MathsCBSE Class 8 · Chapter 13

Algebra Play Class 8: Number Tricks, Number Pyramids, Calendar Grids and Algebra Puzzles

Q: Where is Algebra Play in the Class 8 textbook?

It is Chapter 6 of Ganita Prakash Part II (Chapter 13 in our list), the NCERT Grade 8 Mathematics textbook (Reprint 2026-27). It has six short sections on tricks, pyramids, grids, the largest product and divisibility tricks.

Q: Is this chapter only for fun, or will it be tested?

It is fun, but the skills are examinable: forming expressions, simplifying, solving linear equations and proving statements with algebra. Expect questions that ask you to explain a trick or solve a puzzle step by step.

Q: Who are the Virahanka-Fibonacci numbers named after?

The sequence 1, 2, 3, 5, 8, ..., where each term is the sum of the two before, was described by the Indian scholar Virahanka centuries before Fibonacci. The chapter uses it in number pyramids.

Q: Can I solve the puzzles without algebra?

Often yes, as the book shows for heads and legs: assume all animals are hens and use the extra legs. Algebra is still the most reliable method, so learn both and use the one that is clearer.

Q: How much working should I show for a 'prove the trick' question?

Show the expression after each step, simplify carefully, and finish with a sentence explaining why the result does not depend on the starting number (or why it is always a multiple of 9, 11, and so on).

Updated 15 October 2026 · Fact-checked

Algebra Play is Chapter 6 of Ganita Prakash Part II for Class 8 (Chapter 13 overall). It uses letter-numbers to explain and invent number tricks, solve number pyramids and calendar grids, find the largest product of digits, prove divisibility tricks such as reversing digits, and solve word puzzles about ages, animals and money.

What Algebra Play is about

Algebra is not only for solving equations; it explains why puzzles and tricks work. In this chapter you replace unknown numbers by letter-numbers, simplify, and see the hidden structure.

6.1 Algebra Play: using algebra to model situations and justify claims.
6.2 Thinking about 'Think of a Number' Tricks: tricks that always end in the same number, and a date trick that recovers a birthday from one final answer.
6.3 Number Pyramids: each block is the sum of the two below; finding missing numbers with equations; the top of a 3-row pyramid is $a + 2 b + c$ ; pyramids built from Virahanka-Fibonacci numbers.
6.4 Fun with Grids: calendar blocks whose sum reveals the dates, and algebra grids with shapes.
6.5 The Largest Product: arranging three digits in $□□ \times □$ to get the biggest product, proved with algebra.
6.6 Decoding Divisibility Tricks: why a number minus its reverse is a multiple of 9, why a number plus its reverse is a multiple of 11, cyclic sums and 37, and the $ab c ab c$ trick with 7, 11 and 13; plus word problems on ages, heads and legs, and doubling.

Exams ask you to explain tricks with algebra, fill pyramids and grids, and solve puzzles by forming and solving equations. This page explains each idea with an example and gives exam-level and harder questions with full solutions.

Key concepts

Algebra as a tool for explaining

NCERT section 6.1 Algebra Play

When a pattern works for every number you try, algebra shows why it must always work. Replace the unknown by a letter-number, follow the steps, and simplify.

Three habits run through this chapter:
1. Model: name the unknowns ( $x$ , $a$ , $M$ , $D$ , ...).
2. Translate: write each instruction or condition as an expression or equation.
3. Simplify and interpret: cancel what cancels and read off what is left (a constant, a multiple of 9, an equation to solve).

A check with one or two actual numbers is useful, but only the algebra proves the result for all numbers.

Worked example

Show with algebra that the sum of any three consecutive numbers is 3 times the middle one.

Try it first, then show the solutionSolution

Let the middle number be $n$ ; the numbers are $n - 1$ , $n$ , $n + 1$ .
Sum $= (n - 1) + n + (n + 1) = 3 n$ .
So the sum is always 3 times the middle number (check: $9 + 10 + 11 = 30 = 3 \times 10$ ).

Answer: Sum $= 3 n$ , three times the middle number.

'Think of a number' tricks

NCERT section 6.2 Thinking about ‘Think of a Number’ Tricks

In these tricks the original number cancels out, so everyone ends with the same answer.

Example: think of $x$ ; multiply by 3: $3 x$ ; add 12: $3 x + 12$ ; divide by 3: $x + 4$ ; subtract the original number: $4$ . The answer is always 4.

To change the final answer, change the number added: adding 15 instead of 12 gives 5.

Tricks that reveal the number: if the steps produce a known expression like $10 x$ , you can work backwards. Double, add 5, multiply by 5, subtract 25: $(2 x + 5) \times 5 - 25 = 10 x$ , so dividing the final answer by 10 reveals $x$ .

Design rule: keep track of the expression after each step; to make $x$ disappear, subtract exactly the multiple of $x$ that is left.

Worked example

Think of a number, multiply by 3, add 12, divide by 3, and subtract the number you first thought of. Prove that the result is always 4. How would you change the trick to always end with 7?

Try it first, then show the solutionSolution

Start: $x$ . After $\times 3$ : $3 x$ . After $+ 12$ : $3 x + 12$ .
After $\div 3$ : $x + 4$ . After subtracting $x$ : $4$ .
To end with 7 instead, add 21 in the second step: $\frac{3 x + 21}{3} - x = 7$ .

Answer: Always 4; adding 21 instead of 12 makes it 7.

The date trick

NCERT section 6.2 Thinking about ‘Think of a Number’ Tricks

A trick can hide two numbers at once. Let the month be $M$ and the day $D$ . Suppose the steps are: multiply the month by 4, add 3, multiply by 25, add the day.

$M \to 4 M \to 4 M + 3 \to 100 M + 75 \to 100 M + 75 + D$ .
So **final answer $- 75 = 100 M + D$ **.
Since $D$ is at most 31, it fits in the last two digits, and $M$ is the number in front.

If a friend says 1189: $1189 - 75 = 1114$ , so $M = 11$ , $D = 14$ : the 14th of November.

The book's version (multiply by 5, add 6, multiply by 4, add 9, multiply by 5, add the day) gives $100 M + 165 + D$ , so you subtract 165. Any trick that makes $100 M$ plus a known constant plus $D$ works.

Worked example

Using the steps 'month × 4, + 3, × 25, + day', one friend answers 381 and another answers 375. Find the first date and explain why the second answer must contain a slip.

Try it first, then show the solutionSolution

Every answer equals $100 M + 75 + D$ , so subtract 75.
$381 - 75 = 306 = 100 \times 3 + 6$ : month 3, day 6, the 6th of March.
$375 - 75 = 300 = 100 \times 3 + 0$ : day 0 is not a date, so the second friend made an arithmetic slip.

Answer: 6 March; 375 cannot come from a real date.

Number pyramids

NCERT section 6.3 Number Pyramids

In a number pyramid, each block is the sum of the two blocks directly below it.

General top expressions:

2 rows, bottom $a, b$ : top $a + b$ .
3 rows, bottom $a, b, c$ : middle row $a + b$ , $b + c$ ; top $a + 2 b + c$ .
4 rows, bottom $a, b, c, d$ : top $a + 3 b + 3 c + d$ .

The multipliers 1, 2, 1 and 1, 3, 3, 1 count how many paths lead from each bottom block to the top.

Missing numbers: call the unknown blocks $a$ , $b$ , $c$ , write an equation for each block, and solve. For a 3-row pyramid with top 50 and bottom $9, c, 5$ : $9 + 2 c + 5 = 50$ , so $c = 18$ .

Virahanka-Fibonacci pyramids: if the bottom row is 1, 2, 3, the next row is 3, 5 and the top is 8: every number is a Virahanka-Fibonacci number. With 3, 5, 8, 13 at the bottom of a 4-row pyramid, the top is $3 + 15 + 24 + 13 = 55$ , again in the sequence.

Worked example

A 3-row pyramid has bottom row 9, $c$ , 5 and top number 50. Find $c$ and complete the pyramid.

Try it first, then show the solutionSolution

Top $= 9 + 2 c + 5 = 14 + 2 c$ .
$14 + 2 c = 50$ , so $c = 18$ .
Middle row: $9 + 18 = 27$ and $18 + 5 = 23$ ; check $27 + 23 = 50$ .

Answer: $c = 18$ ; middle row 27, 23; top 50.

Calendar and algebra grids

NCERT section 6.4 Fun with Grids

On a calendar, the number below a date is 7 more, and the number to its right is 1 more.

2 × 2 block with top-left $a$ : $a$ , $a + 1$ , $a + 7$ , $a + 8$ . Sum $= 4 a + 16$ . From the sum you can recover $a$ : sum 36 gives $a = 5$ .

3 × 3 block with top-left $a$ : the nine numbers add to $9 a + 72 = 9 (a + 8)$ , and $a + 8$ is the centre. So the sum is always 9 times the centre date: from a sum of 153 the centre is 17 and the block starts at 9.

Algebra grids: shapes stand for numbers and each row (or column) gives an equation. Solve the row with the fewest unknowns first, then substitute into the others.

Worked example

A friend picks a 3 × 3 block of dates on a calendar page and says the nine dates add up to 153. Find the block.

Try it first, then show the solutionSolution

With top-left $a$ , the sum is $9 a + 72 = 9 (a + 8)$ .
$9 (a + 8) = 153$ , so $a + 8 = 17$ and $a = 9$ .
Rows: 9, 10, 11 / 16, 17, 18 / 23, 24, 25 (centre 17).

Answer: 9–11, 16–18, 23–25, centred on 17.

The largest product

NCERT section 6.5 The Largest Product

Place three different digits in $□□ \times □$ to make the largest product. There are 6 arrangements.

Algebra proof for digits $p < q < r$ :

For each choice of the single-digit multiplier, the bigger tens digit makes the bigger two-digit number, so only $r q \times p$ , $r p \times q$ and $q p \times r$ need comparing.
$r p \times q = 10 r q + pq$ and $r q \times p = 10 r p + pq$ ; since $q > p$ , $10 r q > 10 r p$ , so $r p \times q$ beats $r q \times p$ .
Compare $q p \times r = 10 q r + p r$ with $r p \times q = 10 r q + pq$ : the first parts are equal, and $p r > pq$ because $r > q$ .

So the winner is ** $q p \times r$ **: the largest digit as the multiplier and the other two in decreasing order. For 4, 6, 9: $64 \times 9 = 576$ .

Worked example

Using the digits 4, 6 and 9 once each in $□□ \times □$ , find the largest product and justify it.

Try it first, then show the solutionSolution

By the rule, the largest digit 9 is the multiplier and 6, 4 form 64.
$64 \times 9 = 576$ .
Check the close rival $94 \times 6 = 564$ and $96 \times 4 = 384$ : both smaller.

Answer: $64 \times 9 = 576$

Divisibility tricks with reversed and repeated digits

NCERT section 6.6 Decoding Divisibility Tricks

Write a two-digit number as $10 a + b$ and its reverse as $10 b + a$ .

Difference: $(10 a + b) - (10 b + a) = 9 (a - b)$ : always a multiple of 9, and the quotient is the difference of the digits.
Sum: $(10 a + b) + (10 b + a) = 11 (a + b)$ : always a multiple of 11, and the quotient is the sum of the digits.

Three-digit cycles: $ab c + b c a + c ab = 111 (a + b + c) = 3 \times 37 \times (a + b + c)$ , so the sum is always divisible by 37 and by 3.

Repeating a three-digit number: $ab c ab c = ab c \times 1001$ and $1001 = 7 \times 11 \times 13$ . So dividing $ab c ab c$ by 7, then 11, then 13 gives back $ab c$ .

Three-digit reverse: $ab c - c ba = 99 (a - c)$ , a multiple of 99.

Worked example

Take 527527. Divide it by 7, then by 11, then by 13. Explain the result.

Try it first, then show the solutionSolution

$527527 = 527 \times 1000 + 527 = 527 \times 1001$ .
$1001 = 7 \times 11 \times 13$ .
So dividing by 7, 11 and 13 in turn leaves 527.

Answer: 527, because $ab c ab c = ab c \times 7 \times 11 \times 13$ .

Solving word puzzles with algebra

NCERT section 6.6 Decoding Divisibility Tricks

Many puzzles become simple equations.

Heads and legs: 40 animals (cows and hens) with 112 legs. If $c$ are cows: $4 c + 2 (40 - c) = 112$ , so $2 c = 32$ , $c = 16$ and 24 hens. *Without algebra:* if all 40 were hens there would be 80 legs; the extra 32 legs come from cows having 2 more legs each, so $32 \div 2 = 16$ cows.
Ages: a father is 4 times his son's age; in 5 years he will be 3 times: $4 s + 5 = 3 (s + 5)$ , so $s = 10$ , father 40.
Equal after giving: A has 3 times as many marbles as B; if A gives B 10, they are equal: $3 B - 10 = B + 10$ , $B = 10$ , $A = 30$ .
Doubling and paying: money doubles, then a fixed amount is paid, several times. Work forwards with a letter (or backwards from the end).
Business: profit $=$ (price × quantity) $-$ costs; set it equal to the target and solve.

Worked example

A father is 4 times as old as his son. In 5 years he will be 3 times as old. Find their present ages.

Try it first, then show the solutionSolution

Let the son be $s$ ; the father is $4 s$ .
In 5 years: $4 s + 5 = 3 (s + 5) = 3 s + 15$ .
$s = 10$ ; father $= 40$ . Check: in 5 years, 45 and 15, and $45 = 3 \times 15$ .

Answer: Son 10 years, father 40 years.

Maths in real life: Algebra Play

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Date trick

Suppose you want to guess a friend's birthday at a party using the steps 'month × 4, + 3, × 25, + day'.

Their answer is 1006. When is the birthday?

Try it first, then show the solutionSolution

Subtract 75: $1006 - 75 = 931$ .
931 means month 9, day 31, but September has 30 days, so ask them to recheck; if their answer is 1005, it gives 930: 30 September.

Answer: 1006 is impossible; 1005 would mean 30 September.

Memory hook: Subtract the hidden constant, then split month and day.

Heads and legs

Suppose a parking lot has 30 vehicles, only cars and motorbikes, with 84 wheels in all.

How many cars are there?

Try it first, then show the solutionSolution

If all were bikes: 60 wheels; extra 24 wheels.
Each car adds 2 extra wheels: $24 \div 2 = 12$ cars, 18 bikes.

Answer: 12 cars and 18 bikes.

Memory hook: Assume the smaller kind, then share out the extras.

Profit equation

Suppose a juice stall pays ₹1,000 a day in rent, each glass costs ₹8 to make, and the owner wants ₹2,000 profit selling 125 glasses.

What price should a glass be?

Try it first, then show the solutionSolution

$125 p - 125 \times 8 - 1000 = 2000$ .
$125 p = 4000$ , so $p = ₹32$ .

Answer: ₹32 a glass.

Memory hook: Profit = income − all costs; set it to the target.

Calendar algebra

Suppose a teacher asks you to pick a 2 × 2 block of dates and tell only their sum, 64.

Which dates did you pick?

Try it first, then show the solutionSolution

$4 a + 16 = 64$ , so $a = 12$ .
The block is 12, 13, 19, 20.

Answer: 12, 13, 19 and 20.

Memory hook: Right is +1, down is +7.

Topics in this chapter

Algebra Play formula sheet

Two-digit number: $\overline{ab} = 10 a + b$
Number minus reverse: $\overline{ab} - \overline{ba} = 9 (a - b)$
Number plus reverse: $\overline{ab} + \overline{ba} = 11 (a + b)$
Cyclic sum: $\overline{ab c} + \overline{b c a} + \overline{c ab} = 111 (a + b + c) = 3 \times 37 (a + b + c)$
Repeated three digits: $\overline{ab c ab c} = 1001 \times \overline{ab c} = 7 \times 11 \times 13 \times \overline{ab c}$
3-row pyramid top: $a + 2 b + c$
4-row pyramid top: $a + 3 b + 3 c + d$
Calendar 2 × 2 block: $a + (a + 1) + (a + 7) + (a + 8) = 4 a + 16$
Calendar 3 × 3 block: $sum = 9 (a + 8) = 9 \times centre$
Largest product (p < q < r): $\overline{q p} \times r$

How the school exam asks Algebra Play

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Explain a trick	Show why a 'think of a number' trick always gives the same answer, or how the date trick reveals the date.	Write the expression after every step; simplify; interpret the final expression.
Number pyramids	Find the top without building the pyramid; find missing numbers; Fibonacci pyramids.	Use $a + 2 b + c$ or $a + 3 b + 3 c + d$ ; set up equations for unknown blocks.
Calendar and algebra grids	Recover a block of dates from its sum; find values of shapes.	Express every date in terms of one letter; solve the simplest equation first.
Largest product	Arrange given digits to make the largest product; justify.	Use the largest digit as the multiplier and the others in decreasing order; justify by expanding.
Divisibility tricks	Prove that a number and its reverse give a multiple of 9 or 11; cyclic sums and 37; the 1001 trick.	Write numbers in place-value form and factor out the common number.
Word puzzles	Ages, heads and legs, sharing to make equal, doubling with payments, profit targets.	Choose a letter for one unknown, express the others, form one equation, solve and check.

Tricks to remember Algebra Play

Short memory aids. Say each one aloud once, then use it on a question the same day.

Cancel the thinker
In a fixed-answer trick, the original number must cancel; subtract exactly what multiple of $x$ remains.
Use it when: Designing and explaining tricks.
1-2-1 and 1-3-3-1
Pyramid tops weight the bottom row by 1, 2, 1 (3 rows) or 1, 3, 3, 1 (4 rows).
Use it when: Pyramid shortcuts.
Minus reverse is nines, plus reverse is elevens
Difference gives $9 \times$ digit gap; sum gives $11 \times$ digit total.
Use it when: Reverse-digit tricks.
1001 = 7 × 11 × 13
Repeat any three digits and you have multiplied by 1001.
Use it when: The abcabc trick.
Big digit multiplies
For the largest product, the biggest digit goes alone.
Use it when: Largest product puzzles.

Algebra Play: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Pyramid top · 1 mark · MCQExam level

A 3-row number pyramid has bottom row 4, 7, 2. The top number is
(a) 13
(b) 18
(c) 20
(d) 22

Try it first, then show the solutionSolution

Top $= a + 2 b + c$ .
$4 + 14 + 2 = 20$ .

Answer: (c) 20

Example 2: Reverse and subtract · 1 mark · Assertion-reasonExam level

Assertion (A): $73 - 37$ is divisible by 9.
Reason (R): For digits $a$ and $b$ , $(10 a + b) - (10 b + a) = 9 (a - b)$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

$73 - 37 = 36 = 9 \times 4$ .
R gives $9 (7 - 3) = 36$ , which explains A.

Answer: (a)

Example 3: Fixed-answer trick · 2 marks · Very short answerExam level

Show that 'think of a number, multiply by 4, add 20, divide by 4, subtract the original number' always gives 5.

Try it first, then show the solutionSolution

$x \to 4 x \to 4 x + 20 \to x + 5$ .
$x + 5 - x = 5$ .

Answer: Always 5

Example 4: Decode a date · 3 marks · Short answerHarder exam level

In a trick, a friend multiplies the month by 4, adds 3, multiplies by 25 and adds the day. Their answer is 806. Find the date, and explain why the method always works.

Try it first, then show the solutionSolution

The answer is $100 M + 75 + D$ .
$806 - 75 = 731 = 100 \times 7 + 31$ .
Since $D \leq 31$ fits in two digits, $M = 7$ and $D = 31$ : 31 July.

Answer: 31 July

Example 5: 3 × 3 calendar block · 3 marks · Short answerHarder exam level

The nine dates in a 3 × 3 block of a calendar page add up to 207. Find the centre date and the first date of the block.

Try it first, then show the solutionSolution

With top-left $a$ , the sum is $9 a + 72 = 9 (a + 8)$ .
$a + 8 = 207 \div 9 = 23$ , the centre.
$a = 15$ ; the block is 15–17, 22–24, 29–31.

Answer: Centre 23; first date 15.

Example 6: Largest product · 2 marks · Very short answerHarder exam level

Using 2, 7 and 8 once each, find the largest product $□□ \times □$ and compare it with the two closest rivals.

Try it first, then show the solutionSolution

$72 \times 8 = 576$ (largest digit as multiplier).
Rivals: $82 \times 7 = 574$ , $87 \times 2 = 174$ .

Answer: 576

Example 7: Heads and legs · 5 marks · Long answerHarder exam level

A farm has cows and hens: 40 heads and 112 legs in all. (a) Solve using algebra. (b) Solve without letter-numbers. (c) Check.

Try it first, then show the solutionSolution

(a) Let there be $c$ cows and $40 - c$ hens: $4 c + 2 (40 - c) = 112$ .
$2 c + 80 = 112$ , so $c = 16$ cows and 24 hens.
(b) If all 40 were hens: 80 legs. The extra $112 - 80 = 32$ legs come from cows.
Each cow adds 2 extra legs, so $32 \div 2 = 16$ cows, and 24 hens.
(c) $16 \times 4 + 24 \times 2 = 64 + 48 = 112$ and $16 + 24 = 40$ .

Answer: 16 cows, 24 hens

Example 8: The 101 trick · 3 marks · Short answerHOTS / competency

A 4-digit number has the form $abab$ , such as 4747. Prove that it is always divisible by 101, and find the quotient for 4747.

Try it first, then show the solutionSolution

$\overline{abab} = 1000 a + 100 b + 10 a + b = 1010 a + 101 b$ .
$= 101 (10 a + b) = 101 \times \overline{ab}$ .
So it is always a multiple of 101, and the quotient is the 2-digit number $\overline{ab}$ : $4747 \div 101 = 47$ .

Answer: Because $\overline{abab} = 101 \times \overline{ab}$ ; $4747 \div 101 = 47$ .

Example 9: Doubling and paying · 4 marks · Case-basedHOTS / competency

Each time Ravi walks round a magic tree, his coins double, and then he must pay 16 coins. After three rounds he has no coins left. How many did he start with? What payment per round would make him gain coins?

Try it first, then show the solutionSolution

Start $x$ : after round 1, $2 x - 16$ ; round 2, $4 x - 48$ ; round 3, $8 x - 112$ .
$8 x - 112 = 0$ , so $x = 14$ .
Check: 14 → 28 − 16 = 12 → 24 − 16 = 8 → 16 − 16 = 0.
He gains in a round if $2 x - p > x$ , that is $p < x$ : the payment must be less than the coins he has (less than 14 at the start).

Answer: 14 coins; a payment of less than 14 coins per round.

Example 10: Sum of a number and its reverse · 3 marks · Short answerHOTS / competency

Prove that a two-digit number plus its reverse is always a multiple of 11. Find all two-digit numbers (with non-zero units digit) for which this sum is 121.

Try it first, then show the solutionSolution

$(10 a + b) + (10 b + a) = 11 (a + b)$ .
$11 (a + b) = 121$ gives $a + b = 11$ .
Numbers: 29, 38, 47, 56, 65, 74, 83, 92.

Answer: Always $11 (a + b)$ ; the eight numbers 29, 38, 47, 56, 65, 74, 83, 92.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

A tea stall's prices

Suppose a tea stall pays ₹1,200 a day in rent, each cup of tea costs ₹6 to make, and the owner sells 150 cups a day. He wants a daily profit of ₹1,800.

(i) · 1 mark
Write an expression for the daily profit if each cup sells for ₹ $p$ .
Answer
Answer: $150 p - 2100$
Income $150 p$ ; costs $150 \times 6 + 1200 = 2100$ .
(ii) · 1 mark
What price gives the target profit?
Answer
Answer: ₹26
$150 p - 2100 = 1800$ , so $150 p = 3900$ and $p = 26$ .
(iii) · 2 marks
If customers will pay only ₹20 a cup, how many cups must he sell to reach the same profit? Explain your equation.
Answer
Answer: 215 cups
Profit per cup is $20 - 6 = 14$ , so $14 n - 1200 = 1800$ (1 mark). $14 n = 3000$ , $n \approx 214.3$ , so at least 215 cups (1 mark).

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Think of a number, double it, add 10, halve it, subtract the original number. Prove the result is always the same. (2 marks)

Step	Marks
$x \to 2 x \to 2 x + 10 \to x + 5$	1
$x + 5 - x = 5$ : always 5	1

A 3-row number pyramid has bottom row 7, $c$ , 11 and top 60. Find $c$ and the middle row. (3 marks)

Step	Marks
Top $= 7 + 2 c + 11$	1
$18 + 2 c = 60$ , so $c = 21$	1
Middle row: 28 and 32 (check $28 + 32 = 60$ )	1

A mother is 3 times as old as her son. In 12 years she will be twice as old as her son. (a) Form an equation. (b) Find their ages. (c) Check your answer. (d) How many years ago was the mother 5 times her son's age? (5 marks)

Step	Marks
(a) Son $s$ , mother $3 s$ : $3 s + 12 = 2 (s + 12)$	1
(b) $3 s + 12 = 2 s + 24$ , so $s = 12$ ; mother 36	1
(c) In 12 years: 48 and 24, and $48 = 2 \times 24$	1
(d) $t$ years ago: $36 - t = 5 (12 - t)$ , so $36 - t = 60 - 5 t$	1
$4 t = 24$ , $t = 6$ : 6 years ago (30 and 6)	1

Where students lose marks in Algebra Play

The slips examiners see most often, and what to write instead.

Mistake: Checking a trick with examples only
A few numbers seem to prove it.
Wrong working
5 gives 4 and 10 gives 4, so the trick always gives 4.
Right working
With $x$ : $\frac{3 x + 12}{3} - x = x + 4 - x = 4$ for every $x$ .
Do this: Use a letter for the starting number.
Usually costs: The justification marks.
Mistake: Forgetting to divide every term
Only one term in a sum is divided.
Wrong working
$\frac{3 x + 12}{3} = x + 12$
Right working
$\frac{3 x + 12}{3} = x + 4$
Do this: Divide each term of the expression.
Usually costs: 1 mark.
Mistake: Wrong expressions for calendar neighbours
Below is taken as $+ 1$ instead of $+ 7$ .
Wrong working
2 × 2 block: $a, a + 1, a + 2, a + 3$ .
Right working
$a, a + 1$ on top; $a + 7, a + 8$ below.
Do this: Right is +1, down is +7.
Usually costs: Whole answer.
Mistake: Using the pyramid top as a plain sum of the bottom row
Middle blocks are counted once.
Wrong working
Bottom 4, 7, 2: top $= 4 + 7 + 2 = 13$ .
Right working
Top $= 4 + 2 \times 7 + 2 = 20$ .
Do this: Middle numbers count twice in a 3-row pyramid.
Usually costs: 1 mark.
Mistake: Not checking the answer to a puzzle
Students stop after solving.
Wrong working
$4 c + 2 (40 - c) = 112$ gives $c = 16$ ; done.
Right working
Check: $16 \times 4 + 24 \times 2 = 64 + 48 = 112$ legs and $16 + 24 = 40$ heads.
Do this: Substitute back into every condition.
Usually costs: Can save the whole answer if there was a slip.

NCERT exercise guide: Algebra Play

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Figure it Out (number pyramids)

What it tests: Tops of pyramids without building them, the general top for 4 rows, and pyramids built from Virahanka-Fibonacci numbers.

How to approach it: Use the 1-2-1 and 1-3-3-1 weightings; for Fibonacci bottoms, notice each block is the sum of the two below and so stays in the sequence.

Like the 'top without building' question in Section 6.3

A 4-row pyramid has bottom row 3, 5, 8, 13. Find the top without building the pyramid. Is it a Virahanka-Fibonacci number?

Show the solutionSolution

Top $= a + 3 b + 3 c + d = 3 + 15 + 24 + 13$ .
$= 55$ .
55 is in the sequence 1, 2, 3, 5, 8, 13, 21, 34, 55, ...

Answer: 55, a Virahanka-Fibonacci number.

Figure it Out (largest product)

What it tests: Arranging three given digits in $□□ \times □$ for the largest product.

How to approach it: Largest digit as multiplier, the other two in decreasing order; check the closest rival.

Like the largest-product questions in Section 6.5

Use 2, 5 and 8 once each to make the largest product $□□ \times □$ .

Show the solutionSolution

Multiplier 8; the others in decreasing order: 52.
$52 \times 8 = 416$ .
Rival $82 \times 5 = 410$ is smaller.

Answer: $52 \times 8 = 416$

Figure it Out (divisibility tricks and puzzles)

What it tests: The quotient in the reverse-digit trick, number plus reverse and 11, cyclic sums and 37, the $ab c ab c$ trick, doubling-pond flowers, horses and hens, mother-daughter ages, cowherds sharing cows, a dosa cart's prices, a sequence of fractions of odd numbers, and Karim and the genie.

How to approach it: Write numbers in place-value form; for puzzles choose one letter, express everything in it, form an equation and check.

Like the cowherds puzzle in Section 6.6

Asha has 3 times as many marbles as Bina. If Asha gives Bina 10 marbles, they will have the same number. How many does each have?

Show the solutionSolution

Let Bina have $B$ ; Asha has $3 B$ .
$3 B - 10 = B + 10$ , so $2 B = 20$ and $B = 10$ .
Asha 30, Bina 10; check: $30 - 10 = 20 = 10 + 10$ .

Answer: Asha 30, Bina 10.

Practice questions on Algebra Play

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
For any three-digit number $ab c$ , the sum $ab c + b c a + c ab$ is always divisible by
(a) 7
(b) 11
(c) 37
(d) 13
Hint
$111 (a + b + c)$ .
Answer
(c) 37
Q2 · 1 mark · MCQExam level
The sum of a 2 × 2 block of dates with top-left date $a$ is
(a) $4 a + 4$
(b) $4 a + 8$
(c) $4 a + 16$
(d) $4 a + 32$
Hint
$a, a + 1, a + 7, a + 8$ .
Answer
(c) $4 a + 16$
Q3 · 1 mark · Very short answerExam level
Write the top of a 4-row pyramid with bottom row $a, b, c, d$ .
Hint
Weights 1, 3, 3, 1.
Answer
$a + 3 b + 3 c + d$
Q4 · 2 marks · Very short answerExam level
Find the top of a 4-row pyramid with bottom row 2, 7, 4, 6.
Hint
Use the formula.
Answer
$2 + 21 + 12 + 6 = 41$
Q5 · 2 marks · Very short answerHarder exam level
A 3-row pyramid has bottom row 6, $c$ , 10 and top 46. Find $c$ .
Hint
$6 + 2 c + 10$ .
Answer
$16 + 2 c = 46$ , so $c = 15$
Q6 · 2 marks · Very short answerHarder exam level
Show that 58 plus its reverse is a multiple of 11 and find the quotient. What is the quotient for any $ab$ ?
Hint
$11 (a + b)$ .
Answer
$58 + 85 = 143 = 11 \times 13$ ; the quotient is $a + b$ (here $5 + 8 = 13$ ).
Q7 · 2 marks · Very short answerHarder exam level
Use 3, 6 and 7 once each to make the largest product $□□ \times □$ .
Hint
Largest digit alone.
Answer
$63 \times 7 = 441$ (rival $73 \times 6 = 438$ is just smaller)
Q8 · 3 marks · Short answerHarder exam level
A bhel stall pays ₹900 rent a day, each plate costs ₹15 to make, and 120 plates are sold daily. At what price per plate will the daily profit be ₹1,500?
Hint
Profit = income − costs.
Answer
$120 p - 1800 - 900 = 1500$ , so $120 p = 4200$ and $p = ₹35$ .
Q9 · 3 marks · Short answerHarder exam level
Ravi has 4 times as many stamps as Gita. If Ravi gives Gita 15 stamps, they will have the same number. How many does each have?
Hint
$4 G - 15 = G + 15$ .
Answer
Gita 10, Ravi 40.
Q10 · 5 marks · Long answerHarder exam level
Design a 'think of a number' trick with these steps: double, add 5, multiply by 5, subtract 25. (a) Write the expression after each step. (b) Explain how to find your friend's number from their answer. (c) If the answer is 230, what was the number? (d) Change one step so the answer is always 100 times the number.
Hint
Expand carefully.
Answer
(a) $2 x$ , $2 x + 5$ , $10 x + 25$ , $10 x$ . (b) Divide the answer by 10. (c) 23. (d) Multiply by 50 instead of 5 and subtract 250: $50 (2 x + 5) - 250 = 100 x$ .
Q11 · 3 marks · Short answerHOTS / competency
Prove that a three-digit number minus its reverse (for example $753 - 357$ ) is always a multiple of 99.
Hint
Place values.
Answer
$(100 a + 10 b + c) - (100 c + 10 b + a) = 99 a - 99 c = 99 (a - c)$ . Example: $753 - 357 = 396 = 99 \times 4$ .
Q12 · 3 marks · Short answerHOTS / competency
A trader's money doubles each day, and then he spends ₹600. After 3 days he has nothing left. How much did he start with?
Hint
Track $2 x - 600$ each day.
Answer
After 3 days: $8 x - 600 \times (4 + 2 + 1) = 8 x - 4200 = 0$ , so $x = ₹525$ . Check: 1050 − 600 = 450; 900 − 600 = 300; 600 − 600 = 0.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Why does a 'think of a number' trick give the same answer for everyone?Show answer
Because the starting number cancels out in the algebra.
2.In the date trick 'month × 4, + 3, × 25, + day', what do you subtract from the answer?Show answer
75, leaving $100 M + D$ .
3.What is the top of a 3-row pyramid with bottom $a, b, c$ ?Show answer
$a + 2 b + c$ .
4.What is the top of a 4-row pyramid with bottom $a, b, c, d$ ?Show answer
$a + 3 b + 3 c + d$ .
5.What is the sum of a 2 × 2 calendar block with top-left $a$ ?Show answer
$4 a + 16$ .
6.What is the sum of a 3 × 3 calendar block?Show answer
9 times the centre date.
7.How do you make the largest product $□□ \times □$ from three digits?Show answer
Largest digit as the multiplier; the other two in decreasing order.
8.Why is a two-digit number minus its reverse a multiple of 9?Show answer
$(10 a + b) - (10 b + a) = 9 (a - b)$ .
9.Why is $ab c ab c$ divisible by 7, 11 and 13?Show answer
It equals $ab c \times 1001$ and $1001 = 7 \times 11 \times 13$ .
10.Which numbers are in a pyramid with Virahanka-Fibonacci numbers 1, 2, 3 at the bottom?Show answer
3, 5 and 8 at the top: all Virahanka-Fibonacci numbers.

Best order to study it

Revise forming and simplifying expressions, then explain a simple 'think of a number' trick.
Work through the date trick and invent your own version.
Learn number pyramids: general top expressions and missing values (first Figure it Out set).
Do calendar and algebra grids.
Learn the largest product rule and its algebraic proof (second set).
Prove the reverse-digit, cyclic-sum and 1001 tricks.
Finish with the word puzzles at the end of the chapter.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you prove a fixed-answer trick with algebra?
Can you design a trick that ends with any number you choose?
Can you decode a date from the final answer of the date trick?
Can you write the top of a 3-row and 4-row pyramid in terms of the bottom row?
Can you find a missing number in a pyramid?
Can you recover a calendar block from its sum?
Can you find and justify the largest product for three digits?
Can you prove the reverse-digit tricks for 9 and 11?
Can you explain the $ab c ab c$ trick?
Can you solve heads-and-legs and age puzzles with and without algebra?

Algebra Play: frequently asked questions

Where is Algebra Play in the Class 8 textbook?

It is Chapter 6 of Ganita Prakash Part II (Chapter 13 in our list), the NCERT Grade 8 Mathematics textbook (Reprint 2026-27). It has six short sections on tricks, pyramids, grids, the largest product and divisibility tricks.

Is this chapter only for fun, or will it be tested?

It is fun, but the skills are examinable: forming expressions, simplifying, solving linear equations and proving statements with algebra. Expect questions that ask you to explain a trick or solve a puzzle step by step.

Who are the Virahanka-Fibonacci numbers named after?

The sequence 1, 2, 3, 5, 8, ..., where each term is the sum of the two before, was described by the Indian scholar Virahanka centuries before Fibonacci. The chapter uses it in number pyramids.

Can I solve the puzzles without algebra?

Often yes, as the book shows for heads and legs: assume all animals are hens and use the extra legs. Algebra is still the most reliable method, so learn both and use the one that is clearer.

How much working should I show for a 'prove the trick' question?

Show the expression after each step, simplify carefully, and finish with a sentence explaining why the result does not depend on the starting number (or why it is always a multiple of 9, 11, and so on).

Based on: NCERT Ganita Prakash, Mathematics Textbook for Grade 8, Part II (Reprint 2026-27); NCERT chapter PDF: Algebra Play