Official NCERT 2026-27 Solutions
NCERT Solutions for Class 8 Maths Chapter 13: Algebra Play (Ganita Prakash Part II)
Authentic handwritten solutions in Topper Notebook style. Features digit product maximisation, divisibility trick decodings, 1001 cyclic permutations, the magical pond puzzle, and linear business equations from Pages 143 to 147 of Ganita Prakash Part II.
Quick Chapter Navigation & Blueprint
- Part A: Digit Optimisation & Reversal Tricks (Pages 143–145) — Maximising products with digits (1, 3, 7 and 3, 5, 9), algebraic proof for qp × r, and proving 9|(ba − ab) and 11|(ba + ab).
- Part B: Cyclic Permutations & The Magic 1001 Factor (Page 145) — Proving 37 and 3 divide abc + bca + cab, and why dividing abcabc by 7, 11, and 13 returns abc.
- Part C: Classic Word Puzzles & Linear Systems (Pages 145–146) — The 3 Shrines & Magical Pond puzzle (7 start, 8 offered), horses & hens head-leg method (20 horses, 35 hens), mother-daughter ages (6 & 30), and Gauri-Naina cows (6 & 12).
- Part D: Commercial Break-Even & Karim and the Genie (Pages 146–147) — Dosa cart break-even economics (&inr;80 SP / 175 dosas) and Karim’s 7-coin genie trick reverse engineering.
Official Textbook PDF Reference: Download NCERT Class 8 Maths Chapter 13 (hegp206.pdf)
Part A: Digit Optimisation & Reversals (Pages 143–145)
Questions 1 & 2 (Page 144):
(1) Fill digits 1, 3, and 7 in [ ][ ] × [ ] to make the largest product possible.
(2) Fill digits 3, 5, and 9 in [ ][ ] × [ ] to make the largest product possible.
Prove algebraically why this strategy always works.
(1) Fill digits 1, 3, and 7 in [ ][ ] × [ ] to make the largest product possible.
(2) Fill digits 3, 5, and 9 in [ ][ ] × [ ] to make the largest product possible.
Prove algebraically why this strategy always works.
Algebraic Proof of the Optimal Arrangement:
- Let the three distinct digits be p < q < r.
- To maximise the 2-digit × 1-digit product, the single-digit multiplier must be the largest digit r, or the largest digit must occupy the tens place of the multiplicand.
- Comparing the two top candidates:
• Expression 1: (qp) × r = (10q + p) × r = 10qr + pr.
• Expression 2: (rp) × q = (10r + p) × q = 10rq + pq. - Subtracting Expression 2 from Expression 1:
(10qr + pr) − (10rq + pq) = pr − pq = p(r − q). - Since r > q, (r − q) > 0, and since p is a non-zero digit, p(r − q) > 0!
- Therefore, (qp) × r is strictly the largest possible product!
- Applying to Question 1 (Digits 1, 3, 7 where p=1, q=3, r=7):
Largest Product = 31 × 7 = 217. - Applying to Question 2 (Digits 3, 5, 9 where p=3, q=5, r=9):
Largest Product = 53 × 9 = 477.
Answers:
(1) Digits 1, 3, 7: 31 × 7 = 217.
(2) Digits 3, 5, 9: 53 × 9 = 477.
(1) Digits 1, 3, 7: 31 × 7 = 217.
(2) Digits 3, 5, 9: 53 × 9 = 477.
Mukta’s Divisibility Tricks (Page 145):
(1) In the digit reversal subtraction trick, what is the quotient when divided by 9? Is there a relation with the digits?
(2) Instead of difference, what happens if we add the number and its reverse? Prove algebraically.
(1) In the digit reversal subtraction trick, what is the quotient when divided by 9? Is there a relation with the digits?
(2) Instead of difference, what happens if we add the number and its reverse? Prove algebraically.
Algebraic Proof of Reversals:
- Let the two-digit number be N = 10a + b, with reverse N‘ = 10b + a.
- (1) Difference:
Assuming a > b: N − N‘ = (10a + b) − (10b + a) = 9a − 9b = 9(a − b).
Dividing by 9 gives: Quotient = a − b (the exact difference between the two original digits!). - (2) Sum:
N + N‘ = (10a + b) + (10b + a) = 11a + 11b = 11(a + b).
The sum is always unconditionally divisible by 11, and the quotient is simply the sum of the digits (a + b)!
Summary:
(1) Difference divided by 9 gives quotient equal to (a − b).
(2) Sum is always divisible by 11 with quotient equal to (a + b).
(1) Difference divided by 9 gives quotient equal to (a − b).
(2) Sum is always divisible by 11 with quotient equal to (a + b).
Part B: Cyclic Permutations & The 1001 Mystery (Page 145)
Question 3 (Page 145 – 3-Digit Cyclic Sum):
Take any 3-digit number abc. Form cyclic permutations bca and cab. Prove algebraically that their sum is always divisible by 37 and by 3.
Take any 3-digit number abc. Form cyclic permutations bca and cab. Prove algebraically that their sum is always divisible by 37 and by 3.
Topper’s Expansion & Factorisation:
- Expanding place values:
abc = 100a + 10b + c
bca = 100b + 10c + a
cab = 100c + 10a + b - Adding all three numbers together:
Sum = (100a + 10a + a) + (100b + 10b + b) + (100c + 10c + c)
= 111a + 111b + 111c = 111(a + b + c). - Prime factorising 111:
111 = 3 × 37. - Therefore, Sum = 3 × 37 × (a + b + c).
- Since 37 and 3 are factors of 111, the sum is always divisible by 37, by 3, and by (a + b + c)!
Conclusion: The cyclic sum equals 111(a + b + c), which is divisible by both 37 and 3 for every 3-digit number.
Question 4 (Page 145 – The Repeating Number abcabc):
Take any 3-digit number abc. Write it twice to form a 6-digit number abcabc. Divide successively by 7, 11, and 13. What do you get? Explain why it works.
Take any 3-digit number abc. Write it twice to form a 6-digit number abcabc. Divide successively by 7, 11, and 13. What do you get? Explain why it works.
The Prime Product of 1001:
- A repeated 6-digit number can be factored by place value:
abcabc = (abc × 1000) + abc = abc × (1000 + 1) = abc × 1001. - Now observe the product of the three consecutive prime numbers 7, 11, and 13:
7 × 11 × 13 = 77 × 13 = 1001! - Therefore:
abcabc ÷ (7 × 11 × 13) = (abc × 1001) ÷ 1001 = abc. - Example: 425425 ÷ 7 = 60775 → 60775 ÷ 11 = 5525 → 5525 ÷ 13 = 425!
Final Answer: Dividing abcabc by 7, 11, and 13 always returns the original 3-digit number abc, because 7 × 11 × 13 = 1001.
Part C: Word Problems & Practical Algebra (Pages 145–147)
Question 5 (Page 145 – Shrines & Magical Pond):
A devotee has some flowers. At each of 3 shrines, he first dips all his flowers in a magical pond (which doubles them), and then offers an equal number of flowers at the shrine. After offering flowers at the 3rd shrine, exactly 0 flowers remain. How many flowers did he start with? How many did he offer at each shrine?
A devotee has some flowers. At each of 3 shrines, he first dips all his flowers in a magical pond (which doubles them), and then offers an equal number of flowers at the shrine. After offering flowers at the 3rd shrine, exactly 0 flowers remain. How many flowers did he start with? How many did he offer at each shrine?
Topper’s Backward Induction Method:
- Let the equal number of flowers offered at each shrine be x.
- Working backwards from the end:
• After Shrine 3: 0 flowers left.
• Before offering at Shrine 3: Had x flowers.
• Before Pond 3 (halving): Had x / 2 flowers.
• After Shrine 2: Had x / 2 flowers remaining.
• Before offering at Shrine 2: Had (x / 2) + x = 3x / 2 flowers.
• Before Pond 2 (halving): Had (3x / 2) / 2 = 3x / 4 flowers.
• After Shrine 1: Had 3x / 4 flowers remaining.
• Before offering at Shrine 1: Had (3x / 4) + x = 7x / 4 flowers.
• Initial flowers before Pond 1 (halving): Had (7x / 4) / 2 = 7x / 8 flowers! - For the initial number of flowers to be a whole integer, x must be a multiple of 8.
- Choosing the smallest positive integer x = 8 flowers offered:
Initial flowers = (7 × 8) / 8 = 7 flowers. - Step-by-step verification: Start with 7 → doubles to 14 → offers 8, leaves 6 → doubles to 12 → offers 8, leaves 4 → doubles to 8 → offers 8, leaves 0 ✓!
Final Answer: The devotee started with 7 flowers and placed 8 flowers in each shrine (or any positive multiple: 7k and 8k).
Question 6 (Page 146 – Horses and Hens):
A farm has horses (4 legs) and hens (2 legs). Total heads = 55, total legs = 150. How many horses and hens are on the farm? Solve without algebraic variables.
A farm has horses (4 legs) and hens (2 legs). Total heads = 55, total legs = 150. How many horses and hens are on the farm? Solve without algebraic variables.
Topper’s Assumption / Excess Legs Method:
- Total animals = 55 (each has 1 head).
- Assumption: Suppose ALL 55 animals were hens.
Expected number of legs = 55 × 2 = 110 legs. - Actual number of legs on the farm = 150 legs.
- Excess legs = 150 − 110 = 40 extra legs.
- Each horse has 2 more legs than a hen (4 − 2 = 2 legs).
- Number of horses = Excess legs / 2 = 40 / 2 = 20 horses.
- Number of hens = Total animals − Horses = 55 − 20 = 35 hens.
- Check: (20 × 4) + (35 × 2) = 80 + 70 = 150 legs ✓.
Final Answer: There are 20 horses and 35 hens.
Question 11 (Page 147 – Karim and the Genie):
Karim circles a banyan tree 3 times. Each round, his money doubles, but he must give 8 coins to a genie. After the 3rd round, he has exactly 8 coins left, which he gives to the genie, leaving 0 coins.
(i) How many coins did Karim start with?
(ii) For what fee per round should Karim agree to the deal if he wants to make a profit?
Karim circles a banyan tree 3 times. Each round, his money doubles, but he must give 8 coins to a genie. After the 3rd round, he has exactly 8 coins left, which he gives to the genie, leaving 0 coins.
(i) How many coins did Karim start with?
(ii) For what fee per round should Karim agree to the deal if he wants to make a profit?
Reverse Deduction & Profit Inequality:
- Let initial coins be C.
• Round 1: Coins = 2C − 8.
• Round 2: Coins = 2(2C − 8) − 8 = 4C − 24.
• Round 3: Coins = 2(4C − 24) − 8 = 8C − 56. - After paying the genie 8 coins, 0 remain: 8C − 56 = 0 → 8C = 56 → C = 7 coins!
- (ii) Fee condition for profit:
For money to increase after 1 round: 2C − Fee > C → Fee < C.
Karim must ensure the genie’s fee per round is strictly less than his current number of coins.
Answers:
(i) Karim initially had 7 coins.
(ii) Karim will only make a profit if the fee per round is strictly less than his starting coins (Fee < C).
(i) Karim initially had 7 coins.
(ii) Karim will only make a profit if the fee per round is strictly less than his starting coins (Fee < C).