Reverse Percent Problems on the SAT: Find the Original Value Without Using the Wrong Base

A reverse percent problem gives you the new value after a percentage change and asks you to recover the original value.

The safest method is to use a multiplier.

If a quantity increases by $r\%$:

$$\text{new}=\text{original}(1+r)$$

where $r$ is written as a decimal.

If a quantity decreases by $r\%$:

$$\text{new}=\text{original}(1-r).$$

So the reverse operation is:

$$\boxed{\text{original}=\frac{\text{new}}{\text{multiplier}}}.$$

Example: after a 20% increase, a value is 96.

$$96=1.20(\text{original})$$ $$\text{original}=\frac{96}{1.20}=80.$$

The most common wrong approach is to subtract 20% of 96. That uses the new value as the percent base even though the 20% increase was calculated from the old value.

Why reverse percentages are an SAT skill

College Board's Problem-Solving and Data Analysis domain explicitly includes percentages, ratios, rates, proportional relationships, and units.

Reverse-percent problems can appear through:

• discounts;
• markups;
• tax or fee changes;
• population growth or decline;
• test or survey percentages;
• sales/revenue changes;
• original price/value questions;
• multi-step percentage changes.

For the full parent skill, use SAT Percentages. For the difference between relative percent change and percentage points, see Percent Increase vs Percentage Points.

Forward percent vs reverse percent

Forward problem

Original price:

$$\$80.$$

Increase:

$$20\%.$$

New price:

$$80(1.20)=96.$$

Reverse problem

New price:

$$\$96.$$

It is 20% higher than the original.

Original:

$$\frac{96}{1.20}=80.$$

The same multiplicative relationship works in both directions.

The multiplier table

For reverse problems:

$$\text{original}=\frac{\text{new}}{\text{multiplier}}.$$

Worked example 1: original price before discount

A jacket costs $72 after a 20% discount. What was the original price?

After a 20% discount, the customer pays:

$$80\%$$

of the original.

Multiplier:

$$0.80.$$

Equation:

$$72=0.80P.$$

So:

$$P=\frac{72}{0.80}.$$ $$\boxed{P=90}$$

Original price:

$$\boxed{\$90}.$$

The tempting wrong answer

A student may compute:

$$72+0.20(72)=86.40.$$

But 20% of 72 is not the amount discounted from the original price. The 20% was based on 90.

Worked example 2: original value before an increase

After a 15% increase, a quantity is 230.

Let original be $x$.

$$1.15x=230.$$ $$x=\frac{230}{1.15}.$$ $$\boxed{x=200}.$$

Check:

$$200(1.15)=230.$$

Worked example 3: original value after a decrease

A town's population is 42,500 after a 15% decrease. What was the population before the decrease?

Retained percentage:

$$100\%-15\%=85\%.$$

Multiplier:

$$0.85.$$ $$0.85P=42500.$$ $$P=\frac{42500}{0.85}.$$ $$\boxed{P=50000}.$$

"Percent of" and "percent more than" are different

Suppose:

$$B\text{ is }120\%\text{ of }A.$$

Then:

$$B=1.20A.$$

That means B is:

$$20\%$$

greater than A.

But if:

$$B\text{ is }120\%\text{ greater than }A,$$

then:

$$B=(1+1.20)A=2.20A.$$

That means B is 220% of A.

This wording distinction is a frequent percentage trap.

Worked example 4: "150% of" vs "150% greater"

Case A

B is 150% of A.

$$B=1.5A.$$

If:

$$B=90,$$

then:

$$A=60.$$

Case B

B is 150% greater than A.

$$B=2.5A.$$

If:

$$B=90,$$

then:

$$A=36.$$

Same number 150%. Completely different relationship.

Reverse discount plus sales tax

Suppose an item is discounted by 25%, then 8% sales tax is applied to the discounted price. The final price is $81.

Let original price be $P$.

Discount multiplier:

$$0.75.$$

Tax multiplier:

$$1.08.$$

Combined:

$$81=P(0.75)(1.08).$$ $$81=0.81P.$$

So:

$$\boxed{P=100}.$$

This is why multipliers are powerful: sequential percentage changes become multiplication.

For repeated changes, also read Repeated Percent Change on the SAT.

Reverse markup problem

A retailer sells a product for $156 after applying a 30% markup to its cost.

Let cost be $C$.

$$156=1.30C.$$ $$C=\frac{156}{1.30}.$$ $$\boxed{C=120}.$$

Do not subtract 30% of $156. The markup was 30% of the cost, not the selling price.

Worked example 5: percent less than another value

A is 25% less than B. If:

$$A=90,$$

find B.

"A is 25% less than B" means A is 75% of B:

$$A=0.75B.$$

So:

$$90=0.75B.$$ $$B=120.$$ $$\boxed{120}$$

Why adding 25% to 90 is wrong

$$90(1.25)=112.5.$$

But 25% was measured from B, not from A.

Percent comparisons are reference-value problems.

Worked example 6: percent greater relationship reversed

B is 40% greater than A. If:

$$B=126,$$

find A.

$$B=1.40A.$$ $$126=1.4A.$$ $$A=90.$$ $$\boxed{90}$$

Now ask the reverse question:

What percent less is A than B?

Difference:

$$36.$$

Base is B:

$$\frac{36}{126}\approx28.57\%.$$

So "B is 40% greater than A" does not mean "A is 40% less than B."

The base-value principle

Whenever you see a percentage statement, identify:

Percent of what?

Examples:

"20% of 80"

Base:

$$80.$$

"20% increase from 80"

Base:

$$80.$$

"A is 20% less than B"

Base:

$$B.$$

"B is 20% greater than A"

Base:

$$A.$$

Most hard percentage questions are really about choosing the correct base.

Reverse percentage from a table

Suppose a table shows:

The prompt says 2026 revenue is 10% higher than 2025.

Let 2025 revenue be $R$.

$$264000=1.10R.$$ $$R=240000.$$ $$\boxed{\$240,000}$$

Do not compute 10% of the 2026 value and subtract it.

Reverse percentage in a data context

A survey result is 54%, which is 20% higher than the previous year's proportion.

Let previous proportion be $p$.

$$54=1.20p.$$ $$p=45.$$

So the previous result was:

$$\boxed{45\%}.$$

Notice the phrase "20% higher" is relative change, not 20 percentage points.

See Percent Increase vs Percentage Points.

Worked example 7: recover a value after two changes

A quantity decreases by 20% and then increases by 25%. Its final value is 240.

Let original be $x$.

First factor:

$$0.80.$$

Second:

$$1.25.$$

Combined:

$$0.80(1.25)=1.$$

So:

$$240=x.$$ $$\boxed{240}$$

This particular pair cancels multiplicatively because:

$$0.8\times1.25=1.$$

Do not assume equal percentage increase/decrease always cancels; 20% down and 20% up does not.

Worked example 8: repeated growth reversed

An investment grows by 10% each year for 2 years and is then worth $12,100.

Let starting amount be $P$.

$$12100=P(1.10)^2.$$ $$12100=1.21P.$$ $$P=10000.$$ $$\boxed{\$10,000}$$

This connects reverse percent work to exponential models. See Growth Rate vs Growth Factor.

Reverse percent and linear models are not the same thing

Suppose a price increases by $20.

That is additive:

new=old+20

Suppose a price increases by 20%.

That is multiplicative:

$$\text{new}=1.20(\text{old}).$$

The first is a linear change by a fixed amount. The second is a proportional percentage change.

For model interpretation, see SAT Linear Models in Word Problems.

A ratio approach

Percent statements can also be represented as ratios.

If a value after a 20% discount is 80% of original:

$$\frac{\text{new}}{\text{original}}=\frac{80}{100}=0.8.$$

So:

$$\frac{72}{P}=0.8.$$

Then:

$$P=90.$$

This can be useful if ratio reasoning feels more natural than multiplier language.

Common trap 1: subtracting the percent from the new value

A 20% increase that produces 120 does not mean the original is:

$$120-24=96.$$

Correct:

$$120/1.2=100.$$

Common trap 2: adding the discount back

A 25% discount producing 75 does not mean original is:

$$75+18.75=93.75.$$

Correct:

$$75/0.75=100.$$

Common trap 3: using the wrong multiplier for a decrease

30% decrease means retain:

$$70\%.$$

Multiplier:

$$0.70,$$

not:

$$0.30.$$

Common trap 4: confusing "of" with "greater than"

150% of = factor 1.5.

150% greater than = factor 2.5.

Common trap 5: assuming percent relationships reverse symmetrically

40% greater one way is not 40% less the other way.

Common trap 6: adding sequential percentages

Use multipliers:

$$(1+r_1)(1+r_2)\dots$$

not percentage addition unless the context is explicitly linearized.

Common trap 7: ignoring what the question asks for

The problem may ask for:

• original value;
• percent change;
• amount of change;
• multiplier;
• final value.

Do not stop one step early.

A 20-second reverse-percent routine

1. Identify the original quantity $x$.
2. Translate the percent change into a multiplier.
3. Write: $\text{new}=x(\text{multiplier}).$
4. Divide: $x=\frac{\text{new}}{\text{multiplier}}.$
5. Check by applying the percentage forward.

SAT-style practice

Question 1

A price is $84 after a 30% discount. Original price?

Retained:

$$70\%.$$ $$84=0.70P.$$ $$\boxed{P=120}.$$

Question 2

A population is 55,200 after increasing 15%.

$$55200=1.15P.$$ $$\boxed{P=48000}.$$

Question 3

A value is 35% less than an original and equals 130.

$$130=0.65P.$$ $$\boxed{P=200}.$$

Question 4

B is 25% greater than A. If $B=150$, find A.

$$150=1.25A.$$ $$\boxed{A=120}.$$

Question 5

An item is discounted 20% and then taxed 10%. Final price $88.

$$88=P(0.8)(1.1)$$ $$88=0.88P.$$ $$\boxed{P=100}.$$

FAQ: reverse percentages on the SAT

How do I find the original value after a percent increase?

Divide the new value by:

$$1+r.$$

How do I find the original value after a percent decrease?

Divide the new value by:

$$1-r.$$

Why can't I just subtract the percentage from the final value?

Because the percentage was calculated from the original value, not necessarily the final value.

What multiplier represents a 35% discount?

$$0.65.$$

What multiplier represents a 35% increase?

$$1.35.$$

Is "20% less" the reverse of "20% greater"?

No. The reference base changes.

How do I reverse multiple percentage changes?

Multiply the forward multipliers together, then divide the final value by the combined multiplier.

How to practice this on MastaPrep

MastaPrep starts with a short placement check, then builds a study plan around the SAT Math skills where you are losing the most points. If this one keeps costing you questions, it shows up in your plan rather than getting lost in a random practice set. Every question has a worked explanation, and the ones you miss come back for review a few days later, which is when you find out whether the fix has actually stuck.

See how MastaPrep works or try free SAT practice questions. The free plan needs no credit card.

Related MastaPrep guides

• SAT Percentages
• Percent Increase vs Percentage Points
• Growth Rate vs Growth Factor
• Repeated Percent Change
• Ratios, Rates, and Unit Conversions
• SAT Linear Models in Word Problems

Final takeaway

Reverse percent problems become much easier when you stop asking:

What percent should I add or subtract?

and instead ask:

What fraction of the original became the new value?

Then write:

new=original × multiplier

and solve backward by division.

The essential skill is choosing the correct base for the percentage.

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Official SAT reference:
https://satsuite.collegeboard.org/sat/whats-on-the-test/math/types/problem-solving
https://satsuite.collegeboard.org/practice/student-question-bank/math

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