Obtain award mark for an award 72 and 82. If the average (mean) number of letters per word is 5, find the values of \( x \) and \( y \).

In an objective test marked out of 40, the marks scored by 35 students are given in the table below.

[tex]\[
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
Marks scored & 11-15 & 16-20 & 21-25 & 26-30 & 31-35 & 36-40 \\
\hline
No. of students & 2 & 7 & \( x^2 \) & \( 2x-6 \) & 5 & 3 \\
\hline
\end{tabular}
\][/tex]



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.