Chapter 6
We DISTRIBUTE YET THINGS MULTIPLY
Figure it out Q & Answers
Figure it Out 142
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
2. Expand the following products.
(i) (3 + u) (v – 3)
(ii) 2/3 (15 + 6a) = 10 +4a
(iii) (10a + b) (10c + d)
(iv) (3 – x) (x – 6)
(v) (–5a + b) (c + d)
(vi) (5 + z) (y + 9)
ANSWER:
(i) (3 + u) (v – 3) = 3v – 3u + uv -9
(ii) 2/3 (15 + 6a) = 10 +4a
(iii) (10a + b) (10c + d) = 100ac+10ad+10bc+bd
(iv) (3 – x) (x – 6) = -x² + 9x – 18
(v) (–5a + b) (c + d) = -5ac -5ad + bc + bd
(vi) (5 + z) (y + 9) = 5y+45+yz+9z
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:
Let the numbers be a and b.
Then, ab = (a + 2)(b - 4)
⇒ ab = ab - 4a +2b - 8
=> ab - ab + 4a + 8 = 2b
4a + 8 = 2b [Divide throughout by 2]
2a + 4 = b
For a = 1 , b = 2 * 1 + 4 = 6
ab = 1 * 6 = 6
and (a+2) (b-4)=3×2=6
Hence, ab = (a + 2)(b - 4) Let a = 2 then b = 2 * 2 + 4 = 8
Let a = 3 then b = 2 * 3 + 4 = 10
Three such pairs are 1 and 6, 2 and 8, 3 and 10.
4. Expand (i) (a + ab – 3b²) (4 + b), and (ii) (4y + 7) (y + 11z – 3)
Solution:
(i) (a + ab-3b²) (4+b)
= (4 + b)(a + ab - 3b²)
= 4(a + ab - 3b²) + b(a + ab - 3b²)
= 4a + 4ab - 12b² + ab + ab² - 3b³
= 4a + 5ab - 12b² + ab² - 3b³
(ii) (4y + 7)(y + 11z - 3)
= 4y(y + 11z - 3) + 7(y + 11z - 3)
= 4y² + 44yz - 12y + 7y + 77z - 21
= 4y² + 44yz - 5y + 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(a + b) - b(a + b)
= a² + ab - ab – b²
= a² - b²
(ii) (a - b)(a² + ab + b²)
= a(a² + ab + b²) - b(a² + ab + b²)
= a³ + a²b + ab² - a²b - a * b² - b³
= a³ - b³
(iii) (a - b)(a³ + a²b + ab² + b³)
= a(a³ + a² * b + ab² + b³) -b(a³+a²b+ ab² + b³ )
= π^4 + a³b + a²b² + ab³ - a³b - a²b² -ab³ - π^4
= π^4 - π^4
The pattern follows
(a-b)(π^π + π^(π−1) b+...+ π^π )= π^(π+1) - π^(π+1)
Next identity in the pattern
(a - b)(π^4 + a³b + a²b² + ab³ + π^4) = π^5 - π^5
Figure it Out 149
1.Which is greater: (a – b)² or (b – a)²? Justify your answer.
Solution:
(a - b)² = a² + b² - 2ab ... (1)
and
(b - a)² = b² + a² - 2ba
b² + a²= a² + b²
and
ba = ab
:. (b - a)² = a² + b² - 2ab ... (2)
Comparing (1) and (2), we get (a - b)² = (b - a)²
2.Express 100 as the difference of two squares.
Solution:
a² - b² = 100
(a + b)(a - b) = 100
[100=1×100, 2 × 50, 4 x 25, 5 x 20, 10 × 10]
take any one
Let us take
50 * 2 = 100
Hence, (a + b)(a - b) = 50 * 2
a + b = 50 ... (1)
a - b = 2 ... (2)
Adding (1) and (2)
2a = 52
Substituting a = 26in(1)
⇒ 26 + b = 50
b = 50 - 26 = 24
Let us check
26² - 24²
676 – 576 = 100
Hence
26² - 24² = 100
a = 26
3. Find 406², 72², 145², 1097², and 124² using the identities you have learnt so far.
Solution:
(1) 406² = (400 + 6)² = 400² + 2 * 400 * 6 + 6² = 160000 + 4800 + 36 = 164836
(ii) 72² = (50 + 22)² = 50² + 2 * 50 * 22 + 22² =2500 + 2200 + 484 = 5184
(iii) 145² = (150 - 5)² = 150² - 2 * 150 * 5 + 5² =22500-1500+25 = 21025
(iv) 1097² = (1100-3)² =1100² - 2×1100 × 3 + 3² = 1210000 – 6600 + 9 = 1203409
(Ξ½) 124² = (100 + 24)² = 10000 + 4800 + 576 = 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
Pattern 1 2 (a² + b² ) = (a+b)² + (a – b)²
Case I Let a = 4, b= 2
LHS = 2(4²+2²) = 2x (16+4) = 2 * 20 = 40
RHS = (4 + 2)² + (4 - 2)² = 36 + 4 = 40
LHS = RHS, Pattern 1 holds for counting numbers
Case II Let a = -4, b = -2
LHS = 2((-4)²+(-2)²) = 2x (16+4) = 2 * 20 = 40
RHS = (- 4 + (- 2))² + (- 4 - (- 2))² = (- 4 - 2)² + (- 4 + 2)²
= (- 6)² + (- 2)² =36+4=40
LHS = RHS
Pattern 1 holds for negative integers also.
Case-III Let a =1/2, b = 1/3
LHS
= 2 [ (1/2)² + (1/3)² ]
= 2 [ 1/4 + 1/9 ]
= 2 * 13/36
= 13/18
RHS
(1/2 + 1/3)² + (1/2 - 1/3)²
= (5/6)² + (1/6)²
= 25/36 + 1/36
= 26/36 = 13/18
Pattern 1 holds for fractions also.Pattern 2 (a² - b² ) = ( a + b )(a – b)
Case I Let a = 5, b= 3
LHS 5² - 3² = 25 – 9 = 16
RHS (5+3) (5-3)=8×2=16 .
LHS = RHS Pattern 2 holds for counting numbers.
Case II Let a = - 5, b = - 3
LHS = (-5)² - (-3)² = 25 – 9 = 16
RHS = [(- 5) + (- 3)][(-5)-(-3)] =(-8)(-2) = 16
LHS = RHS
Pattern 2 holds for negative integers also.
Case-III Let a =1/2, b = 1/3
LHS
= [ (1/2)² - (1/3)² ]
= [ 1/4 - 1/9 ]
= 5/36
RHS
(1/2 + 1/3)(1/2 - 1/3)
= (5/6)(1/6) = 5/36
Pattern 2 holds for fractions also.
Figure it Out 154-156
1. Compute these products using the suggested identity.
(i) (46²) using Identity 1A: ((a + b)² = a² + 2ab + b²)
Let (a = 40), (b = 6): [ 46² = (40 + 6)² = 40² + 2*40*6 + 6² = 1600 + 480 + 36 = 2116]
(ii) 397 × 403 using Identity 1C for (a + b) (a – b)
Let (a = 400), (b = 3): [ 397 X 403 = (400 - 3)(400 + 3) = 400² - 3² = 160000 - 9 = 159991]
(iii) 91² using Identity 1B for (a – b)²
Let (a = 100), (b = 9): [ 91² = (100 - 9)² = 100² - 2*100*9 + 9² = 10000 - 1800 + 81 = 8281 ]
(iv) 43 × 45 using Identity 1C for (a + b) (a – b)
Let (a = 44), (b = 1): [ (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) → Identity 1C = p² + 11p - p - 11 = p² + 10p - 11(ii) (3a – 9b) (3a + 9b) → Identity 1C = (3a)² - (9b)² = 9a² - 81b²
(iii) –(2y + 5) (3y + 4) → Distributive = - [6y² + 8y + 15y + 20] = - [6y² + 23y + 20] = [-6y² - 23y - 20 ]
(iv) (6x + 5y)² = → Identity 1A = 36x² + 2 * 6x * 5y + 25y² = 36x² + 60xy + 25y²
(v) (2x – 1/2 )² → Identity 1B = (2x)² - 2*2x *1/2 + (1/2)² = 4x² - 2x + 1/4
(vi) (7p) × (3r) × (p + 2) → Distributive = 21pr(p + 2) = 21p²r + 42pr = 21pr² + 42pr
3. For each statement identify the appropriate algebraic expression(s).
(i) Two more than a square number.
2 + s (s + 2)² s² + 2 ANSWER S² + 4 2s²
(ii) The sum of the squares of two consecutive numbers
m² + n² (m + n)² m² + 1 m² + (m + 1)² ANSWER m² + (m – 1)² (m + (m + 1))² (2m)² + (2m + 1)²
Let the numbers be (m) and (m+1). Then the sum is: [ m² + (m+1)² = m² + m² + 2m + 1 = 2m² + 2m + 1 ]
✅ Correct expression: (m² + (m+1)²) Also valid: (m² + (m-1)²) — if the consecutive pair is (m) and (m-1)
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
Observation: In any 2×2 calendar square:
Diagonal 1: (a *(a + 8))
Diagonal 2: ((a + 1) *(a + 7))
Let’s compute the difference:
[ a(a + 8) - (a + 1)(a + 7) = a² + 8a - [a² + 7a + a + 7] = a² + 8a - (a² + 8a + 7) = -7 ]
✅ The diagonal products always differ by 7 — the number of days in a week!
Why this happens: The calendar is structured in weeks (7 days), so the diagonals always span across a fixed offset. The algebra confirms that the difference between the two diagonal products is always 7, regardless of the starting number (a). Case 1 6 x 14 =84, 13 x 7 = 91, Difference = 91 – 84 = 7
Case 2 9 x 17 = 153, 16 x 10 = 160, Difference = 160 – 153 = 7
(i) (k + 1) (k + 2) – (k + 3) is always 2.
(k+1)(k+2)−(k+3)
Expand: (k + 1)(k + 2) = k² + 3k + 2, k² + 3k + 2 - k - 3 = k² + 2k - 1
❌ Not always 2 — it depends on k. So statement is sometimes true
Example: Let k = 5, then ( 5+ 1) (5+2)-(5+3) = 6 x 7 – 8 =42 – 8 = 34 is a multiple of 2, 4 x 5 -6 = 20 – 6 = 14 is a multiple of 2.
K = 4, then 5 x 6 – 7 = 30 -7 = 23
(ii) (2q + 1) (2q – 3) is a multiple of 4.
Expand: [ = 4q² - 6q + 2q - 3 = 4q² - 4q - 3 ]
This expression is not always divisible by 4 due to the (-3). Verdict: False
Example: Let q = 3, then (6 +1) (6 - 3) = 7 x 3 =21 is not a multiple of 4. so statement is false
(iii) Squares of even numbers are multiples of 4, and squares of odd numbers are 1 more than multiples of 8.
Example: 2² = 4 = 4 x 1, 4² = 16 = 4 x 4, 6² = 36 = 4 x 9 , always true.
Example: 3² = 9 = 8 + 1, 5² = 25 = 8 x 3 + 1, 7² = 49 = 8 x 6 + 1 , always true.
Let (n = 2k): [ (2k)² = 4k² multiple of 4 ]
Squares of odd numbers: Let (n = 2k + 1): [ (2k + 1)² = 4k² + 4k + 1 = multiple of 4 + 1 ]
✅ Always 1 more than a multiple of 8 Verdict: True
(iv) (6n + 2)² – (4n + 3)² is 5 less than a square number.
(iv) ((6n + 2)² − (4n + 3)²)
Use identity: [ a² - b² = (a + b)(a - b) ]
Let (a = 6n + 2,; b = 4n + 3): [ = (10n + 5)(2n - 1) ]
This is a product — not necessarily 5 less than a square. Try (n = 1): [ (6 * 1 + 2)² = 64,; (4 * 1 + 3)² = 49,; 64 - 49 = 15 15 = 20 - 5 = {5 less than } 20 ]
But 20 isn’t a square. Verdict: False
Example: Let n = 2, ( 6 x 2 +2 )² - (4 x 2 + 3)² = 14² - 11² = 196 – 121 = 75 = 80 – 5 but 80 is not a square number. Statement is false.
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?
6. Remainders modulo 7
Let (a ≡ 3 (mod 7),; b ≡ 5 (mod 7)
Sum: (a + b ≡ 3 + 5 = 8 ≡ 1 (mod 7)
Difference: (a - b ≡ 3 - 5 = -2 ≡ 5 (mod 7)
Product: (a * b ≡ 3 *5 = 15 ≡ 1 (mod 7)
✅ Remainders:
Sum → 1
Difference → 5
Product → 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.
7. Three consecutive numbers
Let them be (n - 1,; n,; n + 1)
Square the middle: (n²)
Product of others: ((n - 1)(n + 1) = n² - 1)
Subtract: n² - (n² - 1) = 1
✅ Always 1 — a beautiful identity!
Statement and pattern
Choose: ((n-1),, n,, (n+1)).
Operation: Square the middle and subtract the product of the other two.
[ n²- (n-1)(n+1) = n²- (n²- 1) = 1 ]
Conclusion: The result is always (1), an identity:
[ n²- (n-1)(n+1) \equiv 1 ]
Classroom hook: Link this to the difference of squares ((n+1)(n-1) = n²- 1) and let students test several triples to “feel” the invariance.
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.
Step 1: a + b
Step 2: (a + b) * 1/2 * (a + b)
= (a + b) * 1/2 * (a + b)
= 1/2 * (a + b)²
Product with half of the sum expression
Description and proof
Let: Two numbers be (a) and (b).
Instruction: Add them, then multiply by half their sum.
[ (a+b) * 1/2 (a+b) = 1/2 (a+b)² = 1/2 (a² + 2ab + b² ]
Conclusion: The result is half the square of the sum.
9. Which is larger? Find out without fully computing the product. (i) 14 × 26 or 16 × 24 (ii) 25 × 75 or 26 × 74
Solution: (1) Let p = 14 * 26 p’ = 16 * 24 =(14 + 2)(26 - 2)
p' = 14 * 26 + 2 * 26 - 14 * 2 - 2 * 2
= 14 * 26 + 2(26 - 14 - 2)
= 14 * 26 + 2 * 10
p' = p + 2 * 10
:. p' > p or 16 * 24 > 14 * 26
Solution:
(ii) Let p = 25 * 75
p' = 26 * 74 = (25 + 1)(75 - 1)
= 25 * 75 + 75 * 1 - 25 * 1 - 1 * 1
= p + (75 - 25 - 1) = p + 49
:. p'> p or 26 × 74 > 25 × 75
Comparing products without full computation
Principle
Key idea: For a fixed sum, the product is larger when the pair is closer together (by AM-GM or completing a square).
(i) (14 * 26) vs (16 * 24)
[ 14 *26 = (20-6)(20+6) = 20²- 6²= 400 - 36 = 364 ]
[ 16 *24 = (20-4)(20+4) = 20²- 4²= 400 - 16 = 384 ]
Verdict: (16 *24) is larger.
(ii) (25 *75) vs (26 *74)
[ 25 *75 = (50-25)(50+25) = 50²- 25²= 2500 - 625 = 1875 ]
[ 26 *74 = (50-24)(50+24) = 50²- 24²= 2500 - 576 = 1924 ]
Verdict: (26 *74) is larger.
Quick heuristic: Same sum ⇒ use (S²- d² with mean (S/2) and deviation (d); smaller (d) means larger product.
10. A tiny park is coming up in Dhauli. w ft. 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: Interpretation : the two green square plots each have side length g. The walkway of width www surrounds and separates them so that the overall outer rectangle has horizontal length 2g+3w and vertical length g+2w. (This matches the usual arrangement with a path of width w on both outer sides and between the two plots.)
Total outer rectangle area =(2g+3w)(g+2w).
Area of the two green squares =2g².
So area to be tiled (walking path) = total area −2g²
(2g+3w)(g+2w)−2g².
You can expand if you want:
(2g+3w)(g+2w)−2g² = (2g²+4gw+3gw+6w²)−2g² = 7gw+6w²=w(7g+6w)
Tiny park problem (two square plots of area g² each and walkway width w)
Park tiling area with two g² plots and path width w
The tiling area equals “total outer area” minus “green cover.” The exact total area depends on whether there’s a path between the two squares in addition to the outer border. Here are the two natural interpretations:
Case A: Path only around the outside (no path between squares)
Layout: Two (g*g) squares placed side-by-side with a border path of width (w) around the combined rectangle.
Outer dimensions: Width (= 2g + 2w), Height (= g + 2w).
Tiled area:
[ A tiled = (2g + 2w)(g + 2w) - 2g²]
Case B: Path around the outside and between the squares
Layout: Side-by-side (g*g) squares, a path of width (w) between them, and a border path of width (w) around.
Outer dimensions: Width (= 2g + 3w), Height (= g + 2w).
Tiled area:
[ A tiled = (2g + 3w)(g + 2w) - 2g² ]
If your figure shows a strip between the squares, use Case B; if it’s only the outer border, use Case A. Share the exact placement and I’ll give a clean diagram or simplify the expression further.
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.
Step 1: 2 vertical strips of 3 units each + 1 vertical strip of 3 units
= 3 strips of 3 units each = 9 units squares = (1 + 2)²| unit squares
Step 2: 4 strips of 4 units each = 16 units squares = (2 + 2)²unit squares
Step 3: 5 strips of 5 units each =25 units squares =(3+2)² unit squares
Step 4: (1) 6 strips of 6 units each = 2 are vertical and 4 are horizontal
(ii) Number of unit squares in step 10 = (10 + 2)²= 144
(iii) Number of unit squares in step y = (y + 2)²11. For each pattern shown below,
(b) Number of unit squares in step 1 = 5 = 2²+ 1
(b) Number of unit squares in step 1 = 5 = 2²+ 1
Number of unit squares in step 2 = = 11 = 3²+ 2
Number of unit squares in step 3 = 19 = 4²+ 3
(ii) Step 1 has (1 + 1)²+ 1 or 5 squares
Step 2 has (2 + 1)²+ 2 or 11 squares
Step 3 has (3 + 1)²+ 3 or 19 squares
step 10 has
(10 + 1)²+ 10 or 131 squares
(iii) Step y has [(y + 1)²+ y] squares
Yellow Pattern Step 1: Analyze the pattern
The number of basic units in each step is: Step 1: 5 units Step 2: 9 units Step 3: 13 units
The pattern is an arithmetic sequence where each subsequent step adds 4 units. The common difference is 4.
Step 2: Draw the next figure (i)
10. Tiny park problem (two square plots of area g² each and walkway width w)
The next figure (Step 4) will have 13+4=17 units.
It is formed by adding 4 units to the top-most vertical part of Step 3, extending it upwards. Step 3: Calculate units in Step 10
(ii) The number of units in Step 10 can be found using the formula for the n-th term of an arithmetic sequence: There are 41 basic units in Step 10.
Step 4: Write an expression for Step y (iii) Using the same formula with n = y The expression to describe the number of basic units in Step y is 4y + 1 
Blue Pattern
Step 1: Analyze the pattern
The number of basic units in each step is:
Step 1: 4 units
Step 2: 9 units
Step 3: 16 units
The pattern is a sequence of perfect squares 2², 3², 4². The number of units in Step n is (n+1)²
Step 2: Draw the next figure (i) The next figure (Step 4) will have (4+1)² = 25+4=29 units = 5 x 5 square+4units
Step 3: Calculate units in Step 10 (ii)
The number of units in Step 10 is (10 + 1)² = 121 units. There are 121+10 = 131 basic units in Step 10.
Step 4: Write an expression for Step y (iii)
he expression to describe the number of basic units in Step y is ( y + 1)²+y
