---
title: "NCERT Solutions for Class 8 Maths Chapter 13 Algebra Play"
url: https://www.swavid.com/maths/class/8/chapter/algebra-play/ncert-solutions
dateModified: 2026-10-07T15:38:49+00:00
---

# NCERT Solutions for Class 8 Maths Chapter 13 Algebra Play

This chapter's questions cover algebraic thinking through number tricks, date puzzles, number pyramids, calendar grids, digit placement for products, and divisibility rules.

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## 6.2 Thinking about 'Think of a Number' Tricks

### Question 1

*3 marks · Short answer*

How would you change this game to make the final answer 3? What about 5?

**Solution**

1. Let the original number be $x$.
2. Following the steps: 1. Think of $x$. 2. Double it: $2x$. 3. Add $4$: $2x + 4$. 4. Divide by $2$: $x + 2$.
3. To get a final answer of $3$, we need $x + 3$ instead of $x + 2$, so we should add $6$ instead of $4$ in step 3 ($2x + 6$ divided by $2$ gives $x + 3$, minus $x$ gives $3$).
4. To get a final answer of $5$, we should add $10$ in step 3 ($2x + 10$ divided by $2$ gives $x + 5$, minus $x$ gives $5$).

**Answer:** To get final answer $3$, add $6$ in step 3. To get $5$, add $10$ in step 3.

> Common mistake: Adding the target number directly instead of doubling the addition amount.

### Question 2

*3 marks · Short answer*

Can you come up with more complicated steps that always lead to the same final value?

**Solution**

1. We can design a trick by taking a starting number $x$.
2. Step 1: Think of $x$. Step 2: Multiply by $3$ ($3x$). Step 3: Add $12$ ($3x + 12$). Step 4: Divide by $3$ ($x + 4$). Step 5: Subtract the original number and add the desired constant, or subtract $x$ and $2$, leaving a constant.
3. Alternatively: Think of $x$, multiply by $4$, add $8$, divide by $2$, subtract $2x$, and the result is always $4$.

**Answer:** Example trick: Think of $x$, multiply by $4$, add $8$, divide by $2$, and subtract $2x$, always resulting in $4$.

> Common mistake: Creating steps where the variable $x$ does not cancel out completely.

### Question 3

*3 marks · Short answer*

How did Shubham figure out the date chosen by Mukta?

**Solution**

1. Let the month be $M$ and the day be $D$.
2. Following the steps algebraically: Multiply $M$ by $5$ ($5M$), add $6$ ($5M + 6$), multiply by $4$ ($20M + 24$), add $9$ ($20M + 33$), multiply by $5$ ($100M + 165$), and add $D$ ($100M + 165 + D$).
3. Subtracting the constant $165$ from the final answer leaves $100M + D$, where the hundreds digits represent the month $M$ and the last two digits represent the day $D$.

**Answer:** By simplifying the algebraic expression to $100M + 165 + D$, subtracting $165$ gives $100M + D$, separating the month and day.

> Common mistake: Forgetting the order of operations while expanding the algebraic expressions.

### Question 4

*3 marks · Short answer*

Mukta thinks of another date, follows the same steps, and reports her answer as 1390. What date did Mukta start with this time?

**Solution**

1. The final expression obtained from the trick is $100M + 165 + D = 1390$.
2. Subtracting $165$ from $1390$ gives $1390 - 165 = 1225$.
3. The expression $1225$ corresponds to $100M + D$, where $M = 12$ (December) and $D = 25$ (25th).

**Answer:** 25th December

> Common mistake: Subtracting an incorrect fixed number instead of 165.

### Question 5

*4 marks · Case-based*

Find the dates if the final answers are the following:
(i) 1269
(ii) 394
(iii) 296

**Part (i)**

1. Subtract 165 from the final answer 1269: $1269 - 165 = 1104$.
2. The result represents $100M + D$, so the month $M = 11$ (November) and the day $D = 4$.

Answer (i): 4th November

**Part (ii)**

1. Subtract 165 from the final answer 394: $394 - 165 = 229$.
2. The result represents $100M + D$, so the month $M = 2$ (February) and the day $D = 29$.

Answer (ii): 29th February

**Part (iii)**

1. Subtract 165 from the final answer 296: $296 - 165 = 131$.
2. The result represents $100M + D$, so the month $M = 1$ (January) and the day $D = 31$.

Answer (iii): 31st January

**Answer:** The dates are (i) 4th November, (ii) 29th February, and (iii) 31st January.

> Common mistake: Confusing the month and day parts by not taking the last two digits as the day and the remaining digits as the month.

### Question 6

*3 marks · Short answer*

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.

**Solution**

1. Yes, we can change the steps. The final constant to subtract depends entirely on the numbers added and multiplied during the steps.
2. For instance, if we change 'Add 6' to 'Add 5', the expression becomes $5(4(5M + 5) + 9) + D$, leading to a different constant to subtract at the end.
3. By working backwards algebraically, we can always find the exact constant to subtract to isolate $100M + D$.

**Answer:** Yes, by reworking the algebraic expression, a new constant corresponding to the new steps can be subtracted.

> Common mistake: Failing to distribute multiplication correctly across all terms when changing steps.

### Question 7

*3 marks · Short answer*

Try to devise your own 'Think of a Number' trick.

**Solution**

1. Think of a number: $x$
2. Multiply it by 3: $3x$
3. Add 6: $3x + 6$
4. Divide by 3: $x + 2$
5. Subtract the original number: $(x + 2) - x = 2$

**Answer:** A trick where you think of a number, multiply by 3, add 6, divide by 3, and subtract the original number, always resulting in 2.

> Common mistake: Making algebraic errors while simplifying the steps.

## 6.3 Number Pyramids

### Question 1

*3 marks · Short answer*

Use the same rule to fill these pyramids:

**Solution**

1. Recall the rule that each number in a number pyramid is the sum of the two numbers directly below it.
2. For the first pyramid with bottom row 6 and 2, the top number is $6 + 2 = 8$.
3. For the second pyramid, bottom row 3, 4, 3 gives middle row $3+4=7$ and $4+3=7$, and top number $7+7=14$.
4. For the third pyramid, bottom row 5, 4, 5, 0 gives middle row $5+4=9$, $4+5=9$, $5+0=5$, and top row $9+9=18$, $9+5=14$, with topmost number $18+14=32$.  Completed pyramids have top values 8, 14, and 32 respectively.

**Answer:** The top numbers of the three pyramids are 8, 14, and 32 respectively.

> Common mistake: Adding the wrong adjacent pairs in multi-row pyramids.

### Question 2

*3 marks · Short answer*

How do we fill this pyramid?

**Solution**

1. State the rule of the number pyramid where each number is the sum of the two numbers directly below it.
2. Use subtraction from top to bottom: for the middle row, the left entry is $10 - 4 = 6$, and the right entry is found by working from the bottom where $4 - 1 = 3$ and then $6 - 3 = 3$.
3. Complete the pyramid by placing $6$ in the middle row and $3, 3$ in the bottom row.

**Answer:** The middle row has $6$, and the bottom row has $3$ and $3$.

> Common mistake: Trying to add numbers upwards instead of using subtraction when working downwards from the top.

### Question 3

*3 marks · Short answer*

What about filling in the numbers in this pyramid? Where do we start?

**Solution**

1. Start by filling the empty boxes in the pyramid with letter-numbers such as $a, b, c$.
2. Form linear equations based on the rule that each number is the sum of the two numbers directly below it.
3. Solve the system of equations for the unknowns to completely fill the pyramid.

**Answer:** Start by assigning letter-numbers to the empty boxes and use the adjacent sum rule to set up and solve linear equations.

> Common mistake: Guessing numbers randomly without setting up algebraic equations.

### Question 4

*3 marks · Short answer*

Fill the following pyramids:

**Solution**

1. For the first pyramid with top 50, middle 22, bottom 4 and 6, let the missing bottom middle cell be $c$.
2. Using the rule, we set up equations and solve to find the missing cells: bottom middle is 18, middle left is 22, middle right is 28.
3. For the second and third pyramids, apply the same downward subtraction and upward addition rules systematically to fill all empty boxes.

**Answer:** The completed pyramids are filled by solving for the unknown boxes using adjacent sums and differences.

> Common mistake: Arithmethic errors in subtraction steps.

### Question 5

*3 marks · Short answer*

What is the relationship between the numbers in the bottom row and the number at the top?

**Solution**

1. Consider a simplest pyramid with bottom row $a, b$ and top $a + b$.
2. For a three-row pyramid with bottom row $a, b, c$, the top row expression is $a + 2b + c$.
3. The number at the top is a linear combination of the bottom row elements with coefficients forming rows of Pascal's triangle.

**Answer:** The number at the top is a linear sum of the bottom row elements weighted by coefficients corresponding to rows of Pascal's triangle.

> Common mistake: Forgetting to apply the correct weighting coefficients for larger pyramids.

### Question 6

*3 marks · Short answer*

What about a pyramid with three rows?

**Solution**

1. Let the bottom row of a three-row pyramid be $a, b,$ and $c$.
2. The middle row consists of $a + b$ and $b + c$.
3. The top row is the sum of the middle row entries: $(a + b) + (b + c) = a + 2b + c$.

**Answer:** The expression for the topmost row of a 3-row pyramid is $a + 2b + c$.

> Common mistake: Incorrectly combining like terms when adding the middle row expressions.

## Figure it Out

### Question 1

*3 marks · Short answer*

Without building the entire pyramid, find the number in the topmost row given the bottom row in each of these cases.

**Part (i)**

1. For a 3-row pyramid with bottom row $a, b, c$, the top number is given by the expression $a + 2b + c$.
2. Substitute the bottom row $(4, 13, 8)$ into the expression: $4 + 2(13) + 8$.
3. Calculate the sum: $4 + 26 + 8 = 38$.

Answer (i): 38

**Part (ii)**

1. For a 3-row pyramid with bottom row $a, b, c$, the top number is given by the expression $a + 2b + c$.
2. Substitute the bottom row $(7, 11, 3)$ into the expression: $7 + 2(11) + 3$.
3. Calculate the sum: $7 + 22 + 3 = 32$.

Answer (ii): 32

**Part (iii)**

1. For a 3-row pyramid with bottom row $a, b, c$, the top number is given by the expression $a + 2b + c$.
2. Substitute the bottom row $(10, 14, 25)$ into the expression: $10 + 2(14) + 25$.
3. Calculate the sum: $10 + 28 + 25 = 63$.

Answer (iii): 63

**Answer:** The topmost numbers are 38, 32, and 63.

> Common mistake: Adding the numbers incorrectly or misapplying the coefficients $(1, 2, 1)$ for the three rows.

### Question 2

*3 marks · Short answer*

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

**Solution**

1. Let the bottom row of 4 numbers be $a, b, c, d$.
2. The second row from the bottom is formed by the sums of adjacent pairs: $a+b$, $b+c$, and $c+d$.
3. The third row is formed by the sums of adjacent pairs from the second row: $(a+b) + (b+c) = a + 2b + c$ and $(b+c) + (c+d) = b + 2c + d$.
4. The topmost row (fourth row) is the sum of the two numbers in the third row: $(a + 2b + c) + (b + 2c + d) = a + 3b + 3c + d$.

**Answer:** $a + 3b + 3c + d$

> Common mistake: Incorrectly applying Pascal's triangle coefficients for the 4th row expression.

### Question 3

*3 marks · Short answer*

Without building the entire pyramid, find the number in the topmost row given the bottom row in each of these cases.

**Part (i)**

1. For a 4-row pyramid with bottom row $a, b, c, d$, the topmost number is given by the formula $a + 3b + 3c + d$.
2. Substitute the values from the first case: $8, 19, 21, 13$.
3. Calculate the sum: $8 + 3(19) + 3(21) + 13 = 8 + 57 + 63 + 13 = 141$.

Answer (i): 141

**Part (ii)**

1. Use the formula for the topmost number of a 4-row pyramid: $a + 3b + 3c + d$.
2. Substitute the values from the second case: $7, 18, 19, 6$.
3. Calculate the sum: $7 + 3(18) + 3(19) + 6 = 7 + 54 + 57 + 6 = 124$.

Answer (ii): 124

**Part (iii)**

1. Use the formula for the topmost number of a 4-row pyramid: $a + 3b + 3c + d$.
2. Substitute the values from the third case: $9, 7, 5, 11$.
3. Calculate the sum: $9 + 3(7) + 3(5) + 11 = 9 + 21 + 15 + 11 = 56$.

Answer (iii): 56

**Answer:** The numbers in the topmost row for the three cases are (i) 141, (ii) 124, and (iii) 56.

> Common mistake: Making arithmetic errors while multiplying the middle coefficients by 3.

### Question 4

*3 marks · Short answer*

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**

1. The first three Virahāṅka-Fibonacci numbers are $1, 2, 3$.
2. Write them in the bottom row: $1, 2, 3$. The second row is $1+2=3$ and $2+3=5$.
3. The top number is $3+5=8$.
4. The numbers appearing in the grid are $1, 2, 3, 3, 5, 8$. All of these are Virahāṅka-Fibonacci numbers.

**Answer:** The grid contains 1, 2, 3, 3, 5, 8. The top number is 8. All numbers in the grid are Virahāṅka-Fibonacci numbers.

> Common mistake: Stating that not all numbers in the grid are Fibonacci numbers.

### Question 5

*3 marks · Short answer*

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.

**Part (i)**

1. The first four Virahāṅka-Fibonacci numbers are $1, 2, 3, 5$.
2. Building the 4-row pyramid, the second row is $3, 5, 8$, the third row is $8, 13$, and the top number is $21$.
3. All numbers appearing in the pyramid are Virahāṅka-Fibonacci numbers.

Answer (i): All numbers in the pyramid, including the top number 21, are Virahāṅka-Fibonacci numbers.

**Part (ii)**

1. When the first $n$ Virahāṅka-Fibonacci numbers are placed in the bottom row of an $n$-row pyramid, every entry generated above is also a Virahāṅka-Fibonacci number.
2. Therefore, all numbers in the pyramid and the top number are Virahāṅka-Fibonacci numbers.

Answer (ii): All numbers in the pyramid and the top number are Virahāṅka-Fibonacci numbers.

**Answer:** All numbers in the pyramid and the top number are Virahāṅka-Fibonacci numbers in both cases.

> Common mistake: Assuming numbers outside the standard sequence will be generated.

### Question 6

*3 marks · Short answer*

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

**Solution**

1. If the bottom row of an $n$-row pyramid contains the first $n$ Virahāṅka-Fibonacci numbers, the addition rule of the pyramid matches the recurrence relation of the Fibonacci sequence.
2. Consequently, every number generated in the intermediate rows of the pyramid belongs to the Virahāṅka-Fibonacci sequence.
3. The number at the top of the pyramid is also a Virahāṅka-Fibonacci number.

**Answer:** All numbers in the pyramid, including the number at the top, are Virahāṅka-Fibonacci numbers.

> Common mistake: Failing to connect the pyramid addition rule with the Fibonacci recurrence relation.

## 6.4 Fun with Grids

### Question 1

*3 marks · Short answer*

Can we find the 4 numbers in the grid from just knowing this sum?

**Solution**

1. Let the top-left number of a $2 \times 2$ grid on a calendar be $a$.
2. The other three numbers in the grid are $a + 1$, $a + 7$, and $a + 8$.
3. The sum of these four numbers is $a + (a + 1) + (a + 7) + (a + 8) = 4a + 16$, so we can uniquely determine the top-left number $a$ and the whole grid if we know the sum.

**Answer:** Yes, by letting the top-left number be $a$, the sum of the $2 \times 2$ grid is $4a + 16$, allowing us to uniquely find $a$ and all four numbers.

> Common mistake: Forgetting the relationship of calendar grid numbers where moving down adds 7 and moving right adds 1.

### Question 2

*3 marks · Short answer*

Suppose you are told that the sum is 36. Can you find the 4 numbers in the grid?

**Solution**

1. We are given that the sum of the $2 \times 2$ grid is $36$.
2. Using algebra, the sum is $4a + 16 = 36$, which gives $4a = 20$ after subtracting 16 from both sides.
3. Dividing both sides by 4, we get $a = 5$, so the four numbers in the grid are $5, 6, 12, \text{ and } 13$.

**Answer:** The 4 numbers in the grid are 5, 6, 12, and 13.

> Common mistake: Errors in arithmetic while subtracting or dividing the sum equation.

### Question 3

*3 marks · Short answer*

Create your own calendar trick. For instance, choose a grid of a different size and shape.

**Solution**

1. Choose a $3 \times 3$ square grid on the calendar with top-left number $a$.
2. The 9 numbers in the grid are $a, a+1, a+2, a+7, a+8, a+9, a+14, a+15, \text{ and } a+16$.
3. The sum of all 9 numbers is $9a + 72$, which can be used to perform a calendar number trick.

**Answer:** For a $3 \times 3$ grid with top-left number $a$, the sum of the 9 numbers is $9a + 72$.

> Common mistake: Incorrectly adding the offsets for a different grid size.

### Question 4

*3 marks · Short answer*

In the following grids, find the values of the shapes and fill in the empty squares:

**Solution**

1. In each row, the last column is the sum of the values to its left, allowing us to form linear equations for each shape.
2. Solve the simultaneous linear equations for the shapes step by step.
3. Substitute the values of the shapes back into the empty squares to complete the grid.

**Answer:** The shape values and empty squares are found by equating each row's sum to the last column and solving the resulting equations.

> Common mistake: Misinterpreting the sum rule where the last column represents the sum of the preceding entries in that row.

## 6.5 The Largest Product

### Question 1

*3 marks · Short answer*

Fill the digits 2, 3, and 5 in $\square\square \times \square$, using each digit once. What is the largest product possible?

**Solution**

1. Let the three digits be 2, 3, and 5.
2. To get the largest product, we place the largest digit as the multiplier and arrange the other two digits in decreasing order to form the multiplicand.
3. Thus, the largest product is formed by the expression $52 \times 3 = 156$.

**Answer:** $156$

> Common mistake: Multiplying the largest digits together or placing the smallest digit as the multiplier.

### Question 2

*3 marks · Short answer*

How do we find the largest product among these six options?

**Solution**

1. We can group the six possible products into pairs where the multiplier is the same.
2. In each pair, the combination with the larger multiplicand gives the larger product.
3. By expanding and comparing the remaining expressions, we find the one that yields the maximum value.

**Answer:** Group the options by the multiplier, compare the multiplicands within each pair, and expand the remaining expressions to find the maximum.

> Common mistake: Randomly guessing combinations instead of grouping by multipliers and comparing systematically.

## Figure it Out

### Question 1

*3 marks · Short answer*

Fill the digits 1, 3, and 7 in $\square\square \times \square$ to make the largest product possible.

**Solution**

1. Let the three given digits be $1, 3,$ and $7$, where $1 < 3 < 7$.
2. To make the largest product in the form $\square\square \times \square$, the largest digit should be the multiplier and the other two digits should be arranged in decreasing order to form the multiplicand.
3. Thus, the largest product is formed by $71 \times 3 = 213$.

**Answer:** $71 \times 3 = 213$

> Common mistake: Placing the largest digit in the multiplicand instead of the multiplier.

### Question 2

*3 marks · Short answer*

Fill the digits 3, 5, and 9 in $\square\square \times \square$ to make the largest product possible.

**Solution**

1. Let the three given digits be $3, 5,$ and $9$, where $3 < 5 < 9$.
2. To make the largest product in the form $\square\square \times \square$, the largest digit should be the multiplier and the other two digits should be arranged in decreasing order to form the multiplicand.
3. Thus, the largest product is formed by $53 \times 9 = 477$.

**Answer:** $53 \times 9 = 477$

> Common mistake: Placing the digits in ascending order for the multiplicand.

## 6.6 Decoding Divisibility Tricks

### Question 1

*3 marks · Short answer*

If we choose other 2-digit numbers, and follow the steps, will there always be no remainder?

**Solution**

1. Let the two-digit number be $ab$, represented algebraically as $10a + b$, where $a$ is the tens digit and $b$ is the units digit.
2. When the digits are reversed, the new number is $ba$, represented as $10b + a$.
3. If $b > a$, the difference is $(10b + a) - (10a + b) = 9b - 9a = 9(b - a)$, which is always a multiple of $9$ and leaves no remainder when divided by $9$.

**Answer:** Yes, there will always be no remainder because the difference is a multiple of 9.

> Common mistake: Confusing the place value representation, writing $ab$ as $a \times b$ instead of $10a + b$.

### Question 2

*3 marks · Short answer*

Can you work out what happens if $a > b$?

**Solution**

1. Let the two-digit number be $ab$, represented as $10a + b$, and the reversed number be $ba$, represented as $10b + a$.
2. If $a > b$, then $ab > ba$, so we find the difference as $(10a + b) - (10b + a)$.
3. Simplifying the expression gives $10a - a + b - 10b = 9a - 9b = 9(a - b)$, which is again a multiple of $9$ and leaves no remainder when divided by $9$.

**Answer:** The difference is $9(a - b)$, which is still divisible by 9 with no remainder.

> Common mistake: Subtracting the smaller number from the larger incorrectly without adjusting the order of place values.

## Figure it Out

### Question 1

*3 marks · Short answer*

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**

1. Let the two-digit number be $\overline{ab} = 10a + b$ and the reversed number be $\overline{ba} = 10b + a$ with $b > a$.
2. The difference between the two numbers is given by $(10b + a) - (10a + b) = 9b - 9a = 9(b - a)$.
3. Dividing the difference by $9$ gives the quotient $b - a$, which is simply the difference between the digits of the original number.

**Answer:** The quotient is the difference between the tens and units digits of the original number.

> Common mistake: Confusing the quotient with the digits themselves instead of their difference.

### Question 2

*3 marks · Short answer*

In the trick given above, instead of finding the difference of the two 2-digit numbers, find their sum. What will happen?

**Solution**

1. Let the two-digit number be $\overline{ab} = 10a + b$ and its reverse be $\overline{ba} = 10b + a$.
2. Adding the two numbers gives $(10a + b) + (10b + a) = 11a + 11b = 11(a + b)$.
3. Thus, the sum of a two-digit number and its reverse is always a multiple of $11$, and the quotient when divided by $11$ is the sum of the digits $a + b$.

**Answer:** The sum is always divisible by 11, and the quotient is the sum of the digits of the original number.

> Common mistake: Subtracting the digits instead of adding them when finding the sum.

### Question 3

*3 marks · Short answer*

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**

1. Consider the three-digit number $abc = 100a + 10b + c$, along with its cyclic permutations $bca = 100b + 10c + a$ and $cab = 100c + 10a + b$.
2. Adding the three numbers gives $(100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b) = 111a + 111b + 111c$.
3. Factoring out $111$ gives $111(a + b + c) = 37 \times 3(a + b + c)$, which shows the sum is always divisible by $37$ and also by $3$ since $3$ is a factor.

**Answer:** The sum is $111(a + b + c) = 37 \times 3(a + b + c)$, so it is always divisible by 37 and also by 3.

> Common mistake: Incorrectly expanding the cyclic permutations of the three-digit number.

### Question 4

*3 marks · Short answer*

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**

1. Represent the six-digit number formed by repeating $abc$, which is $\overline{abcabc}$, as $100000a + 10000b + 1000c + 100a + 10b + c$.
2. Factor out $1001$ to get $1001(100a + 10b + c) = 1001 \times abc$.
3. Since $1001 = 7 \times 11 \times 13$, dividing the number successively by $7$, $11$, and $13$ leaves the original three-digit number $abc$.

**Answer:** Dividing by 7, 11, and 13 successively results in the original 3-digit number abc because $1001 = 7 \times 11 \times 13$.

> Common mistake: Failing to recognize that repeating a three-digit number to form a six-digit number multiplies it by 1001.

### Question 5

*3 marks · Short answer*

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 has some flowers. He dips them all 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**

1. Let the initial number of flowers be $x$. After the first pond, there are $2x$ flowers, and he places $y$ flowers in shrine 1, leaving $2x - y$ flowers.
2. After the second pond, there are $2(2x - y) = 4x - 2y$ flowers, and he places $y$ flowers in shrine 2, leaving $4x - 3y$ flowers.
3. After the third pond, there are $2(4x - 3y) = 8x - 6y$ flowers, and he places all of them in shrine 3, which equals $y$ flowers since an equal number $y$ is placed in each shrine.
4. Equating $8x - 6y = y$ gives $8x = 7y$, and choosing the smallest positive integer values gives $x = 7$ and $y = 8$, meaning he started with 7 flowers and placed 8 flowers in each shrine.

**Answer:** He started with 7 flowers and placed 8 flowers in each shrine.

> Common mistake: Assuming the number of flowers placed in each shrine is equal to the initial number.

### Question 6

*3 marks · Short answer*

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?

**Solution**

1. Let the number of horses be $h$ and the number of hens be $n$. From the number of heads, we have $h + n = 55$.
2. From the number of legs, since horses have 4 legs and hens have 2 legs, we have $4h + 2n = 150$.
3. Substitute $n = 55 - h$ into the leg equation: $4h + 2(55 - h) = 150$, which simplifies to $4h + 110 - 2h = 150$.
4. Solving for $h$ gives $2h = 40$, so $h = 20$ horses and $n = 55 - 20 = 35$ hens.

**Answer:** There are 20 horses and 35 hens on the farm.

> Common mistake: Assigning 4 legs to hens and 2 legs to horses by mistake.

### Question 7

*3 marks · Short answer*

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**

1. Let the daughter's present age be $x$ years. Then the mother's present age is $5x$ years.
2. In 6 years' time, the daughter's age will be $x + 6$ years and the mother's age will be $5x + 6$ years.
3. According to the question, the mother will be 3 times her daughter's age: $5x + 6 = 3(x + 6)$.
4. Solving the equation: $5x + 6 = 3x + 18 \Rightarrow 2x = 12 \Rightarrow x = 6$.
5. Therefore, the daughter is 6 years old now.

**Answer:** 6 years

> Common mistake: Adding 6 to only one of the ages instead of both.

### Question 8

*3 marks · Short answer*

Two friends, Gauri and Naina, are cowherds. One day, they pass each other on the road with their cows. 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**

1. Let the number of cows Gauri has be $x$. Then Naina has $2x$ cows.
2. If Naina gives 3 cows to Gauri, Gauri will have $x + 3$ cows and Naina will have $2x - 3$ cows.
3. According to the question, they would each have the same number of cows: $x + 3 = 2x - 3$.
4. Solving the equation: $2x - x = 3 + 3 \Rightarrow x = 6$.
5. Gauri has 6 cows and Naina has $2 \times 6 = 12$ cows.

**Answer:** Gauri has 6 cows and Naina has 12 cows.

> Common mistake: Forgetting to subtract 3 from Naina's cows when adding 3 to Gauri's cows.

### Question 9

*4 marks · Case-based*

I run a small dosa cart and my expenses are as follows:
• Rent for the dosa cart is `5000 per day.
• The cost of making one dosa (including all the ingredients and fuel) 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?

**Part (i)**

1. Let the selling price of one dosa be $S$.
2. Total cost for 100 dosas = Rent + Cost of making 100 dosas = $5000 + (100 \times 10) = 5000 + 1000 = `6000$.
3. Total revenue from selling 100 dosas = $100S$.
4. Profit = Revenue - Total Cost = $100S - 6000 = 2000 \Rightarrow 100S = 8000 \Rightarrow S = 80$.
5. Therefore, the selling price of one dosa should be `80.

Answer (i): `80

**Part (ii)**

1. Let the number of dosas sold be $n$.
2. Selling price of one dosa = `50, so total revenue = $50n$.
3. Total cost for $n$ dosas = $5000 + 10n$.
4. Profit = $50n - (5000 + 10n) = 2000 \Rightarrow 40n - 5000 = 2000 \Rightarrow 40n = 7000 \Rightarrow n = 175$.
5. Therefore, the number of dosas to sell is 175.

Answer (ii): 175 dosas

**Answer:** Selling price is `70 and number of dosas is 140.

> Common mistake: Forgetting to include the fixed daily rent in the total expenses.

### Question 10

*3 marks · Short answer*

Evaluate the following sequence of fractions:
$\frac{1}{3}$, $\frac{1 + 3}{5 + 7}$, $\frac{1 + 3 + 5}{7 + 9 + 11}$
What do you observe? Can you explain why this happens?

**Solution**

1. Evaluate the given fractions: $\frac{1}{3} = \frac{1}{3}$, $\frac{1 + 3}{5 + 7} = \frac{4}{12} = \frac{1}{3}$, $\frac{1 + 3 + 5}{7 + 9 + 11} = \frac{9}{27} = \frac{1}{3}$.
2. Observation: Every fraction in the sequence simplifies to $\frac{1}{3}$.
3. Explanation: The sum of the first $n$ odd numbers is $n^2$. The numerator has $n$ odd numbers starting from 1, giving $n^2$. The denominator has $n$ odd numbers starting from $2n - 1$, and its sum can be shown to equal $3n^2$, making the ratio always $\frac{n^2}{3n^2} = \frac{1}{3}$.

**Answer:** Each fraction evaluates to $\frac{1}{3}$ because the ratio of the sum of the numerator's odd numbers to the denominator's odd numbers is always $\frac{1}{3}$.

> Common mistake: Not simplifying the fractions completely to notice the constant value.

### Question 11

*4 marks · Case-based*

Karim and the Genie
(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?

**Part (i)**

1. Let the initial number of coins Karim had be $x$.
2. After round 1, he has $2x - 8$ coins.
3. After round 2, he has $2(2x - 8) - 8 = 4x - 24$ coins.
4. After round 3, he has $2(4x - 24) - 8 = 8x - 56$ coins.
5. Given that he is left with 8 coins, we have $8x - 56 = 8$, which gives $8x = 64$, so $x = 7$.

Answer (i): Karim initially had 7 coins.

**Part (ii)**

1. Let the number of coins before a round be $C$ and the cost per round be $K$.
2. After doubling and paying the cost, the remaining coins are $2C - K$.
3. To increase the number of coins, we must have $2C - K > C$, which implies $C > K$.
4. Therefore, Karim should agree to a deal where the cost per round is less than half of the coins he possesses at the start of that round.

Answer (ii): The cost per round should be less than half of his current coins.

**Part (iii)**

1. For Karim to be left with nothing or just enough to pay the exact cost, the number of coins after the round must equal the cost, or he must end up with zero gain.
2. To get all of Karim's coins such that he has no coins left after paying the genie, the cost per round must equal the number of coins he has after doubling.
3. Since the coins double to $2C$, the cost set by the genie must be exactly half of the doubled amount, which is $\frac{2C}{2} = C$, meaning the genie must set the cost to half the number of coins Karim has before doubling (or equal to his coins before the round).

Answer (iii): The genie should set the cost per round equal to the number of coins Karim has before doubling (or half the coins after doubling).

**Answer:** Karim initially had 7 coins; the cost per round should be less than half of his coins; the genie should set the cost per round to exactly half the number of coins Karim has.

> Common mistake: Setting up the reverse equations incorrectly without accounting for the order of doubling and subtracting the cost.

## Frequently asked questions

### How many questions are there in NCERT Solutions for Class 8 Maths Chapter 13 Algebra Play?

This chapter contains a total of 40 questions spread across various sections like Thinking about Think of a Number Tricks, Number Pyramids, Fun with Grids, and Decoding Divisibility Tricks. You can find step-by-step solutions for all these questions in the free PDF available on this page.

### Which topics are covered in Class 8 Maths Chapter 13 Algebra Play?

The chapter covers fascinating algebraic puzzles including calendar and date tricks, number pyramids, algebra grids with shapes, maximizing products using given digits, and decoding divisibility tricks. These concepts help students understand how algebra is used in everyday mathematical patterns and number games.

### Which is the hardest question type in this chapter and how should we approach it?

The Figure it Out sections featuring algebraic modelling of puzzles, simultaneous linear equations for animals, and multi-row Virahāṅka-Fibonacci number pyramids are generally considered challenging. To approach them, break down the given word problems into systematic linear equations or identify underlying number patterns before solving.

### How do I write answers to get full marks in Class 8 Maths Chapter 13?

To secure full marks, you must clearly define your variables, write out every algebraic expression step by step, and show the logical progression for tricks involving concepts like the difference of a 2-digit number and its reverse. Referring to the detailed solutions in our free PDF on this page will help you structure your answers correctly.

### Is the free PDF for Class 8 Maths Chapter 13 Algebra Play available?

Yes, the complete free PDF containing accurate solutions for the new NCERT book sessions is available right here on this page. These solutions are tailored according to the latest syllabus guidelines to help students prepare effectively for their exams.

## Related pages

- [Class 8 Maths chapters](https://www.swavid.com/maths/class/8)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student