class8-Part 2 chapter 6 - Algebra Play- solutions

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Class 8 Ganita Prakash · Part 2 · Chapter 6
Algebra Play

Page-wise textbook solutions (pages 135–147, Part 2) · Every step · Figures · Common mistakes · Tips

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🗝️ Key ideas of this chapter

• 'Think of a number' tricks work because the unknown cancels out.
• Date trick: result = 100M + 165 + D.
• Pyramid tops: 3 rows \(a + 2b + c\); 4 rows \(a + 3b + 3c + d\).
• Calendar 2 × 2 block: \(4a + 16\).
• Number − reverse = \(9(a - b)\); number + reverse = \(11(a + b)\); abc + bca + cab = \(111(a + b + c)\); abcabc = abc × 1001.
• Largest □□ × □: biggest digit as multiplier, others in decreasing order.

📖 Textbook Page 135

📖 Page 135Think

Double, add 4, halve, subtract the number — why is the answer always 2?

1x → 2x → 2x + 4 → x + 2 → (x + 2) − x = 2

✅ Answer: The x cancels, so the answer is always 2.

📖 Textbook Page 136

📖 Page 136Think

Change the trick so the answer is 3, or 5. Make a longer trick with a fixed answer.

1Answer 3: add 6 instead of 4 (2x + 6 → x + 3 → 3). Answer 5: add 10.

2Longer: think of x, multiply by 3 (3x), add 12 (3x + 12), divide by 3 (x + 4), add 6 (x + 10), subtract x → 10.

✅ Answer: Add twice the answer you want, then halve and subtract x.

📖 Textbook Page 137

📖 Page 137Date trick

Month × 5, + 6, × 4, + 9, × 5, + day. Why does subtracting 165 reveal the date? Find dates for 1269, 394, 296.

1Result = 100M + 165 + D, so result − 165 = 100M + D: month in front, day in the last two digits.

2(i) 1269 − 165 = 1104 → 4 November

3(ii) 394 − 165 = 229 → 29 February

4(iii) 296 − 165 = 131 → 31 January

✅ Answer: (i) 4/11 (ii) 29/02 (iii) 31/01

📖 Page 137Math Talk

Change the steps and still find the date.

1Example: M × 5, + 3, × 20, + D → 100M + 60 + D → subtract 60.

2Any steps giving 100M + (constant) + D work.

✅ Answer: Subtract the new constant.

📖 Textbook Page 138

📖 Page 138Pyramids

Fill the pyramids with bottom rows 6, 2 · 3, 4, 3 · 5, 4, 5, 0.

1Add neighbours to get the box above.

✅ Answer: Tops: 8, 14, 32

📖 Page 138Pyramid 10

Top 10, middle-left 4, bottom-left 1. Fill it.

110 − 4 = 6

24 − 1 = 3

36 − 3 = 3

✅ Answer: Bottom 1, 3, 3; middle 4, 6

📖 Textbook Page 139

📖 Page 139Pyramid 60

Top 60, bottom 12, c, 8. Find c.

1a = 12 + c, b = c + 8, a + b = 60

220 + 2c = 60 → c = 20

✅ Answer: c = 20; middle 32, 28

📖 Page 139Fill pyramids

Fill: (1) top 50, row 3 right 22, bottom 4, _, 6, _ (2) row 3 left 40, row 2 right 9, bottom 5, _, 7, _ (3) top 35, row 2 right 7, bottom 3, 5, _, _

1(1) Bottom 4, x, 6, y. Row 3: 10 + 2x and x + y + 12 = 22 → x + y = 10. Top: 10 + 2x + 22 = 50 → x = 9, y = 1.

2(2) 7 + y = 9 → y = 2; row 3 left (5 + x) + (x + 7) = 40 → x = 14; top = 40 + 30 = 70.

3(3) Bottom 3, 5, x, y with x + y = 7; top 18 + 3x + y = 35 → 3x + y = 17 → x = 5, y = 2.

✅ Answer: Bottoms: 4, 9, 6, 1 · 5, 14, 7, 2 · 3, 5, 5, 2

📖 Textbook Page 140

📖 Page 140Figure it Out · Q1

Top of 3-row pyramids with bottoms 4, 13, 8 · 7, 11, 3 · 10, 14, 25.

1Top = a + 2b + c

24 + 26 + 8; 7 + 22 + 3; 10 + 28 + 25

✅ Answer: 38, 32, 63

📖 Page 140Figure it Out · Q2

Top of a 4-row pyramid with bottom a, b, c, d.

1Row 2: a + b, b + c, c + d

2Row 3: a + 2b + c, b + 2c + d

3Top: a + 3b + 3c + d

✅ Answer: a + 3b + 3c + d (coefficients 1, 3, 3, 1)

💡 Tip: The coefficients come from Pascal's / Meru-prastāra triangle: 1, 1-1, 1-2-1, 1-3-3-1 …

📖 Page 140Figure it Out · Q3

Tops for bottoms 8, 19, 21, 13 · 7, 18, 19, 6 · 9, 7, 5, 11.

1a + 3b + 3c + d

28 + 57 + 63 + 13 = 141

37 + 54 + 57 + 6 = 124

49 + 21 + 15 + 11 = 56

✅ Answer: 141, 124, 56

📖 Page 140Figure it Out · Q4

Bottom row 1, 2, 3 (first three Virahāṅka-Fibonacci numbers). Fill it.

1Row 2: 3, 5

2Top: 8

✅ Answer: Numbers 1, 2, 3, 3, 5, 8 — all are Virahāṅka-Fibonacci numbers; top 8.

📖 Page 140Figure it Out · Q5

(i) Bottom 1, 2, 3, 5 in 4 rows (ii) first 29 numbers in 29 rows.

1(i) Rows: 3, 5, 8 → 8, 13 → 21. All are Virahāṅka-Fibonacci numbers.

2Adding two neighbours F(k) + F(k+1) = F(k+2), so each row is again consecutive Fibonacci numbers, shifted by 2.

3(ii) Each row up shifts by 2 places; 28 steps up → the top is the 1 + 56 = 57th number.

✅ Answer: (i) top 21 (ii) every number is a Virahāṅka-Fibonacci number; top = the 57th one

📖 Page 140Figure it Out · Q6

Bottom row = first n Virahāṅka-Fibonacci numbers. What about the pyramid and the top?

1Every row is a run of consecutive Fibonacci numbers.

2Top = the (2n − 1)th number: n = 3 → 5th = 8; n = 4 → 7th = 21.

✅ Answer: All entries are Virahāṅka-Fibonacci numbers; top = the (2n − 1)th one.

📖 Textbook Page 141

📖 Page 141Calendar magic

A 2 × 2 block of a calendar has sum 36. Find the four numbers. Check 6, 7, 13, 14.

1Block = a, a + 1, a + 7, a + 8 → sum 4a + 16.

24a + 16 = 36 → a = 5

3Check: 4 × 6 + 16 = 40 ✔

August 2025: a 2 × 2 block is a, a + 1, a + 7, a + 8 → sum 4a + 16. Sum 36 → a = 5.

✅ Answer: 5, 6, 12, 13

📖 Textbook Page 142

📖 Page 142Math Talk

Create your own calendar trick (another size or shape).

13 × 3 block: centre m → the 9 numbers add to 9m. So sum ÷ 9 = centre.

2Vertical strip of 3: a, a + 7, a + 14 → 3a + 21.

✅ Answer: e.g. 3 × 3 block: sum = 9 × middle number.

📖 Page 142Algebra grids

Find the shape values and fill the empty squares.

1Grid 1: 2■ + ● = 27 and 2● + ■ = 21 → adding: 3(■ + ●) = 48 → ■ + ● = 16 → ■ = 11, ● = 5. Row 3 (●■●) = 21.

2Grid 2: ● + 2◆ = 18 and ◆ + 2● = 15 → ● + ◆ = 11 → ◆ = 7, ● = 4. Row 3 = 15.

3Grid 2 column totals: 4 + 7 + 7 = 18, 7 + 4 + 4 = 15, 15, grand total 48.

✅ Answer: ■ = 11, ● = 5 (row 3: 21) · ● = 4, ◆ = 7 (row 3: 15; columns 18, 15, 15; total 48)

📖 Textbook Page 143

📖 Page 143Largest product

Use 2, 3, 5 in □□ × □ for the largest product. Is there a general rule?

1Six products; for each multiplier the bigger 2-digit number wins: 53 × 2 = 106, 52 × 3 = 156, 32 × 5 = 160.

2General p < q < r: compare qp × r = 10qr + pr with rp × q = 10qr + pq; pr > pq.

✅ Answer: 32 × 5 = 160. Rule: largest digit is the multiplier; the other two in decreasing order.

📖 Textbook Page 144

📖 Page 144Figure it Out · Q1

Digits 1, 3, 7: largest □□ × □.

1Largest digit 7 as multiplier, 31 as the number.

✅ Answer: 31 × 7 = 217

📖 Page 144Figure it Out · Q2

Digits 3, 5, 9: largest □□ × □.

153 × 9

✅ Answer: 53 × 9 = 477

📖 Textbook Page 145

📖 Page 145Think

Why is (number − reversed number) always divisible by 9? What if a > b?

1(10a + b) − (10b + a) = 9a − 9b = 9(a − b)

✅ Answer: Always 9 × (difference of the digits).

📖 Page 145Figure it Out · Q1

What is the quotient when dividing by 9?

19(b − a) ÷ 9 = b − a

✅ Answer: The quotient is the difference of the two digits (74 − 47 = 27, 27 ÷ 9 = 3 = 7 − 4).

📖 Page 145Figure it Out · Q2

Number + reversed number: always divisible by 11?

1(10a + b) + (10b + a) = 11a + 11b = 11(a + b)

✅ Answer: Yes, always — it is 11 × (sum of digits).

📖 Page 145Figure it Out · Q3

abc + bca + cab is divisible by 37? By 3?

1Each digit appears once in each place: 100(a + b + c) + 10(a + b + c) + (a + b + c) = 111(a + b + c)

2111 = 3 × 37

✅ Answer: Yes — divisible by 37 and by 3 (and by a + b + c).

📖 Page 145Figure it Out · Q4

abcabc ÷ 7 ÷ 11 ÷ 13 gives?

1abcabc = abc × 1000 + abc = abc × 1001

21001 = 7 × 11 × 13

✅ Answer: You get back abc!

📖 Textbook Page 146

📖 Page 146Figure it Out · Q5

Flowers double in each pond; equal numbers go to each of 3 shrines; nothing left. Starting number?

1Start with x flowers; put k in each shrine.

2After pond 1 and shrine 1: 2x − k left. After pond 2 and shrine 2: 2(2x − k) − k = 4x − 3k left.

3Pond 3 doubles this and ALL of it goes to shrine 3: 2(4x − 3k) = k → 8x = 7k.

4Smallest whole numbers: x = 7, k = 8. Check: 7 → 14 − 8 = 6 → 12 − 8 = 4 → 8 ✔

✅ Answer: 7 flowers; 8 in each shrine (or any multiple: 14 and 16, …)

📖 Page 146Figure it Out · Q6

55 heads and 150 legs of horses and hens. How many of each?

1If all were hens: 110 legs. Extra 40 legs → each horse adds 2 → 20 horses.

2Hens = 55 − 20 = 35. Check: 80 + 70 = 150 ✔

✅ Answer: 20 horses, 35 hens

📖 Page 146Figure it Out · Q7

Mother is 5 times her daughter's age; in 6 years she will be 3 times. Daughter's age?

15d + 6 = 3(d + 6) → 2d = 12

✅ Answer: Daughter 6, mother 30

📖 Page 146Figure it Out · Q8

Naina has twice Gauri's cows; if Naina gives 3, they are equal.

1N = 2G and N − 3 = G + 3 → 2G − 3 = G + 3 → G = 6

✅ Answer: Gauri 6, Naina 12

📖 Page 146Figure it Out · Q9

Dosa cart: rent ₹5000/day, cost ₹10 per dosa. (i) Price for profit ₹2000 selling 100 (ii) dosas at ₹50 for profit ₹2000?

1(i) Costs = 5000 + 1000 = 6000. Need 8000 from 100 dosas → ₹80.

2(ii) Profit = 50n − 10n − 5000 = 2000 → 40n = 7000 → n = 175

✅ Answer: (i) ₹80 (ii) 175 dosas

📖 Page 146Figure it Out · Q10

Evaluate 1/3, (1 + 3)/(5 + 7), (1 + 3 + 5)/(7 + 9 + 11). Why?

11/3, 4/12 = 1/3, 9/27 = 1/3

2First n odd numbers add to n²; the next n odd numbers make the first 2n, minus the first n: (2n)² − n² = 3n².

3n² ÷ 3n² = 1/3

✅ Answer: Always 1/3

📖 Textbook Page 147

📖 Page 147Figure it Out · Q11

Karim's coins double each round and he pays 8. After the 3rd round he has exactly 8 left to pay. (i) Start? (ii) When is the deal good? (iii) Genie's cost to take everything?

1(i) Work backwards: before paying the last 8 he had 8 → before 3rd doubling 4 → +8 = 12 before 2nd payment → 6 → +8 = 14 → 7.

2Algebra: 2(2(2x − 8) − 8) = 8 → 8x − 48 = 8 → x = 7.

3(ii) After a round he has 2x − c. It grows only if 2x − c > x, i.e. c < x — the cost must be less than the coins he has (less than 7).

4(iii) Over 3 rounds: 8x − 7c = 0 → c = 8x ÷ 7 (for x = 7, c = 8).

✅ Answer: (i) 7 coins (ii) cost less than his coins (iii) c = 8x/7

🌍 Where do we use this?

• Puzzles and magic tricks, checking calculations, pricing and profit, word problems.

🚀 Link to higher classes

• Class 9–10: linear equations in two variables, Pascal's triangle and binomial coefficients, proofs of divisibility.

⬆ Top 🏠 Class 8 Hub

🔑 keytoenjoylearningmaths.blogspot.com · Solutions written in our own words, based on NCERT Ganita Prakash Class 8 (Part 2)