A school decided to award prizes to the most punctual and disciplined students. The sum of two prizes is Rs.95 and their product is Rs. 2250. Calculate the prize money of the two prizes.

Options

A. Rs. 50, Rs. 45
B. Rs. 60, Rs. 35
C. Rs. 70, Rs. 25
D. None of these

Answer

To solve the problem, we need to find two numbers (prizes) whose sum is Rs. 95 and whose product is Rs. 2250. Let's denote the two prizes as \( x \) and \( y \).

### Step 1: Set up the equations
From the problem statement, we have:
1. \( x + y = 95 \) (Equation 1)
2. \( xy = 2250 \) (Equation 2)

### Step 2: Express one variable in terms of the other
From Equation 1, we can express \( y \) in terms of \( x \):
\[ y = 95 - x \]

### Step 3: Substitute into the second equation
Now, substitute \( y \) in Equation 2:
\[ x(95 - x) = 2250 \]
Expanding this gives:
\[ 95x - x^2 = 2250 \]

### Step 4: Rearrange the equation
Rearranging the equation leads to:
\[ x^2 - 95x + 2250 = 0 \]

### Step 5: Solve the quadratic equation
Now, we can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -95 \), and \( c = 2250 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-95)^2 - 4 \cdot 1 \cdot 2250 \]
\[ = 9025 - 9000 = 25 \]

Now substituting back into the quadratic formula:
\[ x = \frac{95 \pm \sqrt{25}}{2 \cdot 1} \]
\[ = \frac{95 \pm 5}{2} \]

Calculating the two possible values for \( x \):
1. \( x = \frac{100}{2} = 50 \)
2. \( x = \frac{90}{2} = 45 \)

### Step 6: Find the corresponding values of \( y \)
Using \( y = 95 - x \):
- If \( x = 50 \), then \( y = 95 - 50 = 45 \).
- If \( x = 45 \), then \( y = 95 - 45 = 50 \).

### Conclusion
The two prizes are Rs. 50 and Rs. 45.