Percentage Applications (CAT)
This is where percentages pay off in CAT: the "product is constant" family and exam-style word problems. When two quantities multiply to a fixed product (price × consumption = expenditure, speed × time = distance, length × breadth = area), a rise of p% in one forces a fall of [p/(100+p)]×100% in the other to keep the product fixed. Example: if price rises 25%, consumption must fall 25/125 = 20% to keep spending the same. The other staples are: passing marks ("scored x%, failed by m marks, pass mark is p%"), election problems (winner’s margin as a % of votes), and expenditure/income splits. The trick is to assign the unknown whole a value of 100 (or the LCM of the denominators) so every percentage becomes a clean number.
✅ Solved examples
✏️ Practice — try these, take hints as needed
📝 Topic test — 8 questions
Auto-graded with full solutions; saved to your dashboard. Use the calculator and formula sheet (top-right) any time.
Formula Reference Sheet
Core conversions & change
| Percentage of a number | x% of N = (x/100) × N |
|---|---|
| Value as a percentage | (Part / Whole) × 100 % |
| Percentage change | (New − Old) / Old × 100 % |
| Increase by x% | N × (1 + x/100) |
| Decrease by x% | N × (1 − x/100) |
CAT power-tools
| Successive change a% then b% | net = a + b + ab/100 (%) |
|---|---|
| Reverse percentage | Original = Final / (1 ± x/100) |
| A is x% more than B | B is [x/(100+x)]×100 % less than A |
| A is x% less than B | B is [x/(100−x)]×100 % more than A |
| Product constant (price↑ p% ⇒ consumption↓) | [p/(100+p)]×100 % |