Number Play Class 7 Solutions RBSE Maths Ganita Prakash Chapter 6

Practicing Ganita Prakash Class 7 Solutions and RBSE Class 7 Maths Chapter 6 Number Play Solutions Question Answer helps develop logical thinking and accuracy.

RBSE Class 7 Maths Chapter 6 Number Play Solutions

Ganita Prakash Class 7 Chapter 6 Solutions

Class 7 Maths Ganita Prakash Part 1 Chapter 6 Solutions

In-text Questions
Page 129

Question 1.
Can you figure out which 5 cards add to 30? Is it possible?

Solution:
No, it is not possible
Since the sum of 5 odd numbers is always odd and 30 is an even number.

Page 130

Question 1.
Explore what happens to the sum of
(a) 4 odd numbers
(b) 5 odd numbers and
(c) 6 odd numbers
Solution:
Let the number cards are 1, 3, 5, 7, 11 and 15.
(a) Sum of 4 odd numbers = 1 + 3 + 5 + 7 = 16 (even) these can be arranged in pairs.
(b) Sum of 5 odd numbers = 1 + 3 + 5 + 7 + 11 = 27 (odd) these cannot be arranged in pairs.
(c) Sum of 6 odd numbers= 1 + 3 + 5 + 7 + 11 + 15 = 42 (even) these can be arranged in pairs.

Question 2.
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:
No, it is not possible.
Martin’s and Maria’s ages are consecutive numbers, they cannot add upto 112 as 112 is an even number.
The resulting sum of an even number and an odd number is always an odd number.

Page 132

Question 1.
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
Since 27 and 13 both are odd numbers, and odd × odd = odd
So, the parity of the number of small squares is odd.

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

(c) 135 × 654
Since 135 is odd and 654 is even, and odd × even = even
So, the parity of the number of small squares is even.

Question 2.
Come up with expressions that always have odd parity.
Solution:
4n + 3, 6n + 5, 2n + 11, 2n – 1 etc.

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

Question 4.
Are there expressions using which we can list all the even numbers?
Hint: All even numbers have a factor 2.
Solution:
The expression 2n evaluates to 2,4,6,8, 10, for n = 1,2,3,4, 5, respectively.
So required expression to represent all the even numbers = 2n.

Question 5.
Are there expressions using which we can list all odd numbers?
Solution:
Yes,
(2n – 1) evaluates to 1, 3, 5 for n = 1,2 respectively. So expression (2n – 1) can list all odd numbers.

Question 6.
What is the 100th odd number?
Solution:
100^th odd number = 2 × 100 – 1
= 200 – 1 = 199

Page 133

Question 1.
The numbers in the circles are the sums of the corresponding rows and columns.
Fill the grids below based on the rule mentioned above:

Solution:

Page 137

Question 1.
Choose any magic square that you have made so far using consecutive numbers.

	m

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:
Consider a magic square with center m = 5

8	1	6
3	5	7
4	9	2

Now on taking ‘m’ at the center and expressing other numbers related to m:

m + 3	m – 4	m +  1
m – 2	m	m + 2
m – 1	m + 4	m – 3

Now this generalized form can be used to get other magic squares.

Page 141

Question 1.
Use the systematic method to write down all 6-beat rhythms, i.e., write 6 as the sum of l’s and 2’s in all possible ways. Did you get 13 ways?
Solution:
Yes, there are total 13 ways.
(1) 1 + 1 + 1 + 1 + 1 + 1
(2) 1 + 1 + 1 + 1 + 2
(3) 1 + 1 + 1 + 2 + 1
(4) 1 + 1 + 2 + 1 + 1
(5) 1 + 2 + 1 + 1 + 1
(6) 2 + 1 + 1 + 1 + 1
(7) 1 + 1 + 2 + 2
(8) 1 + 2 + 2 + 1
(9) 2 + 2 + 1 + 1
(10) 2+ 1 + 2 + 1
(11) 1 + 2+ 1 + 2
(12) 2+ 1 + 1 + 2
(13) 2 + 2 + 2

Page 142

Question 1.
Write the next number in the sequence after 55.
1, 2, 3, 5, 8,13, 21, 34, 55,…
Solution:
34 + 55 = 89
So, required number in the sequence is 89.

Question 2.
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 3 numbers in the sequence:
55 + 89 = 144
89 + 144 = 233
144 + 233 =377
Now sequence would become:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, (144), (233), (377),…
One more number = 233 + 377 = odd + odd = even
By following parity cycle, the next term in the sequence after 377 will be an even number.

Question 3.
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.
Therefore, next number (after 377) will be even.

Page 143

Question 1.
Let us look at one more example K2 shown on the right. Here K2 means + K2 that the number is a 2-digit number HMM 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.
\(\begin{array}{r}
\text { K2 } \\
+\quad \text { K2 } \\
\hline \text { HMM }
\end{array}\)
What digit should the letter M correspond to?

\(\begin{array}{r}
72 \\
+\quad 72 \\
\hline 144
\end{array}\)
Both the tens place and the units place of the sum have the same digit.
Solution:
Here, H = 1, M = 4 and K = 7, H = 2 or 3 is not possible because they make a 2-digit number, which is invalid.

Question 2.
Find out what each letter stands for.

Solution:
(i) \(\begin{array}{r}
\text { YY } \\
+\quad \text { Z } \\
\hline \text { ZOO }
\end{array}\)
YY is a 1 two-digit number where both digits are the same.
So it can be 99, 88,….
\(\begin{array}{r}
99 \\
+\quad 1 \\
\hline 100
\end{array}\)
But, Z is a 1 digit number and ZOO is a 3-digit number.
So Y = 9, Z = 1 and O = 0

(ii) \(\begin{array}{r}
\mathrm{B} 5 \\
+\quad 3 \mathrm{D} \\
\hline \mathrm{ED} 5
\end{array}\)
5 + D= 5 ⇒ D = 0
Now, B + 3 = E0 ⇒ B = 7
If B = 7, then E = 1
\(\begin{array}{r}
75 \\
+\quad 30 \\
\hline 105 \\
\hline
\end{array}\)

(iii) \(\begin{array}{r}
\text { KP } \\
+\quad \text { KP } \\
\hline \text { PRR }
\end{array}\)
Here, KP is a 2 digit number and PRR is a 3 digit number.
PRR So, 2 × (KP)= PRR. So P = 1
If P = 1, then R = 2
So K = 6
⇒ \(\begin{array}{r}
61 \\
+\quad 61 \\
\hline 122
\end{array}\)

(iv) \(\begin{array}{r}
\mathrm{Cl} \\
+\mathrm{C} \\
\hline 1 \mathrm{FF}
\end{array} \Rightarrow \begin{array}{r}
91 \\
+9 \\
\hline 100
\end{array}\)
Here, C = 9 and F = 0

Class 7 Maths Ganita Prakash Chapter 6 Solutions

Figure it Out (Page 128)

Question 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) 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

Question 2.
For each of the statements given below, think and identify if it is Always True, Only Sometimes True, or Sever 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 sometimes true.
Reason : 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 front or in a position where no one taller is ahead of them. So, given statement is sometimes true.

(b) Always true.
Reason: If a person is the tallest then no one is taller than him, so he will always say ‘0’. So given statement is always true.

(c) Always true.
Reason : Anumber is assigned to each person that represents how many taller people are ahead of him. Since no one is ahead of the given first person, so given statement is always true.

(d) Only sometimes true.
Reason : ‘0’ can be assigned to a person standing in between provided that—no taller people is ahead of him.

(e) Only sometimes true.
Reason : A person who call out the largest number has many taller people in front but he may not be the shortest one.
For example: If shortest person is standing at the front, he will call out ‘O’. However at the same time, the second shortest person could be at the back and might call out the largest number.

(f) If there are 8 people in a group, then shortest person will see 7 taller people. So maximum the largest possible number will be 7.

Figure it Out (Page 131)

Question 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.
Since even + even = Even and
odd + odd= Even, the total sum is even + even = even.
Therefore, the parity of the sum of 2 even numbers and 2 odd numbers is even.
For example : 2 + 4 + 3 + 5 = 6 + 8 = 14 (even)

(b) Sum of 2 odd numbers and 3 even numbers : Since odd + odd = even and
even + even + even = even,
Now sum = even + even = even
Therefore, the parity of sum of 2 odd numbers and 3 even numbers is even.
For example : 3 + 5 + 2 + 4 + 6 = 8 + 12 = 20 (even)

(c) Sum of 5 even numbers.
Since even number is a complete pair, so on adding any number of even numbers keeps the sum as pairs.
Therefore, the parity of sum of 5 even numbers is even.
For example : 2 + 4 + 6 + 8 + 10 = 30 (even)

(d) Sum of 8 odd numbers: odd + odd = even We have 4 such pairs.
On adding 4 such pairs, we get
even + even + even + even = even
Therefore, the parity of sum of 8 odd numbers is even.
For example : 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64 (even)

Question 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:
Since an odd number of ₹ 1 coins gives an odd amount, an odd number of ₹ 5 coins gives an odd multiple of 5 (which is odd), and an even number of ₹ 10 coins gives an even amount.
∴ odd + odd + even = even
Since ₹ 205 is an odd number, so total of ₹ 205 is not possible with an odd number of ₹ 1 coins and ₹ 5 coins and an even number of ₹ 10 coins. Therefore, Lakpa made a mistake.

Question 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) even – even = ?
For example, 8 – 2 = 6(even)
10 – 6 = 4(even)
Parity of result = even
∴ even – even = even

(e) odd – odd = ?
For example, 7 – 5 = 2(even)
15 – 11 = 4(even)
Parity of result = even
∴ odd – odd = even

(f) even – odd =?
For example, 14 – 7 = 7(odd)
16 – 5 = 11(odd)
Parity of result = odd
∴ even – odd odd

(g) odd – even = ?
For example, 11 – 8 = 3(odd)
17 – 12 = 5(odd)
odd – even = odd

Figure it Out (Page 136)

Question 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.
However there are 8 distinct (3 × 3) magic squares can be made using 1-9 considering rotations and reflections as equivalent.

8	1	6
3	5	7
4	9	2

Question 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:
Here we can see that numbers 2 – 10 are consecutive numbers just like 1-9, but increased by 1.
Magic square using 1 – 9 :

8	1	6
3	5	7
4	9	2

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

9	2	7
4	6	8
5	10	3

Both squares are structurally identical. The magic square having number 1 – 9 has 15 as magic sum and the magic square having number 2-10 has 18 as magic sum.

Question 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:

8	1	6
3	5	7
4	9	2

(a) When each number is increased by 1, we will get a new magic square with the magic sum 18.
On adding a constant to every number, the magic sum increases by 3 × that constant.

9	2	7
4	6	8
5	10	3

(b) When each number is doubled, we will get a new magic square with the magic sum 30.

16	2	12
6	10	14
8	18	4

On multiplying each by a constant, the magic sum is multiplied by that constant.

Question 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.

Question 5.
Discuss ways of creating a magic square using any set of 9 consecutive numbers (like 2 – 10, 3 – 11, 9 – 17, etc.).
Solution:
For magic square with numbers 2-10, see sol. of Q. 2.
For magic square with numbers 3-11, add 2 to each number of the magic square 1-9.

10	3	8
5	7	9
6	11	4

For magic square with numbers 9-17, add 8 of each number of the magic square 1-9.

16	9	14
11	13	15
12	17	10

Figure it Out (Page 137)

Question 1.
Using this generalised form, find a magic square if the centre number is 25.
Solution:
Using the generalised form discussed above,
Let m = 25 as center value Following magic square is obtained:

Question 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 (leading diagonal) = 28 + 25 + 22 = 75
Expression = 3 × m
Where m is the letter number written in the center of the magic square.

Question 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 :

m + 3	m – 4	m + 1
m – 2	m	m + 2
m – 1	m + 4	m – 3

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

m + 4	m – 3	m + 2
m – 1	m + 1	m + 3
m	m + 5	m – 2

(b) Doubling every term in the generalised form

2m + 6	2m – 8	2m + 2
2m – 4	2m	2m + 4
2m – 2	2m + 8	2m – 6

Question 4.
Create a magic square whose magic sum is 60.
Solution:
Generalised magic square

m + 3	m – 4	m + 1
m – 2	m	m + 2
m – 1	m + 4	m – 3

Sum of any row = (m – 2 + m + m + 2) = 3 m
i. e. magic sum 3m = 60
⇒ m = 20
So magic square having center as 20

23	16	21
18	20	22
19	24	17

Question 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.

Two magic squares with magic sum 45.

Figure it Out (Pages 143-144)

Question 1.
A light bulb is ON. Dorjee toggles its switch 77 times. Will the bulb be on or off Why
Solution:
Intially light bulb is ON.
Starting from the ON state of the bulb:
1^st toggle → OFF
2^nd toggle → ON
3^rd toggle → OFF and so on.
So the odd number of toggles ilI 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.

Question 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 number. It means that each sheet will give even + odd = odd number
Since there are total 50 sheets and we know that on adding odd + odd = even Odd number × 50 will give rise to an even number. Since 6000 is an even number, so it is possible for the sum of page numbers of loose pages to be 6000.

Question 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:
Let us consider

Row 1 = A + B + C = odd number
Row 2 = D + E + F = even number
Column 1 = A + D = even number
Column 2 = B + E = even number
Column 3 = C + F = odd number
Now, Row 1 : A = o, B = c, C = e then o + e + e = odd number
Column 1 : A and D must be paired 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
If C = e, F = o, then e + o = odd number
Now, 6 boxes with 3 evens and 3 odds number can be filled as:

Question 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:
We need to make a 3 × 3 magic square with 0 as the magic sum.
Negative numbers can be used.
Using numbers from – 4 to 4 to create a magic square whose magic sum is 0.

-3	2	1
4	0	-4
-1	-2	3

Question 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) Sum of an odd number of even numbers is even.
(b) Sum of an even number of odd numbers is even.
(c) Sum of an even number of even numbers is even.
(d) Sum of an odd number of odd numbers is odd.

Question 6.
What is the parity of the sum of the numbers from 1 to 100?
Solution:
There are total 100 numbers from 1 to 100.
Parity : odd, even, odd, even,
There are exactly 50 odd numbers and 50 even numbers.
On adding 50 odd numbers obtained sum is even and on adding 50 even numbers obtained sum is even.
Total sum: even + even = even Hence, the parity of the sum of numbers from 1 to 100 is even.

Question 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:
In the Virahanka sequence, each number is the sum of two preceding numbers.
So, next two numbers are
987 + 1597 = 2584
1597 + 2584 = 4181 Previous numbers are
1597 – 987 = 610 and
987 – 610 = 377
So sequence is…., 377, 610, 987, 1597, 2584, 4181,….

Question 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:
Different ways by which Angaan can climb the 8 – step staircase by taking either 1 step or 2 steps at a time.

	Different ways	Number of ways
No 2s	1+ 1 + 1 + 1 + 1 + 1 + 1 + 1	1
One 2s	2 + 1 + 1 + 1 +  1 + 1 + 1 1 + 2 +1 + 1 + 1 + 1 + 1 1 +1 + 2 + 1 + 1 + 1 + 1 1 + 1+ 1 + 2+ 1 + 1 + 1 1 + 1 + 1 + 1 + 2 + 1 + 1 1 + 1 + 1+ 1 +  1 + 2 + 1 1 + 1 + 1 + 1 + 1 + 1 + 2	7
Two 2s	2 +2 + 1 +  1+  1 + 1 2 + 1 + 2 + 1 + 1 + 1 1 +2 + 2 + 1 + 1 + 1 2 + 1 + 1 + 2 + 1 + 1 1 + 2 + 1 + 2 + 1 + 1 1 + 1 + 2 + 2 + 1 + 1 2 + 1 +1  + 1 + 2 + 1 1 + 2 + 1 + 1 + 2 + 1 1 + 1 + 2 + 1 +  2 + 1 1 + 1 + 1 +  2 + 2 + 1 2 + 1 + 1 + 1 + 1 + 2 1 + 2 + 1 + 1 + 1 + 2 1 + 1 + 2 + 1 + 1 + 2 1 + 1 + 1 + 2 + 1 + 2 1 + 1 + 1 + 1 + 2 + 2
Three 2s	1 + 1+ 2 + 2 + 2 1 + 2 + 2 + 2 + 1 2 + 2 + 2  + 1 + 1 1 + 2 + 1 + 2 + 2 2 + 1 + 1  + 2 + 2 2 + 1 + 2 +  2 + 1 1 + 2 + 2 + 1 + 2 2 + 1 + 2 + 1 + 2 2 + 2 + 1 +1  +2 2 + 1+ 1 +  2 + 2	10
Four 2s	2 + 2 + 2 + 2	1

So, total ways in which Aangaan reaches the top = 1 + 7 + 15 + 10 + 1 = 34 ways.

Question 9.
What is the parity of the 20th term of the Virahanka sequence?
Solution:
Virahanka sequence :
1, 2, 3, 5, 8, 13, 21, 34, 55, 89,….
1 (odd), 2 (even), 3 (odd), 5 (odd), 8 (even), 13 (odd), 21 (odd), 34 (even), 55 (odd), 89 (odd),…
Parity Cycle:
odd, even, odd, odd, even, odd, odd, even, ………….
So parity cycle odd, even, odd repeats every 3 terms. So the 20^th term of the Virahanka sequence is even.

Question 10.
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.
Solution:
(a) Putting m = 1 in 4m – 1;
4 × 1 – 1 = 3 (odd)
Putting m = 2 in 4m – 1
4 × 2 – 1 = 7 (odd)
Hence, the expression 4m – 1 always gives odd number.
So given statement is true.

(b) Putting j = 1 in 6j – 4
6 ×1 – 4 = 2 (even);
Putting j = 2 in 6j – 4;
6 × 2 – 4 = 8 (even);
Putting j = 3 in 6j – 4;
6 × 3 – 4 = 14 (even);
Since some even numbers can he expressed in the form (6j – 4) but not all, so given statement is false.

(c) For 2p + 1
Putting p = 1 2 × 1 + 1 = 3 (odd)
Putting p = 2 2 × 2 + 1 = 5 (odd)
Putting p = 3 2 × 3 + 1 = 7 (odd)
Now for 2q – 1
Putting q = 1 2 × 1 – 1 = 1 (odd)
Putting q = 2 2 × 2 – 1 = 3 (odd)
Putting q = 3 2 × 3 – 1 = 5 (odd)
Since All odd numbers can be expressed with the help of given expressions, so given statement is true.

(d) Putting f= 1 in 2f + 3
2 × 1 + 3 = 5 (odd)
Putting f = 2 in 2f + 3
2 × 2 + 3 = 7 (odd)
Putting f = 3 in 2f + 3
2 × 3 + 3 = 9 (odd)
Since there is always an odd number, so given statement is false.

Question 11.
Solve the cryptarithm:
\(\begin{array}{r}
\text { UT } \\
+\quad \text { TA } \\
\hline \text { TAT }
\end{array}\)
Solution:
Here, T is at hundreds place,
So T = 1
⇒ A = 0 and U = 9
⇒ \(\begin{array}{r}
\text { UT } \\
+\quad \text { TA } \\
\hline \text { TAT }
\end{array}\) ⇒ \(\begin{array}{r}
91 \\
+\quad 10 \\
\hline 101 \\
\hline
\end{array}\)