Inverse Proportion

Statement

Two quantities are said to be in inverse proportion (or inverse variation) if one increases while the other decreases in such a way that their product is constant. Mathematically:

\[ y \propto \tfrac{1}{x} \quad \Rightarrow \quad y = \tfrac{k}{x} \]

where \(k\) is the constant of proportionality.

Why it’s true

• If doubling one quantity halves the other, the product remains unchanged.
• For example, if speed increases, the time taken for a fixed journey decreases, but speed × time = distance stays constant.
• This constant product is what defines an inverse proportion.

Recipe (how to use it)

1. Write the relationship as \(y = k/x\).
2. Find the constant \(k\) using known values of \(x\) and \(y\).
3. Use this \(k\) to find new values of \(y\) for given values of \(x\).
4. Always check: \(x \times y = k\) should hold.

Spotting it

Inverse proportion appears when a problem says “one value gets smaller as the other gets bigger” with their product constant — e.g., “5 workers finish in 12 days, 10 workers finish in 6 days.”

Common pairings

• Speed, distance, and time problems.
• Work problems (workers vs days).
• Physics: pressure and volume (Boyle’s law).

Mini examples

1. Given: \(y\) varies inversely with \(x\). If \(y=8\) when \(x=3\), find \(y\) when \(x=12\). Solution: \(k=3×8=24\). For \(x=12\), \(y=24/12=2\).
2. Given: Time taken to complete a job varies inversely with the number of workers. If 4 workers take 15 days, how many days for 10 workers? Solution: \(k=4×15=60\). For 10 workers: time=60/10=6 days.

Pitfalls

• Mixing up with direct proportion: Direct uses \(y=kx\); inverse uses \(y=k/x\).
• Forgetting to keep product constant: Always calculate \(k=x×y\) first.
• Zero values: \(x=0\) makes no sense here since division by zero is undefined.

Exam strategy

• Step 1: Write down the product \(x×y=k\).
• Step 2: Use the given pair to find \(k\).
• Step 3: Apply it to find the missing value.
• Check: does the product stay the same?

Summary

Inverse proportion means as one quantity increases, the other decreases, keeping the product constant. The formula is \(y=k/x\). It is used in real-life problems such as speed-time-distance and workers-days jobs.