Question

If x is inversely proportional to y and y is proportional to z then xz

• A. Is proportional to y
• B. Is inversely proportional to y
• C. Is a constant
• D. Is proportional to \(y^2\)

Hint:

When combining proportionalities, substitute the variables directly:
Since \(x \propto 1/y\) and \(y \propto z\), we have \(x \propto 1/z\).
If \(x\) is inversely proportional to \(z\), their product \(x \cdot z\) must be a constant.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Proportionality describes the mathematical relationship between two variables.
- Direct proportionality (\(a \propto b\)) means \(a = k \cdot b\), where \(k\) is a non-zero constant.
- Inverse proportionality (\(a \propto \frac{1}{b}\)) means \(a = \frac{k}{b}\), where \(k\) is a non-zero constant.

Step 2: Detailed Explanation:
Let us translate the proportionalities into algebraic equations:
1. Given that \(x\) is inversely proportional to \(y\):
\[ x \propto \frac{1}{y} \implies x = \frac{k_1}{y} \quad \text{--- (Equation 1)} \]
Where \(k_1\) is a non-zero constant of proportionality.
2. Given that \(y\) is directly proportional to \(z\):
\[ y \propto z \implies y = k_2 \cdot z \quad \text{--- (Equation 2)} \]
Where \(k_2\) is another non-zero constant of proportionality.
3. Rearrange Equation 2 to express \(z\) in terms of \(y\):
\[ z = \frac{y}{k_2} \quad \text{--- (Equation 3)} \]
4. Now, find the product of \(x\) and \(z\) (\(xz\)) by multiplying Equation 1 and Equation 3:
\[ xz = \left(\frac{k_1}{y}\right) \times \left(\frac{y}{k_2}\right) \]
The variable \(y\) in the numerator and denominator cancels out:
\[ xz = \frac{k_1}{k_2} \]
Since both \(k_1\) and \(k_2\) are constants, their quotient \(\frac{k_1}{k_2}\) must also be a constant.
Therefore, the product \(xz\) is a constant value and does not depend on \(y\) or \(z\).
This matches Option C.

Step 3: Final Answer:
The product xz is a constant.