Let the mean and standard deviation of marks of class `A` of `100` students be respectively `40` and `alpha(>0)`,and the mean and standard deviation of marks of class `B` of `n` students be respectively `55` and `30-alpha`.If the mean and variance of the marks of the combined class of `100+n` students are respectively `50` and `350`,then the sum of variances of class `A` and `B` is

Options

A. `500`
B. `450`
C. `650`
D. `900`

Answer

To solve the problem step by step, we will use the given information about the means and standard deviations of classes A and B, as well as the combined class.

### Step 1: Understand the given data
- Class A:
- Number of students, \( n_1 = 100 \)
- Mean, \( \bar{x_1} = 40 \)
- Standard deviation, \( \sigma_1 = \alpha \)

- Class B:
- Number of students, \( n_2 = n \)
- Mean, \( \bar{x_2} = 55 \)
- Standard deviation, \( \sigma_2 = 30 - \alpha \)

- Combined class:
- Total number of students, \( n_1 + n_2 = 100 + n \)
- Mean, \( \bar{x} = 50 \)
- Variance, \( \sigma^2 = 350 \)

### Step 2: Calculate the mean of the combined class
The mean of the combined class can be expressed as:
\[ \bar{x} = \frac{n_1 \bar{x_1} + n_2 \bar{x_2}}{n_1 + n_2} \]
Substituting the known values:
\[ 50 = \frac{100 \cdot 40 + n \cdot 55}{100 + n} \]
This simplifies to:
\[ 50(100 + n) = 4000 + 55n \]
Expanding and rearranging gives:
\[ 5000 + 50n = 4000 + 55n \]
\[ 1000 = 5n \implies n = 200 \]

### Step 3: Calculate the variance of the combined class
The variance of the combined class is given by:
\[ \sigma^2 = \frac{n_1 \sigma_1^2 + n_2 \sigma_2^2}{n_1 + n_2} - \bar{x}^2 \]
Substituting the known values:
\[ 350 = \frac{100 \cdot \alpha^2 + 200 \cdot (30 - \alpha)^2}{300} - 50^2 \]
Calculating \( 50^2 \):
\[ 50^2 = 2500 \]
Thus, we have:
\[ 350 + 2500 = \frac{100 \alpha^2 + 200 (30 - \alpha)^2}{300} \]
This simplifies to:
\[ 2850 = \frac{100 \alpha^2 + 200 (900 - 60\alpha + \alpha^2)}{300} \]
\[ 2850 = \frac{100 \alpha^2 + 180000 - 12000\alpha + 200\alpha^2}{300} \]
Combining terms:
\[ 2850 = \frac{300\alpha^2 - 12000\alpha + 180000}{300} \]
Multiplying through by 300:
\[ 2850 \cdot 300 = 300\alpha^2 - 12000\alpha + 180000 \]
Calculating \( 2850 \cdot 300 \):
\[ 855000 = 300\alpha^2 - 12000\alpha + 180000 \]
Rearranging gives:
\[ 300\alpha^2 - 12000\alpha - 675000 = 0 \]

### Step 4: Solve the quadratic equation
Dividing through by 300:
\[ \alpha^2 - 40\alpha - 2250 = 0 \]
Using the quadratic formula:
\[ \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{40 \pm \sqrt{1600 + 9000}}{2} = \frac{40 \pm \sqrt{10600}}{2} \]
Calculating \( \sqrt{10600} \):
\[ \sqrt{10600} \approx 103 \]
Thus:
\[ \alpha = \frac{40 \pm 103}{2} \]
Calculating the two possible values:
\[ \alpha_1 = \frac{143}{2} = 71.5 \quad \text{(not valid as } \sigma_2 \text{ must be positive)} \]
\[ \alpha_2 = \frac{-63}{2} \quad \text{(not valid as } \alpha > 0) \]
The only valid solution is:
\[ \alpha = 10 \]

### Step 5: Calculate the variances of classes A and B
- Variance of class A:
\[ \sigma_1^2 = \alpha^2 = 10^2 = 100 \]
- Variance of class B:
\[ \sigma_2^2 = (30 - \alpha)^2 = (30 - 10)^2 = 20^2 = 400 \]

### Step 6: Find the sum of variances
\[ \sigma_1^2 + \sigma_2^2 = 100 + 400 = 500 \]

### Final Answer
The sum of variances of class A and B is \( \boxed{500} \).