If y exceeds x by 10% then x is less than y by what percent?

Concept

A reverse-percentage comparison is NOT symmetric: if a quantity is p% more than a base, the base is less than that quantity by a DIFFERENT percentage, because the two comparisons use different denominators. The drop is measured against the LARGER value, so it is always smaller than the original increase.

Rule: if A is less than B, then “A is less than B by” = ((B − A) / B) × 100 — the difference is divided by B (the value we compare against), not by A.

Application

1. Let x be the base. y is 10% more than x, so y = x + 10% of x = 110x/100 = 11x/10.

2. We want how much x is below y, so the comparison is against y: required % = ((y − x) / y) × 100.

3. Difference: y − x = 11x/10 − x = x/10.

4. Substitute: ((x/10) / (11x/10)) × 100 = (1/11) × 100 = 100/11 = 9 1/11%.

Cross-check

Take x = 100. Then y = 110. x is below y by 110 − 100 = 10, and 10/110 = 1/11 = 9 1/11% ≈ 9.09%. Note this is smaller than the original 10% rise, exactly as the asymmetry predicts.