NCERT Solutions for Ch 6 Number Play Class 7 Maths
Book Solutions1
Answer
In the figure, the children have numbers above their heads: 0, 1, 0, 2, 2, 2, 1.These numbers show how many taller people are there in front of that particular person.
2
Answer
In the figure, each number shows how many taller people are standing in front of a person.For example, the number above the second person is 0, which means no one taller is standing in front of him.
3
Answer
Do it Yourself!4
(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
Answer
(a) The required arrangement is FCBGADE.
(b) The required arrangement is AECGBDF.
(c) The required arrangement is FDBGCEA.
(d) The required arrangement is EAGCDBF.
(e) The required arrangement is FAECGBD.
(f) The required arrangement is BDFAECG.
5
(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 says ‘0’ when they see no one taller than themselves. The tallest person will always say ‘0’, but a shorter person can also say ‘0’ if they are at the front or in a position where no one taller is ahead of them. Thus, the given statement is only sometimes true.
(b) Always True: If a person is the tallest, then no one is taller than them, so they will always say ‘0’. So, the given statement is always true.
(c) Always True: Each person is assigned a number that represents how many taller people are ahead of them. Since there is no one ahead of the first person, their number will always be ‘0’. Hence, the given statement is always true.
(d) Only Sometimes True. A person in between can say 0 if no one ahead is taller (e.g., second tallest in second position).
(e) Only Sometimes True: The statement is only sometimes true. A person who calls out the largest number has many taller people in front but may not be the shortest overall. For example, if the shortest person is standing at the front, they will call out ‘0’. Meanwhile, the second shortest person could be at the back and might call out the largest number.
(f) If there are 8 people, then the shortest person will see 7 taller people. So the maximum number someone can say is 7.
6
Can you figure out which 5 cards add to 30? Is it possible? Are there many ways of choosing 5 cards from this collection? Is there a way to find a solution without checking all possibilities?
Answer
No, it is not possible, as the sum of 5 odd numbers is always odd, but 30 is an even number.7
Answer
When even numbers are added together, the sum will always be an even number, regardless of how many numbers are added.
For example, adding 2 + 4 gives 6, an even number. Similarly, adding 8 + 2 + 4 results in 14, which is still an even number.
Thus, the sum of any even numbers will always be even, no matter how many numbers are added.
8
Answer
When odd numbers are added together, the sum will always be odd if an odd number of odd numbers are added. If an even number of odd numbers are added, the sum will be even.
For example:
- 3 + 5 = 8 (even)
- 3 + 5 + 7 = 15 (odd)
So, adding odd numbers together gives an odd or even result based on the number of odd numbers being added.
9
Answer
The sum of 3 odd numbers is odd and cannot be arranged in pairs (e.g., 1 + 3 + 5 = 9).10
Answer
Based on the given examples for number cards 1, 3, 5, 7, 9, 11, 13.(a) Sum of 4 odd numbers = 1 + 3 + 5 + 7 = 16 (even), can be arranged in pairs.
(b) Sum of 5 odd numbers = 1 + 3 + 5 + 7 + 9 = 25 (odd), cannot be arranged in pairs.
(c) Sum of 6 odd numbers = 1 + 3 + 5 + 7 + 9 + 11 = 36 (even), can be arranged in pairs.
11
(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
Answer
(a) Even + Even = Even and Odd + Odd = Even.
Adding the two results, we get Even + Even = Even.
The parity of the result is even.
Example: 2 + 4 + 3 + 5 = 6 + 8 = 14 (Even)
(b) Two odd numbers: Their sum is even (e.g., 3 + 5 = 8, even), as the extra ones pair up.
Three even numbers: Each is even, so their sum is even (e.g., 2 + 4 + 6 = 12, even).
Sum: (even) + (even) = even.
The sum of 2 odd numbers and 3 even numbers is even.
(c) Each even number is a complete pair. Adding any number of even numbers keeps the sum as pairs.
Example: 2 + 4 + 6 + 8 + 10 = 30, even.
The sum of 5 even numbers is even.
(d) Odd + Odd = Even (4 such pairs).
Adding the 4 such results, we get
Even + Even + Even + Even = Even
The parity of the result is even.
Example: 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64 (Even)
12
Answer
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.
13
(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 =
Answer
(d) Example:
6 – 2 = 4 → even
8 – 4 = 4 → even
Parity of result = even
∴ even – even = even
(e) Example:
7 – 3 = 4 → even
9 – 5 = 4 → even
Parity of result = even
∴ odd – odd = even
(f) Example:
8 – 3 = 5
12 – 5 = 7
Parity of result = odd
∴ even – odd = odd
(g) Example:
7 – 2 = 5
9 – 6 = 3
Parity of result = odd
∴ odd – even = odd
14
Small Squares in Grids
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.
Q: Given the dimensions of a grid, can you tell the parity of the number of small squares without calculating the product?
Answer
Yes, we can determine the parity of the number of small squares in a grid without directly calculating the full product, simply by observing the parity of the dimensions.
Rule: The product of two numbers is:
- Even if at least one of the numbers is even.
- Odd if both numbers are odd.
15
(a) 27 × 13
(b) 42 × 78
(c) 135 × 654
Answer
(a) Both 27 and 13 are odd numbers, and Odd × Odd = Odd.
So, the parity of the number of small squares is odd.
(b) Both 42 and 78 are even numbers, and Even × Even = Even.
So, the parity of the number of small squares is even.
Since 3276 is even, the answer is even.
(c) 135 is odd, 654 is even, and Odd × Even = Even.
So, the parity of the number of small squares is even.
16
Parity of Expressions
Consider the algebraic expression: 3n + 4. For different values of n, the expression has different parity:
Q: Come up with an expression that always has even parity.
Answer
Examples: 2n, 4n + 2, 6n - 4. Any expression of the form 2k or 2k + even.17
Answer
Examples: 2n + 1, 2n - 1, 4n + 3. Any expression of the form 2k + 1.18
Answer
Examples: 3n, n + 5, 5n + 2. Any expression where the coefficient of n is odd and the constant term affects parity based on n.19
Answer
Yes, 2n (n = 1, 2, 3, ...) lists all even numbers (2, 4, 6, ...).
20
Answer
Yes, 2n - 1 (n = 1, 2, 3, ...) lists all odd numbers (1, 3, 5, ...).21
Answer
The nth even number is 2n.22
Answer
The 100th odd number is 2 × 100 - 1 = 199.23
Answer
The nth odd number is 2n - 1.24
Some Explorations in Grids
Observe this 3 × 3 grid. It is filled following a simple rule— use numbers from 1 – 9 without repeating any of them. There are circled numbers outside the grid.
Q: Are you able to see what the circled numbers represent?
Answer
The circled numbers represent the sums of the corresponding rows and columns in the 3×3 grid.25
Answer
26
Answer
Do it Yourself!27
Answer
The magic sum cannot be just any number. It is fixed for a 3 x 3 magic square and depends on the numbers being used.For the numbers 1 to 9, the magic sum is 45.
28
Answer
Yes, 1 can be in a corner position. In the given example, 1 is in the top-left corner, and we have 1 + 5 + 9 = 15 along the diagonal. Let's check other ways:
- Row: 1 + (second number) + (third number) = 15
1 + 5 + (third number) = 15
1 + 5 = 6, so third number = 15 - 6 = 9. This matches the given grid. - Column: 1 + (second number) + (third number) = 15
1 + (second number) + 9 = 15
1 + 9 = 10, so second number = 15 - 10 = 5. This also matches.
Thus, it is possible for 1 to be in a corner.
29
Answer
Do it Yourself!30
Answer
31
Answer
There are 8 distinct 3×3 magic squares using 1-9 (considering rotations and reflections as equivalent).32
Answer
The numbers 2-10 are 9 consecutive numbers, just like 1-9, but increased by 1.
Strategy: Start with the classic 1-9 magic square, and add 1 to each number.
Original: After adding 1 to each:
The magic square for numbers 2-10 would have a different magic sum (18) compared to numbers 1-9 (15). The structure remains similar, but the values are shifted up by 1.
33
(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?
Answer
(a) After increasing each number by 1:
This is still a magic square.
New magic sum = 18
(b) After doubling each number
Still a magic square
New magic sum = 30
In case (a), adding a constant to every number, → magic sum increases by 3 × that constant.
In case (b), multiplying all by a constant → magic sum multiplied by that constant.
34
Answer
Add a constant, multiply by a constant, or reflect/rotate the grid.
35
Answer
The magic square for numbers 2-10 can be seen in solution 2 above.
For the magic square using numbers 3-11, add 2 to each number in the original, and we get the adjoining magic square.
Here, magic sum = 21
For the magic square using numbers 9-17, add 8 to each number in the original, and we get a square as shown below.
Here, magic sum = 39
36
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].
Answer
For a magic square with centre m = 5:
Relative to m:
37
Answer
Do it Yourself!38
Answer
Magic sum = 3 × 25 = 75. Example:
39
Answer
Row sum (1st row) = 28 + 21 + 26 = 75Column sum (1st column) = 28 + 23 + 24 = 75
Diagonal sum (1st column) = 28 + 25 + 22 = 75
The expression obtained = 3 × m
where m is the letter-number representing the number in the centre.
40
Q3: Write the result obtained by-
(a) adding 1 to every term in the generalised form.
(b) doubling every term in the generalised form
Answer
(a)
(b)
41
Answer
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 of 60 is:
42
Answer
Yes, it is possible.Justification: Let us consider the two magic squares with a magic sum 45.
43
The First-ever 4 × 4 Magic Square
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. 
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?
Answer
Yes, we can find different combinations of 4 numbers that add up 34 in the given square.
- Sum of 4 corner numbers: 7 + 14 + 4 + 9 = 34
- Sum of 4 central numbers: 13 + 8 + 10 + 3 = 34
- Sum of 4 numbers in any 2 × 2 squares:
- For example, top-left square: 7 + 12 + 13 + 2 = 34
44
Answer
Here n = 6Write a ‘1+’ in front of all rhythms having 5 beats and then a ‘2+’ in front of all rhythms having 4 beats. This gives us all the rhythms having 6 beats.
Yes, we get a total of 13 ways.
45
Answer
55 + 34 = 89.46
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, _______,_______,_______,...
Answer
The next 3 terms in the sequence are:55 + 89 = 144
89 + 144 = 233
144 + 233 = 377
47
Answer
To determine if the next number after 377 is odd or even without adding the previous terms, let’s examine the parity of the sequence.Parity pattern: 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).
48
Answer
Here, the parity alternates as follows:odd, odd, even i.e., two odd numbers are followed by one even number.
So, the next number (after 377) will be even, as per the repeating parity cycle.
The pattern of parities: Repeats every 3 terms as Odd, Odd, Even.
49
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?
Answer
Both the tens place and the units place of the sum have the same digit.
50
Answer
The possible value of two 2-digit numbers 72 with unit digit 2 is 72 + 72 = 144.M corresponds to 4 and H corresponds to 1 in 144.
So, H cannot be 2 or 3.
51
Share how you thought about each question with your classmates; you may find some new approaches.
These types of questions are called ‘cryptarithms’ or ‘alphametics’.
Answer
(i)
Here, Y = 9, Z = 1, and O = 0.
(ii)
Here, D = 0, B = 7 and E = 1.
(iii)
Here, K=6, P=1, R=2.
(iv)
Here, C=9, F=0.
52
Answer
Dorjee toggles the switch 77 times.Each toggle changes the state of the bulb (ON to OFF or OFF to ON).
Starting from ON:
An odd number of toggles will leave the bulb OFF, and an even number of toggles will leave the bulb ON.
Since 77 is odd, after 77 toggles, the bulb will be OFF.
53
Answer
Total sheets = 50Each 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.
54
Answer
Let’s label the cells as:
So the constraints are:
- Row 1 (A, B, C): sum is odd
- Row 2 (D, E, F): sum is even
- Column 1 (A, D): sum is even
- Column 2 (B, E): sum is even
- Column 3 (C, F): sum is odd
We’ll track parities only (o or e), not actual numbers.
Row 1: A = o, B = e, C = e, then o + e + e = odd
Column 1 (A, D) = e means A must be paired with D to get the sum as even.
So, if A = o, D = o, then o + o = even
Similarly, if B = e, E = e, then e + e = even
Again, if C = e, F = o, then e + o = odd
So, the 6 boxes with 3 odd numbers and 3 even numbers are as follows:
55
Answer
It is given that
- The magic square is 3 × 3.
- The magic sum is 0.
- All numbers in the square cannot be zero; we can use negative numbers as needed.
So, we will use the numbers (-4) to 4 to create a magic square whose magic sum is 0.
56
(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 ______
Answer
(a) Sum of odd number of even numbers is even.(b) Sum of even number of odd numbers is even.
(c) Sum of even number of even numbers is even.
(d) Sum of odd number of odd numbers is odd.
57
Answer
Sum = 1 + 2 + ... + 100 = 100 × 101 / 2 = 5050. Even (100 is even).
58
Answer
The given numbers are 987 and 1597.In the Virahanka sequence, each number is the sum of the two preceding numbers.
The next two numbers are:
987 + 1597 = 2584
1597 + 2584 = 4181
The previous two numbers are:
1597 – 987 = 610
987 – 610 = 377
The sequence is …..,377, 610, 987, 1597, 2584, 4181,…..
59
Answer
Ways in which Angaan can climb 8 steps with 1 or 2 steps are as follows:
For n = 8
So, the total ways in which Angaan reaches the top = 1 + 7 + 15 + 10 + 1 = 34 ways.
60
Answer
Consider the Virahanka sequence given below:1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987,…..
Let us observe the pattern of odd/even in Virahanka numbers given above:
Here 1 → odd
2 → even
3 → odd
5 → odd
8 → even
13 → odd
21 → odd
34 → even
So parity cycle is: odd, even, odd, (repeats every 3 terms)
So the 20th term of the Virahanka sequence is even.
61
(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.
Answer
(a) Substituting m = 1 in 4m – 1 = 4 × 1 – 1 = 3 (odd).
Substituting m = 2 in 4m – 1= 4 × 2 – 1 = 7 (odd).
Thus, the expression 4m – 1 always gives odd numbers.
This statement is true.
(b) Substituting j = 1 in 6j – 4 = 6 × 1 – 4 = 2 (even).
Substituting j = 2 in 6j – 4 = 6 × 2 – 4 = 8 (even).
This expression produces even numbers, but it does not produce all even numbers (for e.g., it skips 4 and 6).
This statement is false.
(c) Substituting p = 1, 2, 3,….. in 2p + 1, we get 3, 5, 7,…..
Substituting q = 1, 2, 3,…. in 2q + 1, we get 1, 3, 5, 7,…….
Here, 2q – 1 describes all the odd numbers but 2p + 1 does not describe 1.
Thus, this statement is false.
(d) Substituting f = 1, 2f + 3 = 2 × 1 + 3 = 5 (odd).
Substituting f = 2, 2f + 3 = 2 × 2 + 3 = 7 (odd).
The expression 2f + 3 always gives odd numbers because 2f is even and adding 3 makes it odd.
This statement is false.
62
Answer
Here, T is at hundreds place, so T = 1⇒ A = 0 and U = 9.
So, we have U = 9, T = 1, and A = 0.
