NCERT Solutions for Class 7 Maths Ganita Prakash Chapter 6 PDF Download

NCERT Solutions for Class 7 Maths Ganita Prakash Chapter 6 provide students with clear and step-by-step explanations of all important topics in the chapter Number Play. These solutions help learners understand how to work with number sequences, find the parity of sums and differences, solve magic squares and explore grids and patterns.

By using these NCERT Solutions for Class 7 Maths Chapter 6 students can easily practice exercises, check their answers, and strengthen their problem-solving skills. These solutions help prepare for exams and gain a strong foundation in mathematical reasoning.

NCERT Solutions for Class 7 Maths Chapter 6 Number Play

Class 7 Math Number Play is about exploring numbers in fun and interesting ways. Students learn to find patterns in numbers, work with sequences and identify whether numbers or sums are even or odd. 

The chapter also introduces puzzles like magic squares, grids and cryptarithms, helping students practice logical thinking and problem-solving. By understanding these concepts, learners develop a strong foundation in mathematics while enjoying number games and challenges.

Class 7 Math Chapter 6 Question Answer

Here are the Class 7 Maths Chapter 6 Question and Answer solutions from Ganita Prakash. 

Figure It Out

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

Bookshelves

(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

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 peopl

Solution:
(a) It’s only sometimes true because a tall person might say ‘0’ if there’s no one taller in front of them, but a short person can also say ‘0’ if they are at the front or if no one taller is ahead of them.

(b) It’s always true because the tallest person never has anyone taller in front of them and would always say ‘0’.

(c) It’s always true because there is no one taller in front of the first person, so they will always say ‘0’.

(d) It’s only sometimes true because a person standing somewhere in the middle of the line can say ‘0’ if no one taller is in front of them.

(e) It’s only sometimes true because the shortest person can even call out ‘0’ when they are standing first in the line, and the second shortest person could be at the back and still call out the largest number.

(f) The largest possible number in a group of 8 people is 7 because the shortest person, standing at the end of the line, can see 7 taller people in front of them.

Page 129

Q. Kishor has some number cards and is working on a puzzle: There are 5 boxes, and each box should contain exactly 1 number card. The numbers in the boxes should sum to 30. Can you help him find a way to do it? Can you figure out which 5 cards add to 30? Is it possible?

Solution:
Adding odd numbers for an odd number of times gives odd numbers.
Since 30 is an even number, it is impossible to add five odd numbers to get an even number.

Q. Add a few even numbers together. What kind of number do you get? Does it matter how many numbers are added?
Solution:
Even numbers, when added together, always result in an even number. It doesn’t matter how many even numbers are added—their sum will always be even.

Page 131

Figure It Out

 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:
(a) Sum of 2 even numbers and 2 odd numbers
Even + Even = Even, and Odd + Odd = Even.
Since Even + Even = Even, the total sum is Even.
Therefore, the parity of the sum of 2 even numbers and 2 odd numbers is even.
Example: 2 + 4 + 1 + 3 = 6 + 4 = 10 (even).

(b) Sum of 2 odd numbers and 3 even numbers
Odd + Odd = Even, and Even + Even + Even = Even.
Even + Even = Even, so the total sum is Even.
Therefore, the parity of the sum of 2 odd numbers and 3 even numbers is even.
Example: 1 + 3 + 2 + 4 + 6 = 4 + 12 = 16 (even).

(c) Sum of 5 even numbers
The sum of any number of even numbers is always an even number.
Therefore, the parity of the sum of 5 even numbers is even.
Example: 2 + 4 + 6 + 8 + 10 = 30 (even).

(d) Sum of 8 odd numbers
When an even number of odd numbers are added, the result is even.
Therefore, the parity of the sum of 8 odd numbers is even.
Example: 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64 (even).

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:

The total value of an odd number of ₹1 coins is odd.
The total value of an odd number of ₹5 coins is odd.
The total value of an even number of ₹10 coins is even.
Adding the values of all the coins:
Odd + Odd + Even = Even + Even = Even.
Therefore, the parity of the sum of ₹1 coins, ₹5 coins, and ₹10 coins is even.
But Lakpa calculated a total of ₹205, which is odd.
Therefore, Lakpa must have made a mistake!
The total can never be ₹205 with the given coin counts.

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:
(d) Since 8 – 4 = 4 (even)
10 – 4 = 6 (even)
Therefore,
Even – even = ( even ).

(e) Since 11 – 7 = 4 (even)
13 – 5 = 8 (even)
Therefore,
Odd – odd = ( even ).

(f) Since 8 – 5 = 3 (odd)
10 – 5 = 5 (odd)
Therefore,
Even – odd = ▁( odd ).

(g) Since 9 – 4 = 5 (odd)
11 – 6 = 5 (odd)
Therefore,
Odd – even = ( odd ).

Small Squares in Grids

Q. In a 3 × 3 grid, there are 9 small squares, which is an odd number. Meanwhile, in a 3 × 4 grid, there are 12 small squares, which is an even number.
Given the dimensions of a grid, can you tell the parity of the number of small squares without calculating the product?

Solution:

The parity of the number of small squares in a grid (with m rows and n columns) is:
(i) Even if at least one of the numbers (either m or n) is even.
(ii) Odd if both numbers (m and n) are odd.
(iii) Even if both numbers (m and n) are even.

Page 132

Q. Find the parity of the number of small squares in these grids:
(a) 27 × 13
(b) 42 × 78
(c) 135 × 654
Solution:
(a) 27 × 13
27 and 13 are both odd numbers.
Since, Odd × Odd = Odd
Therefore, the parity of the small squares in 27 × 13 grid is odd.

(b) 42 × 78
42 and 78 are both even numbers.
Since, Even × Even = Even.
Therefore, the parity of the small squares in 42 × 78 grid is even.

(c) 135 × 654
135 is odd and 654 is even.
Since, Odd × Even = Even.
Therefore, the parity of the small squares in 135 × 654 grid is even.

Parity of Expressions

 Q. Consider the algebraic expression: 3n + 4. For different values of n, the expression has different parity:

Come up with an expression that always has even parity.
Solution:
4n, 8n + 6, 6n + 2 etc.

Q. Come up with expressions that always have odd parity.
Solution:
4n + 3, 6n + 5, 2n + 9 etc.

Q. Come up with other expressions, like 3n + 4, which could have either odd or even parity.
Solution:
5n + 2, 9n + 4, 7n + 8 etc.

Q. The expression 6k + 2 evaluates to 8, 14, 20,… (for k = 1, 2, 3,…) — many even numbers are missing. Are there expressions using which we can list all the even numbers?
Solution:
The expression 2n evaluates to 2, 4, 6, 8, 10,…… (for n = 1, 2, 3, 4, 5,….). Therefore, 2n is the expression that can list all even numbers.

Q. Are there expressions using which we can list all odd numbers?
Solution:
The expression 2n -1 evaluates to 1, 3, 5, 7, 9,…. (for n = 1, 2, 3, 4, 5,….). Therefore, 2n – 1 is the expression that can list all odd numbers.

Q. What would be the nth term for multiples of 2? Or, what is the nth even number?
Solution:
The nth term for multiples of 2 or the nth even number is 2n.

Page 133

6.3 Some Explorations in Grids

Page 136

Figure It Out

1. How many different magic squares can be made using the numbers 1 – 9?
Solution:
There is exactly one unique magic square with numbers 1-9, if we ignore rotations and reflections.

 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:
Magic square made using numbers 1-9:

 Now, if we add 1 to each number of the magic square 1-9, we will get a magic square with numbers 2-10.

 Both squares are structurally identical. The only difference is that the magic square with numbers 2–10 has 18 as the magic sum, while the magic square with numbers 1–9 has 15 as the magic sum.

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:
Magic square:

 (a) After increasing each number by 1, we will get a new magic square with the magic sum 18.

 (b) After doubling each number, we will get a new magic square with the magic sum 30.

 4. What other operations can be performed on a magic square to yield another magic square?
Solution:
Operations like Addition, subtraction, multiplication and division can be performed on a magic square to yield another magic square.

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 with numbers 1-9:

Magic square with numbers 2–10: It is created by adding 1 to each number of the magic square 1–9.

 Magic square with numbers 3-11: It is created by adding 2 to each number of the magic square 1-9.

 Magic square with numbers 9-17: It is created by adding 8 to each number of the magic square 1-9.

 Page 137

Generalising a 3 × 3 Magic Square

Q. 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 × 2 grid of a calendar month in the Algebraic Expressions chapter].

 Solution:
Considering a magic square with numbers 2-10:

 Taking ‘m’ as the letter-number of the number in the centre and expressing other numbers in relation to ‘m’.

 Figure It Out

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

 A magic square with the center value 25, where other numbers in the grid are expressed in relation to 25.

 2. What is the expression obtained by adding the 3 terms of any row, column or diagonal?
Solution:
Row Sum (1st row): 28 + 21 + 26 = 75.
Column Sum (1st column): 28 + 23 + 24 = 75.
Diagonal sum (1st diagonal): 28 + 25 + 22 = 75.
Expression = 3m (where m is the letter-number of the number in the centre).

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:

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

(b) Doubling every term in the generalised form:

4. Create a magic square whose magic sum is 60.
Solution:
Magic square with numbers 1-9 has a magic sum of 15.

 To get a magic square with a magic sum of 60, we need to multiply each number of the magic square 1-9 by 4. Therefore, a magic square with a magic sum 60 is:

 5. Is it possible to get a magic square by filling nine non-consecutive numbers?
Solution:
Yes, it is possible to get a magic square with non-consecutive numbers.

The First-ever 4 × 4 Magic Square

Q. The first ever recorded 4 × 4 magic square is found in a 10th century inscription at the Pārśhvanath Jain temple in Khajuraho, India, and is known as the Chautīsā Yantra.

 Chau̐ tīs means 34. Why do you think they called it the Chautīsā Yantra?
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:
Different patterns of four numbers in the square that add up to 34 are:
(i) Sum of four corner numbers 7 + 14 + 4 + 9 is 34.
(ii) Sum of four central numbers 13 + 8 + 10 + 3 is 34.
(iii) Sum of four number in any 2×2 square is 34.

Page 141

Q. 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:
Yes, we got a total of 13 ways.

 Page 142

Q. Write the next 3 numbers in the sequence:
1, 2, 3, 5, 8, 13, 21, 34 , 55, 89, , , __, …
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:
The next number is: 55 + 89 = 144.
After that: 89 + 144 = 233.
And then: 144 + 233 = 377.
The sequence would become:
1, 2, 3, 5, 8, 13, 21, 34 , 55, 89, ( 144 ), ( 233 ), ( 377 ),……
By examining the parity of the sequence:
1 (odd), 2 (even), 3 (odd), 5 (odd), 8 (even), 13 (odd), 21 (odd), 34 (even), 55 (odd), 89 (odd), 144 (even), 233 (odd), 377 (odd).
Therefore, the next term in the sequence after 377 will be an even number following the parity cycle.

Q. What is the parity of each number in the sequence? Do you notice any pattern in the sequence of parities?
Solution:
The parity of the sequence:
1 (odd), 2 (even), 3 (odd), 5 (odd), 8 (even), 13 (odd), 21 (odd), 34 (even), 55 (odd), 89 (odd), 144 (even), 233 (odd), 377 (odd).
Pattern in the sequence of parities: odd, odd, even i.e., two odd numbers are followed by an even number.

Figure It Out

1. A light bulb is ON. Dorjee toggles its switch 77 times. Will the bulb be on or off? Why?
Solution:
Starting from the initial ON state of the bulb:
1st toggle → OFF
2nd toggle → ON
3rd toggle → OFF
…and so on.
Thus, an odd number of toggles will make the bulb ‘OFF’ and even number of toggles will make the bulb ‘ON’.
Since 77 is an odd number, the bulb will end up in the OFF state.

Page 144

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.
The sum of the even numbers is even. (because adding any number of even numbers is even).
The sum of the odd numbers is even. (because adding an even number of odd numbers is 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.

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:

4. Make a 3 × 3 magic square with 0 as the magic sum. All numbers cannot be zero. Use negative numbers, as needed.
Solution:
Using numbers from -4 to 4 to create a 3 ´ 3 magic square with 0 as the magic sum.

Rows:
(−3) + 2 + 1 = 0
4 + 0 + (−4) = 0
(−1) + (−2) + 3 = 0
Columns:
(−3) + 4 + (−1) = 0
2 + 0 + (−2) = 0
1 + (−4) + 3 = 0
Diagonals:
(−3) + 0 + 3 = 0
1 + 0 + (−1) = 0

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:
(a) Even
(b) Even
(c) Even
(d) Odd

6. What is the parity of the sum of the numbers from 1 to 100?
Solution:
There are 100 numbers from 1 to 100.
Parity: odd, even, odd, even…
There are exactly 50 odd numbers and 50 even numbers.
Adding 50 odd numbers gives an even sum (because an even number of odd numbers sums to even).
Adding 50 even numbers gives an even sum.
Total sum: even + even = even.
Therefore, the parity of the sum of numbers from 1 to 100 is even.

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:
In the Virahanka sequence, the next number is obtained by adding the two previous numbers.
Given that two consecutive numbers in sequence are 987 and 1597.
The next two numbers are:
987 + 1597 = 2584.
1597 + 2584 = 4181.
The previous two numbers are:
1597 – 987 = 610.
987 – 610 = 377.

Number Play Class 7 Maths Chapter 6 PDF Download

Students can now easily access NCERT Solutions for Class 7 Maths Ganita Prakash Chapter 6 in PDF format. Students can download the PDF from the link below and use it to practice questions, check answers, and prepare effectively for exams.

Benefits of Using NCERT Solutions for Class 7 Maths Ganita Prakash Chapter 6

Using NCERT Solutions for Class 7 Maths Ganita Prakash Chapter 6 can make learning easier and more effective. These solutions help students understand the chapter clearly and improve their problem-solving skills.

• Provides step-by-step solutions for all exercises in Chapter 6 – Number Play.

• Helps students understand number patterns, sequences, and parity easily.

• Makes it easier to practice questions, check answers, and attempt PYQs effectively.

• Strengthens problem-solving and logical thinking skills.

• Builds confidence in mathematics for exams and higher studies.

• Saves time by giving clear and accurate explanations.

• Helps with sample papers and other practice tests for better exam readiness.