Percent Increase and Decrease - SSAT Upper Level Quantitative

1. A car's value depreciates by $18\%$ each year. After two years, the car is worth $13,448. What was the car's original value?
   Answer: $20,000 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 $x$. 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: $x \times 0.82 \times 0.82 = 13,448$, or $x \times (0.82)^2 = 13,448$. Calculating: $(0.82)^2 = 0.6724$, so $x \times 0.6724 = 13,448$. Solving for $x$: $x = \frac{13,448}{0.6724} = 20,000$. 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.
2. A store marks up the wholesale cost of an item by $40\%$ to set the retail price, then offers a $25\%$ discount during a sale. If the sale price is $63, what was the wholesale cost?
   Answer: $60.00 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 $w$. The store first marks up by 40%, making the retail price $w \times 1.40 = 1.4w$. Then they apply a 25% discount, so the sale price becomes $1.4w \times 0.75 = 1.05w$. Since the sale price is $63, we have $1.05w = 63$. 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 $63 \times 0.75 = 47.25$, not $63. Choice (B) $48.00 leads to: $48 \times 1.4 = 67.20$, then $67.20 \times 0.75 = 50.40$. Choice (C) $52.50 gives: $52.50 \times 1.4 = 73.50$, then $73.50 \times 0.75 = 55.13$. Only choice (D) $60.00 produces the correct sale price: $60 \times 1.4 = 84$, 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.
3. The number of students in a school increased by $15\%$ from Year 1 to Year 2, then decreased by $10\%$ from Year 2 to Year 3. If there were $1,242$ students in Year 3, how many students were there in Year 1?
   Answer: 1,200 students Let Year 1 population be $x$. After 15% increase: $1.15x$. After 10% decrease: $1.15x \times 0.9 = 1.035x = 1,242$. So $x = 1,200$. 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.
4. A store increases the price of a jacket by $25\%$, then offers a $20\%$ discount on the new price. If the final price is $96, what was the original price of the jacket?
   Answer: $96 Let the original price be $x$. After a 25% increase the price is $1.25x$, and a 20% discount on that leaves $0.8 \times 1.25x = 1.00x$. The two changes cancel exactly, so the final price equals the original price: the jacket originally cost $96. A common mistake is to treat the changes as a net 5% increase and divide 96 by 1.2, giving $80. Dividing 96 by 0.8 undoes only the discount and gives $120. Assuming the price ended up 4% lower and dividing 96 by 0.96 gives $100.
5. A recipe calls for $20\%$ more flour than sugar. If the recipe uses $360$ grams of flour, and the baker wants to reduce the sugar by $15\%$, how many grams of sugar will be used?
   Answer: 255 grams 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.
6. Company A's profits increased by $25\%$ while Company B's profits decreased by $20\%$. If both companies now have the same profit of $150,000, what was the difference between their original profits?
   Answer: $67,500 Company A's original profit: $150,000 ÷ 1.25 = 120,000$. Company B's original profit: $150,000 ÷ 0.80 = 187,500$. The difference is $187,500 - 120,000 = 67,500$. Choice B results from calculation errors. Choice C results from incorrectly computing the percentage changes. Choice D results from approximation errors.
7. The population of a town decreased by $15\%$ in the first year and then increased by $20\%$ in the second year. If the population at the end of two years is $10,200$, what was the population at the beginning of the first year?
   Answer: 10,000 people Let the original population be $x$. After 15% decrease: $0.85x$. After 20% increase: $0.85x \times 1.2 = 1.02x = 10,200$. So $x = 10,000$. Choice B results from incorrectly calculating the net change as $-15\% + 20\% = 5\%$. Choice C results from calculation errors. Choice D results from reversing the order of operations.
8. Maria's salary increased by $12\%$ and then by another $8\%$. What single percent increase would have the same effect as these two consecutive increases?
   Answer: 20.96% Two consecutive increases of 12% and 8% result in a total multiplier of $1.12 \times 1.08 = 1.2096$, 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.
9. A stock's value decreased by $30\%$ on Monday. By what percent must it increase on Tuesday to return to its original value?
   Answer: 42.86% If the stock decreases by 30%, it's worth 70% of its original value. To return to 100%, it must increase by $\frac{30}{70} \times 100\% = \frac{3}{7} \times 100\% = 42.86\%$. 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.
10. The price of gasoline increased by $12\%$ in January, decreased by $8\%$ in February, then increased by $5\%$ in March. If the price at the end of March is $3.36 per gallon, what was the price at the beginning of January, to the nearest cent?
    Answer: $3.11 per gallon Let the January starting price be $x$. Each change multiplies the price: a 12% increase by $1.12$, an 8% decrease by $0.92$, and a 5% increase by $1.05$. So $x \times 1.12 \times 0.92 \times 1.05 = 3.36$. The combined multiplier is $1.12 \times 0.92 = 1.0304$, and $1.0304 \times 1.05 = 1.08192$, so $x = 3.36 \div 1.08192 \approx 3.106$, which is $3.11 per gallon to the nearest cent. A price of $3.20 comes from undoing only the March increase, since $3.36 \div 1.05 = 3.20$. Starting at $3.15 would end near $3.41, and starting at $3.25 would end near $3.52, so both are too high.

Start this quiz