

Steps only

−3p(−5p+2q)=15p²−6pq.

Answer

15p²−6pq.

Question 2

Check the simplification of ½(x−1)+3(x+4). Decide whether it is correct; if not, write the correct result.

Steps only

½(x−1)+3(x+4)=x/2−1/2+3x+12=7x/2+23/2.

Answer

7x/2+23/2.

Question 3

Check the simplification of y+2(y+2). Decide whether it is correct; if not, write the correct result.

Steps only

y+2(y+2)=y+2y+4=3y+4.

Answer

3y+4.

Question 4

Check the simplification of (5m+6n)². Decide whether it is correct; if not, write the correct result.

Steps only

(5m+6n)²=25m²+60mn+36n².

Answer

25m²+60mn+36n².

Question 5

Check the simplification of (−q+2)². Decide whether it is correct; if not, write the correct result.

Steps only

(−q+2)²=(−q)²+2(−q)(2)+2²=q²−4q+4.

Answer

Correct: q²−4q+4.

Question 6

Check the simplification of 3a(2b+3c). Decide whether it is correct; if not, write the correct result.

Steps only

3a(2b+3c)=(3a)(2b)+(3a)(3c)=6ab+9ac.

Answer

6ab+9ac.

Question 7

Check the simplification of ½(10s−6)+3. Decide whether it is correct; if not, write the correct result.

Steps only

½(10s−6)+3=5s−3+3=5s.

Answer

Correct: 5s.

Question 8

Check the simplification of 5w²+6w. Decide whether it is correct; if not, write the correct result.

Steps only

5w²+6w stays as 5w²+6w.

Answer

5w²+6w.

Question 9

Check the simplification of 2a³+3a³+6a²b+6ab². Decide whether it is correct; if not, write the correct result.

Steps only

2a³+3a³+6a²b+6ab²=5a³+6a²b+6ab².

Answer

5a³+6a²b+6ab².

Question 10

Check the simplification of (a+2)(b+4). Decide whether it is correct; if not, write the correct result.

Steps only

(a+2)(b+4)=ab+4a+2b+8.

Answer

ab+4a+2b+8.

Question 11

Check the simplification of ab²+a²b+a²b². Decide whether it is correct; if not, write the correct result.

Steps only

ab²+a²b+a²b²=ab(b+a+ab).

Answer

Correct: ab(a+b+ab).

Question 12

Check the simplification of (x+2)(x+5). Decide whether it is correct; if not, write the correct result.

Steps only

(x+2)(x+5)=x²+5x+2x+10=x²+7x+10.

Answer

Correct: x²+7x+10.

Patterns, areas and applications

Textbook pages 151–158

Question tile-pattern-count

A pattern has (n+2)²−n² tiles in Step n. Find the counts in Steps 1, 2, 3, 4 and 10.

Steps only

Simplify: (n+2)²−n²=4n+4.

Substitute n=1,2,3,4,10.

Answer

8, 12, 16, 20 and 44 tiles, respectively.

Question tile-pattern-general

If Step n is a square frame with outside side n+2 and inside side n, write its tile count and simplify.

Steps only

Count by area: (n+2)²−n².

Expand: n²+4n+4−n².

Answer

4n+4 tiles (equivalently (n+2)²−n²).

Question dashed-area

Write the area of the dashed region in two ways. Then substitute p=6, r=3.5 and s=9.

Steps only

(p−r)(s−r)=ps−pr−sr+r².

(6−3.5)(9−3.5)=2.5×5.5=13.75.

Answer

The area is 13.75 square units.

Question calendar-diagonal-products

For a 2×2 block of consecutive calendar dates, how do the diagonal products compare?

Steps only

Write the block as a, a+1 on the top row and a+7, a+8 below.

Diagonal products are a(a+8) and (a+1)(a+7).

Their difference is (a+1)(a+7)−a(a+8)=7.

Answer

The product on the diagonal with the two middle dates is 7 greater.

Question verify-k1

(k+1)(k+2)−(k+3) is always 2.

Steps only

Expand or factor the expression and simplify.

Answer

False. It simplifies to k²+2k−1, which depends on k.

Question verify-odd-product4

(2q+1)(2q−3) is a multiple of 4.

Steps only

Expand or factor the expression and simplify.

Answer

False. It equals 4q(q−1)−3, leaving remainder 1 modulo 4.

Question verify-odd-square8

An odd square is 1 more than a multiple of 8.

Steps only

Expand or factor the expression and simplify.

Answer

True. For odd n=2k+1, n²=4k(k+1)+1; k(k+1) is even, so 4k(k+1) is divisible by 8.

Question verify-square-minus5

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

Steps only

Expand or factor the expression and simplify.

Answer

False. It simplifies to 20n²−5; this is not generally one less than a square.

Question remainders-product-7

One number leaves remainder 3 and another remainder 5 on division by 7. Find the remainders of sum, second minus first, and product.

Steps only

Sum: 3+5=8 leaves 1.

Difference: 5−3=2.

Product: 3×5=15 leaves 1.

Answer

Remainders: 1, 2 and 1.

Question middle-square-minus-neighbours

Choose n−1,n,n+1. What is n²−(n−1)(n+1)?

Steps only

(n−1)(n+1)=n²−1.

n²−(n²−1)=1.

Answer

The result is always 1.

Question half-square-sum

Write the result algebraically for numbers a and b.

Steps only

Their sum is a+b.

Half the sum is (a+b)/2.

Multiply: (a+b)×(a+b)/2.

Answer

(a+b)²/2, which is half the square of their sum.

Question compare-14x26

Which product is larger?

Steps only

14×26=(16−2)(24+2).

Expanding gives 16×24+32−48−4=16×24−20.

Answer

16×24 is larger by 20.

Question compare-25x75

Which product is larger?

Steps only

25×75=(26−1)(74+1).

Expanding gives 26×74+26−74−1=26×74−49.

Answer

26×74 is larger by 49.

Question park-path-area

A rectangular park surrounds two g×g square plots with a path of width w as shown. Find the area to be tiled.

Steps only

The outer rectangle is (2g+4w) by (g+2w).

Subtract the two green squares: (2g+4w)(g+2w)−2g².

Expand and simplify.

Answer

8w(w+g) square feet.

Question square-pattern-next

A pattern’s Step 1, 2 and 3 counts are (2+1)², (2+2)², (2+3)². Find Step 4 and Step 10.

Steps only

The count at Step k is (k+2)².

Step 4: 6²=36. Step 10: 12²=144.

9 small squares

16 small squares

25 small squares

36 small squares

Original stacked squares diagrams, generated from the mathematical rule.

Answer

36 units in Step 4 and 144 in Step 10.

Question circle-pattern-count

A circle pattern has k²+2k circles at Step k. Find the counts at Steps 4, 10 and 15, and give the next count after Step 3.

Steps only

Step 4: 4²+2×4=24.||Step 10: 10²+2×10=120.||Step 15: 15²+2×15=255.||Step 3 has 15 circles; Step 4 has 24.

Answer

Steps 4, 10 and 15 have 24, 120 and 255 circles.

Question interior-shaded-square-area

Four equal rectangles, each with area mn, surround a shaded square inside a larger square with side m+n. Find the shaded area in two ways and verify the expressions are equal.

Steps only

Outer square area minus rectangles: (m+n)²−4mn.||Inner square side n−m: (n−m)².||(m+n)²−4mn=m²+2mn+n²−4mn=m²−2mn+n²=(n−m)².

Answer

The shaded area is (m+n)²−4mn=(n−m)².

Question slanted-shaded-area-equivalent-methods

Three equal rectangles have dimensions x by y. The shaded region is described by x²−xy, x(x+2y)−3xy, and x(x−y). Show the expressions agree and evaluate them for x=8, y=3.

Steps only

x²−xy=x(x−y).||x(x+2y)−3xy=x²+2xy−3xy=x²−xy.||For x=8,y=3: 64−24=40.

Answer

All three methods give area x²−xy; for x=8 and y=3, the area is 40 square units.

Question coin-conjoin-triangle

A triangular arrangement has 10 coins, and the task is to turn it upside down by moving one coin at a time. How many moves are needed? What is the minimum for the next triangular arrangement of 15 coins? Give the general rule for T coins.

Steps only

10 coins: 3 moves.

15 coins: ⌊15/3⌋=5 moves.

For T coins in a triangular arrangement: ⌊T/3⌋ moves.

Answer

The 10-coin arrangement takes 3 moves; the 15-coin arrangement takes 5 moves. For a triangular arrangement with T coins, the rule is ⌊T/3⌋ moves.

6.4 Figure it Out: Questions 1–3

Textbook pages 154–155

Question Q1(i)

Find 46² using the square-of-a-sum identity.

Steps only

46²=(40+6)²=40²+2(40)(6)+6².

=1600+480+36=2116.

Answer

46²=2,116.

Question Q1(ii)

Find 397×403 using the difference-of-squares identity.

Steps only

397×403=(400−3)(400+3).

=400²−3²=160000−9=159991.

Answer

397×403=159,991.

Question Q1(iii)

Find 91² using the square-of-a-difference identity.

Steps only

91²=(100−9)²=100²−2(100)(9)+9².

=10000−1800+81=8281.

Answer

91²=8,281.

Question Q1(iv)

Find 43×45 using a difference of squares.

Steps only

43×45=(44−1)(44+1).

=44²−1²=1936−1=1935.

Answer

43×45=1,935.

Question Q2(i)

Expand (p−1)(p+11).

Steps only

p²+11p−p−11=p²+10p−11.

Answer

p²+10p−11.

Question Q2(ii)

Expand (3a−9b)(3a+9b).

Steps only

(3a)²−(9b)²=9a²−81b².

Answer

9a²−81b².

Question Q2(iii)

Expand −(2y+5)(3y+4).

Steps only

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

−(2y+5)(3y+4)=−6y²−23y−20.

Answer

−6y²−23y−20.

Question Q2(iv)

Expand (6x+5y)².

Steps only

(6x)²+2(6x)(5y)+(5y)²=36x²+60xy+25y².

Answer

36x²+60xy+25y².

Question Q2(v)

Expand (2x−1/2)².

Steps only

(2x)²−2(2x)(1/2)+(1/2)².

=4x²−2x+1/4.

Answer

4x²−2x+1/4.

Question Q2(vi)

Expand (7p)(3r)(p+2).

Steps only

(7p)(3r)=21pr.

21pr(p+2)=21p²r+42pr.

Answer

21p²r+42pr.

Question Q3

Write expressions for (i) two more than a square number and (ii) the sum of the squares of two consecutive integers.

Steps only

(i) If the integer is s: s²+2.

(ii) If the consecutive integers are m and m+1: m²+(m+1)²=2m²+2m+1.

Answer

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

Question Q11(b)

For the second basic-unit pattern, continue the sequence, find the count at Steps 4 and 10, and write a rule for Step y.

Steps only

Step 4: (4+1)²+4=29.

Step 10: (10+1)²+10=131.

Step y: (y+1)²+y.

Answer

Step 4 has 29 units; Step 10 has 131 units; Step y has (y+1)²+y units.

Questions follow NCERT’s Ganita Prakash Part I, Chapter 6 ↗, checked on 2026-10-05. Use our original question summaries and worked answers alongside the questions in your textbook.

Diagrams are drawn from mathematical rules. Learnivelle is not affiliated with NCERT.

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