Class 7 Maths Chapter 6 Number Play – NCERT Solutions, PDF & Key Tips

NCERT Solutions for Class 7 Maths Chapter 6 Number Play (2025-26)

6.1 Numbers Tell us Things

Figure it Out

1. (a) Arrange the stick figure cutouts given at the end of the book or draw a height arrangement such that the sequence reads:

   • 0, 1, 1, 2, 4, 1, 5
   • 0, 0, 0, 0, 0, 0, 0
   • 0, 1, 2, 3, 4, 5, 6
   • 0, 1, 0, 1, 0, 1, 0
   • 0, 1, 1, 1, 1, 1, 1
   • 0, 0, 0, 3, 3, 3, 3

   Answer: For each arrangement, place children in such a way that the number each calls out is equal to the number of children in front of them who are taller, matching the sequence given. For example, in sequence (b) all numbers are 0—so the tallest child stands at the front, and all others are lined up decreasing in height, so none has anyone taller in front.

   
   

2. For each of the statements below, 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?

   Answer:

   • (a) Only Sometimes True. A person can say ‘0’ if there is no one taller in front—even if not the tallest overall.
   • (b) Always True. The tallest person will have no one taller in front, so their number is always ‘0’.
   • (c) Always True. The first person has no one in front, so their number is ‘0’.
   • (d) Only Sometimes True. A person in between can still say ‘0’ if there are no taller people in front.
   • (e) Only Sometimes True. The shortest person may have the largest number if all taller children are in front, but not always.
   • (f) 7. In a group of 8 people, the shortest can have all 7 taller children in front, so the largest number is 7.

6.2 Picking Parity

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 (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
   • (a) Even (Two odds make even, sum with two evens is still even)
   • (b) Even (Pair the odds, get even, total with 3 evens is even)
   • (c) Even (Sum of any number of even numbers is even)
   • (d) Even (Sum of even number of odds is even; 8 is 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?

   Answer: Yes, Lakpa made a mistake. The sum of an odd number of ₹1 coins (odd), odd number of ₹5 coins (odd), and even number of ₹10 coins (even) gives an even total. But 205 is odd, so this is not possible.

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 = even
   • (e) odd – odd = even
   • (f) even – odd = odd
   • (g) odd – even = odd

Small Squares in Grids

• (a) 27 × 13 → Odd × Odd = Odd (Parity: Odd)
• (b) 42 × 78 → Even × Even = Even (Parity: Even)
• (c) 135 × 654 → Odd × Even = Even (Parity: Even)

Parity of Expressions

• Expression that always has even parity: 2n, 4n, 100p, 48w – 2, etc.
• Expressions that always have odd parity: 2n – 1, 2k + 1, 4m – 1, etc.
• Expressions like 3n + 4 (can be even or odd): 3n + 4, n + 1, n + 3, etc.

To list all even numbers: Expressions like 2n, where n = 1, 2, 3,...
To list all odd numbers: Expressions like 2n – 1 or 2n + 1, where n = 1, 2, 3,...

Formula for the nth odd number: 2n – 1

6.3 Some Explorations in Grids

Figure it Out

1. There are 8 possible 3 × 3 magic squares (up to rotation and reflection) made using the numbers 1–9.
2. To create a magic square using 2–10, add 1 to every number in the 1–9 magic square; the new magic sum will be 18 (since 15 + 3 × 1 = 18).
3. Take a magic square, and
   • (a) When you increase each number by 1, the resulting grid is still a magic square, and the magic sum increases by 3.
   • (b) When you double each number, the resulting grid is a magic square and the magic sum doubles.
4. Other operations: multiplying all numbers by a constant, or adding a constant to all numbers will also result in a magic square.
5. To create a magic square with any set of 9 consecutive numbers, start with a standard magic square and add/subtract the same value to all entries as needed.

Generalising a 3 × 3 Magic Square

1. If the centre of the magic square is m, other entries can be expressed as m ± numbers so that all rows, columns, and diagonals sum to 3m.
2. Sum of any row/column/diagonal = 3m
3. (a) Adding 1 to every entry: sum becomes 3(m + 1) = 3m + 3
   (b) Doubling every entry: sum becomes 3 × (2m) = 6m
4. To create a magic square with sum 60, set m = 20.
5. It is not possible to fill using nine non-consecutive numbers and retain the magic property for all rows, columns, and diagonals.

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

Write the next number in the sequence, after 55.

Answer: 89 (since 34 + 55 = 89)

Write the next 3 numbers in the sequence:

Answer: 144, 233, 377 (since 55 + 89 = 144, 89 + 144 = 233, 144 + 233 = 377)

6.5 Digits in Disguise

Cryptarithms Examples

• T + T + T = UT: T = 5, UT = 15
• K2 + K2 = HMM: K = 6, HMM = 124 (since 62 + 62 = 124, M = 4, H = 1)

Figure it Out

1. The bulb will be OFF. Since starting from ON, an odd (77) number of toggles flips the bulb to OFF.
2. No, it is not possible. Since pages are in pairs (even sums), the total sum of page numbers from 50 double-sided leaves cannot be 6000 (not possible as 6000/50 = 120, page numbers are sequential, but total will not match).
3. Place 3 odd and 3 even numbers in the 2×3 grid so that each row and column sum matches the required parity as given in the question.
4. A 3 × 3 magic square with sum 0: Example: 2, -1, -1; -1, 2, -1; -1, -1, 2.
5. (a) odd
   (b) even
   (c) even
   (d) odd
6. Sum of numbers from 1 to 100 = 5050 (even).
7. Next two after 987 and 1597: 2584, 4181. Previous two: 610, 377.
8. Ways to climb 8-step staircase (using steps of 1 or 2): 34 ways (8th Fibonacci number).
9. Parity of the 20th term of the Virahāṅka sequence: Even.
10. True statements:
    • (a) True
    • (b) False
    • (c) True
    • (d) True
11. Solve this cryptarithm:
    UT + TA = TAT
    Answer: U = 6, T = 7, A = 5 (since 67 + 75 = 142, which does not match; try U = 9, T = 9, A = 5, but 99 + 95 = 194, not 999. Check the next. U = 1, T = 1, A = 0, 11 + 10 = 21 = 121. The possible solution is T = 1, U = 1, TA = 10, 11 + 10 = 21. But TAT = 121 which is not possible. This cryptarithm might have no valid solution in conventional digits. Since the book asks to "try and think", this can be left as: No solution found with standard digits. If you have found a different solution, check your working carefully!

Mastering Number Play for Class 7

Understanding NCERT Solutions for Class 7 Maths Chapter 6 Number Play is key to acing your exams. Dive deep into concepts like parity, magic squares, and the Fibonacci sequence to develop strong problem-solving skills.

Practice regularly using chapter-wise solutions to boost your confidence. This chapter connects mathematics to logic and creativity, making maths both fun and meaningful. Strengthening these topics will help in competitive exams ahead.

Focus on understanding how patterns and logical rules work for better concept clarity. Complete the NCERT exercises, review important formulas, and test your knowledge with similar questions for effective exam preparation.