Answer :
To determine the values of \( x \) and \( y \) in the provided question, we need to start by using the given data to form an equation and then solve it step-by-step.
Here's the information provided:
- The total number of students is 35.
- The numbers of students in each mark category are:
- Marks 11-15: 2 students
- Marks 16-20: 7 students
- Marks 21-25: \( x^2 \) students
- Marks 26-30: \( 2x - 6 \) students
- Marks 31-35: 5 students
- Marks 36-40: 3 students
Since the total number of students is 35, we can write an equation summing up the number of students in all categories:
[tex]\[ 2 + 7 + x^2 + (2x - 6) + 5 + 3 = 35 \][/tex]
Let's simplify this equation step-by-step:
1. Combine the constants:
[tex]\[ 2 + 7 + 5 + 3 = 17 \][/tex]
2. Plug the combined constant into the equation:
[tex]\[ 17 + x^2 + 2x - 6 = 35 \][/tex]
3. Simplify the constants on the left-hand side:
[tex]\[ 11 + x^2 + 2x = 35 \][/tex]
4. Subtract 11 from both sides to isolate the quadratic equation:
[tex]\[ x^2 + 2x + 11 - 11 = 35 - 11 \][/tex]
[tex]\[ x^2 + 2x = 24 \][/tex]
5. Rewrite the quadratic equation:
[tex]\[ x^2 + 2x - 24 = 0 \][/tex]
Now we solve this quadratic equation for \( x \).
The quadratic equation \( x^2 + 2x - 24 = 0 \) has solutions. We can use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 2 \), and \( c = -24 \).
However, the results yielded from solving this equation are \( x = -6 \) and \( x = 4 \). Since \( x \) must be a positive value (it represents a number of students), we choose \( x = 4 \).
Next, let's determine \( y \) when \( x = 4 \). Given \( y = 2x - 6 \):
[tex]\[ y = 2(4) - 6 = 8 - 6 = 2 \][/tex]
So, the values are:
- \( x = 4 \)
- \( y = 2 \)
Thus, with [tex]\( x = 4 \)[/tex], the number of students in each category is fully described.
Here's the information provided:
- The total number of students is 35.
- The numbers of students in each mark category are:
- Marks 11-15: 2 students
- Marks 16-20: 7 students
- Marks 21-25: \( x^2 \) students
- Marks 26-30: \( 2x - 6 \) students
- Marks 31-35: 5 students
- Marks 36-40: 3 students
Since the total number of students is 35, we can write an equation summing up the number of students in all categories:
[tex]\[ 2 + 7 + x^2 + (2x - 6) + 5 + 3 = 35 \][/tex]
Let's simplify this equation step-by-step:
1. Combine the constants:
[tex]\[ 2 + 7 + 5 + 3 = 17 \][/tex]
2. Plug the combined constant into the equation:
[tex]\[ 17 + x^2 + 2x - 6 = 35 \][/tex]
3. Simplify the constants on the left-hand side:
[tex]\[ 11 + x^2 + 2x = 35 \][/tex]
4. Subtract 11 from both sides to isolate the quadratic equation:
[tex]\[ x^2 + 2x + 11 - 11 = 35 - 11 \][/tex]
[tex]\[ x^2 + 2x = 24 \][/tex]
5. Rewrite the quadratic equation:
[tex]\[ x^2 + 2x - 24 = 0 \][/tex]
Now we solve this quadratic equation for \( x \).
The quadratic equation \( x^2 + 2x - 24 = 0 \) has solutions. We can use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 2 \), and \( c = -24 \).
However, the results yielded from solving this equation are \( x = -6 \) and \( x = 4 \). Since \( x \) must be a positive value (it represents a number of students), we choose \( x = 4 \).
Next, let's determine \( y \) when \( x = 4 \). Given \( y = 2x - 6 \):
[tex]\[ y = 2(4) - 6 = 8 - 6 = 2 \][/tex]
So, the values are:
- \( x = 4 \)
- \( y = 2 \)
Thus, with [tex]\( x = 4 \)[/tex], the number of students in each category is fully described.
