More Problems Involving Percent Increase and Decrease

Want more practice with percents and related concepts?

Here, you will practice solving more problems involving percent increase and decrease. You may use a calculator for these exercises.

Examples

Question: Suppose the price of an item increases by $\,19\%\,,$ and then decreases by $\,30\%\,.$ What is the resulting percent increase or decrease?

Solution:

$$ \begin{gather} \cssId{s9}{(0.7)(1.19)x} \cssId{s10}{= 0.83x} \cssId{s11}{= (1 - 0.17)x}\cr \cssId{s12}{\text{$17\%\,$ decrease}} \end{gather} $$

Why? As discussed in Problems Involving Percent Increase and Decrease, a price $\,x\,$ changes to $\,1.19x\,$ after the $\,19\%\,$ increase. After the subsequent $\,30\%\,$ decrease, only $\,70\%\,$ of this remains:

$$\cssId{s17}{(1-0.3)(1.19x)} \cssId{s18}{= (0.7)(1.19)x} \cssId{s19}{= 0.83x} $$

The price started at $\,x\,.$ It ended at $\,0.83x\,.$ So, the overall change was a decrease (note that $\,0.83 \lt 1\,$).

How much of a decrease was there in going from $\,x = 1x\,$ to $\,0.83x\,$? Answer:

$$\cssId{s26}{\,1x - 0.83x = 0.17x}$$

That is, $\,17\%\,$ of $\,x\,$ was ‘lost’ in the process. The combined effect of the back-to-back increase/decrease was a $\,17\%\,$ decrease.

Question: Suppose the price of an item decreases by $\,40\%\,,$ and then increases by $\,40\%\,.$ What is the resulting percent increase or decrease?

Solution:

$$ \begin{align} &\cssId{s33}{(1 + 0.4)(1 - 0.4)x}\cr &\qquad \cssId{s34}{= (1.4)(0.6)x}\cr &\qquad \cssId{s35}{= 0.84x}\cr &\qquad \cssId{s36}{= (1-0.16)x} \end{align} $$ $$ \cssId{s37}{\text{$16\%\,$ decrease}} $$

Pause for a moment and appreciate the power in renaming an expression! There are four names for the same expression given above, and each has its strength:

$(1 + 0.4)(1 - 0.4)x$	This name makes it clear that we're doing a $\,40\%\,$ decrease (the $\,1 - 0.4\,$) and a $\,40\%\,$ increase (the $\,1 + 0.4\,$).
$(1.4)(0.6)x$	This name is a whole lot easier to plug into a calculator.
$0.84x$	This name, as compared to the original $\,1x\,,$ shows that the overall effect was a decrease.
$(1 - 0.16)x$	This name shows that it was a $\,16\%\,$ decrease.

Question: Suppose the price of an item increases by $\,50\%\,,$ and then decreases by $\,50\%\,.$ What is the resulting percent increase or decrease?

Solution:

$$ \begin{align} &\cssId{s53}{(1 - 0.5)(1 + 0.5)x}\cr &\qquad \cssId{s54}{= (0.5)(1.5)x}\cr &\qquad \cssId{s55}{= 0.75x}\cr &\qquad \cssId{s56}{= (1 - 0.25)x} \end{align} $$ $$ \cssId{s57}{\text{$25\%\,$ decrease}} $$

Question: Suppose the price of an item increases by $\,30\%\,,$ and then decreases by $\,10\%\,.$ What is the resulting percent increase or decrease?

Solution:

$$ \begin{align} &\cssId{s62}{(1 - 0.1)(1 + 0.3)x}\cr &\qquad \cssId{s63}{= (0.9)(1.3)x}\cr &\qquad \cssId{s64}{= 1.17x}\cr &\qquad \cssId{s65}{= (1 + 0.17)x} \end{align} $$ $$ \cssId{s66}{\text{$17\%\,$ increase}} $$

Question: Suppose the price of an item increases by $\,50\%\,,$ and then increases by $\,50\%\,$ again. What is the resulting percent increase or decrease?

Solution:

$$ \begin{align} &\cssId{s71}{(1 + 0.5)(1 + 0.5)x}\cr &\qquad \cssId{s72}{= (1.5)(1.5)x}\cr &\qquad \cssId{s73}{= 2.25x}\cr &\qquad \cssId{s74}{= (1 + 1.25)x} \end{align} $$ $$ \cssId{s75}{\text{$125\%\,$ increase}} $$

Question: Suppose an item costs $\,\$50\,.$ The price increases by $\,20\%\,,$ and then decreases by $\,70\%\,.$ What is the resulting percent increase or decrease?

Solution: There are two good approaches. You choose!

First approach: Compute new price, then compute percent change:

New price is: $$\cssId{s86}{(0.3)(1.2)(\$50) = \$18}$$ It was an overall decrease.
The percent decrease is: $$ \cssId{s89}{\frac{50-18}{50}} \cssId{s90}{= 0.64} \cssId{s91}{= 64\%} $$

Second approach: You don't need the original price at all! Just denote it by $\,x\,$:

$$ \begin{gather} \cssId{s95}{(0.3)(1.2)x} \cssId{s96}{= 0.36x} \cssId{s97}{= (1 - 0.64)x}\cr \cssId{s98}{\text{$64\%\,$ decrease}} \end{gather} $$

Concept Practice

All answers are rounded to two decimal places.