Algebra Play | CBSE Class 8 Maths Notes

This note covers number tricks, date decoding, number pyramids, calendar and shape grids, the largest product from three digits, divisibility tricks, equations for everyday situations, doubling puzzles, and patterns in fractions.

How does algebra explain a number trick?

Algebra uses letters to represent numbers and describe relationships. A letter-number, or variable, stands for a number whose value may be unknown or allowed to change. Following a trick with a letter shows what happens to every permitted starting value.

An expression combines numbers, letters and operations, such as addition or multiplication. An equation states that two expressions have equal values. The sign = means “is equal to”, × means multiplication, and − means subtraction.

What happens to the starting number?

Let x represent the number first chosen. The notation 2x means 2 multiplied by x. Keep using the same letter for that original number throughout the trick; the changing intermediate answers have their own expressions.

Worked example 1. Think of a number, double it, add four, divide by two, and subtract the original number.

Answer: The successive expressions are x, 2x, 2x + 4, x + 2, and x + 2 − x. The final value is 2, regardless of the starting number.

The final subtraction removes the original number because x − x = 0. Before that, dividing by two acts on the whole expression: (2x + 4)/2 = x + 2. Here, the slash means division, and brackets group the quantity being divided.

How can the final answer change?

To obtain 3 instead, replace “add four” with “add six”. To obtain 5, replace it with “add ten”. These changes follow the same reasoning: doubling is undone by division, while half the added amount remains after the original number is subtracted.

How can a calculation reveal a chosen date?

A date contains two quantities: its month number and its day number. Let M be the month number and D be the day within that month. The notation 5M means five times the month number.

The trick multiplies the month by 5, adds 6, multiplies by 4, adds 9, multiplies by 5, and finally adds the day. Follow this order carefully: changing when an addition occurs changes the later expression.

Why does subtracting 165 work?

Subtracting 165 leaves 100M + D. A digit is a single symbol used to write a number. Since the day is at most 31, it fits into the final two digit places. Read those two places as the day; the digits before them give the month. Retain a leading zero when needed for the day.

Worked example 2. Mukta reports 291 after following the date trick. Recover her date.

Answer: 291 − 165 = 126. The final two digits give D = 26, and the preceding digit gives M = 1. Her date is 26 January.

The calculation for that date runs through 5, 11, 44, 53 and 265 before adding 26 to obtain 291. A second reported answer, 1390, gives 1390 − 165 = 1225, so the chosen date is 25 December.

Subtracting the added fixed number exposes the month and day. If the instructions are changed, derive the new final expression before deciding which fixed number to remove.

How do we fill missing numbers in a number pyramid?

Definition: A number pyramid is an arrangement in which each number above the bottom row equals the sum of the two numbers directly below it.

When both lower entries are known, add them. When an upper entry and one of its two lower entries are known, subtract to find the other lower entry. Choose a place where the available numbers determine a missing value.

The bottom row reads 1, 9, 4 from left to right. The middle row contains 10 and 13, and the top box contains 23. Each upper box sits over the pair whose sum it contains.

Reference: NCERT Class 8, page 137

How does working downwards help?

In a pyramid with top 10, middle-left entry 4 and bottom-left entry 1, the missing middle-right entry is 10 − 4 = 6. The bottom-middle entry is 4 − 1 = 3. The bottom-right entry is then 6 − 3 = 3.

Some pyramids do not offer an immediate subtraction. In these, give letters to the missing values and use the pyramid rule to connect them. A shared lower entry can affect more than one number above it.

Worked example 3. A three-row pyramid has top 60 and bottom row 12, an unknown number, 8. Find all missing entries.

Answer: Let c be the bottom-middle entry, and let a and b be the middle-left and middle-right entries. Then a = 12 + c, b = c + 8, and a + b = 60.

Substitution gives (12 + c) + (c + 8) = 60. Thus 20 + 2c = 60, 2c = 40, and c = 20. Hence a = 32 and b = 28.

Substitution means replacing a letter by its value or by an equal expression. Check the completed pyramid locally: 12 + 20 = 32 and 20 + 8 = 28. Finally, 32 + 28 = 60 confirms the required top number.

How can we find a pyramid's top without filling every box?

Letter-numbers reveal how each bottom entry contributes to the top. For a two-row pyramid with bottom entries a and b, where these letters represent the two chosen numbers, the top is a + b.

Property: The middle entry contributes twice in three rows

For a three-row pyramid, let a, b and c represent the bottom entries from left to right. The next row contains a + b and b + c. Adding them gives the top expression a + 2b + c.

The middle entry contributes twice because it belongs to both adjacent pairs. The outer entries each belong to one pair. This explains the coefficient 2; a coefficient is the number multiplying a letter. An algebraic term is a part of an expression separated by addition or subtraction.

Worked example 4. Find the top of a three-row pyramid whose bottom row is 4, 13, 8, without completing the middle row.

Answer: Substitute a = 4, b = 13 and c = 8. The top is a + 2b + c = 4 + 2 × 13 + 8 = 38.

Property: The four-row top has coefficients 1, 3, 3, 1

For four rows, let d be the fourth bottom entry, after a, b and c. The row above the bottom contains a + b, b + c, c + d. The next contains a + 2b + c and b + 2c + d.

Adding these gives a + 3b + 3c + d. This expression follows by repeated use of the same addition rule. With bottom row 8, 19, 21, 13, the top is 8 + 3 × 19 + 3 × 21 + 13 = 141.

What happens with Virahāṅka-Fibonacci numbers?

The Virahāṅka-Fibonacci sequence begins 1, 2, 3, 5, with each later number equal to the sum of the previous two. A sequence is an ordered list of numbers; each number in it is called a term.

A pyramid with bottom row 1, 2, 3 has middle row 3, 5 and top 8. Every entry belongs to that sequence. For bottom row 1, 2, 3, 5, the higher rows are 3, 5, 8; then 8, 13; then 21.

Let n be the number of rows when the bottom contains the first n sequence terms. Each step upwards starts two terms later and shortens the row by one. The top is term number 2n − 1. For 29 rows, it is the 57th term.

How does calendar magic recover four hidden numbers?

A 2 × 2 grid has two rows and two columns. In a calendar block of this size, moving one column right adds 1, and moving one row down adds 7. Choose a block containing four dates within the calendar.

Let a now represent the top-left date. The top-right date is a + 1, the bottom-left date is a + 7, and the bottom-right date is a + 8. The same letter identifies the starting entry throughout this calculation.

The August 2025 calendar is accompanied by a square block with 6 and 7 in the top row and 13 and 14 in the bottom row. Their displayed sum is 40. A second block labels the positions a, a + 1, a + 7 and a + 8.

Reference: NCERT Class 8, page 141

Property: The calendar block sums to 4a + 16

Adding all four positions gives a + (a + 1) + (a + 7) + (a + 8) = 4a + 16. Thus a reported sum lets us find a by subtracting 16 and dividing the result by 4.

Worked example 5. Four dates in a 2 × 2 calendar block sum to 36. Find the dates.

Answer: 4a + 16 = 36. Subtracting 16 from both sides gives 4a = 20. Dividing both sides by 4 gives a = 5. The block therefore has top row 5, 6 and bottom row 12, 13.

The four values add back to 36. The lower row begins seven after the upper-left entry. The calendar's arrangement determines the expressions.

How do shapes in algebra grids represent unknown numbers?

In an algebra grid, each kind of shape represents a number. Repeated copies of the same shape have the same value. The final column in the given rows shows the sum of the shape values to its left.

Which row should we solve first?

Start with a row containing only one kind of shape when such a row is available. Its total can be shared equally among the repeated shapes. Then substitute that value into another row to find the next shape.

The first row shows three blue squares with total 27. The second shows two red circles followed by one blue square with total 19. The accompanying calculation assigns 9 to the square and 5 to the circle.

Reference: NCERT Class 8, page 142

Worked example 6. Three equal squares total 27. Two equal circles and one of those squares total 19. Find the value of each shape.

Answer: Each square has value 27/3 = 9. Replacing the square in the second row leaves two circles totalling 19 − 9 = 10. Each circle therefore has value 10/2 = 5.

Let s represent a square's value and c a circle's value in this grid. The equations are 3s = 27 and 2c + s = 19. These describe the shape rows.

After solving, check every given row: 9 + 9 + 9 = 27 and 5 + 5 + 9 = 19. A value that fits just one row is insufficient if the same shape also appears in another row with a stated total.

How do three digits form the largest possible product?

The challenge is to use the digits 2, 3 and 5 once each to form a two-digit number multiplied by a one-digit number. The product is the answer to multiplication. Here, call the two-digit number the multiplicand and the one-digit number the multiplier.

The six arrangements are 23 × 5, 25 × 3, 32 × 5, 35 × 2, 52 × 3 and 53 × 2. Grouping them by multiplier reduces the comparison: keep the larger two-digit number in each pair.

Worked example 7. Find the largest product obtainable from 2, 3 and 5 in the required arrangement.

Answer: The remaining candidates are 53 × 2, 52 × 3 and 32 × 5. Their products are 106, 156 and 160. Therefore, 32 × 5 = 160 is the largest.

Why should the largest digit be the multiplier?

Let p, q and r be three distinct positive digits in increasing order, written p < q < r. The sign < means “is less than”. Arrange the two remaining digits in decreasing order when forming each two-digit candidate.

In digit notation, qp means the two-digit number with tens digit q and units digit p, so qp = 10q + p. It does not mean q multiplied by p here. Similarly, rp = 10r + p and rq = 10r + q.

Compare (10r + q) × p with (10r + p) × q. Subtracting the first from the second gives 10 × r × (q − p). This is positive, so the second candidate is larger.

Now compare (10q + p) × r and (10r + p) × q. Both contain the same amount 10 × q × r. Their remaining amounts are p × r and p × q. Since r is larger than q, the first is larger.

Thus the largest digit is the multiplier, and the other two digits form the multiplicand in decreasing order. The remaining practice digit sets give 31 × 7 = 217 and 53 × 9 = 477 by the same reasoning.

Why do reversing-digit tricks involve 9 and 11?

Divisibility means that division by the stated number leaves no remainder. The remainder is the amount left over after division into whole-number groups; the quotient is the result of the division.

For a two-digit number, let a be its tens digit and b its units digit. The number is 10a + b. Reversing the digits gives 10b + a. Writing place values explicitly prevents confusion with multiplying the digits together.

Property: The reversed-number difference is divisible by 9

When b is greater than a, subtract the original number from its reversal. The difference is (10b + a) − (10a + b) = 9b − 9a = 9(b − a). Here, brackets after 9 mean multiplication by the enclosed difference.

The answer is nine times the difference between the digits, so dividing it by 9 leaves no remainder. If a is greater than b, subtract the reversal from the original instead; the corresponding expression is 9(a − b).

Worked example 8. Reverse 47, subtract the smaller number from the larger, and divide by 9.

Answer: The reversal is 74. The difference is 74 − 47 = 27, and 27/9 = 3. This quotient equals the difference between the digits, 7 − 4.

What changes if we add the two numbers?

The sum is (10a + b) + (10b + a) = 11a + 11b = 11(a + b). It is therefore divisible by 11. In this case, the quotient is the sum of the digits, rather than their difference.

The examples 31 + 13 = 44, 28 + 82 = 110 and 12 + 21 = 33 illustrate this property. Algebra explains their shared factor; a factor is a number that divides another number without leaving a remainder.

Why do cycling or repeating three digits create divisibility patterns?

Let a, b and c now be the hundreds, tens and units digits of a three-digit number. In digit notation, abc represents 100a + 10b + c. The letters record positions; abc here is not the product of three letters.

How does cycling the digits work?

Cycling means moving the digits round while retaining their cyclic order. The three arrangements are abc, bca and cab. Their place-value expressions are 100a + 10b + c, 100b + 10c + a, and 100c + 10a + b.

In their sum, each digit appears once in the hundreds place, once in the tens place and once in the units place. Collecting these contributions gives 111(a + b + c). This is the reason the same divisibility result applies across the choices.

Worked example 9. Show that the sum of a three-digit number and its two cyclic arrangements is divisible by 37 and by 3.

Answer: The sum is 111(a + b + c). Since 111 = 3 × 37, it is both 37 × 3(a + b + c) and 3 × 37(a + b + c). Both divisions therefore leave no remainder.

What happens when the whole number is repeated?

Let N represent the original three-digit number. Repeating its three digits forms a six-digit number: the first copy contributes 1000N and the second contributes N. The resulting number is 1001N.

Because 1001 = 7 × 11 × 13, dividing successively by 7, then 11, then 13 returns N. The repeated digit block creates the factor 1001. The trick therefore depends on the place values of the two copies.

How do equations solve animal, age and cow puzzles?

A model translates a situation into mathematical relationships. First identify what is unknown, then express related quantities using it. An equation must retain all the conditions in the problem. Finally, check that the answer fits the original situation.

How can heads and legs determine animal numbers?

Worked example 10. A farm has only horses and hens, with 55 heads and 150 legs. Taking one head per animal, four legs per horse and two per hen, find the number of each.

Answer: Let h be the number of horses. There are 55 − h hens. The leg equation is 4h + 2(55 − h) = 150, giving 2h + 110 = 150. Hence h = 20 and there are 35 hens.

There is also a solution without letters. If all 55 animals were hens, they would have 110 legs. The actual total has 40 extra legs. Each horse replacing a hen adds two legs, so there are 40/2 = 20 horses.

Why must both ages increase?

A mother is five times her daughter's age, and in six years she will be three times her daughter's age. Let d be the daughter's present age in years. The mother's present age is 5d.

Six years later, the ages are d + 6 and 5d + 6. Thus 5d + 6 = 3(d + 6), so 5d + 6 = 3d + 18. This gives 2d = 12 and d = 6. The mother is 30.

How does transferring cows affect both friends?

Let g be Gauri's number of cows. Naina has twice as many, so she has 2g. If Naina gives Gauri three cows, their new numbers are 2g − 3 and g + 3. Equal new totals give 2g − 3 = g + 3.

Therefore g = 6: Gauri has 6 cows and Naina has 12. After the transfer each has 9. Check both the original doubling relationship and the equality after the transfer.

How does algebra connect costs, sales and profit?

Cost is the money spent, sales income is the money received from selling, and profit is sales income minus total cost. The symbol ₹ means rupees. Include every stated expense when turning a sales problem into an equation.

The dosa cart has rent of ₹5000 per day. Making each dosa, including ingredients and fuel, costs ₹10. These expenses behave differently: the daily rent is given as one amount, while the making cost increases with the number of dosas.

What selling price produces the required profit?

Worked example 11. The cart sells 100 dosas in a day. With daily rent ₹5000 and making cost ₹10 per dosa, find the selling price per dosa required for a ₹2000 profit.

Answer: Making 100 dosas costs ₹1000. Total daily cost is ₹5000 + ₹1000 = ₹6000. The required sales income is ₹6000 + ₹2000 = ₹8000. Dividing by 100 gives a selling price of ₹80 per dosa.

What if the selling price is fixed?

Customers are willing to pay ₹50 per dosa. Let n now represent the number of dosas made and sold in one day. Income is 50n rupees, and total cost is 5000 + 10n rupees.

The required profit gives 50n − (5000 + 10n) = 2000. Simplifying gives 40n − 5000 = 2000, then 40n = 7000. Thus n = 175 dosas.

Each sale leaves ₹50 − ₹10 = ₹40 after its making cost. That amount must collectively cover both the ₹5000 rent and the ₹2000 desired profit. This gives the same calculation, ₹7000 divided by ₹40, and explains why rent cannot be omitted.

How can we solve puzzles involving repeated doubling?

Both the magical-pond puzzle and Karim's genie puzzle combine doubling with removal of a fixed amount. The order of operations matters here: first double the current amount, then subtract the offering or payment. Repeat with what remains.

What does the flower puzzle determine?

There are three shrines, each preceded by a pond that doubles the flowers dipped into it. The person dips all remaining flowers before each offering, leaves an equal number at each shrine, and has no flowers after the third offering.

Let x be the starting number of flowers and y the equal offering at each shrine. After the first offering, 2x − y remain. After the second, 2(2x − y) − y = 4x − 3y remain.

After the third, 2(4x − 3y) − y = 8x − 7y remain. Setting this to zero gives 8x = 7y. The given conditions specify a relationship, so they do not determine one unique numerical starting amount.

For positive whole-number counts, write x = 7k and y = 8k, where k is any positive whole number. The smallest such solution starts with 7 flowers and offers 8 at each shrine. The successive remaining counts are 6, 4 and 0.

How does working backwards reveal Karim's coins?

Worked example 12. Karim's coins double each time he circles the tree, after which he pays the genie 8 coins. After three rounds and payments he has none. Find his initial number of coins.

Answer: Immediately before the last payment he has 8 coins, so before the last doubling he has 4. Before the second payment he therefore had 4 + 8 = 12, and before that doubling he had 6. Before the first payment he had 6 + 8 = 14, so he started with 7.

For the general payment question, let x now be the starting coins and f the fixed fee per round. One round leaves 2x − f. This exceeds x precisely when f is less than x. A lower fee lets the coin total grow through subsequent rounds.

After three rounds the remaining amount is 8x − 7f. To take all the coins through this three-round arrangement, the fee must satisfy 7f = 8x. If coins must remain whole, the fee and intermediate amounts must also meet that requirement.

Why do the given fractions have the same value?

A fraction represents division: its numerator is the number above the fraction line and its denominator is the non-zero number below it. Consider the sequence 1/3, (1 + 3)/(5 + 7), and (1 + 3 + 5)/(7 + 9 + 11).

The first value is 1/3. The second is 4/12 = 1/3, and the third is 9/27 = 1/3. An odd number is a whole number not divisible by 2; the sums here are built from consecutive positive odd numbers.

How does the sum of odd numbers explain the pattern?

Let n represent the number of odd-number terms in the numerator. The sum of the first n positive odd numbers is n², meaning n multiplied by itself. In the displayed pattern, the denominator uses n consecutive odd numbers starting with 2n + 1.

Those denominator terms follow the first n odd numbers and complete the first 2n odd numbers. Their sum is therefore (2n)² − n² = 4n² − n² = 3n². Subtracting removes the earlier terms that do not belong in the denominator.

For positive n, the resulting fraction is n²/(3n²) = 1/3. This argument explains the equal values using the relationship between the two sums. Continuing the same rule therefore preserves the value, without separately adding every numerator and denominator.

Glossary

• Algebra — The use of letters to represent numbers and express mathematical relationships.
• Letter-number — A letter representing a number that is unknown or allowed to vary.
• Expression — A combination of numbers, letters and operations representing a mathematical value.
• Equation — A statement that the expressions on its two sides have equal values.
• Substitution — Replacing a letter with its numerical value or with an equal expression.
• Number pyramid — An arrangement where each upper entry sums the two entries directly below it.
• Coefficient — The numerical multiplier of a letter within a term of an expression.
• Sequence — An ordered list of numbers, each of which is called a term.
• Multiplicand — The number being multiplied, here the two-digit number in the digit puzzle.
• Multiplier — The number by which another number is multiplied to obtain a product.
• Divisibility — The property of giving no remainder when divided by the stated number.
• Quotient — The result obtained by dividing one number by another non-zero number.
• Profit — Sales income remaining after all the stated costs have been subtracted.
• Numerator — The number above a fraction line, divided by the number below it.
• Denominator — The non-zero number below a fraction line, by which the numerator is divided.

Common errors and misconceptions

• Misconception: Dividing 2x + 4 by 2 gives x + 4. Correct: Divide both parts by 2 to get x + 2 before subtracting the original x.
• Misconception: The reported date-trick answer already gives the date. Correct: First subtract 165; then read the last two digit places as the day and the preceding digits as the month.
• Misconception: The top of a three-row pyramid is the sum of its bottom entries. Correct: For bottom entries a, b, c, it is a + 2b + c because b enters both middle boxes.
• Misconception: A calendar block contains four consecutive numbers. Correct: If its top-left entry is a, the four entries are a, a + 1, a + 7 and a + 8.
• Misconception: Adjacent letters always mean multiplication in digit puzzles. Correct: When a and b denote tens and units digits, the digit notation ab means 10a + b.
• Misconception: Only the mother's age increases in the future-age equation. Correct: After six years, add six to both the mother's and the daughter's present ages before applying the new relationship.
• Misconception: The dosa-making expense is the entire daily cost. Correct: Add the ₹5000 daily rent to ₹10 for each dosa before calculating the profit.
• Misconception: The equal-offering flower puzzle fixes one starting count. Correct: It gives 8x = 7y, where x is the starting flowers and y the offering; multiples of the smallest solution also work.

Exam-style questions with model answers

Q1. Let x be a chosen number. Double it, add four, divide the result by two, and subtract x. Show why the final answer is 2. [2 marks]

1. After doubling and adding four, the expression is 2x + 4. Dividing the whole expression by two gives x + 2.
2. Subtracting the original x gives x + 2 − x = 2, so the starting value disappears from the answer.

Q2. A date trick multiplies the month by 5, adds 6, multiplies by 4, adds 9, multiplies by 5, then adds the day. Months run from 1 for January to 12 for December; days are at most 31. The reported answer is 1390. Derive the decoding rule and find the date. [4 marks]

1. Let M represent the month number and D the day number. Multiplying M by 5, adding 6 and multiplying by 4 gives 20M + 24.
2. Adding 9 gives 20M + 33. Multiplying this by 5 gives 100M + 165; adding the day gives 100M + 165 + D.
3. Subtract 165 from the reported answer: 1390 − 165 = 1225. This equals 100M + D.
4. A day is at most 31, so the last two digits give D = 25 and the preceding digits give M = 12. The date is 25 December.

Q3. In a three-row number pyramid, every upper entry is the sum of the two directly below it. The top is 60 and the bottom row is 12, an unknown entry, 8. Find all three missing entries and check the result. [3 marks]

1. Let c represent the unknown bottom-middle entry. The two middle entries are 12 + c and c + 8, so their sum gives (12 + c) + (c + 8) = 60.
2. Combining the terms gives 20 + 2c = 60. Subtract 20 and divide by 2 to obtain c = 20.
3. The middle entries are 12 + 20 = 32 and 20 + 8 = 28. Their sum is 60, which verifies the given top entry.

Q4. A calendar has seven days in each row. Moving right increases a date by 1 and moving down increases it by 7. Four dates in a 2 × 2 block sum to 36. Find them algebraically. [3 marks]

1. Let a be the top-left date. The other entries are a + 1 to its right, a + 7 below it, and a + 8 diagonally below and right.
2. The sum is a + (a + 1) + (a + 7) + (a + 8) = 4a + 16. Thus 4a + 16 = 36, giving a = 5.
3. The top row is 5, 6 and the bottom row is 12, 13. Checking gives 5 + 6 + 12 + 13 = 36, as required.

Q5. Use the digits 2, 3 and 5 exactly once to form a two-digit number multiplied by a one-digit number. Find and justify the largest product by grouping the possible arrangements. [5 marks]

1. There are six arrangements: 23 × 5, 25 × 3, 32 × 5, 35 × 2, 52 × 3 and 53 × 2. Each uses all three digits once.
2. For multiplier 2, compare 35 × 2 with 53 × 2. Since 53 is larger than 35, retain 53 × 2.
3. For multiplier 3, retain 52 × 3 rather than 25 × 3. For multiplier 5, retain 32 × 5 rather than 23 × 5.
4. The surviving products are 53 × 2 = 106, 52 × 3 = 156 and 32 × 5 = 160. Compare these three values.
5. The largest is 160, obtained from 32 × 5. Here the largest digit is the multiplier, and the other digits form the two-digit number in decreasing order.

Q6. A two-digit number has tens digit a and units digit b, with b greater than a. Reverse the digits. Show that the reversal minus the original is divisible by 9, identify the quotient, and check using 47. [4 marks]

1. The original number has value 10a + b. Its reversal has value 10b + a and is larger because b is greater than a.
2. Subtracting gives (10b + a) − (10a + b) = 9b − 9a = 9(b − a).
3. Division by 9 therefore leaves no remainder, and the quotient is b − a, the difference between the two digits.
4. For 47, the reversal is 74. Their difference is 27, and 27/9 = 3, equal to 7 − 4.

Q7. A cart pays ₹5000 daily rent, and each dosa costs ₹10 to make, including ingredients and fuel. It makes and sells 100 dosas at one fixed price in one day. What selling price per dosa gives a ₹2000 daily profit? Show the cost and income calculations. [5 marks]

1. The making cost depends on the number of dosas. For 100 dosas at ₹10 each, it is 100 × ₹10 = ₹1000.
2. The cart must also pay the stated daily rent. Adding ₹5000 rent to ₹1000 making cost gives total daily expenses of ₹6000.
3. Profit is sales income minus expenses. To retain ₹2000 after paying expenses of ₹6000, the required daily sales income is ₹8000.
4. All 100 dosas are sold at the same price. Dividing the required ₹8000 income by 100 gives a selling price of ₹80 per dosa.
5. Check the result: selling 100 dosas at ₹80 gives ₹8000. Subtracting ₹5000 rent and ₹1000 making cost leaves the required ₹2000 profit.

Q8. Karim's coins double whenever he circles a tree, and he then pays the genie 8 coins. After three such rounds and payments, he has no coins left. Work backwards to find his initial coins and check all three rounds. [5 marks]

1. After the third payment Karim has zero coins. Immediately before paying the final 8 coins, he must therefore have had exactly 8 coins.
2. The third doubling produced those 8 coins. Before that doubling, and therefore after the second payment, he had 8/2 = 4 coins.
3. Before the second payment he had 4 + 8 = 12 coins. Before the second doubling, he had 12/2 = 6 coins.
4. Before the first payment he had 6 + 8 = 14 coins. Before the first doubling, he therefore had 14/2 = 7 coins.
5. The forward check is 7 × 2 − 8 = 6, then 6 × 2 − 8 = 4, then 4 × 2 − 8 = 0. He began with 7 coins.

Key takeaways

• Letters let us follow a calculation for an unknown starting number and explain why a number trick works.
• The date trick produces 100M + 165 + D, where M is the month number and D is the day number.
• Every upper pyramid entry adds the two below it; a three-row pyramid with bottom a, b, c has top a + 2b + c.
• A calendar block with top-left entry a contains a, a + 1, a + 7 and a + 8, totalling 4a + 16.
• For three distinct positive digits, use the largest as multiplier and arrange the other two in decreasing order.
• Reversing two digits creates a difference divisible by 9; adding the original and reversed numbers gives a multiple of 11.
• Writing place values explicitly explains why cycling three digits gives divisibility by 37 and repeating a block creates the factor 1001.
• Translate every condition into the equation, preserve the stated operation order, and check the solution against the original quantities.

Test yourself

What addition replaces “add four” if the double, add, halve, subtract-original trick must finish at 5?

Add ten: halving leaves the original number plus 5, and subtracting the original leaves 5.

A date trick gives 291 and has final expression 100M + 165 + D, where M is the month and D the day. What date was chosen?

Subtracting 165 gives 126. The month is 1 and the day is 26, so the date is 26 January.

A three-row pyramid adds adjacent entries upwards. Its bottom row is 7, 11, 3. What is the top?

The top is 7 + 2 × 11 + 3 = 32, because the middle bottom entry contributes twice.

Three equal squares total 27. Two equal circles and one square total 19. What is one circle worth?

Each square is 9, leaving 10 for two circles. Each circle therefore has value 5.

Use 1, 3 and 7 once each in a two-digit number times a one-digit number. What is the largest product?

Use 7 as multiplier and 31 as multiplicand. The largest product is 31 × 7 = 217.

For a two-digit number with tens digit a and units digit b, why is its sum with its reversal divisible by 11?

The sum is (10a + b) + (10b + a) = 11(a + b), which has 11 as a factor.

A three-digit number N is repeated to form a six-digit number. Why does division successively by 7, 11 and 13 return N?

The repeated number is 1001N, and 1001 = 7 × 11 × 13. Dividing by these three factors leaves N.

A mother is five times her daughter's age now and three times her daughter's age in six years. How old is the daughter now?

Let d be her present age. Then 5d + 6 = 3(d + 6), giving 2d = 12 and d = 6 years.