Number Play Class 7 Solutions Maths Ganita Prakash Chapter 6

By using Ganita Prakash Class 7 Solutions and Part 1 Chapter 6 Number Play Class 7 Question Answer, students can improve their problem-solving skills.

Class 7 Maths Chapter 6 Number Play Solutions

Ganita Prakash Class 7 Chapter 6 Solutions

Class 7 Maths Ganita Prakash Chapter 6 Solutions Number Play

6.1. Numbers tell us Things

Page: 127

Question 1.
What do the numbers in the figure below tell us?

Answer:
The students are calling a number each of his/her choice.

Question 2.
What do you think these numbers mean?

Answer:
Students have arranged themselves in a row based on the numbers they have called.

Question 3.
Could you figure out what these numbers convey? Observe and try to find out.
Answer:
Here the rule is: Each child calls out the number of children in front of them who are taller than them.

Page: 128

Question 4.
Write down the number each child should say based on this rule for the arrangement shown below.

Solution:

Page: 128

Question 5.
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)

(b)

(c)

(d)

(e)

(f)

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:
(a) Only sometime true.
If a person is at the extreme left i.e., at the beginning of the row then also he or she will say ‘0’.
(b) Always true.
If a person is the tallest, than there is no other person taller than him. Hence, his or her number is ‘0’.
(c) Always true.
As there is no person in front of the first person. So, the first person’s number is ‘0’.
(d) Never true.
Tallest person will always say ‘0’ irrespective of his or her position.
(e) Only sometimes true.
If the shortest person stands at the beginning then he will not say the largest number.
(f) If 8 people are standing, then the largest number possible is 7.

6.2. Picking Parity

Page: 129

Question 6.
Add a few even numbers together. What kind of number do you get? Does it matter how many numbers are added?
Solution:
Taking two even numbers :
2 + 6 = 8 (even number)
Taking three even numbers :
4 + 10 + 20 = 34 (even number)
Taking four even numbers :
14 + 100 + 204 = 318 (even number)
In all the above cases, we get an even number as the outcome.
It does not matter, how many numbers are added, we always get even number if any number of even numbers are added.

Question 7.
Now, add a few odd numbers together. What kind of number do you get? Does it matter how many odd numbers are added?
Solution:
Taking two odd numbers :
3 + 7 = 10 (even)
Taking three odd numbers :
3+7+9 = 19 (odd)
Taking four odd numbers :
3 + 7 + 9 + 11 = 30 (even)
Taking five odd numbers :
3 + 11 + 15 + 17 + 23 = 69 (odd)
We five both parity – even as well as odd.
But we can notice here that if we add even number of odd number then we obtain even number and if we obtain odd number of odd numbers we get an odd number. So, it does not matter how many odd number are added.

Page: 130

Question 8.
What about adding 3 odd numbers? Can the resulting sum be arranged in pairs? No.
Solution:
If we add 3 add numbers, then we get an odd number. Since only even numbers can be arranged in pairs. Hence, resulting sum in the above case cannot be arranged in pairs.

Question 9.
Explore what happens to the sum of (a) 4 odd numbers, (b) 5 odd numbers, and (c) 6 odd numbers.
Solution:
(a) 4 odd numbers :
3 + 7 + 11 + 15 = 36 (even)
We get an even number.

(b) 5 odd numbers :
7 + 11 + 19 + 23 + 29 = 89 (odd)
We get an odd number.

(c) 6 odd numbers :
5 + 7 + 9 + 21 + 23 + 31 = 96 (even)
We get an even number.

Question 10.
Two siblings, Martin and Maria, were born exactly one year apart. Today they are celebrating their birthday. Maria exclaims that the sum of their ages is 112. Is this possible? Why or why not?
Solution:
It is not possible that the sum of their ages is 112.
Since, they were born exactly one year apart. Hence, their ages will be represented by two consecutive numbers. Now, we know that the sum of two consecutive numbers can never be an even number.
Thus, the sum of their ages can never be 112.

Page: 131

Question 11.
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) even
(b) even
(c) even
(d) 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:
Yes. He made a mistake as sum of ₹1 coins an odd number of times will gives an odd number, sum of ₹5 coins an odd number of time will give an odd number like- 5,15,25, 35, etc. ₹10 added for an even number of times will give us an even number.
Now if we sum up the sums of these numbers we will be adding two odd numbers and an even number. As a result we will obtain an even number so we cannot get ₹205 which is an odd number.

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 = ______
(f) odd – even = ______
Solution:
(d) even
(e) even
(f) odd
(g) odd

Question 12.
Given the dimensions of a grid, can you tell the parity of the number of small squares without calculating the product?
Solution:
Yes we can tell the parity of the number of small squares without calculating the product. Our answers will be based on the following facts :
(a) Product of an odd number and an even number is an even number.
(b) Product of two even numbers is an even number.
(c) Product of two odd numbers is an odd number.

Page: 132

Question 13.
Find the parity of the number of small squares in these grids :
(a) 27 × 13
(b) 42 × 78
(c) 135 × 654
Solution:
(a) Here, 27 × 13
= odd × odd
= odd

So, the parity of the number of small squares is ‘odd’.

(b) Here, 42 × 78
= even × even
= even
So, the parity of the number of small squares is ‘even’.

(c) Here, 135 × 654
= odd × even
= even
So, the parity of the number of small square is ‘even’.

Question 14.
Come up with an expression that always has even parity.
Some examples are : 100 p and 48 w-2. Try to find more.
Solution:
Some of the expressions that always has even parity are :
(i) 20 p
(ii) 10 p
(iii) 2 p
(iv) 32 p
(v) 36w + 10
(vi) 26 w+12
(vii) 8 w-6
(viii) 52 w , etc.
Note: Students have to write only one such expression.

Question 15.
Come up with expressions that always have odd parity.
Solution:
Some of the expressions that always has odd parity are :
(i) 3p
(ii) 2p + 1
(iii) 8p + 7
(iv) 10p – 3
(v) 7w
(vi) 10w + 13
(vii) 6w + 1
(viii) 3w – 7, etc.

Question 16.
Come up with other expressions, like 3n + 4, which could have either odd or even parity.
Solution:
Some of the expressions which could have either odd or even parity are:
(i) 5 x-2, it will have even parity if ‘x’ is even and it will have odd parity if ‘x’ is odd.
(ii) 7 x-3, it will have odd parity if ‘x’ is even and even parity if ‘ x ‘ is odd.

Question 17.
The expression 6k+2 evaluates to 8, 14, 20,… (for k+1,2,3,….)- many even numbers are missing.
Solution:
The expression 6k+2 will give two consecutive values which will differ by 6 as it is got by multiplying by 6 and then adding 2.
Hence, there will be many even numbers which will not come in the sequence if we put k = 1,2,3,4,…. etc.

Question 18.
Are there expressions using which we can list all the even numbers?
Solution:
By using the expression 2x , where x = 1,2, 3,…., we can list all the even numbers.
Note: To include all integers we need to take x = 0, ± 1, ± 2, ± 3,….

Question 19.
Are there expressions using which can list all odd numbers?
Solution:
By using the expression 2x + 1, where x = 1,2, 3,.., we can list all the odd numbers.
Note: To include all integers we need to take x=0, ± 1, ± 2, ± 3,….

Question 20.
What would be the n^th term for multiples of 2? Or, what is the n^th even number?
Solution:
n^th term for multiples of 2 = ‘2n’
So, n^th even number is ‘2n’.
Note: In this way we can conclude that 10^th even number is 20 and 32^nd even number is 64 etc.

Question 21.
What is the 100th odd number?
Solution:
Let is write an expression which will generate sequence of odd numbers :
The expressions is: ‘2 k-1’, where k = 1, 2, 3,…
To get the 100^th odd number we put k-100 in the expression.
So, we get
100^th odd number = 2 × 100 – 1
= 200 – 1
= 199

Page: 133

Question 22.
What is the 100^th even number?
Solution:
Sequence of even numbers are given by ‘2k’, k = 1,2,3,….
So, to get the 100^th even number we put k = 100
Thus, 100^th even number = 2 × 10
= 200

Question 23.
Write a formula to find n^th odd number.
Solution:
Formula for n^th odd number can be written as : ‘2n – 1’, n = 1,2,3,…. etc.

6.3. Some Explorations in Grids

Question 24.
Are you able to see what the circled numbers represent?

Solution:

Page: 134

Question 25.
Make a couple of questions like this on your own and challenge your peers.
Solution:
Fill the grids below based on the following rule :
Use numbers from 1-9 without repeating any of them. There are circled numbers outside the grid. You cannot change the numbers already filled in the grid if any.

Page: 136

Question 26.
Figure it Out

1. How many different magic squares can be made using the numbers 1-9?
Solution:
There are 8 different magic squares which can be made using 1-9 :

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:
We have to use 2-10 : My strategy will be : First of all we will fill the centre most square by 6.

We fill 10 and 2 in the middle position.

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:
Let us consider a magic square :

(a) We increase each number by 1.

Yes, it is a magic square. Magic sum in this case is 18 which is 3 more than the previous magic sum.
(b)

Above grid is obtained after doubling each number of the magic square we have considered.
Here the magic sum is 30 which is double the magic sum of the initial magic square we have considered.

4. What other operations can be performed on a magic square to yield another magic square?
Solution:
We can do the following operations on a magic square to yield another magic square :
(a) decrease each number by some fixed value.
(b) divide each number by some fixed number.

5. Discuss ways of creating a magic square using any set of 9 consecutive numbers (like 2-10, 3-11, 9-17, etc.)
Solution:
We can create a magic square using numbers 3-11 as follows :
(i) Fill the centre square by the number 7.
(ii) Put the smallest number 3 in the central row. Write the number in the central row so that the sum is 21 , i.e., write the number 11.

(iii) Write 8 in the top left corner. To get the sum 21, write in the bottom row of I^{\text {st column the number 10.
(iv) Write 4 in the top corner of the third column.
(v) Write 6 in the bottom row of the third column so that the sum of numbers in the third column also becomes 21.
(vi) Now in the middle of the top row fill a number so that the sum of numbers becomes 21.
(vii) Fill in the last two numbers at the bottom row keeping the magic sum in mind. So, we fill 5 in the middle of the bottom row and 6 in the corner of the bottom row.
Here the magic sum = 21.
By using similar strategy we can create another magic square by using numbers 9-17 as follows :

Here the magic sum is ’39’.

Page: 137

Question 27.
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.
Solution:
We can make the following arrangement in the squares :

Note: Many such arrangements are possible.

Question 28.
Once the generalised form is obtained, share your observations with the class.
Solution:
The generalised form can be :

Note: There can be more generalised form.

Question 29.
Figure it Out

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

For the centre number to be 25, we take m = 25. By putting m = 25 in the above generalised form, we get the following magic square.

Magic sum = 75.

2. What is the expression obtained by adding the 3 terms of any row, column or diagonal?
Solution:
The expression obtained by adding the 3 terms of any row, column or diagonal is : ‘3m’.

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:
The generalised magic box is :

(a) adding 1 to every term in the generalised form, we get

(b) doubling every term in the generalised form we obtain.

4. Create a magic square whose magic sum is 60.
Solution:
To create a magic square whose magic sum is 60 , we need to fill the centre most square by 20, the central row by 24, 20, 16
Upper row by 19, 18, 23
Lower row by 17, 22, 21
Therefore, we get the magic square by

5. Is it possible to get a magic square by filling nine non-consecutive numbers?
Solution:
Yes. There are many magic squares with nine non-consecutive numbers.
Let us consider the following magic square with magic number =30 and involving numbers 3,6,7,9,10,11,13,14 and 17, which are not consecutive numbers, obviously!

Note: The generalised form for the above magic square is :

Hey! Many other such magic square are possible. Think!!!

6.4. Nature’s Favourite Sequence: The Virahanka-Fibonacci Numbers

Page: 141

Question 30.
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:
Following are the different possible systematic method to write down all 6 -beat rhythms :
(i) 6 = 1 + 1 + 1 + 1 + 1 + 1
(ii) 6 = 1 + 1 + 1 + 1 + 2
(iii) 6 = 1 + 1 + 1 + 2 + 1
(iv) 6 = 1 + 1 + 2 + 1 + 1
(v) 6 = 1 + 2 + 1 + 1 + 1
(vi) 6 = 2 + 1 + 1 + 1 + 1
(vii) 6 = 2 + 2 + 1 + 1
(viii) 6 = 2 + 1 + 2 + 1
(ix) 6 = 2 + 1 + 1 + 2
(x) 6 = 1 + 2 + 1 + 2
(xi) 6 = 1 + 2 + 2 + 1
(xii) 6 = 1 + 1 + 2 + 2
(xiii) 6 = 2 + 2 + 2
Yes! there are 13 different ways.

Page: 142

Question 31.
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:
Give sequence is :
1,2,3,5,8,13,21,34,55,89,
______, ______, ______, _____
Next three numbers will be :
55 + 89 = 144
89 + 144 = 233
144 + 233 = 377
It will be an even number as the next two numbers row will be odd numbers and the sum of two odd numbers results in an ‘even number’.

Question 32.
What is the parity of each number in the sequence? Do you notice any pattern in the sequence of parties?
Solution:
In the sequence, 1,2,3,5,8,13,21,34,55,89,144,233,377,…. we can easily notice that there are an even number after two odd numbers. So, parity of each number in the sequence is : odd, even, odd, odd, even, odd, odd, even, odd, odd, even, odd, odd,… This gives us a sequence of “odd, odd, even” if we consider from the third term or it gives us a sequence of “even, odd, odd”, if we consider from the second term.

6.5. Digits in Disguise

Page: 143

Question 33.
What could U and T be? Can T be 2? Can it be 3?

Solution:
Taking T = 5, we get

Comparing with, we get U = 1
T can not be 2 as

so there will not be any value for U. Similarly, T can not be 3 as for T = 3, we get

Again there will not be any value for U.

Question 34.
What about H? Can it be 2? Can it be 3?

Solution:
Here sum of two equal two-digit numbers is a three-digit number with unit’s and ten’s digit equal.
So, we can take K = 7, so that

We get H = 1, M = 4
Maximum value of digit can be 9, so we will have

which does not contain 2 in place of H, moreover unit’s digit and ten’s digit are not same.
With the same reason H can not be 3.

Page: 143-144

Question 35.
Figure it Out :

1. A light bulb is ON. Dorjee toggles its switch 77 times. Will the bulb be on or off? Why?
Solution:
Here, parity for ON is even and parity for OFF is odd.
Now, Dorjee has toggled its switch 77 times, whose parity is odd and odd parity is for off.
So, the bulb will be off.

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:
We have 50 pages printed on both sides of the pages.
∴ One even and one odd number will be printed on each page.
So, the sum of page numbers on any page will be an odd number.
We know that sum of 50 consecutive odd numbers can not 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:

Note: There can be other possible ways.

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:

Magic sum = 0.
Note: There can be other possible magic squares with magic = 0.

5. Fill in the following blanks with ‘odd’ or ‘eve’:
(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 numbers 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:
Sum of the numbers from 1 to 100 is 5050 which is even.
Hence, the parity of the sum of the numbers from 1 to 100 is even.

7. Two consecutive numbers in the Virahanka sequence are 987 and 1597. What are the next 2 numbers in the sequence? What are the previous 2 numbers in the sequence?
Solution:
Here two consecutive numbers in the Virahanka sequence are 987 and 1597.
So, the next number will be 987 + 1597 = 2584
The next number to 2584 will be 1597 + 2584 = 4181
Therefore, the next 2584 and 4181.
Just previous number to the 987 will be 1597 – 987 = 610
Again the previous number to 610 will be 987 – 610 = 377.
So, the previous 2 numbers in the sequence are 610 and 377.

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:
Angaan can reach the top in the following ways:

9. What is the parity of the 20^th term of the Virahanka sequence?
Solution:
We know that the parity of terms in the Virahanka sequence are:
odd, even, odd, odd, even, odd, odd, even, odd, odd, even, odd, odd, even, odd, odd, even, odd, odd, even.
If we continue this sequence of parity the 20^th term of Virahanka sequence is ‘even’.

10. Identify the statements that are true.
(a) The expression 4 m-1 always gives odd numbers.
(b) All even numbers can be expressed as 6j – 4.
(c) Both expressions 2p+1 and 2p-1 describe all odd numbers.
(d) The expression 2f + 3 gives both even and odd numbers.
Solution:
(a) True.
By multiplying any number by 4 we get an even number.
If we subtract 1 from an even number we get an odd number.
(b) Not true.
6j-4 cannot give us 0 for any integer ‘ j ‘.
(c) True.
By taking different values of p in 2p+1 and 2p-1, we can get all odd numbers.
(d) Not true.
2f+3 will always give odd number.

11. Solve this cryptarithm :

Solution:
Take U = 9, T = 1 and A = 0
We get