The Logic Behind the Math: Multipliers

The most efficient way to handle percentages in competitive exams is the Decimal Multiplier.

• To increase a value by $x\%$, multiply by $(1 + \frac{x}{100})$.
• To decrease a value by $x\%$, multiply by $(1 - \frac{x}{100})$.

Example: To increase $500$ by $20\%$, simply do $500 \times 1.2 = 600$. This is much faster than finding $20\%$ and adding it back.

Step-by-Step Solved Example (Successive Change)

Problem: The price of a book increases by $20\%$ and then decreases by $10\%$. What is the net percentage change?

• Step 1: Identify Initial Multiplier. $20\%$ increase $\to 1.2$.
• Step 2: Identify Second Multiplier. $10\%$ decrease $\to 0.9$.
• Step 3: Calculate Net Multiplier. $1.2 \times 0.9 = 1.08$.
• Step 4: Interpret the Result. Since $1.08 > 1$, it is an increase.$1.08 - 1 = 0.08$.$0.08 \times 100 = 8\%$.
• Conclusion: The net change is an $8\%$ increase.

Alternative Methods: The $x + y + \frac{xy}{100}$ Formula

For two successive changes $x$ and $y$, the net change is given by this formula.For the example above: $+20 + (-10) + \frac{20 \times -10}{100} = 10 - 2 = 8\%$.This is a legendary shortcut for JEE/NEET students.

Exam Trap Alert: The "Reversing" Fallacy

If a value decreases by $25\%$, it must increase by more than $25\%$ to get back to the original value.

Logic Trap: If you lose $25\%$ of ₹100, you have ₹75. To get back to ₹100, you need to add ₹25. But ₹25 is 33.33% of ₹75, not $25\%$. In competitive exams, "25%" will always be Option A. Don't fall for it!

Practice Problem (JEE Physics Error Logic)

Question: If the radius of a sphere is increased by $2\%$, what is the percentage increase in its volume?Hint: $V = \frac{4}{3}\pi r^3$. The multiplier is $(1.02)^3$. Use the binomial approximation $(1+x)^n \approx 1+nx$.