Class 8 Maths Ganita Prakash Chapter 13 Solutions (NCERT 2026–27) – Algebra Play

These Class 8 Maths Ganita Prakash Chapter 13 solutions cover Algebra Play from the new NCF-2023 textbook (2026–27). This is Chapter 6 of Ganita Prakash Part II (the 13th chapter of the Class 8 course). Every “Figure it Out” question, Math Talk prompt and reflective task is solved step by step, with each algebraic step checked, so you can master tricks, number pyramids, grids and divisibility proofs and revise the whole chapter quickly.

Class: 8 Subject: Mathematics Book: Ganita Prakash (Part II) Chapter: 13 (Part II Ch 6) Exercises: Figure it Out (3 sets), Math Talk Session: 2026–27
On this page

Chapter 13 Overview

Chapter 6 of Ganita Prakash Part II, Algebra Play, shows how algebra explains why number tricks and puzzles always work. By letting a letter-number stand for the “unknown” starting value, you can model a “Think of a Number” trick, a calendar/date trick, a number pyramid, an algebra grid, a digit-arrangement problem or a divisibility puzzle — and then prove what happens for every input. The chapter also revisits the Virahāṅka–Fibonacci sequence, builds and solves simple linear equations, and uses generalisation (writing the top of a pyramid in terms of the bottom row) to spot patterns. These Class 8 Maths Ganita Prakash Chapter 13 solutions work through every part step by step.

Key Concepts & Ideas

Modelling with a letter-number: represent the unknown starting value by a letter (say x). Carry out each step of the trick on the expression to see the guaranteed result.

Place value in algebra: a two-digit number with tens digit a and units digit b equals 10a + b; a three-digit number abc equals 100a + 10b + c.

Number pyramid rule: every cell is the sum of the two cells directly below it; the bottom row determines the whole pyramid.

Forming & solving simple equations: set up a linear equation from the situation, then isolate the unknown by doing the same operation to both sides.

Generalising patterns: replace the bottom-row numbers with letters to find a formula for the top cell, then read off the coefficients (the binomial pattern 1, 2, 1 / 1, 3, 3, 1).

Virahāṅka–Fibonacci sequence: 1, 2, 3, 5, 8, 13, 21, … where each term is the sum of the two before it.

Important Formulas & Relations

Two-digit number: ab = 10a + b  •  reversed: ba = 10b + a

Difference of a 2-digit number and its reverse: (10a + b) − (10b + a) = 9(a − b) → divisible by 9

Sum of a 2-digit number and its reverse: (10a + b) + (10b + a) = 11(a + b) → divisible by 11

3-row pyramid with bottom a, b, c: top = a + 2b + c

4-row pyramid with bottom a, b, c, d: top = a + 3b + 3c + d

2 × 2 calendar grid with top-left a: numbers are a, a+1, a+7, a+8; sum = 4a + 16

6.2 Thinking about ‘Think of a Number’ Tricks (Math Talk)

The date trick: think of a date as month M and day D; multiply M by 5, add 6, multiply by 4, add 9, multiply by 5, then add the day D. This gives 100M + 165 + D, so subtracting 165 from the final answer leaves a number whose last two digits are D and whose leading digits are M.

Math Talk How would you change this game to make the final answer 3? What about 5? Answer. The original trick (double, add 4, halve, subtract the number) gives x + 2 − x = 2 because of the “add 4 then halve” step, which contributes +2. To end on 3, add 6 instead of 4: 2x + 6 → halve → x + 3 → subtract x → 3. To end on 5, add 10: 2x + 10 → halve → x + 5 → subtract x → 5. In general, to finish on the number k, add 2k after doubling.
Math Talk Can you come up with more complicated steps that always lead to the same final value? Answer. Yes — any sequence of steps in which the letter-number x finally cancels out works. Example: Think of x; triple it (3x); add 12 (3x + 12); divide by 3 (x + 4); subtract the original number (4). The result is always 4. As long as the multiply and divide undo each other and the “subtract the original number” step removes x, only the constant survives.
Math Talk Find the dates if the final answers are the following: (i) 1269   (ii) 394   (iii) 296 Answer. Subtract 165 from each answer; the last two digits are the day D and what comes before is the month M. (i) 1269 − 165 = 1104 → M = 11, D = 04 → 4th November. (ii) 394 − 165 = 229 → M = 2, D = 29 → 29th February. (iii) 296 − 165 = 131 → M = 1, D = 31 → 31st January.
Math Talk Can you change the steps in this trick and still find the original date? Instead of subtracting 165 from the final answer, you might have to subtract some other number. Answer. Yes. The number you subtract is just the constant that builds up from the “add” steps. If you change those numbers, recompute the constant. For instance, if instead of “add 6 … add 9” you used “add 7 … add 8”, the expression becomes 5M·4 = 20M; 20M + 28 from +7; ×5 = 100M + 140; +8 (instead of +9 it folds in) — in general work out the new constant K so the answer is 100M + K + D, then subtract K. The structure (last two digits = day, leading digits = month) stays the same.
Math Talk Try to devise your own ‘Think of a Number’ trick. Answer (one example). Think of a number x. Add 5 (x + 5). Double it (2x + 10). Subtract 4 (2x + 6). Halve it (x + 3). Subtract the number you thought of. The result is always 3, because x + 3 − x = 3 for every starting value. You can change the final constant by adjusting the “add” step.

6.3 Number Pyramids — Figure it Out

In a number pyramid, each number is the sum of the two numbers directly below it. For a 3-row pyramid with bottom row a, b, c, the middle row is (a + b), (b + c) and the top is (a + b) + (b + c) = a + 2b + c.

1. Without building the entire pyramid, find the number in the topmost row given the bottom row in each of these cases. (a) 4   13   8     (b) 7   11   3     (c) 10   14   25

SOLUTION For a 3-row pyramid, top = a + 2b + c. (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.

2. Write an expression for the topmost row of a pyramid with 4 rows in terms of the values in the bottom row.

SOLUTION Let the bottom row be a, b, c, d. Row 2: (a + b), (b + c), (c + d). Row 3: (a + 2b + c), (b + 2c + d). Top = (a + 2b + c) + (b + 2c + d) = a + 3b + 3c + d. (The coefficients 1, 3, 3, 1 are the binomial numbers, just as 1, 2, 1 appear for a 3-row pyramid.)

3. Without building the entire pyramid, find the number in the topmost row given the bottom row in each of these cases. (a) 8   19   21   13     (b) 7   18   19   6     (c) 9   7   5   11

SOLUTION For a 4-row pyramid, top = a + 3b + 3c + d. (a) 8 + 3(19) + 3(21) + 13 = 8 + 57 + 63 + 13 = 141. (b) 7 + 3(18) + 3(19) + 6 = 7 + 54 + 57 + 6 = 124. (c) 9 + 3(7) + 3(5) + 11 = 9 + 21 + 15 + 11 = 56.

4. If the first three Virahāṅka–Fibonacci numbers are written in the bottom row of a number pyramid with three rows, fill in the rest of the pyramid. What numbers appear in the grid? What is the number at the top? Are they all Virahāṅka–Fibonacci numbers?

SOLUTION The first three Virahāṅka–Fibonacci numbers are 1, 2, 3. Middle row: (1 + 2) = 3 and (2 + 3) = 5. Top: 3 + 5 = 8. Numbers in the grid: bottom 1, 2, 3; middle 3, 5; top 8. The full sequence is 1, 2, 3, 5, 8, … so every number in the pyramid (1, 2, 3, 3, 5, 8) is a Virahāṅka–Fibonacci number, and so is the top number 8.

5. What can you say about the numbers in the pyramid and the number at the top in the following cases? (i) The first four Virahāṅka–Fibonacci numbers are written in the bottom row of a four row pyramid. (ii) The first 29 Virahāṅka–Fibonacci numbers are written in the bottom row of a 29 row pyramid.

SOLUTION (i) Bottom row 1, 2, 3, 5. Middle: (1+2)=3, (2+3)=5, (3+5)=8. Next: (3+5)=8, (5+8)=13. Top: 8 + 13 = 21. Every entry (1, 2, 3, 5, 3, 5, 8, 8, 13, 21) is a Virahāṅka–Fibonacci number, because adding two consecutive terms of the sequence gives the next term. The top number 21 is also a Virahāṅka–Fibonacci number. (ii) The same pattern continues: when consecutive Virahāṅka–Fibonacci numbers sit side by side, their sums are again terms of the sequence. So all numbers in the pyramid are Virahāṅka–Fibonacci numbers, and the top number is also a Virahāṅka–Fibonacci number.

6. If the bottom row of an n row pyramid contains the first n Virahāṅka–Fibonacci numbers, what can we say about the numbers in the pyramid? What can we say about the number at the top?

SOLUTION Because each cell is the sum of two cells directly below it, and the sum of two consecutive Virahāṅka–Fibonacci numbers is the next Virahāṅka–Fibonacci number, every number that appears anywhere in the pyramid is a Virahāṅka–Fibonacci number. In particular, the number at the top is also a Virahāṅka–Fibonacci number — the pattern holds for every value of n.

6.4 Fun with Grids (Math Talk)

Calendar magic: in a 2 × 2 grid on a calendar, if a is the top-left number the others are a + 1, a + 7 and a + 8, so the four-number sum is 4a + 16. From the sum you can recover a, and hence all four numbers.

Math Talk — Algebra Grids In the following grids, find the values of the shapes and fill in the empty squares. Grid A: Row 1 has three blue squares summing to 27; Row 2 has two red circles and a blue square summing to 21; Row 3 has red circle, blue square, red circle. Grid B: Row 1 blue circle + two purple diamonds = 18; Row 2 purple diamond + blue circle + blue circle = 15; Row 3 purple diamond + two blue circles. Answer (Grid A). Let the blue square = S and the red circle = C. The last column is the sum of the values to its left in each row. Row 1: S + S + S = 27 ⇒ 3S = 27 ⇒ S = 9. Row 2: C + C + S = 21 ⇒ 2C + 9 = 21 ⇒ 2C = 12 ⇒ C = 6. Row 3 sum = C + S + C = 6 + 9 + 6 = 21, so the empty last square in Row 3 is 21. Answer (Grid B). Let the blue circle = B and the purple diamond = D. Row 1: B + D + D = 18 ⇒ B + 2D = 18. Row 2: D + B + B = 15 ⇒ 2B + D = 15. From the first, B = 18 − 2D. Substitute: 2(18 − 2D) + D = 15 ⇒ 36 − 4D + D = 15 ⇒ 36 − 3D = 15 ⇒ 3D = 21 ⇒ D = 7; then B = 18 − 14 = 4. Row 3 sum = D + B + B = 7 + 4 + 4 = 15, so the empty last square in Row 3 is 15. Check: Grid A Row 2: 6 + 6 + 9 = 21 ✓. Grid B Row 1: 4 + 7 + 7 = 18 ✓.

6.5 The Largest Product — Figure it Out

From the chapter: to get the largest product when filling a 2-digit × 1-digit form, use the largest digit as the single multiplier, and arrange the other two digits in decreasing order to form the 2-digit multiplicand.

1. Fill the digits 1, 3, and 7 in (  )(  ) × (  ) to make the largest product possible.

SOLUTION Use the largest digit, 7, as the multiplier; arrange 3 and 1 in decreasing order as the multiplicand: 31. Largest product = 31 × 7 = 217. Check other options: 73 × 1 = 73, 71 × 3 = 213, 37 × 1 = 37, 17 × 3 = 51, 13 × 7 = 91. The greatest is indeed 31 × 7 = 217.

2. Fill the digits 3, 5, and 9 in (  )(  ) × (  ) to make the largest product possible.

SOLUTION Use the largest digit, 9, as the multiplier; arrange 5 and 3 in decreasing order as the multiplicand: 53. Largest product = 53 × 9 = 477. Check other options: 95 × 3 = 285, 93 × 5 = 465, 59 × 3 = 177, 39 × 5 = 195, 35 × 9 = 315. The greatest is indeed 53 × 9 = 477.

6.6 Decoding Divisibility Tricks — Figure it Out

Trick: choose a 2-digit number of different digits, reverse the digits, find the difference, and divide by 9 — there is no remainder. Writing the number as 10a + b and its reverse as 10b + a, the difference is 9(b − a) (when b > a), which is divisible by 9.

1. In the trick given above, what is the quotient when you divide by 9? Is there a relationship between the two numbers and the quotient?

SOLUTION When b > a, the difference is (10b + a) − (10a + b) = 9b − 9a = 9(b − a). Dividing by 9 gives the quotient = b − a, i.e. the difference between the two digits of the chosen number. For example, with 74 and 47: 74 − 47 = 27, and 27 ÷ 9 = 3, which equals 7 − 4 = 3. So the quotient always equals the (positive) difference of the digits.

2. In the trick given above, instead of finding the difference of the two 2-digit numbers, find their sum. What will happen? For example, 31 + 13 = 44, 28 + 82 = 110, 12 + 21 = 33. Observe that all these numbers are divisible by 11. Is this always true? Can we justify this claim using algebra?

SOLUTION Let the number be 10a + b; its reverse is 10b + a. Sum = (10a + b) + (10b + a) = 11a + 11b = 11(a + b). Since 11(a + b) is a multiple of 11, the sum is always divisible by 11 — so yes, the pattern is always true. Check: 31 + 13 = 44 = 11 × 4; 28 + 82 = 110 = 11 × 10; 12 + 21 = 33 = 11 × 3. ✓

3. Consider any 3-digit number, say abc (100a + 10b + c). Make two other 3-digit numbers from these digits by cycling these digits around, yielding bca and cab. Now add the three numbers. Using algebra, justify that the sum is always divisible by 37. Will it also always be divisible by 3?

SOLUTION abc = 100a + 10b + c; bca = 100b + 10c + a; cab = 100c + 10a + b. Sum = (100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b) = (100 + 10 + 1)a + (100 + 10 + 1)b + (100 + 10 + 1)c = 111a + 111b + 111c = 111(a + b + c). Now 111 = 3 × 37, so the sum = 3 × 37 × (a + b + c). Hence the sum is always divisible by 37, and since it is also a multiple of 3, it is always divisible by 3 too.

4. Consider any 3-digit number, say abc. Make it a 6-digit number by repeating the digits, that is abcabc. Divide this number by 7, then by 11, and finally by 13. What do you get? Try this with other numbers. Figure out why it works.

SOLUTION Let the 3-digit number be N = abc (so N = 100a + 10b + c). The 6-digit number abcabc places N in the thousands and again in the units: abcabc = N × 1000 + N = N × 1001. Now 1001 = 7 × 11 × 13. So abcabc = N × 7 × 11 × 13. Dividing abcabc by 7, then by 11, then by 13 removes all three factors and leaves the original 3-digit number N (abc), with no remainder at any stage. Check (N = 256): 256256 ÷ 7 = 36608; ÷ 11 = 3328; ÷ 13 = 256 ✓.

5. There are 3 shrines, each with a magical pond in the front. If anyone dips flowers into these magical ponds, the number of flowers doubles. A person dips all his flowers in the first pond and then places some flowers in shrine 1. Next, he dips the remaining flowers in the second pond and places some flowers in shrine 2. Finally, he dips the remaining flowers in the third pond and then places them all in shrine 3. If he placed an equal number of flowers in each shrine, how many flowers did he start with? How many flowers did he place in each shrine?

SOLUTION Let him start with x flowers and place s flowers at each shrine. After pond 1: 2x; place s → remaining 2x − s. After pond 2: 2(2x − s) = 4x − 2s; place s → remaining 4x − 3s. After pond 3: 2(4x − 3s) = 8x − 6s; he places all of these at shrine 3, and this also equals s: 8x − 6s = s ⇒ 8x = 7s ⇒ s = 8x/7. For whole numbers, the smallest solution is x = 7, giving s = 8. So he started with 7 flowers and placed 8 flowers in each shrine. Check: 7 → double 14, place 8 → 6; double 12, place 8 → 4; double 8, place 8 → 0. Each shrine got 8. ✓

6. A farm has some horses and hens. The total number of heads of these animals is 55 and the total number of legs is 150. How many horses and how many hens are on the farm? Can you solve this without letter-numbers?

SOLUTION With algebra: let horses = r and hens = h. Then r + h = 55 and 4r + 2h = 150. From the first equation h = 55 − r. Substitute: 4r + 2(55 − r) = 150 ⇒ 4r + 110 − 2r = 150 ⇒ 2r = 40 ⇒ r = 20. So horses = 20 and hens = 55 − 20 = 35. Without letter-numbers (the hint): if all 55 animals were hens, legs = 2 × 55 = 110. The actual count is 150, so there are 150 − 110 = 40 extra legs. Each horse has 2 legs more than a hen, so number of horses = 40 ÷ 2 = 20, and hens = 35. Check: 20 horses × 4 + 35 hens × 2 = 80 + 70 = 150 legs; heads = 55. ✓

7. A mother is 5 times her daughter’s age. In 6 years’ time, the mother will be 3 times her daughter’s age. How old is the daughter now?

SOLUTION Let the daughter’s present age = d years; then the mother’s age = 5d. In 6 years: mother = 5d + 6, daughter = d + 6, and mother = 3 × daughter: 5d + 6 = 3(d + 6) = 3d + 18 ⇒ 5d − 3d = 18 − 6 ⇒ 2d = 12 ⇒ d = 6. So the daughter is now 6 years old (and the mother is 30). Check: in 6 years, mother = 36, daughter = 12, and 36 = 3 × 12. ✓

8. Two friends, Gauri and Naina, are cowherds. Gauri says to Naina, “You have twice as many cows as I do”. Naina says, “That’s true, but if I gave you three of my cows, we would each have the same number of cows”. How many cows do Gauri and Naina have?

SOLUTION Let Gauri have g cows; then Naina has 2g cows. If Naina gives 3 cows to Gauri: Gauri has g + 3, Naina has 2g − 3, and these are equal: g + 3 = 2g − 3 ⇒ 3 + 3 = 2g − g ⇒ g = 6. So Gauri has 6 cows and Naina has 2 × 6 = 12 cows. Check: after the gift, Gauri = 9 and Naina = 9 — equal. ✓

9. I run a small dosa cart and my expenses are: rent for the dosa cart is ₹5000 per day; the cost of making one dosa is ₹10. (i) If I can sell 100 dosas a day, what should be the selling price of my dosa to make a profit of ₹2000? (ii) If my customers are willing to pay only ₹50 for a dosa, how many dosas should I aim to sell in a day to make a profit of ₹2000?

SOLUTION (i) Total cost for 100 dosas = rent + making cost = 5000 + 100 × 10 = 5000 + 1000 = ₹6000. For a profit of ₹2000, the takings must be 6000 + 2000 = ₹8000. Selling price per dosa = 8000 ÷ 100 = ₹80. (ii) Let the number of dosas sold be n. Revenue = 50n; total cost = 5000 + 10n; profit = revenue − cost: 50n − (5000 + 10n) = 2000 ⇒ 40n − 5000 = 2000 ⇒ 40n = 7000 ⇒ n = 175. So I should aim to sell 175 dosas a day. Check (ii): 175 × 50 = 8750; cost = 5000 + 1750 = 6750; profit = 8750 − 6750 = 2000. ✓

10. Evaluate the following sequence of fractions: (1/3), (1 + 3)/(5 + 7), (1 + 3 + 5)/(7 + 9 + 11). What do you observe? Can you explain why this happens?

SOLUTION First fraction: 1/3. Second: (1 + 3)/(5 + 7) = 4/12 = 1/3. Third: (1 + 3 + 5)/(7 + 9 + 11) = 9/27 = 1/3. Observation: every fraction equals 1/3. Why: the sum of the first n odd numbers is n2, so the numerator is n2. The denominator is the sum of the next n odd numbers, which is (sum of first 2n odds) − (sum of first n odds) = (2n)2 − n2 = 4n2 − n2 = 3n2. So each fraction = n2 ÷ 3n2 = 1/3.

11. Karim and the Genie. Each time Karim goes around the tree his coins double, and he must give the genie 8 coins. After the third round he is left with exactly 8 coins, the amount he owes, leaving him with nothing. (i) How many coins did Karim initially have? (ii) For what cost per round should Karim agree to the deal, if he wants to increase the number of coins he has? (iii) Through its magical powers, the genie knows the number of coins that Karim has. How should the genie set the cost per round so that it gets all of Karim’s coins?

SOLUTION (i) Let Karim start with x coins. Each round: double, then subtract 8. After round 1: 2x − 8. After round 2: 2(2x − 8) − 8 = 4x − 24. After round 3: 2(4x − 24) − 8 = 8x − 56. He ends with 0 (he gives away the last 8): 8x − 56 = 0 ⇒ x = 7. So Karim initially had 7 coins. (ii) After one round with cost c, Karim has 2x − c. For his coins to grow we need 2x − c > x, i.e. c < x. So the cost per round must be less than the number of coins he currently has. Since he starts with 7, any fixed cost of 6 coins or fewer would let his money increase each round. (iii) To wipe him out in a single round the genie needs 2x − c = 0, i.e. c = 2x. So the genie should set the cost equal to twice the number of coins Karim currently has (for 7 coins, demand 14). Then after doubling and paying, Karim is left with nothing.

Common Mistakes to Avoid

Watch out for these

  • Writing a two-digit number as a × b instead of 10a + b when setting up a divisibility proof.
  • In a 3-row pyramid, forgetting that the middle term is doubled — the top is a + 2b + c, not a + b + c.
  • Mixing up the 4-row coefficients; they are 1, 3, 3, 1 (not 1, 2, 2, 1).
  • In the “equal flowers” and Karim problems, forgetting to double first, then subtract (order of operations matters).
  • In word problems, not doing the same operation to both sides of the equation, or losing a sign while transposing.
  • Forgetting to verify by substituting the answer back into the original statement (heads & legs, ages, cows).
  • Assuming the largest digit goes in the multiplicand — for the largest product the largest digit is the single multiplier.

Practice MCQs & Assertion–Reason

1. A two-digit number with tens digit a and units digit b is written in algebra as:

(a) a + b    (b) ab    (c) 10a + b    (d) 10b + a

2. For a 3-row number pyramid with bottom row a, b, c, the top number is:

(a) a + b + c    (b) a + 2b + c    (c) 2a + b + 2c    (d) a + 3b + c

3. The top number of a 3-row pyramid with bottom row 4, 13, 8 is:

(a) 25    (b) 34    (c) 38    (d) 42

4. The difference between a 2-digit number and the number formed by reversing its digits is always divisible by:

(a) 7    (b) 9    (c) 11    (d) 13

5. The sum of a 2-digit number and its reverse equals 11(a + b), so it is always divisible by:

(a) 9    (b) 10    (c) 11    (d) 37

6. Using the digits 1, 3 and 7 once each in ( )( ) × ( ), the largest product is:

(a) 213    (b) 217    (c) 91    (d) 73

7. The 6-digit number abcabc is always equal to abc multiplied by:

(a) 111    (b) 1001    (c) 1010    (d) 1111

8. A farm has 55 heads and 150 legs of horses and hens. The number of horses is:

(a) 15    (b) 20    (c) 25    (d) 35

9. A mother is 5 times her daughter’s age; in 6 years she will be 3 times as old. The daughter’s present age is:

(a) 5    (b) 6    (c) 8    (d) 10

10. The value of (1 + 3 + 5)/(7 + 9 + 11) is:

(a) 1/2    (b) 1/3    (c) 2/3    (d) 1/4

Answer key: 1-(c), 2-(b), 3-(c), 4-(b), 5-(c), 6-(b), 7-(b), 8-(b), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The sum of any 2-digit number and the number obtained by reversing its digits is divisible by 11.

Reason: (10a + b) + (10b + a) = 11(a + b).

A-R 2. Assertion: When you subtract a 2-digit number from its reverse and divide by 9, the quotient equals the difference of the two digits.

Reason: The difference (10b + a) − (10a + b) equals 9(b − a).

A-R 3. Assertion: The number abcabc is always divisible by 7, 11 and 13.

Reason: abcabc = abc × 1001 and 1001 = 7 × 11 × 13.

A-R 4. Assertion: In a 3-row number pyramid the middle bottom number is added twice to get the top.

Reason: The top equals a + b + c for a bottom row a, b, c.

A-R 5. Assertion: To make the largest product with three given digits in the form (two-digit) × (one-digit), the largest digit should be the one-digit multiplier.

Reason: Expanding the products shows the leading terms match and the larger second term comes from using the largest digit as the multiplier.

Answer key: 1-(A), 2-(A), 3-(A), 4-(C), 5-(A).

Quick Revision Summary

  • Algebra explains why number tricks work: model the unknown by a letter and simplify — if the letter cancels, the result is fixed for every input.
  • A 2-digit number is 10a + b; a 3-digit number is 100a + 10b + c. Reverse, cycle or repeat the digits and simplify to expose divisibility.
  • Difference of a number and its reverse = 9(a − b) → divisible by 9; sum = 11(a + b) → divisible by 11.
  • For abc + bca + cab = 111(a + b + c) = 3 × 37 × (a + b + c); and abcabc = abc × 7 × 11 × 13.
  • Number pyramid: top of 3 rows = a + 2b + c; top of 4 rows = a + 3b + 3c + d (binomial coefficients 1,2,1 and 1,3,3,1).
  • If the bottom row holds consecutive Virahāṅka–Fibonacci numbers, every entry — including the top — is also a Virahāṅka–Fibonacci number.
  • For the largest product, use the biggest digit as the multiplier and the rest in decreasing order as the multiplicand.
  • Word problems (heads & legs, ages, cows, dosa cart, flowers, Karim) become simple linear equations — always substitute the answer back to check.

How to score full marks in this chapter

For every “trick” question, first define the letter-number (e.g. “let the number be 10a + b”), then show each algebraic step so the cancellation or factor is visible. In word problems, write the two relations clearly, solve neatly, and finish with a one-line verification. Keep fractions reduced (4/12 = 1/3) and state the rule you are using (pyramid rule, place value, sum of odd numbers = n2) before you compute.

Frequently Asked Questions

What is Class 8 Maths Ganita Prakash Chapter 13 about?

Chapter 13 (Chapter 6 of Part II), Algebra Play, uses algebra to explain why number tricks, calendar/date tricks, number pyramids, algebra grids, the largest-product puzzle and divisibility tricks always work. You model an unknown with a letter, simplify, and prove the result for every input.

Why is this called Chapter 13 if the book says Chapter 6?

It is Chapter 6 of Ganita Prakash Part II, but it is the 13th chapter of the full Class 8 course (Part I and Part II combined), which is how ClearStudy numbers it for easy navigation.

Why is the sum of a 2-digit number and its reverse divisible by 11?

If the number is 10a + b, its reverse is 10b + a. Their sum is 11a + 11b = 11(a + b), which is always a multiple of 11.

Are these Class 8 Maths Ganita Prakash Chapter 13 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Prakash (Part II) textbook for 2026–27, with every algebraic step verified.

← Chapter 12 All Ganita Prakash Chapters Chapter 14 →
Scroll to Top