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). Explore number tricks, completed pyramids, calendar and shape grids, divisibility proofs and word problems with worked explanations and original diagrams.

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
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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. The solutions below show the equations, calculations and checks.

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

Here the digit strings ab, abc and abcabc denote place-value numerals, not products of letters. In ordinary algebra, ab means a × b. A leading zero after reversing or cycling is read by its numerical value; the divisibility identities still hold.

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.
Algebra flow shows x becoming 2x, 2x plus 4, x plus 2 and finally 2.
Algebra flow shows x becoming 2x, 2x plus 4, x plus 2 and finally 2.
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, a valid date only in a leap year. (iii) 296 − 165 = 131 → M = 1, D = 31 → 31st January.
The date trick yields 100M plus 165 plus D. Subtracting 165 from 1269 gives month 11 and day 04.
The date trick yields 100M plus 165 plus D. Subtracting 165 from 1269 gives month 11 and day 04.
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. Replacing the two additions by +7 and +8 gives M → 5M → 5M+7 → 20M+28 → 20M+36 → 100M+180 → 100M+180+D. Subtract 180 to obtain 100M+D. More generally, additions +A and +B give constant K=20A+5B. Use month 1–12 and a day valid for that month; read the last two digits as the day.
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.

A symbolic three-row pyramid has bottom a b c, middle a plus b and b plus c, and top a plus 2b plus c.
A symbolic three-row pyramid has bottom a b c, middle a plus b and b plus c, and top a plus 2b plus c.

Complete the introductory pyramids (textbook pages 138–139)

Add neighbouring cells to work upwards; subtract a known child from its parent to work backwards.

  • Bottom 6, 2: top 8.
  • Bottom 3, 4, 3: next row 7, 7; top 14.
  • Bottom 5, 4, 5, 0: rows 9, 9, 5; then 18, 14; top 32.
  • Backward examples: bottom 1, 3, 3 gives 4, 6 and top 10; bottom 12, 20, 8 gives 32, 28 and top 60.

The three partially filled four-row pyramids on page 139 have the completed values below.

Completed pyramids with bottom rows 4, 9, 6, 1; 5, 14, 7, 2; 3, 5, 5, 2.
Completed pyramids with bottom rows 4, 9, 6, 1; 5, 14, 7, 2; 3, 5, 5, 2.

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.
Completed pyramids with bottom rows 4, 13, 8; 7, 11, 3; 10, 14, 25.
Completed pyramids with bottom rows 4, 13, 8; 7, 11, 3; 10, 14, 25.

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.)
A four-row algebra pyramid expands to a plus 3b plus 3c plus d at the top.
A four-row algebra pyramid expands to a plus 3b plus 3c plus d at the top.

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.
Completed pyramids with bottom rows 8, 19, 21, 13; 7, 18, 19, 6; 9, 7, 5, 11.
Completed pyramids with bottom rows 8, 19, 21, 13; 7, 18, 19, 6; 9, 7, 5, 11.

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 1, 2, 3, 5 gives rows 3, 5, 8; then 8, 13; then 21. Every entry is a Virahāṅka–Fibonacci number.(ii) Write the sequence V1=1, V2=2, V3=3, …. Each new row starts two sequence positions later: V1, then V3, V5, …. A 29-row pyramid has 28 upward steps, so its top is V57 = 591,286,729,879. All its entries are sequence terms.

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?

SOLUTIONLet V1=1 and V2=2. The bottom entries are V1, …, Vn. Since Vj+Vj+1=Vj+2, each upward row consists of consecutive sequence terms starting two positions later.After r upward steps, its jth entry is Vj+2r. At the top r=n−1 and j=1, so top = V2n−1. This includes n=1. Every cell is a sequence term.
Three- and four-row Fibonacci pyramids have tops 8 and 21. The 29-row top is the 57th term, 591286729879.
Three- and four-row Fibonacci pyramids have tops 8 and 21. The 29-row top is the 57th term, 591286729879.

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.

A symbolic calendar block a a+1 a+7 a+8 and a worked block 5 6 12 13 whose sum is 36.
A symbolic calendar block a a+1 a+7 a+8 and a worked block 5 6 12 13 whose sum is 36.

Worked calendar example: if the 2 × 2 sum is 36, a=(36−16)/4=5. The entries are 5, 6, 12, 13. For your own trick, a complete 3 × 3 calendar block has sum 9a+72=9(a+8), so divide its sum by 9 to find its centre. Use complete blocks within one month, without wrapping across the end of a week.

Math Talk — Algebra Grids

Grid A: The first row contains two squares and one circle. Let square=S and circle=C. Then 2S+C=27 and S+2C=21. Doubling the second equation and subtracting the first gives 3C=15, so C=5 and S=11. The last row totals 5+11+5=21. Check the first row: 11+11+5=27.

Grid B: Let blue circle=B and diamond=D. B+2D=18 and 2B+D=15 give B=4 and D=7. Row totals are 18, 15, 15. Column totals are also 18, 15, 15, and the bottom-right grand total is 48.

Correct algebra grids: Grid A has two squares and one circle in its first row, so square 11 and circle 5. Grid B has circle 4 and diamond 7; column totals are 18, 15, 15 and grand total 48.
Correct algebra grids: Grid A has two squares and one circle in its first row, so square 11 and circle 5. Grid B has circle 4 and diamond 7; column totals are 18, 15, 15 and grand total 48.

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.

Why the largest-product rule works

For distinct positive digits p<q<r, choose the greater possible two-digit number for each multiplier. The best candidates are (10q+p)r, (10r+p)q and (10r+q)p. The first exceeds the second by p(r−q)>0; the second exceeds the third by 10r(q−p)>0. Thus the first is greatest. A zero digit can produce a tie: 30 × 7 = 70 × 3 = 210.

Comparing candidate products leaves difference p times r minus q, positive for three distinct positive digits.
Comparing candidate products leaves difference p times r minus q, positive for three distinct positive digits.

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.
Cyclic digit arrangements show that every digit contributes hundreds, tens and ones, making a sum divisible by 3 and 37.
Cyclic digit arrangements show that every digit contributes hundreds, tens and ones, making a sum divisible by 3 and 37.

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 ✓.
Repeating a three-digit number N yields 1001N, so division by 7, 11 and 13 returns N.
Repeating a three-digit number N yields 1001N, so division by 7, 11 and 13 returns N.

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. All positive whole-number solutions have x=7k and s=8k, where k is any positive integer. The smallest is 7 flowers initially and 8 at each shrine; 14 and 16 also work, so the answer is not unique unless the smallest is requested. Check: 7 → double 14, place 8 → 6; double 12, place 8 → 4; double 8, place 8 → 0. Each shrine got 8. ✓
The flower sequence starts with 7, then after doubling and offering 8 leaves 6, 4 and finally 0.
The flower sequence starts with 7, then after doubling and offering 8 leaves 6, 4 and finally 0.

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. ✓
Starting with 55 hens gives 110 legs. Forty extra legs correspond to 20 horses, leaving 35 hens.
Starting with 55 hens gives 110 legs. Forty extra legs correspond to 20 horses, leaving 35 hens.

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. ✓
The age relation changes from mother 5d and daughter d to 5d+6 and d+6, giving daughter age 6.
The age relation changes from mother 5d and daughter d to 5d+6 and d+6, giving daughter age 6.

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. This textbook model assumes every dosa made is sold and the stated rent and making cost are the only costs. Check (ii): 175 × 50 = 8750; cost = 5000 + 1750 = 6750; profit = 8750 − 6750 = 2000. ✓
The dosa model separates rent 5000, variable cost 10n, and revenue. Selling 175 at 50 each gives profit 2000.
The dosa model separates rent 5000, variable cost 10n, and revenue. Selling 175 at 50 each gives profit 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.
A 6 by 6 square contains a blue 3 by 3 square of area 9 and a remaining gold area of 27, illustrating the constant fraction one third.
A 6 by 6 square contains a blue 3 by 3 square of area 9 and a remaining gold area of 27, illustrating the constant fraction one third.

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 distinct positive 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 worked explanations and original teaching diagrams.

Textbook reference: NCERT Ganita Prakash, Chapter 13 (official PDF). ClearStudy diagrams and worked explanations are original teaching aids.

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