Question

If H.C.F. of \(135x^2\) and \(189x^3\) is 108, then find the value of x.

Hint:

If you are unsure of division like \(\frac{108}{27}\), simplify it by dividing both numerator and denominator by common factors like 9 first:
\[ \frac{108}{27} = \frac{12}{3} = 4 \]
This reduces chances of division errors.

Answer 1

Correct Answer: 2

Step 1: Prime Factorise 135 and 189.
\[ 135 = 3 \times 45 = 3 \times 3 \times 15 = 3 \times 3 \times 3 \times 5 = 3^3 \times 5 \] \[ 189 = 3 \times 63 = 3 \times 3 \times 21 = 3 \times 3 \times 3 \times 7 = 3^3 \times 7 \]
Step 2: Find the HCF of the Numerical Coefficients.
$\text{HCF}(135, 189) = 3^3 = 27$ (taking the lowest power of common prime factors).
Step 3: Find the HCF of the Algebraic Terms.
For $135x^2$ and $189x^3$: \[ \text{HCF} = \text{HCF}(135, 189) \times x^{\min(2,3)} = 27 \times x^2 = 27x^2 \]
Step 4: Set Up the Equation.
We are told that $\text{HCF}(135x^2, 189x^3) = 108$. So: \[ 27x^2 = 108 \]
Step 5: Solve for x.
\[ x^2 = \frac{108}{27} = 4 \] \[ x = \sqrt{4} = 2 \quad (\text{taking positive value}) \]
Step 6: Verify the Answer.
Check: $135 \times 2^2 = 135 \times 4 = 540$ and $189 \times 2^3 = 189 \times 8 = 1512$. $\text{HCF}(540, 1512)$: $540 = 2^2 \times 3^3 \times 5$ and $1512 = 2^3 \times 3^3 \times 7$. HCF $= 2^2 \times 3^3 = 4 \times 27 = 108$. Confirmed! \[ \boxed{x = 2} \]