Important Questions Class 8 Maths Part 2 Chapter 6 Algebra Play 2026-2027

Algebra Play teaches students how letter-numbers can explain number tricks, puzzles, grids, pyramids and real-life problems. It shows that algebra helps prove why a trick works every time.

Important Questions Class 8 Maths Part 2 Chapter 6 help students practise think-of-a-number tricks, date tricks, number pyramids, calendar grids, algebra grids, digit puzzles, divisibility tricks and word problems. This chapter builds reasoning skills because students must form expressions before solving.

Class 8 Maths Part 2 Chapter 6 focuses on algebra through games and puzzles. Students learn that algebra is useful when a number is unknown, when a pattern repeats, or when a trick gives the same result for every starting number.

The chapter begins with simple number tricks and then moves to pyramids, calendar grids and divisibility puzzles. Each activity asks students to think like problem solvers. Instead of guessing answers, they use variables such as (x), (a), (b), (M) and (D) to explain the logic behind the result.

Key Takeaways from Class 8 Maths Part 2 Chapter 6

Important Questions Class 8 Maths Part 2 Chapter 6 with Answers

These questions cover the main ideas of the chapter. Students should show the algebraic steps, not only the final answer.

Important Questions Class 8 Maths Part 2 Chapter 6: Algebra Play Basics

Q1. What is the main idea of Algebra Play Class 8?
The main idea of Algebra Play is to use algebra to understand tricks, puzzles and real-life number situations.

The chapter shows how letter-numbers help represent unknown values. Once the unknown is written as a variable, students can form equations and explain why a result always works.

Q2. Why do we use letter-numbers in algebra tricks?
We use letter-numbers because the starting number can be anything.

For example, if a trick begins with “Think of a number”, we can call that number (x). This helps us prove the trick for every possible number.

Q3. Explain this trick using algebra: Think of a number, double it, add 4, divide by 2, subtract the original number.
Let the number be (x).

Double it:

[2x]

Add 4:

[2x + 4]

Divide by 2:

[x + 2]

Subtract the original number:

[x + 2 - x = 2]

So, the final answer is always 2.

Q4. How can the same trick give 5 as the final answer?
Change the “add 4” step to “add 10”.

Let the number be (x).

Double it:

[2x]

Add 10:

[2x + 10]

Divide by 2:

[x + 5]

Subtract the original number:

[x + 5 - x = 5]

So, the final answer becomes 5.

Q5. Why does algebra help in puzzles?
Algebra helps because it converts a puzzle into expressions and equations.

For example, a number pyramid may have missing boxes. Once we call one missing number (c), the other boxes can be written in terms of (c). Then one equation gives the answer.

Number Tricks Class 8 Maths Questions

Number tricks class 8 maths questions are important because they show why algebra is useful. Students should not only perform the trick. They should prove why the answer always comes out the same.

Think of a Number Tricks Class 8

Q1. Think of a number, triple it, add 6, divide by 3, subtract the original number. What is the answer?
Let the number be (x).

Triple it:

[3x]

Add 6:

[3x + 6]

Divide by 3:

[x + 2]

Subtract the original number:

[x + 2 - x = 2]

So, the answer is always 2.

Q2. Change the trick above so that the final answer becomes 4.
Use “add 12” instead of “add 6”.

Let the number be (x).

[3x + 12]

Divide by 3:

[x + 4]

Subtract the original number:

[x + 4 - x = 4]

So, the final answer becomes 4.

Q3. Why does the original number disappear in these tricks?
The original number disappears because the last step subtracts the same unknown number.

For example, (x + 2 - x = 2). The (x) terms cancel, so the answer does not depend on the starting number.

Q4. Create a trick where the final answer is 7.
Use this trick: think of a number, double it, add 14, divide by 2, subtract the original number.

Let the number be (x).

[2x + 14]

Divide by 2:

[x + 7]

Subtract (x):

[x + 7 - x = 7]

So, the final answer is 7.

Q5. Why should students use variables in number tricks?
Variables help prove the trick for every starting number.

Testing with one number only proves that the trick works once. Algebra proves that it works for all possible numbers.

Class 8 Maths Part 2 Chapter 6 Question Answer on Date Tricks

Date tricks use place value. The month and day can be recovered because the final expression separates the month and day.

Q1. In the date trick, how does the final answer reveal the date?
Let the month be (M) and the day be (D).

The steps give:

[100M + 165 + D]

So, when the final answer is given, subtract 165.

[\text{Final answer} - 165 = 100M + D]

The last two digits give the day. The remaining digits give the month.

Q2. Mukta’s final answer is 1390. What date did she choose?
Subtract 165:

[1390 - 165 = 1225]

This equals:

[100M + D]

So, (M = 12) and (D = 25).

The date is 25th December.

Q3. Find the date if the final answer is 1269.
Subtract 165:

[1269 - 165 = 1104]

So:

[100M + D = 1104]

This gives (M = 11) and (D = 04).

The date is 4th November.

Q4. Find the date if the final answer is 394.
Subtract 165:

[394 - 165 = 229]

This can be written as:

[029]

So, (M = 2) and (D = 29).

The date is 29th February.

Q5. Find the date if the final answer is 296.
Subtract 165:

[296 - 165 = 131]

This means:

[100M + D = 131]

So, (M = 1) and (D = 31).

The date is 31st January.

Number Pyramids Class 8 Important Questions

In a number pyramid, each box contains the sum of the two boxes directly below it. Algebra helps when the missing numbers are not obvious.

Q1. What is the rule for a number pyramid?
The rule is simple: each number is the sum of the two numbers directly below it.

For example, if the bottom row has (a) and (b), the top number is:

[a + b]

Q2. Write the top expression for a three-row number pyramid with bottom row (a, b, c).
The second row has:

[a + b]

and

[b + c]

The top number is:

[(a + b) + (b + c)]

[= a + 2b + c]

So, the top expression is:

[a + 2b + c]

Q3. Find the top number if the bottom row is 4, 13 and 8.
Use:

[a + 2b + c]

Here, (a = 4), (b = 13), (c = 8).

[4 + 2(13) + 8 = 4 + 26 + 8 = 38]

So, the top number is 38.

Q4. Find the top number if the bottom row is 7, 11 and 3.
[a + 2b + c = 7 + 2(11) + 3]

[= 7 + 22 + 3]

[= 32]

So, the top number is 32.

Q5. What is the top expression for a four-row pyramid with bottom row (a, b, c, d)?
First row above bottom:

[a+b,\quad b+c,\quad c+d]

Next row:

[a+2b+c,\quad b+2c+d]

Top:

[(a+2b+c) + (b+2c+d)]

[= a + 3b + 3c + d]

So, the top expression is:

[a + 3b + 3c + d]

Calendar Grid Questions from Class 8 Maths Part 2 Chapter 6

Calendar grids look like number puzzles, but they follow a fixed weekly pattern. The number below any date is 7 more than it.

Q1. In a 2 × 2 calendar grid, write all numbers in terms of the top-left number (a).
The top-left number is (a).

The top-right number is:

[a + 1]

The bottom-left number is:

[a + 7]

The bottom-right number is:

[a + 8]

So, the grid is:

[\begin{matrix}
a & a+1 \
a+7 & a+8
\end{matrix}]

Q2. What is the sum of a 2 × 2 calendar grid in terms of (a)?
Add all four numbers:

[a + (a+1) + (a+7) + (a+8)]

[= 4a + 16]

So, the sum is:

[4a + 16]

Q3. If the sum of a 2 × 2 calendar grid is 36, find the four numbers.
Use:

[4a + 16 = 36]

[4a = 20]

[a = 5]

The four numbers are:

[5,\ 6,\ 12,\ 13]

Q4. If the top-left number is 9, find the sum of the 2 × 2 grid.
Use:

[4a + 16]

Here, (a = 9).

[4(9) + 16 = 36 + 16 = 52]

So, the sum is 52.

Q5. If the sum of a 2 × 2 calendar grid is 80, find the top-left number.
[4a + 16 = 80]

[4a = 64]

[a = 16]

The top-left number is 16.

Algebra Grids Class 8 Maths Questions

Algebra grids class 8 questions use shapes instead of variables. Each shape has a fixed value. Students must use row and column sums to decode them.

Q1. What is an algebra grid?
An algebra grid is a puzzle where symbols or shapes represent numbers.

Students use row sums and column sums to find the value of each shape. It works like solving equations.

Q2. If three identical shapes add to 27, what is the value of one shape?
Let the shape be (x).

[x + x + x = 27]

[3x = 27]

[x = 9]

So, one shape equals 9.

Q3. If two circles and one triangle make 21, and each circle is 6, find the triangle.
Let the triangle be (t).

[6 + 6 + t = 21]

[
12 + t = 21
]

[t = 9]

So, the triangle equals 9.

Q4. If a square plus a square plus a circle equals 18, and the circle is 4, find the square.
Let the square be (s).

[s + s + 4 = 18]

[2s + 4 = 18]

[2s = 14]

[s = 7]

So, the square equals 7.

Q5. Why are algebra grids useful for students?
Algebra grids help students practise equations without making the topic feel abstract.

They show that a symbol can stand for a number. This builds the foundation for variables and equation solving.

Divisibility Tricks Class 8 Maths Important Questions

Divisibility tricks class 8 maths questions show how place value and algebra work together. Students should write two-digit and three-digit numbers correctly before proving the trick.

Q1. Prove that the difference between a two-digit number and its reverse is divisible by 9.
Let the two-digit number be (10a + b).

Its reverse is:

[10b + a]

If (b > a), the difference is:

[(10b + a) - (10a + b)]

[= 9b - 9a]

[= 9(b-a)]

Since it has 9 as a factor, it is divisible by 9.

Q2. What is the quotient when the difference is divided by 9?
From the proof:

[(10b+a) - (10a+b) = 9(b-a)]

When divided by 9, the quotient is:

[b-a]

So, the quotient is the difference between the two digits.

Q3. Prove that the sum of a two-digit number and its reverse is divisible by 11.
Let the number be:

[10a + b]

Its reverse is:

[10b + a]

Their sum is:

[(10a+b) + (10b+a)]

[= 11a + 11b]

[= 11(a+b)]

So, the sum is divisible by 11.

Q4. Show that (abcabc) is divisible by 7, 11 and 13.
Let the three-digit number (abc) be (n).

Repeating it gives:

[abcabc = 1000n + n]

[= 1001n]

Now:

[1001 = 7 \times 11 \times 13]

So, (abcabc) is divisible by 7, 11 and 13.

Q5. Why is algebra useful in divisibility tricks?
Algebra writes numbers using place value.

For example, a two-digit number becomes (10a + b). Once written this way, students can factor expressions and prove divisibility for all numbers.

Class 8 Maths Part 2 Chapter 6 Extra Questions with Answers

Class 8 maths Part 2 Chapter 6 extra questions cover puzzle-based and word-problem-based practice from Algebra Play.

Q1. Fill digits 1, 3 and 7 in (_ _ \times _) to make the largest product.
Use the largest digit as the multiplier. Arrange the other two digits in decreasing order.

So, use:

[31 \times 7]

[31 \times 7 = 217]

The largest product is 217.

Q2. Fill digits 3, 5 and 9 in (_ _ \times _) to make the largest product.
Use the largest digit as the multiplier. Arrange the other two digits in decreasing order.

[53 \times 9 = 477]

The largest product is 477.

Q3. A farm has horses and hens. Total heads are 55, and total legs are 150. Find the number of horses and hens.
Let all 55 animals be hens.

Then total legs would be:

[55 \times 2 = 110]

Actual legs = 150.

Extra legs:

[150 - 110 = 40]

Each horse has 2 extra legs compared to a hen.

Number of horses:

[40 \div 2 = 20]

Number of hens:

[55 - 20 = 35]

So, there are 20 horses and 35 hens.

Q4. A mother is 5 times her daughter’s age. After 6 years, the mother will be 3 times her daughter’s age. Find the daughter’s age now.
Let the daughter’s age be (x).

Mother’s age is:

[5x]

After 6 years:

Daughter’s age:

[x + 6]

Mother’s age:

[5x + 6]

Given:

[5x + 6 = 3(x + 6)]

[5x + 6 = 3x + 18]

[2x = 12]

[x = 6]

The daughter is 6 years old.

Q5. Gauri says Naina has twice as many cows as she does. If Naina gives 3 cows to Gauri, both have equal cows. Find their cows.
Let Gauri have (x) cows.

Naina has:

[2x]

After Naina gives 3 cows:

Gauri has:

[x + 3]

Naina has:

[2x - 3]

Given:

[x + 3 = 2x - 3]

[x = 6]

So, Gauri has 6 cows and Naina has 12 cows.

Algebra Word Problems Class 8 with Answers

Algebra word problems class 8 questions train students to model real situations. The key step is choosing a useful variable.

Q1. A dosa cart has daily rent ₹5000. One dosa costs ₹10 to make. If 100 dosas are sold, what should be the selling price per dosa to make ₹2000 profit?
Total cost:

[5000 + 100(10)]

[= 5000 + 1000 = 6000]

To make ₹2000 profit, total selling amount should be:

[6000 + 2000 = 8000]

Selling price per dosa:

[8000 \div 100 = 80]

So, the selling price should be ₹80 per dosa.

Q2. If customers pay ₹50 per dosa, how many dosas should be sold to make ₹2000 profit?
Let the number of dosas be (x).

Revenue:

[50x]

Cost:

[5000 + 10x]

Profit:

[50x - (5000 + 10x)]

Given profit = ₹2000.

[50x - 5000 - 10x = 2000]

[40x - 5000 = 2000]

[40x = 7000]

[x = 175]

So, the cart should sell 175 dosas.

Q3. Karim goes around a tree three times. Each time his coins double, and he gives 8 coins to the genie. At the end, he has 8 coins. How many coins did he start with?
Work backwards.

Before paying 8 coins after the third round, he had:

[8 + 8 = 16]

Before the third doubling, he had:

[16 \div 2 = 8]

Before paying after the second round, he had:

[8 + 8 = 16]

Before the second doubling, he had:

[16 \div 2 = 8]

Before paying after the first round, he had:

[8 + 8 = 16]

Before the first doubling, he had:

[16 \div 2 = 8]

So, Karim started with 8 coins.

Q4. A person dips flowers into three magical ponds. Each pond doubles the flowers. He places equal flowers at three shrines and ends with none. Find one possible answer.
Let the equal number of flowers placed at each shrine be (x).

Work backwards.

Before placing flowers at shrine 3, he had (x) flowers after doubling.

So, before pond 3, he had:

[\frac{x}{2}]

After shrine 2, he had:

[\frac{x}{2}]

Before placing at shrine 2, after doubling, he had:

[x + \frac{x}{2} = \frac{3x}{2}]

Before pond 2, he had:

[\frac{3x}{4}]

After shrine 1, he had:

[\frac{3x}{4}]

Before placing at shrine 1, after doubling, he had:

[x + \frac{3x}{4} = \frac{7x}{4}]

Before pond 1, he had:

[\frac{7x}{8}]

Choose (x = 8).

Then he started with:

[\frac{7 \times 8}{8} = 7]

So, he can start with 7 flowers and place 8 flowers at each shrine.

Q5. Evaluate the pattern (\frac{1}{3}, \frac{1+3}{5+7}, \frac{1+3+5}{7+9+11}). What do you observe?
First fraction:

[\frac{1}{3}]

Second fraction:

[\frac{1+3}{5+7} = \frac{4}{12} = \frac{1}{3}]

Third fraction:

[\frac{1+3+5}{7+9+11} = \frac{9}{27} = \frac{1}{3}]

The value remains:

[\frac{1}{3}]

This happens because sums of consecutive odd numbers form square patterns.

Class 8 Maths Part 2 Chapter 6 MCQ with Answers

Class 8 Maths Part 2 Chapter 6 MCQ questions test variables, pyramids, grids and digit-value reasoning. These questions are useful for quick revision.

Q1. In algebra, an unknown number can be represented by:
(a) Only 0
(b) A letter
(c) A calendar
(d) A shape only

Answer: (b) A letter
Letters such as (x), (a) and (b) can represent unknown numbers.

Q2. In a number pyramid, each number is:
(a) The product of two numbers below it
(b) The sum of two numbers below it
(c) Always equal to 10
(d) Always a square number

Answer: (b) The sum of two numbers below it
This is the rule used in number pyramids.

Q3. In a 2 × 2 calendar grid, if the top-left number is (a), the bottom-right number is:
(a) (a+1)
(b) (a+6)
(c) (a+7)
(d) (a+8)

Answer: (d) (a+8)
The number below is (a+7), and the next column adds 1.

Q4. The reverse of the two-digit number (10a+b) is:
(a) (10a-b)
(b) (10b+a)
(c) (a+b)
(d) (ab)

Answer: (b) (10b+a)
Reversing swaps the tens and ones digits.

Q5. The sum of a two-digit number and its reverse is always divisible by:
(a) 5
(b) 7
(c) 9
(d) 11

Answer: (d) 11

[(10a+b)+(10b+a)=11(a+b)]

Class 8 Maths Important Questions Chapter-Wise