Net Percentage Change

Cheatsheet Content

Net Percentage Change Net percentage change is the overall percentage change when a quantity undergoes multiple successive percentage changes. It's crucial in finance, economics, and various fields to understand the cumulative effect of changes. 1. Basic Formula If a quantity $Q$ changes by $P_1\%$ and then by $P_2\%$, the final value $Q_f$ can be found by: Factor approach: $Q_f = Q \times (1 + \frac{P_1}{100}) \times (1 + \frac{P_2}{100})$ Net Change Factor: $NCF = (1 + \frac{P_1}{100}) \times (1 + \frac{P_2}{100}) \times ... \times (1 + \frac{P_n}{100})$ Net Percentage Change: $(NCF - 1) \times 100\%$ Note: $P_i$ is positive for an increase and negative for a decrease. 2. Two Successive Changes For two changes, $P_1\%$ and $P_2\%$, the net percentage change ($P_{net}$) is given by: $P_{net} = P_1 + P_2 + \frac{P_1 \times P_2}{100}$ This formula is derived from the factor approach: $(1 + \frac{P_1}{100})(1 + \frac{P_2}{100}) = 1 + \frac{P_1}{100} + \frac{P_2}{100} + \frac{P_1 P_2}{10000}$ So, $NCF - 1 = \frac{P_1}{100} + \frac{P_2}{100} + \frac{P_1 P_2}{10000}$ $P_{net} = (NCF - 1) \times 100 = P_1 + P_2 + \frac{P_1 P_2}{100}$ Example 1: Successive Increases Problem: A stock price increases by $10\%$ on Monday and then by $20\%$ on Tuesday. What is the net percentage change? Using formula: $P_1 = 10$, $P_2 = 20$ $P_{net} = 10 + 20 + \frac{10 \times 20}{100} = 30 + \frac{200}{100} = 30 + 2 = 32\%$ Using factors: Initial price $Q$. After Monday: $Q \times (1 + \frac{10}{100}) = Q \times 1.10$ After Tuesday: $(Q \times 1.10) \times (1 + \frac{20}{100}) = Q \times 1.10 \times 1.20 = Q \times 1.32$ Net change: $1.32 - 1 = 0.32$, so $32\%$ increase. Example 2: Successive Decreases Problem: The value of a car decreases by $10\%$ in the first year and then by $15\%$ in the second year. What is the net percentage change? Using formula: $P_1 = -10$, $P_2 = -15$ $P_{net} = -10 + (-15) + \frac{(-10) \times (-15)}{100} = -25 + \frac{150}{100} = -25 + 1.5 = -23.5\%$ Net decrease of $23.5\%$. Using factors: Initial value $V$. After year 1: $V \times (1 - \frac{10}{100}) = V \times 0.90$ After year 2: $(V \times 0.90) \times (1 - \frac{15}{100}) = V \times 0.90 \times 0.85 = V \times 0.765$ Net change: $0.765 - 1 = -0.235$, so $23.5\%$ decrease. Example 3: Increase followed by Decrease Problem: A product's price is increased by $25\%$, and then later discounted by $20\%$. What is the net percentage change in price? Using formula: $P_1 = 25$, $P_2 = -20$ $P_{net} = 25 + (-20) + \frac{25 \times (-20)}{100} = 5 + \frac{-500}{100} = 5 - 5 = 0\%$ No net change in price. Using factors: Initial price $P$. After increase: $P \times (1 + \frac{25}{100}) = P \times 1.25$ After discount: $(P \times 1.25) \times (1 - \frac{20}{100}) = P \times 1.25 \times 0.80 = P \times 1.00$ Net change: $1.00 - 1 = 0$, so $0\%$ change. Example 4: Decrease followed by Increase Problem: A company's revenue decreased by $20\%$ last quarter, but increased by $25\%$ this quarter. What is the net percentage change over two quarters? Using formula: $P_1 = -20$, $P_2 = 25$ $P_{net} = -20 + 25 + \frac{(-20) \times 25}{100} = 5 + \frac{-500}{100} = 5 - 5 = 0\%$ No net change. 3. Multiple Successive Changes (More than Two) For $n$ successive changes ($P_1, P_2, ..., P_n$), the factor approach is generally easier: $NCF = (1 + \frac{P_1}{100}) \times (1 + \frac{P_2}{100}) \times ... \times (1 + \frac{P_n}{100})$ $P_{net} = (NCF - 1) \times 100\%$ Example 5: Three Changes Problem: A population increases by $5\%$, then by $8\%$, and finally decreases by $10\%$. What is the net percentage change in population? Using factors: $P_1 = 5, P_2 = 8, P_3 = -10$ $NCF = (1 + \frac{5}{100}) \times (1 + \frac{8}{100}) \times (1 - \frac{10}{100})$ $NCF = 1.05 \times 1.08 \times 0.90$ $NCF = 1.134 \times 0.90 = 1.0206$ $P_{net} = (1.0206 - 1) \times 100 = 0.0206 \times 100 = 2.06\%$ Net increase of $2.06\%$. Using two-change formula iteratively: First two changes ($5\%$ and $8\%$): $P_{net12} = 5 + 8 + \frac{5 \times 8}{100} = 13 + \frac{40}{100} = 13 + 0.4 = 13.4\%$ Now combine $13.4\%$ with the third change ($-10\%$): $P_{net} = 13.4 + (-10) + \frac{13.4 \times (-10)}{100} = 3.4 + \frac{-134}{100} = 3.4 - 1.34 = 2.06\%$ Same result. 4. Applications and Key Concepts Equivalent Single Discount When multiple discounts are applied, they are successive percentage decreases. Problem: A shirt is discounted by $20\%$ and then an additional $10\%$ off the discounted price. What is the single equivalent discount? Using formula: $P_1 = -20, P_2 = -10$ $P_{net} = -20 + (-10) + \frac{(-20) \times (-10)}{100} = -30 + \frac{200}{100} = -30 + 2 = -28\%$ Equivalent to a single $28\%$ discount. (Not $30\%$) Area/Volume Changes If the sides of a 2D shape change by percentages, the area changes by a net percentage. Problem: The length of a rectangle increases by $20\%$ and its width decreases by $10\%$. What is the net percentage change in its area? Using formula: $P_L = 20, P_W = -10$ $P_{Area} = 20 + (-10) + \frac{20 \times (-10)}{100} = 10 + \frac{-200}{100} = 10 - 2 = 8\%$ The area increases by $8\%$. Order of Changes The order of percentage changes does NOT affect the net percentage change. $P_1 + P_2 + \frac{P_1 P_2}{100} = P_2 + P_1 + \frac{P_2 P_1}{100}$ Example 3 and 4 showed this: Increasing by $25\%$ then decreasing by $20\%$ gives $0\%$ change, and decreasing by $20\%$ then increasing by $25\%$ also gives $0\%$ change. Finding Original Value After Net Change If a final value $Q_f$ is known after a net percentage change $P_{net}$, and you need to find the original value $Q$. $Q_f = Q \times (1 + \frac{P_{net}}{100})$ $Q = \frac{Q_f}{(1 + \frac{P_{net}}{100})}$ Problem: After a $20\%$ increase, a product costs $ \$120$. What was its original price? $120 = Q \times (1 + \frac{20}{100}) = Q \times 1.20$ $Q = \frac{120}{1.20} = \$100$ Finding a Missing Percentage Change If the net percentage change and one of the successive changes are known, you can find the other. Problem: A price increased by $10\%$, and the net increase after a second change was $21\%$. What was the second percentage change? Let $P_1 = 10$, $P_{net} = 21$. Find $P_2$. $21 = 10 + P_2 + \frac{10 \times P_2}{100}$ $21 = 10 + P_2 + \frac{P_2}{10}$ $11 = P_2 + 0.1 P_2$ $11 = 1.1 P_2$ $P_2 = \frac{11}{1.1} = 10\%$ The second change was a $10\%$ increase.