You use growth factor in the formula below, and solve the calculation or equation that appears. When using the formula, you always insert $x$ for the value you want to find.

Formula

New value = old value $\cdot $ growth factor

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Example 1

The price of a car decreases by $5$ %. What will the price be after the decrease if the price of the car is $$20\phantom{\rule{0.17em}{0ex}}000$?

Here the old price is $$20\phantom{\rule{0.17em}{0ex}}000$, new value $=x$, and the growth factor is as follows:

$$\phantom{\rule{-0.17em}{0ex}}\left(1-\frac{5}{100}\right)=1-0.05=0.95$$ |

The new price will be

$$\begin{array}{llll}\hfill \text{newvalue}& =\text{oldvalue}\cdot \text{growthfactor}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x=\text{\$}20\phantom{\rule{0.17em}{0ex}}000& \cdot 0.95=\text{\$}19\phantom{\rule{0.17em}{0ex}}000\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$Video Crash Courses

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Example 2

iPhone 13 costs $$799$ and is increased by $15$ % from iPhone 12. What did the iPhone 12 cost?

Here is old price $=x$, new value $$799$, and the growth factor becomes

$$\phantom{\rule{-0.17em}{0ex}}\left(1+\frac{15}{100}\right)=1+0.15=1.15$$ |

The old price was thus:

$$\begin{array}{llll}\hfill \text{newvalue}& =\text{oldvalue}\cdot \text{growthfactor}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 799& =x\cdot 1.15\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{799}{1.15}& =\frac{x\cdot 1.15}{1.15}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& \approx 694.78\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$So, an iPhone 12 costs $$694.78$.

Example 3

An item originally costed $$599$. The price was reduced to $$349$, and the store says that the item is lowered by $45$ %. Is it true?

To find $p$, you must first find the growth factor. You do this by using the formula above.

Let’s call the growth factor gf to save some space. $$\begin{array}{llll}\hfill \text{newvalue}& =\text{oldvalue}\cdot \text{gf}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 349& =599\cdot \text{gf}\phantom{\rule{1em}{0ex}}|:599\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \text{gf}& =\frac{349}{599}\approx 0.583\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

$$\begin{array}{llll}\hfill \text{newvalue}& =\text{oldvalue}\cdot \text{growthfactor}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 349& =599\cdot \text{growthfactor}\phantom{\rule{1em}{0ex}}|:599\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \text{growthfactor}& =\frac{349}{599}\approx 0.583\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

Since the value of the growth factor is less than $1$, you get confirmed that it is a drop in price. Then you use the formula for growth factor, with minus, to find $p$ in percentage: $$\begin{array}{llll}\hfill 0.583& =1-\frac{p}{100}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{p}{100}& =1-0.583\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{p}{100}& =0.417\phantom{\rule{2em}{0ex}}|\cdot 100\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill p& =41.7\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$ The price has only been reduced by $41.7$ %. You could have been tricked!