The frequency distribution of marks obtained by 28 students in a test carrying 40 marks is given below:
`{:("Marks":,0-10,10-20,20-30,30-40),("Number of students":,6,x,y,6):}`
If the mean of the above data is 20, then the difference between x and y is

Options

A. 3
B. 2
C. 1
D. 0

Answer

To solve the problem step by step, we will follow the outlined process:

### Step 1: Set up the frequency distribution table
The frequency distribution of marks obtained by 28 students is given as follows:

| Marks | Number of Students (Frequency) |
|----------|-------------------------------|
| 0 - 10 | 6 |
| 10 - 20 | x |
| 20 - 30 | y |
| 30 - 40 | 6 |

### Step 2: Calculate total frequency
The total number of students is given as 28. Therefore, we can write the equation for total frequency:

\[ 6 + x + y + 6 = 28 \]

Simplifying this gives:

\[ x + y + 12 = 28 \]

Thus,

\[ x + y = 16 \quad \text{(Equation 1)} \]

### Step 3: Calculate midpoints and frequency times midpoints
Next, we calculate the midpoints (x_i) for each class interval:

- For 0-10: \( \frac{0 + 10}{2} = 5 \)
- For 10-20: \( \frac{10 + 20}{2} = 15 \)
- For 20-30: \( \frac{20 + 30}{2} = 25 \)
- For 30-40: \( \frac{30 + 40}{2} = 35 \)

Now we calculate \( f_i \times x_i \):

| Marks | Frequency (f_i) | Midpoint (x_i) | \( f_i \times x_i \) |
|----------|-----------------|----------------|-----------------------|
| 0 - 10 | 6 | 5 | \( 6 \times 5 = 30 \) |
| 10 - 20 | x | 15 | \( 15x \) |
| 20 - 30 | y | 25 | \( 25y \) |
| 30 - 40 | 6 | 35 | \( 6 \times 35 = 210 \) |

### Step 4: Calculate the mean
The mean is given as 20. The formula for mean is:

\[ \text{Mean} = \frac{\Sigma (f_i \times x_i)}{\Sigma f_i} \]

Substituting the values we have:

\[ 20 = \frac{30 + 15x + 25y + 210}{28} \]

Multiplying both sides by 28 gives:

\[ 560 = 30 + 15x + 25y + 210 \]

Simplifying this leads to:

\[ 560 = 240 + 15x + 25y \]

Thus,

\[ 15x + 25y = 560 - 240 \]

This simplifies to:

\[ 15x + 25y = 320 \quad \text{(Equation 2)} \]

### Step 5: Solve the system of equations
We now have two equations:

1. \( x + y = 16 \) (Equation 1)
2. \( 15x + 25y = 320 \) (Equation 2)

From Equation 1, we can express \( y \) in terms of \( x \):

\[ y = 16 - x \]

Substituting this into Equation 2:

\[ 15x + 25(16 - x) = 320 \]

Expanding this gives:

\[ 15x + 400 - 25x = 320 \]

Combining like terms results in:

\[ -10x + 400 = 320 \]

Solving for \( x \):

\[ -10x = 320 - 400 \]
\[ -10x = -80 \]
\[ x = 8 \]

Now substituting \( x \) back into Equation 1 to find \( y \):

\[ 8 + y = 16 \]
\[ y = 16 - 8 = 8 \]

### Step 6: Find the difference between x and y
Now we find the difference between \( x \) and \( y \):

\[ \text{Difference} = x - y = 8 - 8 = 0 \]

### Final Answer
The difference between \( x \) and \( y \) is **0**.

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