---
title: "NCERT Solutions for Class 7 Maths Chapter 6 Number Play (2026-27)"
url: https://www.swavid.com/maths/class/7/chapter/number-play/ncert-solutions
dateModified: 2026-10-07T15:05:27+00:00
---

# NCERT Solutions for Class 7 Maths Chapter 6 Number Play (2026-27)

This chapter covers questions related to number patterns, parity, magic squares, and cryptarithms. Students explore relationships between numbers, algebraic expressions, and historical sequences like Virahāṅka–Fibonacci numbers.

Free PDF (33 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-7/swavid-ncert-solutions-class-7-maths-chapter-6-number-play-20c5b7ec35.pdf

## 6.1 Numbers Tell us Things

### Question 1

*2 marks · Very short answer*

What do the numbers in the figure below tell us?

**Solution**

1. Each number in the speech bubble represents the number of children standing in front of that child who are taller than them.

**Answer:** The numbers tell us the count of taller children standing in front of each child.

> Common mistake: Confusing the numbers with the actual height measurements or positions in line.

### Question 2

*2 marks · Very short answer*

What do you think these numbers mean?

**Solution**

1. When children rearrange themselves, the number each child says changes based on their new position relative to taller children in front of them.

**Answer:** The numbers mean the count of children taller than them in their current arrangement.

> Common mistake: Assuming the numbers remain the same regardless of rearrangement.

### Question 3

*2 marks · Very short answer*

Could you figure out what these numbers convey? Observe and try to find out.

**Solution**

1. The numbers convey information about the relative heights of the children without needing their actual height values.

**Answer:** They convey the relative height order of the children in the line.

> Common mistake: Thinking the numbers represent the ages or roll numbers of the children.

### Question 4

*2 marks · Very short answer*

Check if the number each child says matches this rule in both the arrangements.

**Solution**

1. Verify each child's count against the rule by counting how many children taller than them stand ahead in both arrangements.

**Answer:** Yes, the numbers match the rule that each child calls out the number of children in front of them who are taller than them.

> Common mistake: Counting children behind instead of in front.

### Question 5

*2 marks · Very short answer*

Write down the number each child should say based on this rule for the arrangement shown below.

**Solution**

1. Examine the arrangement from left to right and apply the rule for each child.
2. The numbers from left to right are 0, 0, 1, 0, 3, 0, 3.

**Answer:** The numbers each child should say are 0, 0, 1, 0, 3, 0, 3.

> Common mistake: Miscounting the number of taller children standing ahead of a particular child.

## Figure it Out

### Question 1

*3 marks · Short answer*

Arrange the stick figure cutouts given at the end of the book or draw a height arrangement such that the sequence reads:
(a) $0, 1, 1, 2, 4, 1, 5$
(b) $0, 0, 0, 0, 0, 0, 0$
(c) $0, 1, 2, 3, 4, 5, 6$
(d) $0, 1, 0, 1, 0, 1, 0$
(e) $0, 1, 1, 1, 1, 1, 1$
(f) $0, 0, 0, 3, 3, 3, 3$

**Part (a)**

1. The sequence reads 0, 1, 1, 2, 4, 1, 5.
2. Arranging the stick figure cutouts in height order from left to right gives F, C, B, G, A, D, E.

Answer (a): F, C, B, G, A, D, E

**Part (b)**

1. The sequence reads 0, 0, 0, 0, 0, 0, 0.
2. This means every person has zero taller people in front, so heights must be in descending order from left to right.

Answer (b): Heights in descending order from left to right

**Part (c)**

1. The sequence reads 0, 1, 2, 3, 4, 5, 6.
2. This means heights must be in ascending order from left to right.

Answer (c): Heights in ascending order from left to right

**Part (d)**

1. The sequence reads 0, 1, 0, 1, 0, 1, 0.
2. Heights alternate in pairs or alternating step pattern.

Answer (d): Alternating height arrangement

**Part (e)**

1. The sequence reads 0, 1, 1, 1, 1, 1, 1.
2. The first person is shortest and all others are taller than the first person.

Answer (e): Shortest person at the front with all others taller

**Part (f)**

1. The sequence reads 0, 0, 0, 3, 3, 3, 3.
2. Arrangement has three shorter/equal persons followed by four taller persons.

Answer (f): Arrangement corresponding to sequence 0, 0, 0, 3, 3, 3, 3

**Answer:** Arrangements for each sequence given in parts

> Common mistake: Miscounting the number of taller people standing in front of a child.

### Question 2

*3 marks · Short answer*

For each of the statements given below, think and identify if it is Always True, Only Sometimes True, or Never True. Share your reasoning.
(a) If a person says '0', then they are the tallest in the group.
(b) If a person is the tallest, then their number is '0'.
(c) The first person's number is '0'.
(d) If a person is not first or last in line (i.e., if they are standing somewhere in between), then they cannot say '0'.
(e) The person who calls out the largest number is the shortest.
(f) What is the largest number possible in a group of 8 people?

**Part (a)**

1. A person saying '0' has no taller person in front of them.
2. They could be the tallest, or simply taller than everyone ahead of them in line.

Answer (a): Only sometimes true

**Part (b)**

1. The tallest person in the group has no one taller than them anywhere in front.
2. Therefore, the tallest person always says '0'.

Answer (b): Always true

**Part (c)**

1. The first person in line has no one standing in front of them.
2. Hence, the number of people taller than them in front is always 0.

Answer (c): Always true

**Part (d)**

1. A person in between can say '0' if everyone standing in front of them is shorter than them.

Answer (d): Only sometimes true

**Part (e)**

1. The person calling out the largest number has many taller people in front.
2. They are generally near the front, but not necessarily the shortest.

Answer (e): Only sometimes true

**Part (f)**

1. In a group of 8 people, the maximum number of people taller than someone standing at the front is 7.

Answer (f): 7

**Answer:** Reasoning and classifications for statements (a) to (f)

> Common mistake: Confusing 'tallest person' with a person who says '0'.

## 6.2 Picking Parity

### Question 1

*2 marks · Very short answer*

Can you help him find a way to do it?
$\_\_\_ + \_\_\_ + \_\_\_ + \_\_\_ + \_\_\_ = 30$

**Solution**

1. Kishor has 5 empty boxes, which means an odd number of boxes.
2. All the available number cards contain odd numbers, and their sum must equal 30, which is an even number.
3. Since adding any 5 odd numbers always results in an odd number, it is not possible to fill the boxes to add up to 30.

**Answer:** It is not possible to arrange these cards in the boxes to add up to 30.

> Common mistake: Trying out different combinations of cards without realizing the parity rule.

### Question 2

*2 marks · Very short answer*

Can you figure out which 5 cards add to 30? Is it possible?

**Solution**

1. All the number cards in Kishor's collection are odd numbers.
2. Choosing 5 cards means we are adding 5 odd numbers together.
3. The sum of an odd number of odd numbers is always odd, whereas 30 is an even number, so it is impossible.

**Answer:** No, it is not possible because the sum of 5 odd numbers is always odd.

> Common mistake: Assuming we can find a combination by trial and error without checking the sum's parity.

### Question 3

*2 marks · Very short answer*

Add a few even numbers together. What kind of number do you get? Does it matter how many numbers are added?

**Solution**

1. Any even number can be arranged in pairs without any leftovers.
2. When adding any number of even numbers together, the result can still be arranged in pairs without any leftovers.
3. Therefore, the sum will always be an even number, and it does not matter how many even numbers are added.

**Answer:** The sum is always an even number, and the number of terms added does not matter.

> Common mistake: Thinking that adding many even numbers can result in an odd number.

### Question 4

*2 marks · Very short answer*

Now, add a few odd numbers together. What kind of number do you get? Does it matter how many odd numbers are added?

**Solution**

1. An odd number cannot be arranged in pairs and is one more than a collection of pairs.
2. The parity of the sum depends on how many odd numbers are added together.
3. The sum of an even number of odd numbers is even, while the sum of an odd number of odd numbers is odd.

**Answer:** The sum is even if an even count of odd numbers is added, and odd if an odd count of odd numbers is added.

> Common mistake: Believing that the sum of odd numbers is always odd regardless of how many terms are added.

### Question 5

*2 marks · Very short answer*

Can we also think of an odd number as one less than a collection of pairs?

**Solution**

1. Yes, an odd number can be thought of as one less than a collection of pairs.
2. It is represented as a collection of pairs with one extra dot left over.

**Answer:** Yes, an odd number is one less than a collection of complete pairs.

> Common mistake: Stating that odd numbers cannot be related to pairs at all.

### Question 6

*2 marks · Very short answer*

What about adding 3 odd numbers? Can the resulting sum be arranged in pairs? No.

**Solution**

1. Adding two odd numbers gives an even number.
2. Adding a third odd number to that even number results in an odd number.
3. Therefore, the resulting sum cannot be arranged in pairs and is always odd.

**Answer:** No, the resulting sum cannot be arranged in pairs because the sum of 3 odd numbers is always odd.

> Common mistake: Assuming three odd numbers can add up to an even number.

### Question 7

*3 marks · Short answer*

Explore what happens to the sum of (a) 4 odd numbers, (b) 5 odd numbers, and (c) 6 odd numbers.

**Part (a)**

1. Pair up the four odd numbers into two groups of two.
2. Since the sum of two odd numbers is always even, we have even + even.
3. Thus, the sum of 4 odd numbers is an even number.

Answer (a): Even number

**Part (b)**

1. Consider 5 odd numbers as the sum of 4 odd numbers and 1 odd number.
2. We know that 4 odd numbers add up to an even number.
3. Adding one more odd number gives even + odd, which is always odd.

Answer (b): Odd number

**Part (c)**

1. Consider 6 odd numbers as the sum of 5 odd numbers and 1 odd number.
2. The sum of 5 odd numbers is an odd number.
3. Adding one more odd number gives odd + odd, which is always even.

Answer (c): Even number

**Answer:** The sum of 4 odd numbers is even, 5 odd numbers is odd, and 6 odd numbers is even.

> Common mistake: Confusing the parity pattern by not grouping pairs of odd numbers.

### Question 8

*3 marks · Short answer*

Two siblings, Martin and Maria, were born exactly one year apart. Today they are celebrating their birthday. Maria exclaims that the sum of their ages is 112. Is this possible? Why or why not?

**Solution**

1. Martin and Maria were born exactly one year apart, so their ages are two consecutive numbers.
2. In any two consecutive numbers, one is always even and the other is always odd.
3. The sum of an even number and an odd number is always an odd number.
4. Since 112 is an even number, the sum of their ages cannot be 112.

**Answer:** No, it is not possible because the sum of two consecutive ages must always be an odd number, whereas 112 is even.

> Common mistake: Assuming that any two numbers can add up to an even total without checking their parity.

## Figure it Out

### Question 1

*3 marks · Short answer*

Using your understanding of the pictorial representation of odd and even numbers, find out the parity of the following sums:
(a) Sum of 2 even numbers and 2 odd numbers (e.g., even + even + odd + odd)
(b) Sum of 2 odd numbers and 3 even numbers
(c) Sum of 5 even numbers
(d) Sum of 8 odd numbers

**Part (a)**

1. Sum of 2 even numbers is even, and sum of 2 odd numbers is even.
2. Adding an even number and an even number gives an even number.

Answer (a): Even number

**Part (b)**

1. Sum of 2 odd numbers is even.
2. Sum of 3 even numbers is even.
3. Adding an even number and an even number gives an even number.

Answer (b): Even number

**Part (c)**

1. Any number of even numbers added together always results in an even number.

Answer (c): Even number

**Part (d)**

1. An even amount of odd numbers (8 odd numbers) grouped in pairs will all have even sums.

Answer (d): Even number

**Answer:** The parities are (a) Even number (b) Even number (c) Even number (d) Even number

> Common mistake: Confusing how many odd numbers change the parity of a sum.

### Question 2

*3 marks · Short answer*

Lakpa has an odd number of ₹1 coins, an odd number of ₹5 coins and an even number of ₹10 coins in his piggy bank. He calculated the total and got ₹205. Did he make a mistake? If he did, explain why. If he didn't, how many coins of each type could he have?

**Solution**

1. Lakpa has an odd number of ₹1 coins and an odd number of ₹5 coins, whose sum is odd + odd = even.
2. He has an even number of ₹10 coins, which contributes an even amount to the total.
3. The total sum must be even + even = even, but ₹205 is an odd number, so he made a mistake.

**Answer:** Lakpa must have made a mistake because the total amount with the given coin counts can only be an even number, not ₹205.

> Common mistake: Ignoring the parity of the coin counts and trying to find the exact number of coins directly.

### Question 3

*3 marks · Short answer*

Similarly, find out the parity for the scenarios below:
(d) even – even = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
(e) odd – odd = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
(f) even – odd = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
(g) odd – even = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

**Part (d)**

1. Subtracting an even number from an even number leaves no leftovers in pairs.
2. Therefore, even - even = Even.

Answer (d): Even

**Part (e)**

1. Subtracting an odd number from an odd number removes the single leftover from both, leaving a collection of pairs.
2. Therefore, odd - odd = Even.

Answer (e): Even

**Part (f)**

1. Subtracting an odd number from an even number leaves one leftover.
2. Therefore, even - odd = Odd.

Answer (f): Odd

**Part (g)**

1. Subtracting an even number from an odd number leaves the single leftover intact.
2. Therefore, odd - even = Odd.

Answer (g): Odd

**Answer:** The parities are (d) Even, (e) Even, (f) Odd, (g) Odd.

> Common mistake: Confusing subtraction parity rules with addition parity rules.

### Question 4

*2 marks · Very short answer*

Given the dimensions of a grid, can you tell the parity of the number of small squares without calculating the product?

**Solution**

1. Let the grid have $m$ rows and $n$ columns.
2. The total number of small squares is the product $m \times n$.
3. The product is even if at least one dimension is even, and odd if both dimensions are odd.

**Answer:** The number of small squares is odd if both dimensions are odd, and even if at least one dimension is even.

> Common mistake: Thinking that multiplying two numbers always yields an even number.

### Question 5

*2 marks · Very short answer*

Find the parity of the number of small squares in these grids:
(a) $27 \times 13$
(b) $42 \times 78$
(c) $135 \times 654$

**Part (a)**

1. Both 27 and 13 are odd numbers.
2. The product of two odd numbers is odd.

Answer (a): Odd

**Part (b)**

1. 42 and 78 are both even numbers.
2. The product involving an even number is even.

Answer (b): Even

**Part (c)**

1. 654 is an even number.
2. The product involving an even number is even.

Answer (c): Even

**Answer:** Parities are (a) Odd (b) Even (c) Even

> Common mistake: Calculating the full product instead of checking the parity of the factors.

### Question 6

*2 marks · Very short answer*

Come up with an expression that always has even parity.

**Solution**

1. An even number always has a factor of 2.
2. Multiplying any integer by 2 ensures the result is always even.

**Answer:** An expression like $2n$ or $100p$ always has even parity.

> Common mistake: Writing an expression with an added odd constant like $2n + 1$, which makes it odd.

### Question 7

*2 marks · Very short answer*

Come up with expressions that always have odd parity.

**Solution**

1. An expression always has odd parity if it adds or subtracts an odd number to an even expression like $2n$.
2. Examples of expressions with always odd parity are $2n + 1$ or $4m - 1$.

**Answer:** Expressions like $2n + 1$ and $4m - 1$ always have odd parity.

> Common mistake: Giving an expression that results in even numbers for some values of $n$.

### Question 8

*2 marks · Very short answer*

Come up with other expressions, like $3n + 4$, which could have either odd or even parity.

**Solution**

1. An expression can have either odd or even parity depending on whether the variable $n$ takes an even or odd value.
2. For example, in $3n + 4$, if $n$ is odd, the value is odd, and if $n$ is even, the value is even.

**Answer:** Expressions like $3n + 4$ can have either odd or even parity depending on $n$.

> Common mistake: Assuming an expression with a variable always maintains the same parity.

### Question 9

*2 marks · Very short answer*

The expression $6k + 2$ evaluates to $8, 14, 20,...$ (for $k = 1, 2, 3,...$) — many even numbers are missing.

**Solution**

1. The expression $6k + 2$ misses many even numbers because it only gives multiples of 6 shifted by 2, such as 8, 14, 20, etc.
2. It skips intermediate even numbers like 10, 12, 16, 18.

**Answer:** The expression $6k + 2$ misses even numbers like 10, 12, 16, 18 because it increases in steps of 6.

> Common mistake: Thinking an expression containing even numbers lists all even numbers.

### Question 10

*2 marks · Very short answer*

Are there expressions using which we can list all the even numbers?

**Solution**

1. Yes, since all even numbers have a factor of 2, we can list all even numbers using the expression $2n$, where $n = 1, 2, 3, ...$.

**Answer:** Yes, the expression $2n$ can list all even numbers.

> Common mistake: Using an expression that skips some even numbers like $4n$.

### Question 11

*2 marks · Very short answer*

Are there expressions using which we can list all odd numbers?

**Solution**

1. Yes, we can list all odd numbers by subtracting 1 from the $n^{\text{th}}$ even number.
2. Thus, the expression $2n - 1$ lists all odd numbers.

**Answer:** Yes, the expression $2n - 1$ can list all odd numbers.

> Common mistake: Using $2n + 1$ which starts from 3 for $n = 1$ and misses the first odd number 1.

### Question 12

*2 marks · Very short answer*

What would be the $n^{\text{th}}$ term for multiples of 2? Or, what is the $n^{\text{th}}$ even number?

**Solution**

1. To find the $n^{\text{th}}$ multiple of 2, we multiply the position number $n$ by 2.
2. Therefore, the $n^{\text{th}}$ even number is given by $2n$.

**Answer:** The $n^{\text{th}}$ even number is $2n$.

> Common mistake: Writing $n + 2$ instead of $2n$.

### Question 13

*2 marks · Very short answer*

What is the 100th odd number?

**Solution**

1. The formula for the $n^{\text{th}}$ even number is $2n$, so the $100^{\text{th}}$ even number is $2 \times 100 = 200$.
2. The odd number sequence is one less than the even number sequence, so the $100^{\text{th}}$ odd number is $200 - 1 = 199$.

**Answer:** 199

> Common mistake: Writing 200 instead of 199 by forgetting to subtract 1.

### Question 14

*2 marks · Very short answer*

What is the 100th even number?

**Solution**

1. The $n^{\text{th}}$ even number is given by multiplying the position number by 2.
2. For the $100^{\text{th}}$ even number, calculate $2 \times 100 = 200$.

**Answer:** 200

> Common mistake: Subtracting 1 as done for odd numbers.

### Question 15

*2 marks · Very short answer*

Write a formula to find the $n^{\text{th}}$ odd number.

**Solution**

1. First find the even number at position $n$, which is $2n$.
2. Subtract 1 from the even number to get the formula for the $n^{\text{th}}$ odd number as $2n - 1$.

**Answer:** $2n - 1$

> Common mistake: Writing $2n + 1$ instead of $2n - 1$.

## 6.3 Some Explorations in Grids

### Question 1

*2 marks · Very short answer*

Are you able to see what the circled numbers represent?

**Solution**

1. The numbers in the yellow circles outside the grid represent the sums of the corresponding rows and columns.

**Answer:** The circled numbers represent the sums of the corresponding rows and columns.

> Common mistake: Confusing row and column sums with the numbers inside the grid.

### Question 2

*3 marks · Short answer*

Fill the grids below based on the rule mentioned above:

**Solution**

1. Use the numbers from 1 to 9 without repetition such that each row and column adds up to the respective circled sum.
2. For the first grid, the row sums are 13, 14, 18 and column sums are 24, 9, 12 giving the filled grid with rows (9, 1, 3), (8, 2, 4), (7, 6, 5).
3. For the second grid, the row sums are 24, 15, 6 and column sums are 12, 16, 17 giving the filled grid with rows (7, 8, 9), (4, 6, 5), (1, 2, 3).

**Answer:** The grids are filled with numbers from 1 to 9 matching the given row and column sums.

> Common mistake: Repeating a number or missing a number between 1 and 9.

### Question 3

*Activity*

Make a couple of questions like this on your own and challenge your peers.

**Solution**

1. Create custom $3 \times 3$ grids by placing numbers from 1 to 9 and calculating their row and column sums.

**Answer:** This is an activity question for students to design their own grid puzzles.

### Question 4

*2 marks · Very short answer*

You might have realised that it is not possible to find a solution for this grid. Why is this the case?

**Solution**

1. The smallest possible sum of three distinct numbers from 1 to 9 is $1 + 2 + 3 = 6$.
2. The largest possible sum of three distinct numbers from 1 to 9 is $9 + 8 + 7 = 24$.
3. Since the given grid has a sum of 5 and 26 which lie outside this range, it is impossible to find a solution.

**Answer:** It is impossible because the smallest possible row sum is 6 and the largest is 24, whereas the given grid contains sums 5 and 26.

> Common mistake: Not checking the minimum and maximum possible bounds for row and column sums.

### Question 5

*2 marks · Very short answer*

Why should the row sums and column sums always add to 45?

**Solution**

1. All the row sums added together include each number in the $3 \times 3$ grid exactly once.
2. The sum of all numbers from 1 to 9 is $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$.
3. Therefore, the three row sums and three column sums must always add up to 45.

**Answer:** The row sums and column sums add up to 45 because their total equals the sum of the numbers from 1 to 9.

> Common mistake: Forgetting that every number in the grid is counted once in the row sum total.

### Question 6

*2 marks · Very short answer*

1. What can the magic sum be? Can it be any number?

**Solution**

1. In a $3 \times 3$ magic square using numbers 1 to 9, the total of all three row sums is 45.
2. Since all three row sums must be equal in a magic square, each row sum must be $45 \div 3 = 15$.
3. Thus, the magic sum cannot be any number; it must be exactly 15.

**Answer:** The magic sum must be 15 and cannot be any arbitrary number.

> Common mistake: Assuming the magic sum can be chosen freely without dividing the total sum of 45 by 3.

### Question 7

*2 marks · Very short answer*

Using such reasoning, find out which other numbers $1 - 9$ cannot occur at the centre.

**Solution**

1. Testing numbers $1 - 9$ at the centre of a $3 \times 3$ magic square shows that only $5$ can be placed at the centre.
2. Any other number placed at the centre fails to form rows, columns, and diagonals that sum to $15$.

**Answer:** Numbers other than 5, such as 1, 2, 3, 4, 6, 7, 8, and 9, cannot occur at the centre.

> Common mistake: Assuming numbers other than 5 can be placed at the centre by changing the magic sum.

### Question 8

*2 marks · Very short answer*

If yes, then there should exist three ways of adding 1 with two other numbers to give 15. We have $1 + 5 + 9 = 1 + 6 + 8 = 15$. Is any other combination possible?

**Solution**

1. We need pairs of distinct numbers from $1 - 9$ (excluding $1$) whose sum is $15 - 1 = 14$.
2. The available pairs from $1 - 9$ that add up to $14$ are $5 + 9 = 14$ and $6 + 8 = 14$. No other pair of distinct numbers adds up to $14$.

**Answer:** No other combination is possible because only two pairs of numbers from 1 to 9 add up to 14.

> Common mistake: Including numbers greater than 9 or repeating the same number.

### Question 9

*2 marks · Very short answer*

Similarly, can 9 can be placed in a corner position?

**Solution**

1. A corner square in a $3 \times 3$ magic square belongs to two lines (one row and one column).
2. Since $9$ requires three different pairs to sum to $15$ but only two such pairs exist ($1 + 5 + 9$ and $2 + 4 + 9$), $9$ cannot be in a corner.

**Answer:** No, 9 cannot be placed in a corner position because it only has two valid combinations to sum to 15.

> Common mistake: Thinking corners have more than two intersecting lines.

### Question 10

*2 marks · Very short answer*

Can you find the other possible positions for 1 and 9?

**Solution**

1. Since numbers $1$ and $9$ cannot be at the centre or in the corners, they must occupy the middle positions of the boundary rows or columns.
2. Specifically, 1 and 9 must be placed opposite each other in the middle cells of the grid.

**Answer:** The other possible positions for 1 and 9 are the middle cells of the outer edges of the grid.

> Common mistake: Placing 1 and 9 in adjacent middle positions instead of opposite sides.

### Question 11

*2 marks · Very short answer*

Now, we have one full row or column of the magic square! Try completing it! [Hint: First fill the row or columns containing 1 and 9]

**Solution**

1. Place 1 and 9 in opposite middle cells with 5 at the centre.
2. Complete the remaining cells using numbers $1 - 9$ without repetition so that every row, column, and diagonal adds up to $15$.

**Answer:** The magic square can be completed by placing 8, 1, 6 in the first row, 3, 5, 7 in the second row, and 4, 9, 2 in the third row.

> Common mistake: Repeating numbers while filling the rest of the grid.

## Figure it Out

### Question 1

*3 marks · Short answer*

How many different magic squares can be made using the numbers $1 - 9$?

**Solution**

1. There is exactly one unique magic square using the numbers $1 - 9$ if we ignore rotations and reflections.
2. The magic square is arranged such that every row, column, and diagonal adds up to the magic sum of $15$.
3. The unique magic square is: $$\begin{matrix} 8 & 1 & 6 \\ 3 & 5 & 7 \\ 4 & 9 & 2 \end{matrix}$$

**Answer:** There is exactly one unique magic square made using the numbers $1 - 9$.

> Common mistake: Listing rotated or reflected versions as different magic squares.

### Question 2

*3 marks · Short answer*

Create a magic square using the numbers $2 - 10$. What strategy would you use for this? Compare it with the magic squares made using $1 - 9$.

**Solution**

1. To create a magic square using the numbers $2 - 10$, we add $1$ to every number in the standard $1 - 9$ magic square.
2. The strategy works because shifting every number by a constant maintains the equal sums across rows, columns, and diagonals.
3. The resulting magic square is: $$\begin{matrix} 9 & 2 & 7 \\ 4 & 6 & 8 \\ 5 & 10 & 3 \end{matrix}$$

**Answer:** Add $1$ to each number of the $1 - 9$ magic square to get the new magic square with numbers $2 - 10$.

> Common mistake: Trying to recalculate positions from scratch instead of using the addition strategy.

### Question 3

*3 marks · Short answer*

Take a magic square, and
(a) increase each number by 1
(b) double each number
In each case, is the resulting grid also a magic square? How do the magic sums change in each case?

**Solution**

1. Consider the standard magic square with numbers $1 - 9$ having a magic sum of $15$.
2. For part (a), increasing each number by $1$ gives a new magic square where the new magic sum increases by $3 \times 1 = 3$, becoming $18$.
3. For part (b), doubling each number gives a new magic square where the new magic sum is multiplied by $2$, becoming $15 \times 2 = 30$.

**Answer:** The resulting grids are still magic squares; the magic sum increases by $3$ for part (a) and gets doubled for part (b).

> Common mistake: Forgetting that a row has three numbers, so increasing each number by $1$ increases the row sum by $3$.

### Question 4

*3 marks · Short answer*

What other operations can be performed on a magic square to yield another magic square?

**Solution**

1. Operations like addition, subtraction, and multiplication can be performed on every entry of a magic square to yield another magic square.
2. Adding or subtracting a fixed constant to every number shifts the magic sum by three times that constant.
3. Multiplying or dividing every number by a non-zero constant scales the magic sum by that same constant.

**Answer:** Addition, subtraction, and multiplication (or division) by a constant can be performed on each entry to yield another magic square.

> Common mistake: Stating that division always works without restricting the divisor to non-zero numbers or checking if entries remain integers.

### Question 5

*3 marks · Short answer*

Discuss ways of creating a magic square using any set of 9 consecutive numbers (like $2 - 10$, $3 - 11$, $9 - 17$, etc.).

**Solution**

1. To create a magic square using any set of $9$ consecutive numbers, we find the difference between the smallest number of the new set and $1$.
2. We add this difference to each number in the standard $1 - 9$ magic square.
3. For example, for numbers $3 - 11$, the difference from $1$ is $2$, so we add $2$ to every entry of the $1 - 9$ magic square.

**Answer:** Add the difference between the starting number of the new set and $1$ to each entry of the standard $1 - 9$ magic square.

> Common mistake: Recalculating the entire grid structure instead of applying a uniform shift.

### Question 6

*2 marks · Very short answer*

Choose any magic square that you have made so far using consecutive numbers. If $m$ is the letter-number of the number in the centre, express how other numbers are related to $m$, how much more or less than $m$. [Hint: Remember, how we described a $2 \times 2$ grid of a calendar month in the Algebraic Expressions chapter].

**Solution**

1. Let $m$ be the number in the centre of the magic square.
2. Expressing all other entries in terms of $m$, the generalised magic square is: $$\begin{matrix} m + 3 & m - 4 & m + 1 \\ m - 2 & m & m + 2 \\ m - 1 & m + 4 & m - 3 \end{matrix}$$

**Answer:** The generalised form with centre $m$ is given with surrounding terms expressed relative to $m$.

> Common mistake: Incorrectly assigning offsets from the centre number $m$.

### Question 7

*2 marks · Very short answer*

Once the generalised form is obtained, share your observations with the class.

**Solution**

1. In a magic square with centre number $m$, the sum of any row, column, or diagonal is $3m$.
2. The surrounding numbers are symmetrically placed as $m+3, m-4, m+1$ etc., around the centre $m$.

**Answer:** The row, column, and diagonal sums of the generalised magic square are always equal to $3m$.

> Common mistake: Confusing the magic sum with the centre number $m$.

## Figure it Out

### Question 1

*3 marks · Short answer*

Using this generalised form, find a magic square if the centre number is 25.

**Solution**

1. The generalised 3 $\times$ 3 magic square with centre $m$ is given by replacing $m$ with 25.
2. Substitute $m = 25$ into each cell of the generalised form: $m+3$, $m-4$, $m+1$, $m-2$, $m$, $m+2$, $m-1$, $m+4$, $m-3$.
3. The resulting magic square is: row 1: 28, 21, 26; row 2: 23, 25, 27; row 3: 24, 29, 22.

**Answer:** The magic square with centre 25 is:
28 21 26
23 25 27
24 29 22

> Common mistake: Wrong signs while substituting the value of $m$ in the generalised cells.

### Question 2

*3 marks · Short answer*

What is the expression obtained by adding the 3 terms of any row, column or diagonal?

**Solution**

1. Consider any row, column, or diagonal in the generalised form of a $3 \times 3$ magic square with centre $m$.
2. Add the three terms of a row, for example, the first row: $(m + 3) + (m - 4) + (m + 1)$.
3. Simplify the algebraic expression by combining like terms to get $3m$.

**Answer:** $3m$

> Common mistake: Incorrectly adding positive and negative offsets from $m$.

### Question 3

*3 marks · Short answer*

Write the result obtained by—
(a) adding 1 to every term in the generalised form.
(b) doubling every term in the generalised form

**Part (a)**

1. Add 1 to every term in the generalised form: $m+3+1 = m+4$, $m-4+1 = m-3$, etc.
2. The resulting grid is: $m+4 \quad m-3 \quad m+2$ $m-1 \quad m+1 \quad m+3$ $m \quad m+5 \quad m-2$

Answer (a): m+4, m-3, m+2 / m-1, m+1, m+3 / m, m+5, m-2

**Part (b)**

1. Double every term in the generalised form: $2(m+3) = 2m+6$, $2(m-4) = 2m-8$, etc.
2. The resulting grid is: $2m+6 \quad 2m-8 \quad 2m+2$ $2m-4 \quad 2m \quad 2m+4$ $2m-2 \quad 2m+8 \quad 2m-6$

Answer (b): 2m+6, 2m-8, 2m+2 / 2m-4, 2m, 2m+4 / 2m-2, 2m+8, 2m-6

**Answer:** The resulting grids after adding 1 and doubling every term are obtained.

> Common mistake: Multiplying only the variable $m$ by 2 instead of the entire expression in each cell during doubling.

### Question 4

*3 marks · Short answer*

Create a magic square whose magic sum is 60.

**Solution**

1. The magic sum of a 3 $\times$ 3 magic square is 3 times the centre number, so $3 \times m = 60$, which gives $m = 20$.
2. Substitute $m = 20$ into the generalised magic square cells: $m+3 = 23$, $m-4 = 16$, $m+1 = 21$, $m-2 = 18$, $m = 20$, $m+2 = 22$, $m-1 = 19$, $m+4 = 24$, $m-3 = 17$.
3. The resulting magic square with magic sum 60 is obtained.

**Answer:** 23 16 21 / 18 20 22 / 19 24 17

> Common mistake: Dividing the magic sum incorrectly to find the centre number.

### Question 5

*3 marks · Short answer*

Is it possible to get a magic square by filling nine non-consecutive numbers?

**Solution**

1. Yes, it is possible to create a magic square by filling nine non-consecutive numbers.
2. Consider an example grid filled with nine non-consecutive numbers such as 24, 3, 18, 9, 15, 21, 12, 27, 6.
3. Check that each row, column, and diagonal adds up to the same magic sum of 45, confirming it is a valid magic square.

**Answer:** Yes, it is possible.
24 3 18
9 15 21
12 27 6

> Common mistake: Assuming magic squares can only be formed using consecutive integers.

### Question 6

*2 marks · Very short answer*

Chau̐tīs means 34. Why do you think they called it the Chautīsā Yantra?

**Solution**

1. Identify the meaning of the word 'Chautīs', which means $34$ in Hindi.
2. Explain that every row, column, and diagonal in this $4 \times 4$ magic square adds up to $34$.

**Answer:** It is called the Chautīsā Yantra because every row, column, and diagonal in the magic square adds up to $34$ (Chautīs).

> Common mistake: Confusing the magic sum with the number of cells in the grid.

### Question 7

*2 marks · Very short answer*

Every row, column and diagonal in this magic square adds up to 34. Can you find other patterns of four numbers in the square that add up to 34?

**Solution**

1. In the $4 \times 4$ Chautisā Yantra magic square, the four numbers in any $2 \times 2$ sub-grid also add up to 34.
2. Similarly, the four corner numbers, the four central numbers, and the numbers in the four quadrants each sum to 34.

**Answer:** Other patterns of four numbers that add up to 34 include any $2 \times 2$ sub-grid, the four corners, and the four central numbers.

> Common mistake: Listing patterns of four numbers that do not form symmetrical or standard sub-grids of the magic square.

## 6.4 Nature's Favourite Sequence: The Virahāṅka–Fibonacci Numbers!

### Question 1

*2 marks · Very short answer*

Can you find others?

**Solution**

1. Other possibilities of rhythms for 8 beats can be written using combinations of 1-beat short syllables and 2-beat long syllables.
2. Examples include short long short long long short and long short long short short long.

**Answer:** Other valid rhythms include short long short long long short and long short long short short long.

> Common mistake: Listing combinations that do not add up to the correct total number of beats.

### Question 2

*2 marks · Very short answer*

Do you see other ways?

**Solution**

1. Yes, other ways of writing the number as a sum of 1s and 2s can be found by changing the order of terms or combining different numbers of 1s and 2s.
2. For example, for 8, another way is $1 + 2 + 1 + 2 + 2$.

**Answer:** Yes, other ways like $1 + 2 + 1 + 2 + 2$ are possible by rearranging or changing combinations of 1s and 2s.

> Common mistake: Repeating the same arrangement in a different order thinking it is a new combination.

### Question 3

*2 marks · Very short answer*

Try writing the number 5 as a sum of 1s and 2s in all possible ways in your notebook! How many ways did you find? (You should find 8 different ways!) Can you figure out the answer without listing down all the possibilities? Can you try it for $n = 8$?

**Solution**

1. Writing 5 as a sum of 1s and 2s in all possible ways gives 8 different ways.
2. For $n = 8$, the number of ways is given by the 8th element of the Virahāṅka sequence, which is 34.

**Answer:** There are 8 ways for $n = 5$, and 34 ways for $n = 8$.

> Common mistake: Listing possibilities manually for large numbers instead of using the Virahāṅka sequence rule.

### Question 4

*2 marks · Very short answer*

How many 6-beat rhythms are there? By the same reasoning, it will be the number of 5-beat rhythms plus the number of 4-beat rhythms, i.e., $8 + 5 = 13$. Thus, there are 13 rhythms having 6 beats.

**Solution**

1. Every 6-beat rhythm must begin with either '1+' or '2+'.
2. Thus, the number of 6-beat rhythms is the sum of the number of 5-beat rhythms and 4-beat rhythms, which is $8 + 5 = 13$.

**Answer:** There are 13 rhythms having 6 beats.

> Common mistake: Adding incorrect previous terms of the sequence.

### Question 5

*2 marks · Very short answer*

Use the systematic method to write down all 6-beat rhythms, i.e., write 6 as the sum of 1's and 2's in all possible ways. Did you get 13 ways?

**Solution**

1. Writing all possible sums of 1s and 2s that equal 6 using the systematic method gives all combinations starting with 1+ (5-beat rhythms) and 2+ (4-beat rhythms).
2. Counting them all gives exactly 13 different ways.

**Answer:** Yes, writing out all combinations yields exactly 13 ways.

> Common mistake: Missing out some combinations while writing them manually.

### Question 6

*2 marks · Very short answer*

Write the next number in the sequence, after 55.

**Solution**

1. The next number in the Virahāṅka sequence is obtained by adding the two previous numbers.
2. Adding 34 and 55 gives $34 + 55 = 89$.

**Answer:** The next number after 55 is 89.

> Common mistake: Multiplying the terms instead of adding them.

### Question 7

*2 marks · Very short answer*

Write the next 3 numbers in the sequence:
$1, 2, 3, 5, 8, 13, 21, 34, 55, \_\_\_\_\_, \_\_\_\_\_, \_\_\_\_\_, \dots$

**Solution**

1. The next number is found by adding the two previous numbers, which gives $34 + 55 = 89$.
2. Continuing this, the next two numbers are $55 + 89 = 144$ and $89 + 144 = 233$.
3. The next 3 numbers in the sequence are 89, 144, and 233.

**Answer:** 89, 144, 233

> Common mistake: Adding incorrect previous terms or making arithmetic errors while summing large numbers.

### Question 8

*2 marks · Very short answer*

If you have to write one more number in the sequence above, can you tell whether it will be an odd number or an even number (without adding the two previous numbers)?

**Solution**

1. The parities of the Virahanka sequence repeat in a pattern of odd, odd, even.
2. Thus, the next number will be an even number.

**Answer:** Even number.

> Common mistake: Adding the numbers instead of using the repeating parity pattern.

### Question 9

*2 marks · Very short answer*

What is the parity of each number in the sequence? Do you notice any pattern in the sequence of parities?

**Solution**

1. Writing the sequence with parities: 1(Odd), 2(Even), 3(Odd), 5(Odd), 8(Even), 13(Odd), 21(Odd), 34(Even), ...
2. We notice that the sequence of parities follows a repeating pattern of three terms: Odd, Even, Odd, and then Odd, Even, Odd.
3. This happens because the sum of an odd and an even number is odd, and the sum of two odd numbers is even.

**Answer:** The parity pattern is Odd, Even, Odd, which repeats throughout the sequence.

> Common mistake: Missing the repeating group of three parities (Odd, Even, Odd).

### Question 10

*2 marks · Very short answer*

How many petals do you see on each of these flowers?

**Solution**

1. Observe the number of petals on each daisy shown in the figure in the textbook.
2. The first daisy has 13 petals, the second has 21 petals, and the third has 34 petals.
3. All these numbers (13, 21, 34) belong to the Virahāṅka–Fibonacci sequence.

**Answer:** 13 petals, 21 petals, and 34 petals respectively.

> Common mistake: Miscounting the petals on the daisy illustrations.

## 6.5 Digits in Disguise

### Question 1

*2 marks · Very short answer*

What could U and T be? Can T be 2? Can it be 3?

**Solution**

1. Observe the addition sum where a single-digit number T is added to itself twice to give a 2-digit number UT.
2. Testing values for T shows that when T = 5, $5 + 5 + 5 = 15$, which gives U = 1 and T = 5.
3. T cannot be 2 or 3 because $2 + 2 + 2 = 6$ and $3 + 3 + 3 = 9$, which are single-digit numbers, not 2-digit numbers.

**Answer:** T = 5 and U = 1. T cannot be 2 or 3 as their sums are single-digit numbers.

> Common mistake: Trying to test numbers without considering the number of digits in the sum.

### Question 2

*2 marks · Very short answer*

What digit should the letter M correspond to?

**Solution**

1. In the given addition $\text{K}2 + \text{K}2 = \text{HMM}$, the tens digit $\text{K}$ added to itself gives a sum whose units digit is $\text{M}$.
2. Testing digits for $\text{K}$, if $\text{K} = 6$, then $62 + 62 = 124$, where $\text{H} = 1$ and $\text{M} = 2$, matching the condition that the tens and units digits of the sum are the same.

**Answer:** M corresponds to the digit 2.

> Common mistake: Assuming all letters can take any digit without checking the carry-over.

### Question 3

*2 marks · Very short answer*

What about H? Can it be 2? Can it be 3?

**Solution**

1. The letter H cannot be 2 or 3 because adding a two-digit number K2 to itself can yield a maximum of 198, making H at most 1.

**Answer:** No, H cannot be 2 or 3 as it can only be 1.

> Common mistake: Assuming H can be any single digit without checking the maximum possible sum of K2 and K2.

## Figure it Out

### Question 1

*3 marks · Short answer*

A light bulb is ON. Dorjee toggles its switch 77 times. Will the bulb be on or off? Why?

**Solution**

1. The light bulb starts in the ON state.
2. Each toggle changes the state of the bulb from ON to OFF or OFF to ON, alternating the parity of state changes.
3. Since 77 is an odd number of toggles starting from ON, the bulb will be OFF.

**Answer:** The bulb will be OFF.

> Common mistake: Assuming an odd number of toggles leaves the bulb in the ON state because it started ON.

### Question 2

*3 marks · Short answer*

Liswini has a large old encyclopaedia. When she opened it, several loose pages fell out of it. She counted 50 sheets in total, each printed on both sides. Can the sum of the page numbers of the loose sheets be 6000? Why or why not?

**Solution**

1. Each sheet of paper has two pages, one on the front and one on the back, so consecutive page numbers appear together on the same sheet (such as page 1 and page 2).
2. The sum of the two page numbers on any single sheet is always the sum of an odd number and an even number, which is always odd.
3. There are 50 sheets in total, so the sum of all page numbers is the sum of 50 odd numbers, which must be an even number.
4. Since 6000 is an even number, it is possible for the sum of the page numbers of the loose sheets to be 6000.

**Answer:** Yes, it is possible because the sum of 50 odd sheet-sums is an even number, and 6000 is an even number.

> Common mistake: Confusing the sum of individual pages with the sum of sheet totals, or incorrectly assuming the sum of 50 odd numbers is odd.

### Question 3

*3 marks · Short answer*

Here is a $2 \times 3$ grid. For each row and column, the parity of the sum is written in the circle; 'e' for even and 'o' for odd. Fill the 6 boxes with 3 odd numbers ('o') and 3 even numbers ('e') to satisfy the parity of the row and column sums.

**Solution**

1. We are given a $2 \times 3$ grid with 3 odd numbers ('o') and 3 even numbers ('e').
2. We place the numbers such that the row and column parities match the given circles.
3. One valid placement of numbers is placing even numbers in one row and odd numbers in the other, or mixing them appropriately to satisfy the circle conditions.

**Answer:** A valid grid can be filled with 3 odd and 3 even numbers matching the given row and column parities.

> Common mistake: Placing incorrect counts of odd and even numbers.

### Question 4

*3 marks · Short answer*

Make a $3 \times 3$ magic square with 0 as the magic sum. All numbers can not be zero. Use negative numbers, as needed.

**Solution**

1. Take a standard $3 \times 3$ magic square using consecutive integers such as $1$ to $9$, whose magic sum is $15$.
2. Subtract $5$ from every number in the magic square to shift the centre to $0$ and the magic sum to $0$.
3. The resulting $3 \times 3$ magic square with a magic sum of $0$ is $\begin{matrix} 3 & -4 & 1 \\ -2 & 0 & 2 \\ -1 & 4 & -3 \end{matrix}$.

**Answer:** $\begin{matrix} 3 & -4 & 1 \\ -2 & 0 & 2 \\ -1 & 4 & -3 \end{matrix}$

> Common mistake: Forgetting that all rows, columns, and diagonals must sum to $0$ including negative numbers.

### Question 5

*3 marks · Short answer*

Fill in the following blanks with 'odd' or 'even':
(a) Sum of an odd number of even numbers is \_\_\_\_\_\_\_
(b) Sum of an even number of odd numbers is \_\_\_\_\_\_\_
(c) Sum of an even number of even numbers is \_\_\_\_\_\_\_
(d) Sum of an odd number of odd numbers is \_\_\_\_\_\_\_

**Part (a)**

1. An even number added to another even number always results in an even number.
2. Summing any even number of even numbers repeatedly preserves the even parity.

Answer (a): Even

**Part (b)**

1. The sum of two odd numbers is always an even number.
2. Therefore, taking an even number of odd numbers and grouping them in pairs yields an even sum.

Answer (b): Even

**Part (c)**

1. Each even number can be arranged in pairs without any leftovers.
2. Adding an even number of even numbers results in a sum that can still be arranged in pairs without leftovers.

Answer (c): Even

**Part (d)**

1. Pairing up an even number of odd numbers gives pairs with even sums, leaving one unpaired odd number if the count is odd.
2. Thus, the sum of an odd number of odd numbers results in an odd parity.

Answer (d): Odd

**Answer:** The parities are: (a) Even, (b) Even, (c) Even, (d) Odd.

> Common mistake: Confusing the number of terms being added with the values of the terms themselves when determining parity.

### Question 6

*3 marks · Short answer*

What is the parity of the sum of the numbers from 1 to 100?

**Solution**

1. The sum of numbers from 1 to 100 can be found using the formula $\frac{n(n+1)}{2}$.
2. Substitute $n = 100$: $\frac{100 \times 101}{2} = 50 \times 101 = 5050$.
3. Since 5050 is divisible by 2, its parity is even.

**Answer:** Even

> Common mistake: Trying to add all numbers individually instead of using the sum formula or parity property.

### Question 7

*3 marks · Short answer*

Two consecutive numbers in the Virahāṅka sequence are 987 and 1597. What are the next 2 numbers in the sequence? What are the previous 2 numbers in the sequence?

**Solution**

1. The next number is found by adding the two previous numbers, $987 + 1597 = 2584$.
2. The second next number is $1597 + 2584 = 4181$.
3. The previous numbers are found by subtracting, $1597 - 987 = 610$ and $987 - 610 = 377$.
4. Thus, the next two numbers are 2584 and 4181, and the previous two numbers are 610 and 377.

**Answer:** Next two numbers: 2584, 4181; Previous two numbers: 610, 377

> Common mistake: Subtracting incorrectly while finding the previous numbers.

### Question 8

*3 marks · Short answer*

Angaan wants to climb an 8-step staircase. His playful rule is that he can take either 1 step or 2 steps at a time. For example, one of his paths is $1, 2, 2, 1, 2$. In how many different ways can he reach the top?

**Solution**

1. The number of ways to climb an $n$-step staircase taking 1 or 2 steps at a time is given by the $n$-th Virahāṅka number.
2. The sequence of Virahāṅka numbers is 1, 2, 3, 5, 8, 13, 21, 34, ...
3. The 8th term of the Virahāṅka sequence is 34.
4. Thus, Angaan can reach the top in 34 different ways.

**Answer:** 34 ways

> Common mistake: Counting paths manually and missing some combinations.

### Question 9

*3 marks · Short answer*

What is the parity of the 20th term of the Virahāṅka sequence?

**Solution**

1. The Virahāṅka sequence is 1, 2, 3, 5, 8, 13, 21, 34, ...
2. The parities of the terms follow the repeating pattern: Odd, Even, Odd, Odd, Even, Odd, Odd, Even, ...
3. The pattern repeats every 3 terms.
4. Since 20 divided by 3 leaves a remainder of 2 ($20 = 3 \times 6 + 2$), the 20th term has the parity of the 2nd term, which is even.

**Answer:** Even

> Common mistake: Writing out all 20 terms and making a calculation error.

### Question 10

*3 marks · Short answer*

Identify the statements that are true.
(a) The expression $4m - 1$ always gives odd numbers.
(b) All even numbers can be expressed as $6j - 4$.
(c) Both expressions $2p + 1$ and $2q - 1$ describe all odd numbers.
(d) The expression $2f + 3$ gives both even and odd numbers.

**Part (a)**

1. Consider the expression $4m - 1$.
2. Since $4m$ is always an even number for any integer $m$, subtracting $1$ from an even number always gives an odd number.

Answer (a): True

**Part (b)**

1. Consider the expression $6j - 4$.
2. For $j = 1$, $6(1) - 4 = 2$, and for $j = 2$, $6(2) - 4 = 8$, but for $j = 0$, $6(0) - 4 = -4$. This expression misses several even numbers like $0$ and $4$ for positive integer values of $j$.

Answer (b): False

**Part (c)**

1. The expressions $2p + 1$ and $2q - 1$ both describe odd numbers.
2. However, using two different variables $p$ and $q$ independently does not mean both expressions describe the exact same set of all odd numbers simultaneously in a single restricted context, though individually both generate all odd numbers.

Answer (c): False

**Part (d)**

1. Consider the expression $2f + 3$.
2. Since $2f$ is always even, adding $3$ (which is odd) to an even number always results in an odd number, so it cannot give both even and odd numbers.

Answer (d): False

**Answer:** Statements (a) is True, while (b), (c), and (d) are False.

> Common mistake: Assuming that any linear algebraic expression with a constant term can produce both evens and odds without checking the parity of the terms.

### Question 11

*3 marks · Short answer*

Solve this cryptarithm:
  UT
+ TA
-----
 TAT

**Solution**

1. Consider the addition column-wise: $\text{UT} + \text{TA} = \text{TAT}$.
2. From the units column, $T + A = T$, which implies $A = 0$.
3. From the tens column, $U + T = \text{TA}$ (or $10 \times T + A$). Substituting $A = 0$, we get $U + T = 10 \times T$, which means $U = 9T$.
4. Since $U$ and $T$ must be single-digit numbers from $0$ to $9$ and $T \neq 0$, the only possibility is $T = 1$, which gives $U = 9$.

**Answer:** U = 9, T = 1, and A = 0

> Common mistake: Assuming letters can take multi-digit values or zero for the leading digits incorrectly.

## Frequently asked questions

### How many total questions are there in Class 7 Maths Chapter 6 Number Play?

This chapter contains a total of 69 questions spread across various sections and Figure It Out exercises based on the new NCERT book for the 2026-27 session. You can find step-by-step solutions for all of them in the free PDF available on this page.

### Which topics do the questions in Chapter 6 cover?

The questions cover topics like numbers telling things, picking parity, explorations in grids, the Virahāṅka-Fibonacci sequence, and digits in disguise. SwaVid provides detailed answers for each of these concepts to help Class 7 students prepare effectively.

### What are the hardest question types in Number Play and how do we approach them?

Questions involving magic squares, grid filling, and algebraic parity expressions in the Figure It Out sections can be challenging. To approach them, you should carefully analyze the given rules, use step-by-step logical deductions, and refer to the explanations in our free PDF.

### How should I write answers to score full marks in Class 7 Maths Chapter 6?

To score full marks, you need to clearly state the underlying mathematical rules, show every step of calculation, and verify your results like grid sums or parity conditions. Following the structured formats in SwaVid's free solutions page will guide you on how to present your answers properly.

### Is the free PDF for NCERT Solutions of Class 7 Maths Chapter 6 available?

Yes, the complete free PDF containing accurate answers for all 69 questions of this chapter is available right on this SwaVid page. It strictly follows the new NCERT syllabus for the 2026-27 session to support your exam preparation.

## Related pages

- [Class 7 Maths chapters](https://www.swavid.com/maths/class/7)

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