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Unit 3: Ratios, Rates, and Percents

Topic 4

Successive Percent Changes (Compound Percentage)

Successive percent changes multiply factors. If an amount AA changes by p1%p_1\% then by p2%p_2\%, the final value is
A⋅(1±p1)⋅(1±p2),A\cdot (1\pm p_1)\cdot (1\pm p_2),
where each pip_i is written as a decimal and the sign is plus for increase, minus for decrease.
Key facts:
  • Percent changes are not additive. A 20% increase then a 20% decrease does not return to the original.
  • Order does not matter for multiplication of factors. A(1+p)(1−q)=A(1−q)(1+p)A(1+p)(1-q) = A(1-q)(1+p).
  • Repeated changes by the same percent use exponents: A(1+p)nA(1+p)^n or A(1−p)nA(1-p)^n.
  • To undo a single change by p%p\%, divide by the factor: original =new1±p=\dfrac{\text{new}}{1\pm p}. To undo multiple, divide by the product of factors.

Core Skills

  • Convert each percent to a multiplier 1±p1\pm p before computing.
  • Multiply factors in sequence to get the net factor.
  • For repeated yearly or stepwise changes, write a compact exponential model.
  • Distinguish percent points from percent change when reading contexts.

Example 1: Increase then Decrease

A price increases 25% then decreases 20%. Net factor
(1+0.25)(1−0.20)=1.25⋅0.80=1.00.(1+0.25)(1-0.20)=1.25\cdot0.80=1.00.
Final equals original in this special case.

Example 2: Decrease then Decrease

An item is discounted 30% and then an extra 10%.
Final factor=(1−0.30)(1−0.10)=0.70⋅0.90=0.63.\text{Final factor}=(1-0.30)(1-0.10)=0.70\cdot0.90=0.63.
Final price is 63% of the original.

Example 3: Two Increases

Population grows by 8% one year and 5% the next.
Net factor=(1.08)(1.05)=1.134.\text{Net factor}=(1.08)(1.05)=1.134.
Overall increase is 13.4%.

Example 4: Repeated Change Model

A device loses 12% of its value each year. Initial value V0V_0. After nn years
Vn=V0(1−0.12)n=V0(0.88)n.V_n=V_0(1-0.12)^n=V_0(0.88)^n.

Example 5: Reverse After Two Changes

After a 10% increase and then a 15% decrease the final price is $306. Original =306(1.10)(0.85)=3060.935≈$327.27=\dfrac{306}{(1.10)(0.85)}=\dfrac{306}{0.935}\approx \$327.27.

Example 6: Not Additive

A 20% increase then a 20% decrease: factor 1.2⋅0.8=0.961.2\cdot0.8=0.96. Net change is a 4% decrease, not zero.

Key Takeaways

  • Translate every change into a multiplier and multiply.
  • Use powers for repeated identical changes.
  • To recover an original, divide by the product of all change factors.
  • Do not add percent changes. Work with factors to avoid errors.