SSAT Upper Level Quantitative Quiz: Percent Increase And Decrease
10 questions · exam conditions
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Percent Increase And DecreaseQuestion 1 of 10

Maria's salary increased by 12%12\% and then by another 8%8\%. What single percent increase would have the same effect as these two consecutive increases?

20.0%
20.96%
19.84%
21.12%
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SSAT Upper Level Quantitative Quiz

SSAT Upper Level Quantitative Quiz: Percent Increase And Decrease

Practice Percent Increase And Decrease in SSAT Upper Level Quantitative with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Percent Increase And Decrease, giving you a quick way to practice the rules, question types, and explanations that matter most for SSAT Upper Level Quantitative.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Maria's salary increased by 12%12\% and then by another 8%8\%. What single percent increase would have the same effect as these two consecutive increases?

  1. 20.0%
  2. 20.96% (correct answer)
  3. 19.84%
  4. 21.12%
Explanation: Two consecutive increases of 12% and 8% result in a total multiplier of 1.12×1.08=1.20961.12 \times 1.08 = 1.2096, which represents a 20.96% increase. Choice A is the common error of simply adding the percentages. Choice C results from incorrect calculation of compound effects. Choice D results from calculation errors in the multiplication.

Question 2

The price of gasoline increased by 12%12\% in January, decreased by 8%8\% in February, then increased by 5%5\% in March. If the price at the end of March is $3.36 per gallon, what was the price at the beginning of January?

  1. $3.15 per gallon
  2. $3.25 per gallon
  3. $3.20 per gallon (correct answer)
  4. $3.18 per gallon
Explanation: Let the original price be xx. After the three changes: x×1.12×0.92×1.05=x×1.08192=3.36x × 1.12 × 0.92 × 1.05 = x × 1.08192 = 3.36. So x=3.36÷1.08192=3.20x = 3.36 ÷ 1.08192 = 3.20. Choice A results from calculation errors. Choice B results from incorrect handling of the percentage changes. Choice D results from rounding errors in intermediate calculations.

Question 3

Company A's profits increased by 25%25\% while Company B's profits decreased by 20%20\%. If both companies now have the same profit of $150,000, what was the difference between their original profits?

  1. $67,500 (correct answer)
  2. $75,000
  3. $62,500
  4. $70,000
Explanation: Company A's original profit: 150,000÷1.25=120,000150,000 ÷ 1.25 = 120,000. Company B's original profit: 150,000÷0.80=187,500150,000 ÷ 0.80 = 187,500. The difference is 187,500−120,000=67,500187,500 - 120,000 = 67,500. Choice B results from calculation errors. Choice C results from incorrectly computing the percentage changes. Choice D results from approximation errors.

Question 4

The population of a town decreased by 15%15\% in the first year and then increased by 20%20\% in the second year. If the population at the end of two years is 10,20010,200, what was the population at the beginning of the first year?

  1. 10,000 people (correct answer)
  2. 12,000 people
  3. 9,500 people
  4. 11,500 people
Explanation: Let the original population be xx. After 15% decrease: 0.85x0.85x. After 20% increase: 0.85x×1.2=1.02x=10,2000.85x \times 1.2 = 1.02x = 10,200. So x=10,000x = 10,000. Choice B results from incorrectly calculating the net change as −15%+20%=5%-15\% + 20\% = 5\%. Choice C results from calculation errors. Choice D results from reversing the order of operations.

Question 5

The number of students in a school increased by 15%15\% from Year 1 to Year 2, then decreased by 10%10\% from Year 2 to Year 3. If there were 1,2421,242 students in Year 3, how many students were there in Year 1?

  1. 1,200 students (correct answer)
  2. 1,180 students
  3. 1,300 students
  4. 1,150 students
Explanation: Let Year 1 population be xx. After 15% increase: 1.15x1.15x. After 10% decrease: 1.15x×0.9=1.035x=1,2421.15x \times 0.9 = 1.035x = 1,242. So x=1,200x = 1,200. Choice B results from calculation errors. Choice C results from working with the wrong base values. Choice D results from incorrectly handling the percentage calculations.

Question 6

A store increases the price of a jacket by 25%25\%, then offers a 20%20\% discount on the new price. If the final price is $96, what was the original price of the jacket?

  1. $80
  2. $100 (correct answer)
  3. $120
  4. $75
Explanation: Let the original price be xx. After a 25% increase: 1.25x1.25x. After a 20% discount on the new price: 1.25x×0.8=x1.25x \times 0.8 = x. So x=96x = 96, meaning the original price was $100. Choice A results from incorrectly calculating 96÷1.296 ÷ 1.2. Choice C results from incorrectly calculating 96÷0.896 ÷ 0.8. Choice D results from multiple calculation errors.

Question 7

A recipe calls for 20%20\% more flour than sugar. If the recipe uses 360360 grams of flour, and the baker wants to reduce the sugar by 15%15\%, how many grams of sugar will be used?

  1. 300 grams
  2. 270 grams
  3. 255 grams (correct answer)
  4. 285 grams
Explanation: This problem tests your ability to work with percentages in multi-step word problems. When you see "X% more than Y," remember that X = Y + (percentage × Y), which means you need to work backwards to find the original amount. Start by finding how much sugar the original recipe calls for. If flour is 20% more than sugar, then: flour = sugar + 0.20 × sugar = 1.20 × sugar. Since we know there are 360 grams of flour: 360 = 1.20 × sugar, so sugar = 360 ÷ 1.20 = 300 grams. Now apply the 15% reduction to find the final amount of sugar used. A 15% reduction means using 85% of the original amount: 300 × 0.85 = 255 grams. Looking at the wrong answers: Choice A (300 grams) is the original amount of sugar before the reduction—this traps students who find the initial sugar amount but forget to apply the 15% reduction. Choice B (270 grams) likely comes from incorrectly calculating 15% of 300 as 30, then subtracting to get 300 - 30 = 270, but 15% of 300 is actually 45. Choice D (285 grams) might result from miscalculating the percentage relationship or making arithmetic errors in the multi-step process. The key strategy here is to break percentage problems into clear steps: first find the original amount by setting up the percentage relationship correctly, then apply any additional changes. Always double-check that you've answered what the question actually asks for—the final amount after all modifications.

Question 8

A stock's value decreased by 30%30\% on Monday. By what percent must it increase on Tuesday to return to its original value?

  1. 30%
  2. 42.86% (correct answer)
  3. 43.33%
  4. 40.00%
Explanation: If the stock decreases by 30%, it's worth 70% of its original value. To return to 100%, it must increase by 3070×100%=37×100%=42.86%\frac{30}{70} \times 100\% = \frac{3}{7} \times 100\% = 42.86\%. Choice A is the common error of thinking the same percentage works in reverse. Choice C results from rounding errors. Choice D is an arbitrary intermediate value.

Question 9

A store marks up the wholesale cost of an item by 40%40\% to set the retail price, then offers a 25%25\% discount during a sale. If the sale price is $63, what was the wholesale cost?

  1. $45.00
  2. $48.00
  3. $52.50
  4. $60.00 (correct answer)
Explanation: When you encounter multi-step percentage problems, work backwards from the given information to trace through each operation in reverse order. Let's call the wholesale cost ww. The store first marks up by 40%, making the retail price w×1.40=1.4ww \times 1.40 = 1.4w. Then they apply a 25% discount, so the sale price becomes 1.4w×0.75=1.05w1.4w \times 0.75 = 1.05w. Since the sale price is $63, we have $1.05w=631.05w = 63 $. Solving for w : w = \frac{63}{1.05} = 60 So the wholesale cost was $60. Let's check why the other answers don't work. If you chose (A) $45.00, you might have incorrectly calculated the combined effect of the markup and discount. Starting with $45: $$45 \times 1.4 = 63,then, then 63 \times 0.75 = 47.25$$, not $63. Choice (B) $48.00 leads to: $$48 \times 1.4 = 67.20,then, then 67.20 \times 0.75 = 50.40$$. Choice (C) $52.50 gives: $52.50×1.4=73.5052.50 \times 1.4 = 73.50 ,then, then 73.50×0.75=55.1373.50 \times 0.75 = 55.13 $. Only choice (D) $60.00 produces the correct sale price: $$60 \times 1.4 = 84,then, then 84 \times 0.75 = 63$$. Study tip: In sequential percentage problems, always multiply by the decimal form of each percentage change in order, and remember that a 40% markup means multiplying by 1.40, while a 25% discount means multiplying by 0.75.

Question 10

A car's value depreciates by 18%18\% each year. After two years, the car is worth $13,448. What was the car's original value?

  1. $19,500
  2. $16,400
  3. $18,000
  4. $20,000 (correct answer)
Explanation: This problem involves exponential decay, where a value decreases by a fixed percentage each year. When you see depreciation or decay problems, you're working backwards from a final value to find the original amount. Let's call the original value xx. After the first year, the car loses 18% of its value, so it retains 82% or 0.82 of its original worth. After the second year, it again retains 82% of the previous year's value. This gives us the equation: x×0.82×0.82=13,448x \times 0.82 \times 0.82 = 13,448, or x×(0.82)2=13,448x \times (0.82)^2 = 13,448. Calculating: (0.82)2=0.6724(0.82)^2 = 0.6724, so x×0.6724=13,448x \times 0.6724 = 13,448. Solving for xx: x=13,4480.6724=20,000x = \frac{13,448}{0.6724} = 20,000. Choice A ($19,500) would result in a final value of $19,500 \times 0.6724 = \13,112 , which is too low. Choice B ($16,400) gives $16,400 \times 0.6724 = \11,023, far too low. Choice C ($18,000) yields $$18,000 \times 0.6724 = \$12,103, still too low. Only choice D ($20,000) produces the correct final value of $13,448. Strategy tip: In depreciation problems, remember that "loses X%" means "retains (100-X)%." When working backwards from a final value, divide by the retention rate raised to the power of the number of years. Always check your answer by working forward to verify you get the given final amount.