Percentage problems — increase, decrease, successive percentage change

Question

A mobile phone costs Rs 15,000. Its price increases by 10% in the first year and then decreases by 10% in the second year. Is the final price equal to the original price? Find the effective percentage change.

Solution — Step by Step

After first year: $15000 \times 1.10 = \text{Rs } 16500$

After second year: $16500 \times 0.90 = \text{Rs } 14850$

Notice: the final price (Rs 14,850) is NOT the original (Rs 15,000). A 10% increase followed by a 10% decrease does NOT cancel out.

Change $= 15000 - 14850 = \text{Rs } 150$ (decrease)

Effective change $= \frac{150}{15000} \times 100 = \mathbf{1\%}$ decrease

Why This Works

graph TD
    A["Successive Percentage Change"] --> B["Increase of a% then decrease of a%"]
    B --> C["Net effect = -a²/100 percent DECREASE"]
    A --> D["Increase of a% then increase of b%"]
    D --> E["Net effect = a + b + ab/100 percent"]
    A --> F["Common trap: 10% up then 10% down ≠ 0%"]
    F --> G["Actual result: 1% decrease always"]

The key insight: the 10% decrease applies to a LARGER number (16,500, not 15,000). So 10% of 16,500 is more than 10% of 15,000. The decrease removes more than the increase added.

For successive changes of $a\%$ and $b\%$ , the effective change is:

\text{Effective} = a + b + \frac{ab}{100}

For $a = +10, b = -10$ : Effective $= 10 - 10 + \frac{10 \times (-10)}{100} = 0 - 1 = -1\%$

This formula works for any combination of increases and decreases, and saves time over step-by-step calculation.

Alternative Method

Use the multiplier method for any chain of percentage changes. A $p\%$ increase means multiply by $(1 + p/100)$ . A $p\%$ decrease means multiply by $(1 - p/100)$ .

For our problem: $15000 \times 1.1 \times 0.9 = 15000 \times 0.99 = 14850$ .

The combined multiplier $1.1 \times 0.9 = 0.99$ tells you the effective result immediately: $0.99$ means a $1\%$ decrease.

Common Mistake

Assuming equal percentage increase and decrease cancel out. They never do. An $a\%$ increase followed by an $a\%$ decrease always results in a net decrease of $a^2/100$ percent. For 10%, that is $100/100 = 1\%$ decrease. For 20%, it is $400/100 = 4\%$ decrease. The larger the percentage, the bigger the gap. This is tested regularly in CBSE Class 8 exams.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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