Question

The sum of three numbers is 6. Thrice the third number when added to the first number gives 7. On adding three times first number to the sum of second and third number we get 12. The product of these numbers is

• A. $20$
• B. $3$
• C. $\frac{20}{3}$
• D. $\frac{5}{3}$

Hint:

Look for direct subtractions to save time! Subtracting Equation 1 directly from Equation 3 eliminates $y$ and $z$ completely in one go: $(3x + y + z) - (x + y + z) = 12 - 6 \implies 2x = 6 \implies x = 3$. Spotting this pattern avoids lengthy matrix row operations or substitution tracking.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given three word-problem statements describing linear relationships among three unknown numbers. We need to construct a system of linear equations, solve for the individual numbers, and calculate their mathematical product.

Step 2: Key Formula or Approach:
Let the three numbers be represented by variables $x$, $y$, and $z$. 1. Translate the phrases into algebraic equations: "The sum of three numbers is 6" $\implies x + y + z = 6$ "Thrice the third number when added to the first number gives 7" $\implies x + 3z = 7$ "Adding three times first number to the sum of second and third number we get 12" $\implies 3x + y + z = 12$ 2. Solve the linear system using substitution or elimination methods.

Step 3: Detailed Explanation:
Let's write down our system of equations: $$\text{Equation 1: } x + y + z = 6$$ $$\text{Equation 2: } x + 3z = 7$$ $$\text{Equation 3: } 3x + y + z = 12$$ Notice that Equation 1 and Equation 3 both contain the common grouping term $(y + z)$. Let's isolate $(y + z)$ in Equation 1: $$y + z = 6 - x$$ Substitute this group directly into Equation 3: $$3x + (6 - x) = 12$$ $$2x + 6 = 12 \implies 2x = 6 \implies x = 3$$ Now substitute the value of $x = 3$ back into Equation 2 to find $z$: $$3 + 3z = 7 \implies 3z = 4 \implies z = \frac{4}{3}$$ Finally, substitute the values of $x = 3$ and $z = \frac{4}{3}$ back into Equation 1 to find $y$: $$3 + y + \frac{4}{3} = 6$$ $$y = 6 - 3 - \frac{4}{3} = 3 - \frac{4}{3} = \frac{9 - 4}{3} = \frac{5}{3}$$ The problem asks for the product of these three numbers ($x \cdot y \cdot z$): $$\text{Product} = 3 \times \frac{5}{3} \times \frac{4}{3} = \frac{20}{3}$$

Step 4: Final Answer:
The product of the three numbers is $\frac{20}{3}$, which corresponds to option (C).