The product of the ages of Ram and Shyam is 280. If thrice the age of Shyam is more than Ram's age by 2 years, then what is the age of Shyam?

Options

A. 10 years
B. 15 years
C. 12 years
D. 18 years

Answer

To solve the problem, we need to find the ages of Ram and Shyam based on the information given. Let's break it down step by step.

### Step 1: Define Variables
Let Ram's age be \( x \) and Shyam's age be \( y \).

### Step 2: Set Up the Equations
From the problem, we have two pieces of information:
1. The product of their ages is 280:
\[ x \cdot y = 280 \quad \text{(Equation 1)} \]
2. Thrice the age of Shyam is more than Ram's age by 2 years:
\[ 3y = x + 2 \quad \text{(Equation 2)} \]

### Step 3: Express One Variable in Terms of the Other
From Equation 2, we can express \( x \) in terms of \( y \):
\[ x = 3y - 2 \quad \text{(Substituting this into Equation 1)} \]

### Step 4: Substitute into the First Equation
Now substitute \( x \) from the above equation into Equation 1:
\[ (3y - 2) \cdot y = 280 \]
Expanding this gives:
\[ 3y^2 - 2y = 280 \]

### Step 5: Rearrange the Equation
Rearranging the equation to set it to zero:
\[ 3y^2 - 2y - 280 = 0 \]

### Step 6: Factor or Use the Quadratic Formula
To solve this quadratic equation, we can either factor it or use the quadratic formula. Let's factor it:
We need two numbers that multiply to \( 3 \times (-280) = -840 \) and add to \( -2 \). The numbers are \( -30 \) and \( 28 \).
Thus, we can rewrite the equation as:
\[ 3y^2 - 30y + 28y - 280 = 0 \]
Grouping gives:
\[ 3y(y - 10) + 28(y - 10) = 0 \]
Factoring out \( (y - 10) \):
\[ (y - 10)(3y + 28) = 0 \]

### Step 7: Solve for \( y \)
Setting each factor to zero gives:
1. \( y - 10 = 0 \) → \( y = 10 \)
2. \( 3y + 28 = 0 \) → \( y = -\frac{28}{3} \) (not a valid age)

Thus, the only valid solution is:
\[ y = 10 \]

### Conclusion
Shyam's age is \( 10 \) years.