Percentage problems — increase, decrease, successive percentage change

Question

A price increases by 20% and then decreases by 20%. Is the final price the same as the original? If not, find the net percentage change. Also, if a population of 50,000 grows at 5% per year, find the population after 3 years.

(CBSE 7 & 8 — comparing quantities)

Solution — Step by Step

Let original price = Rs 100.

After 20% increase: $100 \times 1.20 = \text{Rs } 120$

After 20% decrease on Rs 120: $120 \times 0.80 = \text{Rs } 96$

Final price = Rs 96, which is LESS than the original. Net change = $-4\%$ .

For successive changes of $+a\%$ and $-a\%$ :

\text{Net change} = -\frac{a^2}{100}\%

Here: $-\frac{20^2}{100} = -\frac{400}{100} = -4\%$

This always results in a net decrease. A 20% increase followed by 20% decrease is NOT zero — it’s a 4% loss.

P = P_0\left(1 + \frac{r}{100}\right)^n = 50000\left(1 + \frac{5}{100}\right)^3

= 50000 \times (1.05)^3 = 50000 \times 1.157625 = \mathbf{57{,}881}

(Rounded to nearest whole number since population must be a whole number.)

Why This Works

Percentages are relative to the current value, not the original. When you increase by 20% and then decrease by 20%, the decrease is applied to the larger number ( $120$ , not $100$ ). So 20% of $120 = 24$ , which is more than the original increase of $20$ . The net effect is always a loss.

graph TD
    A["Percentage Problem"] --> B{"What type?"}
    B -->|"Single change"| C["New = Original × (1 ± p/100)"]
    B -->|"Successive changes"| D["Multiply factors<br/>(1 + p₁/100)(1 + p₂/100)..."]
    B -->|"Compound growth/decay"| E["Final = Initial × (1 ± r/100)^n"]
    D --> F["Equal +a% and -a%"]
    F --> G["Net = -a²/100 %<br/>(always a loss)"]
    B -->|"Find original from final"| H["Original = Final / (1 ± p/100)"]

Alternative Method — Fraction Equivalents

For common percentages, use fractions: 20% = $1/5$ , 25% = $1/4$ , 10% = $1/10$ .

20% increase means multiply by $6/5$ . 20% decrease means multiply by $4/5$ .

Net: $\frac{6}{5} \times \frac{4}{5} = \frac{24}{25}$ . Since $\frac{24}{25} = 0.96$ , the net change is $-4\%$ .

For compound growth, remember: $(1.05)^2 = 1.1025$ and $(1.05)^3 = 1.157625$ . For 10% growth: $(1.1)^2 = 1.21$ and $(1.1)^3 = 1.331$ . Memorising these saves calculation time in CBSE board exams.

Common Mistake

Students assume that a 20% increase followed by a 20% decrease returns to the original value. It doesn’t — there’s always a net loss of $a^2/100$ percent. Similarly, two successive 10% increases don’t give a 20% increase — they give $(1.1)^2 = 1.21$ , i.e., a 21% increase. Never add successive percentage changes; always multiply the factors.

Question

Solution — Step by Step

Why This Works

Alternative Method — Fraction Equivalents

Common Mistake

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