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NCERT Solutions for Class 7 Maths Chapter 6 Number Play

NCERT Solutions for Class 7 Maths Chapter 6 Number Play

Edited By Komal Miglani | Updated on Jun 26, 2025 09:03 AM IST

Number Play shows that math isn’t just about answers — it’s about the joy of exploring patterns. Chapter 6 of the NCERT Class 7 is an integral part of the syllabus and carries significant weightage in the final exam. This chapter briefly introduces patterns, the relationship between numbers, parity, magic square, Virahāṅka–Fibonacci Numbers to students. The aim of these NCERT solutions for Class 7 Maths is to clear students' doubts about this chapter and provide them with an important tool to learn all the necessary concepts.

This Story also Contains
  1. NCERT Solutions for Class 7 Maths Chapter 6 Exercise
  2. Number Play Class 6 Maths Chapter 6 - Topics
  3. NCERT Solutions for Class 7 Maths Chapter Wise
NCERT Solutions for Class 7 Maths Chapter 6 Number Play
NCERT Solutions for Class 7 Maths Chapter 6 Number Play

The Number Play chapter makes maths exciting and new. Curious minds will find patterns, tricks, and functions combined in this chapter. These NCERT solutions for Class 7 are created by experienced Careers360 teachers in a student-friendly manner, allowing students to understand all the solutions in a step-by-step approach and solve similar questions independently in the future. Explore this link for NCERT syllabus outlines, notes, and PDF downloads: NCERT.

NCERT Solutions for Class 7 Maths Chapter 6 Exercise

Page number: 128
Number of Questions: 2

Question 1: 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

Solution:

(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

Question 2: 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?

Solution:

Statement

Answer

Reasoning

(a) If a person says ‘0’, then they are the tallest in the group.

Only Sometimes True

A person saying ‘0’ could indicate they are the tallest, but this depends on how the numbering is assigned, as a short person can also say “0” if they are in front of the line.

(b) If a person is the tallest, then their number is ‘0’.

Always True

If the tallest person is at the front of the line, their number is ‘0’ as there will be no one taller than them.

(c) The first person’s number is ‘0’.

Always True

The first person in line is always numbered as ‘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’.

Only Sometimes True

A person standing somewhere in the middle of the line can say ‘0’ if no one taller is in front of them.

(e) The person who calls out the largest number is the shortest.

Only Sometimes True

In a lineup, the shortest person may call out the largest number, but this is not guaranteed.

(f) What is the largest number possible in a group of 8 people?

7

In a group of 8 people, the largest number assigned is 7 (since numbering starts from 0).

Page number: 131

Number of Questions: 3


Question 1: 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

Solution:

Statement

Parity

Reasoning

(a) Sum of 2 even numbers and 2 odd numbers (e.g., even + even + odd + odd)

Even

Even + Even = Even

Odd + Odd = Even

So, Even + Even = Even

(b) Sum of 2 odd numbers and 3 even numbers

Even

Odd + Odd = Even
Even + Even + Even = Even

So, Even + Even = Even

(c) Sum of 5 even numbers

Even

The sum of any number of even numbers is always even.

(d) Sum of 8 odd numbers

Even

The sum of 2 odd numbers is even. The sum of 8 odd numbers is the same as the sum of 4 pairs of odd numbers, which is even.

Example: 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64

Question 2: 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:

Lakpa has an odd number of ₹1 coins.

1 × odd = odd

He has an odd number of ₹5 coins.

5 × odd = odd

He has an even number of ₹10 coins.

10 × even = Even

We know, Odd + Odd = Even
Then, Even + Even = Even
But 205 is not an even number.
So, it can’t be possible.

Question 3: We know that:

(a) even + even = even

(b) odd + odd = even

(c) even + odd = odd

Similarly, find out the parity for the scenarios below:

(d) even – even = ___________________

(e) odd – odd = ___________________

(f) even – odd = ___________________

(g) odd – even = ___________________

Solution:

Operation

Result Parity

Example

(d) even – even

even

8 - 4 = 4 (even)

(e) odd – odd

even

7 - 3 = 4 (even)

(f) even – odd

odd

6 - 3 = 3 (odd)

(g) odd – even

odd

7 - 4 = 3 (odd)

Page number: 136

Number of Questions: 5

Question 1: How many different magic squares can be made using the numbers 1 – 9?

Solution:

A magic square means all rows, columns, and diagonals sum to the same number.
Here, Magic Sum =1+2+3+4+5+6+7+8+93=453=15

There can be one magic square using the numbers 1 to 9.

8

1

6

3

5

7

4

9

2

Question 2: 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:

Here, Magic Sum =2+3+4+5+6+7+8+9+103=543=18
There can be a magic square possible with numbers 2–10.

9

2

7

4

6

8

5

10

3

Both structures are the same. But the magic sum is different.

Question 3: 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:

(a)

We can take the example of Questions 1 and 2 for this question.
Magic Square of numbers from 1 to 9.

8

1

6

3

5

7

4

9

2

After adding 1 to each number, we will get the following magic square.

9

2

7

4

6

8

5

10

3

(b)

If we double each number, the magic sum will also be doubled.
The magic sum increases from 15 to 30.

New magic square:

16

2

12

6

10

14

8

18

4

Question 4: What other operations can be performed on a magic square to yield another magic square?

Solution:

Here are some operations that can be performed on a magic square to yield another magic square.

Operation

Effect on Magic Square

Scaling (Multiplying by a constant)

Magic square remains the same

Transposition (Swap rows and columns)

Magic square remains the same

Rotation (90°, 180°, 270°)

Magic square remains the same

Reflection (Flip horizontally/vertically)

Magic square remains the same

Adding/Subtracting a constant

Magic square remains the same

Question 5: Discuss ways of creating a magic square using any set of 9 consecutive numbers

(like 2 – 10, 3 – 11, 9 – 17, etc.).

Solution:

Magic square using 2-10

Here, the magic sum = 18

9

2

7

4

6

8

5

10

3


We can add one with each number to get a magic square of 3-11.

Here, magic sum = 21

10

3

8

5

7

9

6

11

4

We have to add 6 to each number to get a magic square of 9-17.

Here, magic sum = 39

16

9

14

11

13

15

12

17

10

Page number: 137

Number of Questions: 5

Question 1: Using this generalised form, find a magic square if the centre number is 25.

Solution:

Generalised form of a magic square:

K + 3

K - 4

K + 1

K - 2

K

K +2

K - 1

K + 4

K - 3

So, if the centre number is 25, the magic square will be:

28

21

26

23

25

27

24

29

22

Question 2: What is the expression obtained by adding the 3 terms of any row, column or diagonal?

Solution:

Let's take the last magic square as an example.

28

21

26

23

25

27

24

29

22

1st row sum = 28 + 21 + 26 = 75

1st column sum = 28 + 23 + 24 = 75

Diagonal sum = 28 + 25 + 22 = 75

As we can see, it gives the magic sum when we add three terms of any row, column, or diagonal.

Question 3: Write the result obtained by—

(a) Adding 1 to every term in the generalised form.

(b) Doubling every term in the generalised form

Solution:

Generalised form of a magic square:

K + 3

K - 4

K + 1

K - 2

K

K + 2

K - 1

K + 4

K - 3

(a)

Adding 1 to every term in the generalised form:

K + 4

K - 3

K + 2

K - 1

K + 1

K + 3

K

K + 5

K - 2

We can see that the magic square still holds, as the new magic sum = 3K + 3

(b)

Doubling every term in the generalised form

2K + 6

2K - 8

2K + 2

2K - 4

2K

2K + 4

2K - 2

2K + 8

2K - 6

We can see that the magic square still holds, as the new magic sum = 2K

Question 4: Create a magic square whose magic sum is 60.

Solution:

In the magic square of 1 to 9, the magic sum is 15.

8

1

6

3

5

7

4

9

2


So, to get a magic sum of 60, we need to multiply every digit of the magic square from 1 to 9 by 4 as
15 × 4 = 60

32

4

24

12

20

28

16

36

8

Here,
Row sums:
32 + 4 + 24 = 60
12 + 20 + 28 = 60

16 + 36 + 8 = 60
Column sums:
32 + 12 + 16 = 60
4 + 20 + 36 = 60
24 + 28 + 8 = 60
Diagonals:

32 + 20 + 8 = 60
24 + 20 + 16 = 60

Question 5: Is it possible to get a magic square by filling nine non-consecutive numbers?

Solution:

Yes, a magic square can be formed with 9 non-consecutive numbers if they still follow an arithmetic progression.
For example, all even numbers, or numbers with a common difference.
Here is a magic square with 3,6,9,12,15,18,21,24,27.
As you can see, the common difference between all the numbers is 3.

24

3

18

9

15

21

12

27

6

Here, the magic sum = 45

But it is impossible to create a magic square if the 9 numbers are just random, non-consecutive numbers with no pattern.

Page number: 143-144

Number of Questions: 11

Question 1: A light bulb is ON. Dorjee toggles its switch 77 times. Will the bulb be on or off? Why?

Solution:

Initially, the light bulb is on.
So, if Dorjee toggles 1 time, then it will be off.
Subsequently, if he toggles two times, the bulb will be on again.
As we can see, after an odd number of times, the bulb is off, and after an even number of times, the bulb is on.
77 is an odd number.
So, the bulb will be off after Dorjee toggles its switch 77 times.

Question 2: 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:

Total sheets = 50

Each sheet has one even and one odd page number.

Thus, the total of 50 sheets consists of 50 even numbers and 50 odd numbers.

We know that the sum of an even number of odd numbers is “Even”.
And the Sum of an even number of even numbers is also “Even”.

Thus, the total sum is even + even = even.=

Since 6000 is an even number, it is possible for the sum of the page numbers of the loose sheets to be 6000.

Question 3: Here is a 2 × 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:

Question 4: Make a 3 × 3 magic square with 0 as the magic sum. All numbers can not be zero. Use negative numbers, as needed.

Solution:

-3

2

1

4

0

-4

-1

-2

3

Question 5: 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 ______

Solution:

Statement

Answer

Example

(a) Sum of an odd number of even numbers

Even

2 + 4 + 6 = 12

(b) Sum of an even number of odd numbers

Even

1 + 3 + 5 + 7 = 16

(c) Sum of an even number of even numbers

Even

2 + 4 + 6 + 8 = 20

(d) Sum of an odd number of odd numbers

Odd

1 + 3 + 5 = 9


Question 6: What is the parity of the sum of the numbers from 1 to 100?

Solution:

In numbers from 1 to 100, there are 50 odd numbers and 50 even numbers.
We know that the sum of an even number of odd numbers is “Even”.
And the Sum of an even number of even numbers is also “Even”.
So, Even + Even = Even.

Hence, the parity of the sum of numbers from 1 to 100 is even.

Question 7: 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:

A Virahāṅka sequence is a sequence where each number is the sum of the two preceding numbers.
So, the next number in the sequence = 987 + 1597 = 2584

After that, the next number in the sequence = 1597 + 2584 = 4181

Let the previous number of 987 in the sequence be x.
So, 1597 = x + 987
⇒ x = 1597 - 987 = 610

Let the previous number of 610 in the sequence be y.
So, 987 = y + 610
⇒ y = 987 - 610 = 377

Thus, the total sequence is:
377, 610, 987, 1597, 2584, 4181

Question 8: 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:

This problem is a variation of the Fibonacci sequence, where:
At each step, Angaan can either take 1 step or 2 steps.
And the number of ways to reach step n is the sum of the ways to reach step n−1 and step n−2.
So, f(n) = f(n-1) + f(n-2)

Step number (n)

How many ways Angaan can reach each step using 1-step or 2-step moves [f(n)]

0

1

1

1

2

1 + 1 = 2

3

1 + 2 = 3

4

2 + 3 = 5

5

3 + 5 = 8

6

5 + 8 = 13

7

8 + 13 = 21

8

13 + 21 = 34

Hence, the number of different ways Angaan can reach the top of the 8-step staircase is 34.

Question 9: What is the parity of the 20th term of the Virahāṅka sequence?

Solution:

Virahāṅka sequence:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181,…

We can see a pattern.
1 (odd), 2 (even), 3 (odd), 5 (odd), 8 (even), 13 (odd), 21 (odd), 34 (even), 55 (odd), 89 (odd),…

This pattern repeats every 3 terms.

So, after 6 × 3 = 18 terms pattern will again be odd, even, odd.
19th term = odd
20th term = even
21st term = odd

Hence, the parity of the 20th term of the Virahāṅka sequence is “even”.

Question 10: 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.

Solution:

Statement

True/False

Example

(a) The expression 4m−1 always gives odd numbers.

True

For m = 0, 4(0) - 1 = -1, which is odd;
For m=1, 4(1)−1 = 3, which is odd;

For m=2, 4(2)−1 = 7, which is odd

(b) All even numbers can be expressed as 6j−4

False

For j=1, 6(1)−4=2(even);

For j=2, 6(2)−4=8 (even)

This expression produces even numbers, but it does not produce all even numbers (for example, it skips 4 and 6).

(c) Both expressions 2p+1 and 2q−1 describe all odd numbers.

True

For p=1, 2(1)+1=3, which is odd;

For q=1, 2(1)−1=1, which is odd

(d) The expression 2f + 3 gives both even and odd numbers.

False

For f=0, 2(0)+3=3(odd);

For f=1, 2(1)+3=5 (odd);

For f=−1, 2(−1)+3=1 (odd)
So, it doesn’t give even numbers.

Question 11: Solve this cryptarithm:

Solution:

Where, U = 9, T = 1 and A = 0

Number Play Class 6 Maths Chapter 6 - Topics


The topics discussed in the NCERT Solutions for class 7, chapter 6, Number Play are:

  • Numbers Tell Us Things
  • Picking Parity
  • Some Explorations in Grids
  • Nature’s Favourite Sequence: The Virahāṅka– Fibonacci Numbers!
  • Digits in Disguise

NCERT Solutions for Class 7 Maths Chapter Wise

The links below allow students to access all the Maths solutions from the NCERT book.

NCERT Solutions for Class 7 Subject Wise

Students can refer to the link below to access the subject-wise solutions for Class 7.

Students can also check the NCERT Books and the NCERT Syllabus here:

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

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Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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