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Percent Increase and Decrease

SAT Math · Problem-Solving & Data Analysis · Updated June 2026

A percent change is always measured against the original amount. Multiplying by a single factor is usually faster than adding and subtracting.

Reading the concept isn't enough. The score comes from practicing the real Bluebook-style format and finding the exact mistakes you keep making.

What the SAT tests here

To increase by $r\%$, multiply by $1 + \frac{r}{100}$; to decrease, multiply by $1 - \frac{r}{100}$. The base is the starting value, not the new one.

How to solve one, step by step

Example: increase $\$80$ by $25\%$.

  1. Multiply by $1.25$: $80 \times 1.25$.
  2. Result: $\$100$.

The mistakes that cost points

Practice questions

Try these the way you would on test day, then open the solution to check your method.

Easy
A jacket is discounted from $\$80$ to $\$k$ after a $25\%$ decrease. What is the value of $k$?
  • A$k = 60$
  • B$k = 25$
  • C$k = 55$
  • D$k = 100$
Show solution
Answer: A, $k = 60$. Decrease $= 25\%$ of $\$80 = 0.25 \times 80 = 20$. New price $= 80 - 20 = \$60$.
Medium
A store raises a price by $k\%$. The original price is $\$50$ and the new price is $\$63$. What is the value of $k$?
  • A$k = 74$
  • B$k = 26$
  • C$k = 20.6$
  • D$k = 13$
Show solution
Answer: B, $k = 26$. $k = \frac{63 - 50}{50} \times 100 = \frac{13}{50} \times 100 = 26\%$.
Hard
A store marks down a price by $k\%$. The new price is $\$51$ and the original price was $\$60$. What is the value of $k$?
  • A$k = 15$
  • B$k = 17.6$
  • C$k = 9$
  • D$k = 85$
Show solution
Answer: A, $k = 15$. $k = \frac{60 - 51}{60} \times 100 = \frac{9}{60} \times 100 = 15\%$.

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