Answer :
Sure, let's tackle each question in detail.
### Question 1:
#### Problem:
A survey was carried out among 300 students. It was found that:
- 125 students like to play cricket.
- 145 students like to play football.
- 90 students like to play tennis.
- 32 students like to play exactly two games out of the three games.
How many students like to play exactly one game?
#### Solution:
We need to find how many students like to play exactly one game.
The number of students who play exactly one game can be calculated using the principle of inclusion-exclusion.
Let:
- [tex]\( C \)[/tex] be the set of students who like cricket.
- [tex]\( F \)[/tex] be the set of students who like football.
- [tex]\( T \)[/tex] be the set of students who like tennis.
Given:
- [tex]\( |C| = 125 \)[/tex]
- [tex]\( |F| = 145 \)[/tex]
- [tex]\( |T| = 90 \)[/tex]
- Students who like exactly two games = 32.
Using the principle of inclusion-exclusion formula, the number of students who like only one game is given by:
[tex]\[ |C \text{ only}| + |F \text{ only}| + |T \text{ only}| = |C| + |F| + |T| - 2(\text{students who like exactly two games}) - 3(\text{students who like all three games}) \][/tex]
Assuming no student likes all three games ([tex]\( 0 \)[/tex] students like exactly three games):
[tex]\[ |C \text{ only}| + |F \text{ only}| + |T \text{ only}| = 125 + 145 + 90 - 2 \times 32 - 3 \times 0 \][/tex]
[tex]\[ = 125 + 145 + 90 - 64 \][/tex]
[tex]\[ = 296 \][/tex]
Thus, 296 students like to play exactly one game.
### Question 2:
#### Problem:
In a class of 188 students:
- 130 students like ice cream.
- 90 students like cake.
- The number who like both is [tex]\( x \)[/tex].
- The number who like neither is [tex]\( 212 - 2x \)[/tex].
Find the value of [tex]\( x \)[/tex].
#### Solution:
Let:
- [tex]\( A \)[/tex] be the set of students who like ice cream.
- [tex]\( B \)[/tex] be the set of students who like cake.
Given:
- [tex]\( |A| = 130 \)[/tex]
- [tex]\( |B| = 90 \)[/tex]
- Total students = 188
- Neither like ice cream nor cake = [tex]\( 212 - 2x \)[/tex]
Using the principle of inclusion-exclusion:
[tex]\[ |A \cup B| = |A| + |B| - |A \cap B| \][/tex]
[tex]\[ |A \cup B| = 188 - (212 - 2x) \][/tex]
[tex]\[ |A \cup B| = 2x - 24 \][/tex]
So, equating the values:
[tex]\[ 130 + 90 - x = 2x - 24 \][/tex]
[tex]\[ 220 - x = 2x - 24 \][/tex]
[tex]\[ 220 + 24 = 3x \][/tex]
[tex]\[ 244 = 3x \][/tex]
[tex]\[ x = 61 \][/tex]
The value of [tex]\( x \)[/tex] is 61.
### Question 3:
#### Problem:
Given equations, determine which of the resulting values for [tex]\( x, y, z, u \)[/tex] are rational or irrational:
(i) [tex]\( x^2=5 \)[/tex]
(ii) [tex]\( y^2=9 \)[/tex]
(iii) [tex]\( z^2=0.04 \)[/tex]
(iv) [tex]\( u^2=\frac{17}{4} \)[/tex]
#### Solution:
For a number to be rational, it must be expressible as a ratio of two integers:
- [tex]\( x^2 = 5 \)[/tex]:
[tex]\[ x = \sqrt{5} \][/tex]
Since [tex]\( \sqrt{5} \)[/tex] is not a perfect square, [tex]\( x \)[/tex] is irrational.
- [tex]\( y^2 = 9 \)[/tex]:
[tex]\[ y = \sqrt{9} = 3 \][/tex]
Since 3 is a whole number, [tex]\( y \)[/tex] is rational.
- [tex]\( z^2 = 0.04 \)[/tex]:
[tex]\[ z = \sqrt{0.04} = 0.2 \][/tex]
Since 0.2 can be expressed as [tex]\( \frac{1}{5} \)[/tex], [tex]\( z \)[/tex] is rational.
- [tex]\( u^2 = \frac{17}{4} \)[/tex]:
[tex]\[ u = \sqrt{\frac{17}{4}} = \frac{\sqrt{17}}{2} \][/tex]
Since [tex]\( \sqrt{17} \)[/tex] is not a perfect square, [tex]\( u \)[/tex] is irrational.
Hence:
- [tex]\( x \)[/tex] is irrational.
- [tex]\( y \)[/tex] is rational.
- [tex]\( z \)[/tex] is rational.
- [tex]\( u \)[/tex] is irrational.
### Question 4:
#### Problem:
Prove that [tex]\( 1 + 3 + 5 + \cdots + (2n-1) = n^2 \)[/tex] for all values of [tex]\( n \geq 1 \)[/tex].
#### Solution:
We need to show that the sum of the first [tex]\( n \)[/tex] odd numbers is equal to [tex]\( n^2 \)[/tex].
Let's confirm this by using mathematical induction:
1. Base Case:
For [tex]\( n = 1 \)[/tex]:
[tex]\[ 1 = 1^2 \][/tex]
This is true.
2. Inductive Step:
Assume for some [tex]\( k \geq 1 \)[/tex]:
[tex]\[ 1 + 3 + 5 + \cdots + (2k-1) = k^2 \][/tex]
We need to prove that:
[tex]\[ 1 + 3 + 5 + \cdots + (2k-1) + (2(k+1) - 1) = (k+1)^2 \][/tex]
From the inductive hypothesis:
[tex]\[ k^2 + (2(k+1) - 1) = k^2 + 2k + 1 = (k + 1)^2 \][/tex]
Since the formula holds for [tex]\( n = 1 \)[/tex] and the inductive step is valid, by the principle of mathematical induction, the sum of the first [tex]\( n \)[/tex] odd numbers is indeed [tex]\( n^2 \)[/tex].
This completes the proof.
### Question 1:
#### Problem:
A survey was carried out among 300 students. It was found that:
- 125 students like to play cricket.
- 145 students like to play football.
- 90 students like to play tennis.
- 32 students like to play exactly two games out of the three games.
How many students like to play exactly one game?
#### Solution:
We need to find how many students like to play exactly one game.
The number of students who play exactly one game can be calculated using the principle of inclusion-exclusion.
Let:
- [tex]\( C \)[/tex] be the set of students who like cricket.
- [tex]\( F \)[/tex] be the set of students who like football.
- [tex]\( T \)[/tex] be the set of students who like tennis.
Given:
- [tex]\( |C| = 125 \)[/tex]
- [tex]\( |F| = 145 \)[/tex]
- [tex]\( |T| = 90 \)[/tex]
- Students who like exactly two games = 32.
Using the principle of inclusion-exclusion formula, the number of students who like only one game is given by:
[tex]\[ |C \text{ only}| + |F \text{ only}| + |T \text{ only}| = |C| + |F| + |T| - 2(\text{students who like exactly two games}) - 3(\text{students who like all three games}) \][/tex]
Assuming no student likes all three games ([tex]\( 0 \)[/tex] students like exactly three games):
[tex]\[ |C \text{ only}| + |F \text{ only}| + |T \text{ only}| = 125 + 145 + 90 - 2 \times 32 - 3 \times 0 \][/tex]
[tex]\[ = 125 + 145 + 90 - 64 \][/tex]
[tex]\[ = 296 \][/tex]
Thus, 296 students like to play exactly one game.
### Question 2:
#### Problem:
In a class of 188 students:
- 130 students like ice cream.
- 90 students like cake.
- The number who like both is [tex]\( x \)[/tex].
- The number who like neither is [tex]\( 212 - 2x \)[/tex].
Find the value of [tex]\( x \)[/tex].
#### Solution:
Let:
- [tex]\( A \)[/tex] be the set of students who like ice cream.
- [tex]\( B \)[/tex] be the set of students who like cake.
Given:
- [tex]\( |A| = 130 \)[/tex]
- [tex]\( |B| = 90 \)[/tex]
- Total students = 188
- Neither like ice cream nor cake = [tex]\( 212 - 2x \)[/tex]
Using the principle of inclusion-exclusion:
[tex]\[ |A \cup B| = |A| + |B| - |A \cap B| \][/tex]
[tex]\[ |A \cup B| = 188 - (212 - 2x) \][/tex]
[tex]\[ |A \cup B| = 2x - 24 \][/tex]
So, equating the values:
[tex]\[ 130 + 90 - x = 2x - 24 \][/tex]
[tex]\[ 220 - x = 2x - 24 \][/tex]
[tex]\[ 220 + 24 = 3x \][/tex]
[tex]\[ 244 = 3x \][/tex]
[tex]\[ x = 61 \][/tex]
The value of [tex]\( x \)[/tex] is 61.
### Question 3:
#### Problem:
Given equations, determine which of the resulting values for [tex]\( x, y, z, u \)[/tex] are rational or irrational:
(i) [tex]\( x^2=5 \)[/tex]
(ii) [tex]\( y^2=9 \)[/tex]
(iii) [tex]\( z^2=0.04 \)[/tex]
(iv) [tex]\( u^2=\frac{17}{4} \)[/tex]
#### Solution:
For a number to be rational, it must be expressible as a ratio of two integers:
- [tex]\( x^2 = 5 \)[/tex]:
[tex]\[ x = \sqrt{5} \][/tex]
Since [tex]\( \sqrt{5} \)[/tex] is not a perfect square, [tex]\( x \)[/tex] is irrational.
- [tex]\( y^2 = 9 \)[/tex]:
[tex]\[ y = \sqrt{9} = 3 \][/tex]
Since 3 is a whole number, [tex]\( y \)[/tex] is rational.
- [tex]\( z^2 = 0.04 \)[/tex]:
[tex]\[ z = \sqrt{0.04} = 0.2 \][/tex]
Since 0.2 can be expressed as [tex]\( \frac{1}{5} \)[/tex], [tex]\( z \)[/tex] is rational.
- [tex]\( u^2 = \frac{17}{4} \)[/tex]:
[tex]\[ u = \sqrt{\frac{17}{4}} = \frac{\sqrt{17}}{2} \][/tex]
Since [tex]\( \sqrt{17} \)[/tex] is not a perfect square, [tex]\( u \)[/tex] is irrational.
Hence:
- [tex]\( x \)[/tex] is irrational.
- [tex]\( y \)[/tex] is rational.
- [tex]\( z \)[/tex] is rational.
- [tex]\( u \)[/tex] is irrational.
### Question 4:
#### Problem:
Prove that [tex]\( 1 + 3 + 5 + \cdots + (2n-1) = n^2 \)[/tex] for all values of [tex]\( n \geq 1 \)[/tex].
#### Solution:
We need to show that the sum of the first [tex]\( n \)[/tex] odd numbers is equal to [tex]\( n^2 \)[/tex].
Let's confirm this by using mathematical induction:
1. Base Case:
For [tex]\( n = 1 \)[/tex]:
[tex]\[ 1 = 1^2 \][/tex]
This is true.
2. Inductive Step:
Assume for some [tex]\( k \geq 1 \)[/tex]:
[tex]\[ 1 + 3 + 5 + \cdots + (2k-1) = k^2 \][/tex]
We need to prove that:
[tex]\[ 1 + 3 + 5 + \cdots + (2k-1) + (2(k+1) - 1) = (k+1)^2 \][/tex]
From the inductive hypothesis:
[tex]\[ k^2 + (2(k+1) - 1) = k^2 + 2k + 1 = (k + 1)^2 \][/tex]
Since the formula holds for [tex]\( n = 1 \)[/tex] and the inductive step is valid, by the principle of mathematical induction, the sum of the first [tex]\( n \)[/tex] odd numbers is indeed [tex]\( n^2 \)[/tex].
This completes the proof.
