SAT · Math · Problem-Solving and Data Analysis

Percentages

Percentages is part of Problem-Solving and Data Analysis in the SAT Math section. The questions below are written in the style of the real test, each with the answer and a worked explanation.

Question types you will see here include find a percent of a quantity, percent increase or decrease, successive percent changes and work back to the original.

Common traps: treating 15% as 1.5; dividing by the part instead of the whole; rounding an intermediate value before the final step.

Sample questions

Question 1
From last year to this year, a store's weekly sales decreased from 250 to 125. By what percent did it change?
- A.
  55%
- B.
  100%
- C.
  50%
- D.
  125%
Answer: C. 50%
Percent change = |new - old|/old × 100 = 125/250 × 100 = 50%.
Question 2
At a fuel station, the price of petrol was 1.50 euros per litre. It rose by 20%, and the next week fell by 10% of the new price. What was the price per litre after both changes?
- A.
  1.35
- B.
  1.50
- C.
  1.62
- D.
  1.65
Answer: C. 1.62
Multiply the factors: 1.50 × 1.20 × 0.90 = 1.62. Choice D adds the changes (a net +10%) and applies it to 1.50; B assumes a rise and a fall cancel; A applies only the 10% fall to the original price. Percent changes compound; they do not add.
Question 3
A coding bootcamp's instalment plan is a $1,200 deposit followed by 24 weekly payments of $250. The plan's total is 20% more than the price for paying the full tuition up front. What is the up-front price?
- A.
  $6,000
- B.
  $7,200
- C.
  $8,640
- D.
  $9,000
Answer: A. $6,000
The plan totals 1,200 + 24(250) = 7,200, which is 120% of the up-front price p: 1.2p = 7,200, so p = 6,000 (A). B is the plan total itself; C adds another 20% instead of removing it; D divides by 0.8, treating 7,200 as if the up-front price were 20% less than it, which uses the wrong base.
Question 4
A sprinter's best 100-meter time decreased by 4% after a season of training. Her new best time is 11.52 seconds. What was her best time, in seconds, before the season?
- A.
  11.06
- B.
  11.98
- C.
  12
- D.
  15.52
Answer: C. 12
The new time is 96% of the old time, so the old time was 11.52 ÷ 0.96 = 12 seconds (C). 11.98 adds 4% of the new time instead of dividing, 11.06 takes another 4% off, and 15.52 adds 4 seconds instead of 4%.
Question 5
A herd of deer decreased by 25% from one year to the next and then increased by 40% the year after that. After both changes the herd has 840 deer. How many deer were in the herd before the decrease?
- A.
  600
- B.
  630
- C.
  800
- D.
  840
Answer: C. 800
The two changes multiply the original by 0.75 × 1.40 = 1.05, so the original is 840 ÷ 1.05 = 800 (C). Choice A undoes only the 40% increase; B takes 25% off the final count; D is the final count, with no reversal at all.

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