CBSE Important Questions Class 7 Maths Chapter 6 Number Play

CBSE Important Questions Class 7 Maths Chapter 6 – Number Play

Class 7 Maths Chapter 6 Number Play is a chapter of the new NCERT (National Council of Educational Research and Training) textbook Ganita Prakash. It is a fun chapter about patterns and puzzles with numbers: odd and even numbers, the Virahāṅka–Fibonacci sequence, children announcing the number of taller people in front of them, cryptarithms where letters stand for digits, and 3×3 magic squares.

These CBSE important questions for Class 7 Maths Chapter 6 include multiple choice questions, puzzle questions with pictures, a cryptarithm, and magic square problems with full reasoning. Every question has a clear step-by-step answer in simple English, so students can practise on their own, parents can check the working, and teachers can use them in class. A free printable PDF of these important questions is also available on this page.

CBSE Important Questions Class 7 Maths Chapter 6 Number Play

Number Play (New NCERT)
Medium
Q.
Write the result obtained by—
(a) Adding 1 to every term in the generalised form.
(b) Doubling every term in the generalised form.
Number Play (New NCERT)
Medium
Q.
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.
Number Play (New NCERT)
Medium
Q.
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.
Number Play (New NCERT)
Medium
Q.
​ What is the parity of the 20th term of the Virahāṅka sequence?​
Number Play (New NCERT)
Medium
Q.
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?
Number Play (New NCERT)
Medium
Q.
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?
Number Play (New NCERT)
Medium
Q.
​What is the parity of the sum of the numbers from 1 to 100?​
Number Play (New NCERT)
Medium
Q.
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 _____
Number Play (New NCERT)
Medium
Q.
​Make a 3 × 3 magic square with 0 as the magic sum. All numbers cannot be zero. Use negative numbers, as needed.​
Number Play (New NCERT)
Medium
Q.
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?
Number Play (New NCERT)
Medium
Q.
​Create a magic square whose magic sum is 60.​
Number Play (New NCERT)
Medium
Q.
​A light bulb is ON. Dorjee toggles its switch 77 times. Will the bulb be on or off? Why?​
Number Play (New NCERT)
Medium
Q.
These types of questions can be interesting and fun to solve! Here are some more questions like this for you to try out. Find out what each letter stands for. Share how you thought about each question with your classmates; you may find some new approaches.
​YY+ZZOOB5+3DED5KP+KPPRRC1+C1FF\begin{array}{r}\mathrm{YY} \\+\quad \mathrm{Z} \\\hline \mathrm{ZOO}\end{array} \quad \begin{array}{r}\mathrm{B5} \\+\quad 3 \mathrm{D} \\\hline \mathrm{ED} 5\end{array} \quad \begin{array}{r}\mathrm{KP} \\+\quad \mathrm{KP} \\\hline \mathrm{PRR}\end{array} \quad \begin{array}{r}\mathrm{C} 1 \\+\quad \mathrm{C} \\\hline 1 \mathrm{FF}\end{array}​​
Number Play (New NCERT)
Medium
Q.
Let us look at one more example shown on the right. Here K2 means that the number is a 2-digit number having the digit ‘2’ in the units place and ‘K’ in the tens place. K2 is added to itself to give a 3-digit sum HMM. What digit should the letter M correspond to? Both the tens place and the units place of the sum have the same digit. What about H? Can it be 2? Can it be 3?
​K2+K2HMM\begin{array}{r} \mathrm{K} 2 \\ +\quad \mathrm{K} 2 \\ \hline \mathrm{HMM} \end{array}​​
Number Play (New NCERT)
Medium
Q.
​What is the parity of each number in the sequence? Do you notice any pattern in the sequence of parities?​
Number Play (New NCERT)
Medium
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)?
Number Play (New NCERT)
Medium
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?​
Number Play (New NCERT)
Medium
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?
Number Play (New NCERT)
Medium
Q.
​ Is it possible to get a magic square by filling nine non-consecutive numbers?​
Number Play (New NCERT)
Medium
Q.
Solve this cryptarithm:
​UT+TATAT\begin{array}{r}\rm { UT } \\+\quad \rm{ TA } \\\hline \rm { TAT }\end{array}​​

CBSE Important Questions Class 7 Maths Chapter 6 Number Play

Q1. Which of the following expressions results in both odd and even numbers for any integer n?
(a) 2n
(b) 2n – 1
(c) 5n
(d) 4n

Answer: (c) 5n
2n and 4n are always even, and 2n – 1 is always odd. But 5n is odd when n is odd (5 × 3 = 15) and even when n is even (5 × 2 = 10). So 5n gives both odd and even numbers.

Q2. Seven children are standing in a line as shown in the picture, each with a different height. If each child reports the number of children taller than them standing in front, which of the following sequences could correctly represent their answers (from left to right)?
(a) 0, 1, 2, 2, 3, 5, 6
(b) 0, 1, 2, 2, 3, 5, 5
(c) 0, 1, 2, 2, 4, 5, 5
(d) 0, 0, 1, 2, 2, 4, 6

Seven children of different heights standing in a line

Answer: (b) 0, 1, 2, 2, 3, 5, 5
The first child has nobody in front, so 0. The second child is shorter than the first, so 1. The third child is shorter than both, so 2. The fourth child is taller than the third, so still 2. The fifth child has 3 taller children in front, the sixth (the smallest) has 5, and the last child is taller than the sixth but shorter than the rest, so 5. The last number can never be 6, because the sixth child is shorter than the seventh.

Q3. Which of these combinations of 7 odd numbers adds up to 77?
(a) 1, 3, 5, 7, 9, 11, 39
(b) 3, 5, 7, 9, 11, 13, 31
(c) 5, 7, 9, 11, 13, 15, 17
(d) 7, 9, 11, 13, 15, 17, 19

Answer: (c) 5, 7, 9, 11, 13, 15, 17
5 + 7 + 9 + 11 + 13 + 15 + 17 = 77. The other options give 75, 79 and 91.

Q4. In a line of children, each child announces the number of taller children standing in front of them. Which of the following does not represent a valid sequence?
(a) 0, 0, 0, 0, 0
(b) 0, 1, 1, 1, 1
(c) 1, 0, 1, 0, 1
(d) 0, 1, 0, 1, 0

Answer: (c) 1, 0, 1, 0, 1
The first child in the line has nobody in front, so the first number must always be 0. Sequence (c) starts with 1, so it is not possible.

Q5. Six friends decide to stand in a single line, each with different heights. To make things interesting, each child announces the number of taller children standing ahead of them in the line. The sequence of numbers they call out is: 0, 1, 0, 2, 1, 5. Construct a possible height arrangement for the 6 children.

Answer: The numbers tell how many taller children are ahead of each child.
Child 1 says 0: nobody is ahead.
Child 2 says 1: child 1 is taller than child 2.
Child 3 says 0: child 3 is taller than both children ahead, so child 3 is the tallest so far.
Child 4 says 2: two of the first three are taller, so child 4 is taller than child 2 but shorter than children 1 and 3.
Child 5 says 1: only one child ahead is taller, so child 5 is taller than children 1, 2 and 4 but shorter than child 3.
Child 6 says 5: all five children ahead are taller, so child 6 is the shortest.
A possible arrangement of heights (shortest to tallest): child 6 < child 2 < child 4 < child 1 < child 5 < child 3.

Six children calling out 0, 1, 0, 2, 1, 5

Q6. In a line of 16 people, each states the number of taller people standing in front of them. What is the largest possible number that can appear in this sequence?
(a) 16
(b) 13
(c) 14
(d) 15

Answer: (d) 15
The last person in the line has 15 people in front. If the last person is the shortest of all, then all 15 people in front are taller. So the largest possible number is 15.

Q7. A tribal musician is arranging small wooden drums in a display for a cultural exhibition. The number of drums in each row follows a specific pattern: 1, 2, 3, 5, 8, ... Describe the pattern and calculate the total number of drums in the 7th, 8th and 9th rows combined. Show your calculations.

Answer: This is the Virahāṅka–Fibonacci pattern: each number is the sum of the two numbers before it.
1st row: 1, 2nd row: 2, 3rd row: 1 + 2 = 3, 4th row: 2 + 3 = 5, 5th row: 3 + 5 = 8, 6th row: 5 + 8 = 13, 7th row: 8 + 13 = 21, 8th row: 13 + 21 = 34, 9th row: 21 + 34 = 55.
Total drums in the 7th, 8th and 9th rows = 21 + 34 + 55 = 110.

Q8. The Virahāṅka–Fibonacci sequence is given as 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... Identify the pattern in this sequence and use it to find the 13th term.

Answer: Each term is the sum of the two terms before it. Continuing the sequence: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, ...
The 11th term is 89 + 55 = 144, the 12th term is 144 + 89 = 233, and the 13th term is 233 + 144 = 377.

Q9. A puzzle enthusiast is attempting to design a 3×3 magic square for a weekend challenge. The rule is simple: use the digits 1 to 9 exactly once so that every row, column and diagonal adds up to 15. Excitedly, the enthusiast fills in a few numbers: the second column gets a 7 in the first row, the second row takes a 6 on the left and a 4 on the right, and the third row has a 3 in the middle. Looking at the half-filled square, can this arrangement be completed into a proper magic square, or is it impossible?

Half-filled 3x3 square with 7, 6, 4 and 3

Answer: In any 3×3 magic square using 1 to 9, the centre must be 5. Then the middle row 6 + 5 + 4 = 15 and the middle column 7 + 5 + 3 = 15 work. The digits used are 3, 4, 5, 6, 7, and the unused digits are 1, 2, 8, 9.
The top row needs a + 7 + c = 15, so a + c = 8. But no two of the digits 1, 2, 8, 9 add up to 8.
Hence the top row cannot be completed, so this arrangement cannot become a magic square with sum 15 using the digits 1–9 without repetition.

Q10. In the cryptarithm below, each letter stands for a unique digit from 0 to 9. Find the digits represented by A and B. Justify your answer.

Cryptarithm: AB + AB + 5B = 97

Answer: Look at the ones place: B + B + B = 3B must end in 7. The only digit for which 3B ends in 7 is B = 9 (3 × 9 = 27). Write 7 and carry 2.
Now the tens place: A + A + 5 + 2 (carry) = 9, so 2A = 2 and A = 1.
So A = 1 and B = 9. Check: 19 + 19 + 59 = 97 ✓

Q11. Generalise a 3×3 magic square where the centre value is an algebraic expression m. Write expressions for all 8 other cells in terms of m such that the magic sum is 3m. Then, if m = 31, compute the full grid and verify that it forms a magic square.

Answer: Using the standard template, the cells are:
Row 1: m – 3, m + 2, m + 1
Row 2: m + 4, m, m – 4
Row 3: m – 1, m – 2, m + 3
Every row, column and diagonal adds up to 3m.
With m = 31: Row 1: 28, 33, 32; Row 2: 35, 31, 27; Row 3: 30, 29, 34.
Each row, column and diagonal adds up to 93 = 3 × 31, so it is a magic square.

Magic square formed with m = 31

Q12. You are given a 3×3 grid with numbers 1 to 9. Only two corner values are given: top-left = 2 and bottom-right = 8. The sum of each row is known to be 15. Can the grid be a magic square? Prove your conclusion.

Answer: Yes, a magic square is possible. Place 5 in the centre (the centre of a 1–9 magic square is always 5). The diagonal 2 + 5 + 8 = 15 works. Now fill the remaining cells so that every row, column and diagonal adds up to 15:
Row 1: 2, 7, 6
Row 2: 9, 5, 1
Row 3: 4, 3, 8
Rows: 15, 15, 15. Columns: 2 + 9 + 4 = 15, 7 + 5 + 3 = 15, 6 + 1 + 8 = 15. Diagonals: 2 + 5 + 8 = 15 and 6 + 5 + 4 = 15. Hence proved.

Completed magic square with 2 and 8 in the corners

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