NCERT Solutions for Class 8 Maths Chapter 13: Algebra Play – Ganita Prakash

Chapter 13 of the Class 8 NCERT Ganita Prakash textbook (Part 2, Chapter 6 in the book’s own numbering) is titled “Algebra Play”. It uses number tricks, number pyramids, grid puzzles and divisibility puzzles to build algebraic reasoning — representing unknowns with letters and justifying number patterns using algebra. These Class 8 Mathematics Chapter 13 solutions are also useful as quick revision notes before exams.

Below are original, independently verified solutions to the chapter’s key questions.

13.1–13.2 “Think of a Number” Tricks

Q. How would you change the “think of a number” game so the final answer is always 3? What about always 5?
Solution: Let the chosen number be x.
For a final answer of 3: triple it (3x), add 9 (3x+9), divide by 3 (x+3), then subtract the original number: (x+3)−x = 3. Example: start with 23 → 69 → 78 → 26 → 26−23 = 3. ✓
For a final answer of 5: double it (2x), add 10 (2x+10), divide by 2 (x+5), then subtract the original number: (x+5)−x = 5.

Q. Can you design more complicated steps that always lead to the same final value?
Solution: Yes. Multiply by 5 (5x), add 25 (5x+25), divide by 5 (x+5), subtract the original number: 5. Example: start with 18 → 90 → 115 → 23 → 23−18 = 5. ✓ The key idea: whatever multiply-and-add steps you use, dividing back by the same multiplier and subtracting the original number always cancels the “x” term, leaving a fixed constant.

Q (Date-guessing trick). A trick produces a number using the rule final = 100 × Month + 165 + Day. Find the date if the final answer is: (i) 1269 (ii) 394 (iii) 296.
Solution: Subtract 165, then split the remainder into (100 × Month) + Day.
(i) 1269 − 165 = 1104 = 100(11) + 4 → 4th November (04/11).
(ii) 394 − 165 = 229 = 100(2) + 29 → 29th February (29/02).
(iii) 296 − 165 = 131 = 100(1) + 31 → 31st January (31/01).

13.3 Number Pyramids

Rule: each number in a pyramid is the sum of the two numbers directly below it (just like Pascal’s triangle).

Q. Without building the entire pyramid, find the topmost number given a 3-row pyramid’s bottom row: (a) 4, 13, 8 (b) 7, 11, 3 (c) 10, 14, 25.
Solution: For a 3-row pyramid with bottom a, b, c, the top = a + 2b + c (the middle number is counted twice, since it feeds into both numbers on the row above).
(a) 4 + 2(13) + 8 = 4 + 26 + 8 = 38
(b) 7 + 2(11) + 3 = 7 + 22 + 3 = 32
(c) 10 + 2(14) + 25 = 10 + 28 + 25 = 63

Q. Write an expression for the top of a 4-row pyramid in terms of the bottom row a, b, c, d.
Solution: Row 3 (from bottom): a+b, b+c, c+d. Row 2: (a+2b+c), (b+2c+d). Top = a+3b+3c+d. (Notice the coefficients 1, 3, 3, 1 — these are exactly Pascal’s triangle row 3, because the pyramid follows the same addition rule.)

Q. Using the formula top = a+3b+3c+d, find the top value given: (a) a=8, b=19, c=21, d=13 (b) a=7, b=18, c=19, d=6.
Solution:
(a) 8 + 3(19) + 3(21) + 13 = 8 + 57 + 63 + 13 = 141
(b) 7 + 3(18) + 3(19) + 6 = 7 + 54 + 57 + 6 = 124
Note: one part of this question in common online answer keys shows an internal inconsistency between the stated bottom-row values and the numbers used in the working. We’ve applied the verified formula (a+3b+3c+d) directly and shown our own worked steps rather than reproduce that error.

Q. The Virahanka-Fibonacci sequence is 1, 2, 3, 5, 8, 13, 21, … (each number is the sum of the two before it). If the first three Virahanka-Fibonacci numbers (1, 2, 3) form the bottom row of a 3-row pyramid, what appears in the rest of the pyramid, and what’s at the top?
Solution: Bottom row: 1, 2, 3. Middle row: 1+2=3, 2+3=5. Top: 3+5=8. Full set of numbers appearing: 1, 2, 3, 3, 5, 8 — all of these are themselves Virahanka-Fibonacci numbers, and the top = 8.

Q. If the first four Virahanka-Fibonacci numbers (1, 2, 3, 5) form the bottom row of a 4-row pyramid, what’s at the top? What pattern do you notice for a pyramid built from the first n Virahanka-Fibonacci numbers?
Solution: Bottom row 1, 2, 3, 5 gives middle row 3, 5, 8 then row above 8, 13, and top = 21. Every number appearing in the pyramid (1, 2, 3, 5, 8, 13, 21) is itself a Virahanka-Fibonacci number. Comparing this (n=4, top=21=7th term) with the earlier example (n=3, top=8=5th term), the pattern is: the top of an n-row pyramid built from the first n Virahanka-Fibonacci numbers is always the (2n−1)th Virahanka-Fibonacci number. For a 29-row pyramid (n=29), the top would be the (2×29−1) = 57th Virahanka-Fibonacci number.

13.4 Fun with Grids

Q. In a plus/cross-shaped 5-cell grid from a number grid (numbers increasing by 1 across, by 7 down), with 1, 7, 8, 9, 15 selected, find the sum and a general rule.
Solution: 1+7+8+9+15 = 40. If a = the topmost value of the cross, the other four cells are a+6, a+7, a+8, a+14, so the sum = 5a+35. Check: a=1 → 5(1)+35 = 40. ✓

Q. For a 3×3 block of the grid starting at 10 (i.e. 10,11,12 / 17,18,19 / 24,25,26), find the sum and a general rule.
Solution: Sum = (10+11+12)+(17+18+19)+(24+25+26) = 33+54+75 = 162. If a = top-left value, sum = 9a+72 = 9(a+8) — note that a+8 is exactly the centre cell, so the sum of a 3×3 block always equals 9 times its centre value. Check: centre = 18, 9×18 = 162. ✓

Q. For a horizontal 1×3 strip starting at 28 (28, 29, 30), find the sum and a general rule.
Solution: Sum = 28+29+30 = 87. If a = the left value, sum = 3a+3 = 3(a+1) — three times the middle value. Check: 3×29 = 87. ✓

13.5 The Largest Product

Q. Using the digits 1, 3 and 7 exactly once (two digits forming a 2-digit number, one digit as a single-digit multiplier), what arrangement gives the largest possible product?
Solution: Checking all six arrangements: 71×3=213, 73×1=73, 31×7=217, 37×1=37, 13×7=91, 17×3=51. The largest is 31 × 7 = 217. The winning strategy: make the single digit the largest available digit (7), and arrange the two remaining digits in descending order to form the 2-digit number (31, not 13).

Q. Repeat using the digits 3, 5 and 9.
Solution: Using the same strategy (largest digit, 9, as the single-digit multiplier; remaining digits 3 and 5 arranged in descending order as 53): 53 × 9 = 477. Checking alternatives confirms this is the maximum: 93×5=465, 95×3=285, 39×5=195, 59×3=177, 35×9=315.

13.6 Decoding Divisibility Tricks

Q1. For a two-digit number “ab” (with b > a), reversing the digits to “ba” and subtracting gives a result always divisible by 9. What’s the quotient?
Solution: ba − ab = (10b+a) − (10a+b) = 9b − 9a = 9(b−a). Dividing by 9 gives quotient (b−a).

Q2. Instead of subtracting, add the original and reversed two-digit numbers (e.g. 31+13=44, 28+82=110, 12+21=33 — all divisible by 11). Is this always true? Justify with algebra.
Solution: Yes. For “ab”: original = 10a+b, reversed = 10b+a. Sum = 11a+11b = 11(a+b), which is always divisible by 11.

Q3. For a 3-digit number “abc”, cycle the digits to form “bca” and “cab” and add all three. Show the sum is always divisible by 37, and check if it’s also always divisible by 3.
Solution: abc+bca+cab = (100a+10b+c)+(100b+10c+a)+(100c+10a+b) = 111a+111b+111c = 111(a+b+c) = 37 × 3 × (a+b+c), always divisible by 37. Since 111 = 3 × 37, the sum is also always divisible by 3. Example: 153 → 531, 315 → sum = 153+531+315 = 999 = 37×27 (divisible by 37) and digit sum 9+9+9=27 (divisible by 3). ✓

Q4. For a 3-digit number “abc”, form the 6-digit number “abcabc” by repeating it. Divide by 7, then 11, then 13. What do you notice, and why?
Solution: abcabc = 100100a+10010b+1001c = 1001 × (100a+10b+c) = 1001 × abc. Since 1001 = 7 × 11 × 13, dividing the 6-digit number successively by 7, 11 and 13 always returns the original 3-digit number exactly. Example: 836 → 836836 = 1001 × 836. ✓

Q5 (Magical shrines). Three shrines each have a pond that doubles any flowers dipped in it. A devotee dips his flowers, places an equal number k at each of the 3 shrines (after doubling each time), and ends with none left. How many flowers did he start with, and how many did he place at each shrine (smallest whole-number solution)?
Solution: Let x = starting flowers, k = flowers placed at each shrine. After shrine 1: 2x−k. After shrine 2: 2(2x−k)−k = 4x−3k. After shrine 3: 2(4x−3k)−k = 8x−7k = 0 → x = 7k/8. Smallest whole-number solution: k=8, x=7. Check: 7→(double)14→(−8)6→(double)12→(−8)4→(double)8→(−8)0. ✓ He started with 7 flowers and placed 8 at each shrine.

Q6 (Horses and hens). A farm has 55 heads and 150 legs total (horses have 4 legs, hens have 2). Find the number of each — algebraically and without algebra.
Solution (algebra): x+y=55, 4x+2y=150 → 2x+y=75. Subtracting: x=20 (horses), y=35 (hens). Check: 20×4+35×2=80+70=150. ✓
Without algebra: If all 55 were hens: 55×2=110 legs, which is 40 short of 150. Each hen swapped for a horse adds 2 legs, so horses = 40÷2 = 20, hens = 55−20 = 35.

Q7 (Mother and daughter’s ages). A mother is currently 5 times her daughter’s age. In 6 years, she’ll be 3 times her daughter’s age. Find the daughter’s present age.
Solution: Daughter = x, mother = 5x. In 6 years: 3(x+6) = 5x+6 → 3x+18 = 5x+6 → 12 = 2x → x = 6 years (mother is currently 30). Check: in 6 years, mother=36, daughter=12, and 36=3×12. ✓

Q8 (Gauri and Naina’s cows). Naina has twice as many cows as Gauri. If Naina gives Gauri 3 cows, they’ll have equal numbers. How many cows does each have?
Solution: Gauri = x, Naina = 2x. After the transfer: x+3 = 2x−3 → 6 = x. Gauri = 6, Naina = 12. Check: Naina gives 3 (12→9), Gauri receives 3 (6→9) — equal. ✓

Q9 (Dosa cart economics). Rent = ₹5000/day, cost per dosa = ₹10. (i) Selling 100 dosas, what price gives ₹2000 profit? (ii) At ₹50/dosa, how many dosas give ₹2000 profit?
Solution (i): Total cost = 100(10)+5000 = ₹6000. Target revenue = 6000+2000 = ₹8000. Price/dosa = 8000/100 = ₹80.
Solution (ii): Let n = dosas. Revenue − cost = profit: 50n − (10n+5000) = 2000 → 40n = 7000 → n = 175 dosas.

Q10. Evaluate 1/3, (1+3)/(5+7), (1+3+5)/(7+9+11). What do you notice, and why?
Solution: Using the fact that the sum of the first n odd numbers = n²: numerators are 1², 2², 3² and denominators are 3×1², 3×2², 3×3² (3, 12, 27). Each fraction simplifies to n²/(3n²) = 1/3 — all three fractions are equivalent to 1/3.

Q11 (Karim and the genie). Karim’s coins double each time he circles a tree, but he must pay the genie 8 coins per round. After 3 rounds he’s left with exactly 8 coins (which he owes). (i) How many coins did he start with? (ii) For what cost-per-round would Karim come out ahead? (iii) How should the genie set the cost (in terms of starting coins n) to take everything?
Solution (i): Let n = starting coins. After 3 rounds of “double then pay 8”: 8n−56 = 0 → n = 7 coins.
Solution (ii): With cost c per round starting from 7 coins, after 3 rounds Karim has 56−7c. For an increase (final > 7): 56−7c > 7 → c < 7. Example: c=6 → final = 56−42 = 14 (his money doubled).
Solution (iii): For the genie to take everything after 3 rounds starting from n coins: 8n−7c = 0 → c = 8n/7 (n must be a multiple of 7 for c to be a whole number).

Why This Chapter Matters (for Boards)

Algebra Play builds the core skill of translating a word problem or number pattern into an equation and solving it — the same skill used throughout linear equations, word problems and proofs in Class 9 and 10. The divisibility-trick proofs (Q1–Q4) are also a gentle first introduction to algebraic proof.

More on This Chapter

See also: Class 8 Maths NCERT Book (Ganita Prakash) and the Class 8 Maths Formulas Handbook.

Related pages: Extra Questions for Class 8 Maths Chapter 13 | Revision Notes for Class 8 Maths Chapter 13

Frequently Asked Questions

Q: Why does the “think of a number” trick always give the same answer no matter what number you start with?
A: Because the multiply-and-divide steps cancel out the unknown x algebraically, leaving only a fixed constant — the trick is really just simplifying an algebraic expression.

Q: What’s the fastest way to solve word problems like the horses-and-hens or ages problems?
A: Assign a letter to the unknown quantity, translate each sentence in the problem into one equation, then solve. For two unknowns, you need two independent equations.