Successive changes and reverse percentages

What you will learn

A percentage change acts on a named base. When a second change acts on the new amount, that base has moved. This lesson shows how to keep track of the movement, undo a known sequence, compare quantities in the reverse direction, and describe a change in a percentage rate precisely.

You should already be able to convert a percentage to a decimal, find a part of a positive whole and solve a simple equation. First check: 12% = 0.12, and retaining 88% means multiplying by 0.88.

Build the multiplier from the meaning

Let the original quantity be x. A rise of p% adds (p/100)x, so the new amount is x + (p/100)x = x(1 + p/100). A fall of p% removes that fraction, leaving x(1 − p/100). The 1 represents the original 100%; it must not disappear.

Change	Multiplier	Meaning
Increase by 15%	1.15	Retain 100% and add 15%
Decrease by 10%	0.90	Retain 90%
No change	1	Retain the whole
Decrease by 100%	0	Nothing remains

These ordinary price, count and measurement models use x > 0. A percentage change from an original value of zero is undefined because its formula divides by that value. A decrease over 100% would produce a negative amount, so it does not fit these nonnegative contexts. A rise over 100% can fit: doubling is a 100% rise, not a 200% rise.

Follow each current base

For factors a and b, the sequence is x → ax → abx. Overall signed percentage change is (ab − 1) × 100%. A positive result means increase; a negative result means decrease. If asked for the size of a decrease, give its positive magnitude and the word decrease.

Multiplying fixed factors in either order gives the same final value when there is no rounding, cap, fixed addition or other intervening rule. The intermediate values can still differ. Do not extend this observation to an operation such as adding a fixed ₹50: multiplication and addition do not generally commute. For example, 100 × 1.10 + 50 = 160 but (100 + 50) × 1.10 = 165.

Equal percentage rise and fall do not cancel. For a decimal rate r, (1 + r)(1 − r) = 1 − r². With 0 < r < 1, that factor is below 1. If percentages are explicitly both calculated from the original base, however, use that wording; it is a different model from successive changes on current amounts.

Reverse the complete factor

If final = original × M and the known combined factor M is nonzero, original = final/M. Dividing is the inverse operation. Applying an equal percentage change in the opposite direction is usually a new change, not an undo command. After recovering the original, run the complete forward sequence to check it.

A 100% fall gives M = 0: many different original amounts become zero. Zero alone cannot tell you which original was used. A final amount without a specified change is also insufficient. Do not invent a factor, and do not divide by zero.

Reverse the comparison base

If A is p% more than B, then A = (1 + p/100)B. The difference is measured against B in that statement. To ask how much B is below A, divide the same difference by A instead. For positive p, the percentage below is 100p/(100 + p).

After a p% loss, where 0 < p < 100, the recovery increase is 100p/(100 − p)%. The denominator is the reduced amount. These expressions are consequences of the base choice; a 100-unit example is often safer than memorising them.

Percentage points and relative change

A rate is itself a ratio. If a rate moves from u% to v%, its change is v − u percentage points. Its relative change is ((v − u)/u) × 100%, provided u is nonzero. Keep the sign or state increase/decrease. A rate moving from 0% to 5% rises by 5 percentage points; a finite relative percentage increase from 0% is not defined.

This distinction also prevents a count error: a higher number of successes does not alone establish a higher success rate if the number of attempts changed. Compare success/attempts for each period first.

Products and compensation

When total spending = unit price × quantity, multiply the price factor and quantity factor. The same reasoning works for any genuinely stated product, such as rectangular area = length × width. It is not permission to assume that every relationship is a product.

To keep a positive product fixed, its factor must be 1. If the first factor is a > 0, the second must be 1/a. Explain which amount is the base when turning that reciprocal back into a percentage. Use this alongside the fixed-distance inverse model from the preceding lesson.

A repeatable decision process

1. Name the original quantity, its unit and every comparison base.
2. Translate each stated change to a factor; preserve the sequence and any non-percentage operation.
3. Multiply forward, or divide by the full known nonzero factor to recover the original.
4. Compare final with original. Use percentage points only for differences between percentage rates.
5. Check direction, units and feasibility, then substitute your answer back into the original story.

Six worked cases

Worked example 1

A price of ₹1200 rises by 15% and then falls by 10% of the raised price. Find the final price and overall percentage change.

1. The first base is ₹1200. After the rise: 1200 × 1.15 = ₹1380.
2. The second base is ₹1380. Its 10% is ₹138, so final = 1380 − 138 = ₹1242.
3. Combined factor = 1.15 × 0.90 = 1.035. The gain is 0.035 of the original, or 3.5%.

Answer: ₹1242; 3.5% increase

Check and trap: 1242 − 1200 = 42 and 42/1200 = 0.035. Subtracting 10 from 15 would incorrectly use one unchanged base.

Worked example 2

After a 20% discount, a device costs ₹560. Find the undiscounted price.

1. The remaining price is 80% of the original x: 0.80x = 560.
2. Divide by 0.80: x = 560/0.80 = ₹700.
3. The original is larger than the discounted amount, as expected.

Answer: ₹700

Check and trap: 700 × 0.80 = 560. Adding 20% of 560 gives 672, and 672 × 0.80 = 537.60; that does not reproduce 560.

Worked example 3

A has 25% more samples than B. By what percentage is B below A?

1. Use a proportional comparison: set B = 100 units and A = 125 units. This fixes the ratio, not the actual sample counts.
2. The gap is 25 units. For “B below A”, the reference is A = 125.
3. Percentage below = 25/125 × 100 = 20%.

Answer: 20%

Check and trap: 125 × 0.80 = 100. Saying 25% below would give 93.75, so it changes the relationship.

Worked example 4

A completion rate changes from 40% to 46%. Report the change in percentage points and the relative percentage increase.

1. The displayed rate difference is 46 − 40 = 6 percentage points.
2. Relative to the old rate, the difference is 6/40 = 0.15, or a 15% increase.
3. These statements are compatible: 40% × 1.15 = 46%.

Answer: 6 percentage points; 15% relative increase

Check and trap: A 6% relative increase would produce 40% × 1.06 = 42.4%, not 46%. Write the label with the number.

Worked example 5

A unit price rises by 8%, while purchased quantity falls by 5%. With no other charges, what is the percentage change in total spending?

1. Let old unit price be P and old quantity be Q; old spending is PQ.
2. New spending = (1.08P)(0.95Q) = 1.026PQ.
3. The factor exceeds 1 by 0.026, so spending increases by 2.6%.

Answer: 2.6% increase

Check and trap: With P = ₹100 and Q = 20, spending changes from ₹2000 to ₹108 × 19 = ₹2052; 52/2000 = 2.6%. The shortcut 8 − 5 = 3% misses the product term.

Worked example 6

A measurement falls by 20%. What percentage increase from the reduced value restores its original value?

1. Normalize the positive original to 100; a 20% fall leaves 80.
2. The required increase is 20, but its base is now 80.
3. Recovery increase = 20/80 × 100 = 25%; equivalently, recovery factor = 1/0.80 = 1.25.

Answer: 25%

Check and trap: 80 × 1.25 = 100. A 20% gain would reach only 96. At a 100% loss the reduced base would be zero, so this reversal would fail.

Ten fresh practice questions

Try these before reading the solutions. Multiple-choice checks explicitly say select all. For numeric answers, give only the requested number in the stated unit.

Practice question 1

A value of 1600 rises by 20%, then falls by 25% of the new value. Which final value and overall change are correct?

A. 1520; 5% decrease

B. 1440; 10% decrease

C. 1600; no change

D. 1920; 20% increase

Practice question 2

A positive measurement is reduced by 15% twice, each time from its current value. What is the overall percentage decrease? Enter the positive percentage number.

Practice question 3

A positive amount is increased by 10% and then decreased by 30%, each on the current amount. Assume exact arithmetic with no fees or other changes. Select all true statements.

A. The final factor is 0.77

B. The overall decrease is 20%

C. Reversing the order of these two percentage factors gives the same final value

D. The 30% decrease is calculated from 1.10 times the original amount

Practice question 4

After an 8% increase, a price is ₹972. What was the original price? Enter the number of rupees.

Practice question 5

A price first rises by 10% and then falls by 10% of the raised price. The final price is ₹1188. What was the original price?

A. ₹1176.12

B. ₹1188

C. ₹1200

D. ₹1320

Practice question 6

B is 40% less than positive A. By what percentage is A greater than B?

A. 40%

B. 60%

C. 66⅔%

D. 150%

Practice question 7

A positive quantity loses 37.5% of its value. What percentage increase from the remaining value restores the original? Enter the positive percentage number.

Practice question 8

A rate falls from 75% to 69%. Which pair gives its decrease in percentage points and its relative percentage decrease?

A. 6 points; 6%

B. 6 points; 8%

C. 8 points; 6%

D. 8 points; 8%

Practice question 9

Unit price increases by 60%. With no other charges, what decrease in purchased quantity keeps total spending unchanged?

A. 60%

B. 40%

C. 37.5%

D. 62.5%

Practice question 10

Select all mathematically justified statements for the ordinary nonnegative quantity models in this lesson.

A. A 100% increase multiplies the original by zero

B. A final zero after a 100% decrease does not determine a unique positive original

C. The relative percentage change from 0 to 8 is undefined

D. A final value of 150 determines the original even if the percentage change is unknown

Practice solutions

Solution 1

Answer: B

The intermediate value is 1600 × 1.20 = 1920. Final = 1920 × 0.75 = 1440. The factor 0.90 gives a 10% fall from the original.

• A: A subtracts the displayed rates and ignores the new second base.
• B: B uses both factors and compares with 1600.
• C: C assumes the changes cancel.
• D: D stops after the first change.

Solution 2

Answer: 27.75

The retained factor is 0.85 × 0.85 = 0.7225. Therefore 1 − 0.7225 = 0.2775, or 27.75%, is lost. From a normalized 100, the stages are 85 and 72.25. The second loss is 12.75, not another 15 of the original.

Solution 3

Answer: A, C, D

Multiply 1.10 × 0.70 = 0.77, so the net decrease is 23%. Fixed factors commute under the stated exact model, but the amount before the second step is 1.10 times the original.

• A: A is the correct product.
• B: B wrongly subtracts 10 from 30.
• C: C follows from 1.10 × 0.70 = 0.70 × 1.10; the qualification matters.
• D: D identifies the second base correctly.

Solution 4

Answer: 900

Let the original be x. Then 1.08x = 972 and x = 972/1.08 = 900. Forward check: 8% of 900 is 72, giving 972. Subtracting 8% of 972 would use the wrong base.

Solution 5

Answer: C

The combined factor is 1.10 × 0.90 = 0.99. Original = 1188/0.99 = 1200. Check the whole chain: 1200 → 1320 → 1188.

• A: A applies the forward factor again instead of dividing.
• B: B treats equal percentages as cancellation.
• C: C reverses the full factor.
• D: D reverses only the last decrease and gives the intermediate value.

Solution 6

Answer: C

Set A = 100 and B = 60. For “A greater than B”, the base is 60. The gap 40 is 40/60 × 100 = 66⅔% of B.

• A: A keeps the old base A.
• B: B is B as a percentage of A, not the requested increase.
• C: C uses B as the denominator.
• D: D would make A 2.5 times B, contradicting A/B = 5/3.

Solution 7

Answer: 60

Of an original 100, 62.5 remains and 37.5 must be restored. The recovery rate is 37.5/62.5 × 100 = 60%. Check: 0.625 × 1.60 = 1.

Solution 8

Answer: B

The difference is 75 − 69 = 6 percentage points. Relative to the old rate, 6/75 × 100 = 8%. Check: 75% × 0.92 = 69%.

• A: A confuses the point difference with the relative percentage.
• B: B labels both quantities correctly.
• C: C swaps the two numbers.
• D: D has the correct relative change but an incorrect point difference.

Solution 9

Answer: C

The price factor is 1.60. Quantity must have factor 1/1.60 = 0.625. Thus 62.5% of the old quantity remains, a 37.5% decrease. Check: 1.60 × 0.625 = 1.

• A: A wrongly matches equal opposite percentages.
• B: B gives factor 0.60 and spending factor 0.96.
• C: C gives the required reciprocal factor.
• D: D confuses the retained percentage with the decrease.

Solution 10

Answer: B, C

A 100% increase has factor 2, but a 100% decrease has factor 0. Multiplication by zero loses the original information. Relative change divides by the old value, so an old value of zero is invalid. With an unknown factor, 150 could be unchanged from 150 or a 50% increase from 100, among other possibilities.

• A: A confuses doubling with erasing.
• B: B follows because every positive original times zero gives zero.
• C: C would require division by zero.
• D: D is underdetermined; two consistent originals have been exhibited.

If a step went wrong

Practice questions 1–3: revisit the multiplier and current-base sections and worked example 1. Questions 4–5: write the full forward equation before dividing. Questions 6–7: label the new comparison base. Question 8: separate percentage points from relative percent. Question 9: set the product factor to 1. Question 10: check whether the required division and information exist.

A rescaling window

Imagine a picture on a screen with a width control. A multiplier rescales the current width: 120% means 1.20 times the current width; a later 75% setting means 0.75 times that new width. To recover the starting width, divide by the combined scale. A slider that sends the width to zero erases information: its output cannot identify the old width.

The analogy tracks a changing base, not a physical law. Real image software may round pixels, preserve aspect ratio or impose minimum sizes; our exact arithmetic assumes those rules are absent. A fixed added border is an addition and is not another percentage multiplier. For rates such as 40% becoming 46%, use the separate percentage-point comparison rather than thinking the slider moved merely “6%”.

Quick reference

• Positive original x; increase p%: x(1 + p/100); decrease p%: x(1 − p/100)
• Successive factors multiply; overall signed change = (combined factor − 1) × 100%
• Recover original = final/combined factor, only when that known factor is nonzero
• A 100% decrease destroys invertibility; an old value of zero makes relative change undefined
• Compare a gap with the quantity named after “than” or “relative to”; reverse comparison changes the denominator
• Rate u% → v%: v − u percentage points; relative change = (v − u)/u × 100%, for u ≠ 0
• Fixed product: the factors must multiply to 1; use a reciprocal for compensation
• Fixed additions, rounding and caps need their own stated rules
• Name the base, keep units, check direction, and replay the forward calculation

Concept reference: OpenStax Prealgebra 2e, section 6.2, percent equations and change from an original base. The multiplier, inverse and product reasoning here is derived explicitly.