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Lesson 10 of 12

Fractions in Disguise · Lesson 10 of 12

Decline and Repeated Percentage Change

“Model depreciation, repeated decline and successive changes with updated percentage bases.”

Learning Objectives

• Interpret depreciation as a reduction in value. • Calculate repeated decreases using the remaining-percentage multiplier. • Combine successive increases and decreases with changing bases. • Handle rounded model quantities carefully and explain why equal opposite percentage changes need not cancel.

A television bought for ₹21,000 loses 5% of its value over the first year in a simple depreciation model. The reduction is ₹1050, leaving ₹19,950. If the same 5% rate applies the following year, should you remove another ₹1050? Only if the reduction were based again on the original purchase price. When the problem says the current value falls by 5% each year, the second reduction must use the smaller ₹19,950 as its base.

Repeated decline uses the same updated-base reasoning as compounding. The multiplier is below one instead of above one. Each step keeps a fixed fraction of the current amount, so the absolute reductions gradually become smaller.

We can also mix increases and decreases. The safest method is to process them in sequence, writing the new amount after every step or multiplying the corresponding factors.

Definition
Depreciation

A reduction in the value of an item over time, often associated with age or use. A fixed percentage depreciation problem is a mathematical model of that change.

A 5% reduction leaves 100% − 5% = 95% of the current value. The remaining-value multiplier is 0.95. For the television, 0.95 × ₹21,000 = ₹19,950. This calculation already includes the subtraction; do not subtract another 5% afterward.

After the second year, multiply the updated value by 0.95 again. The new value is 0.95 × ₹19,950 = ₹18,952.50. The second reduction is ₹997.50, smaller than the first ₹1050, because five per cent acts on a smaller base.

A fixed percentage does not mean a fixed amount. This distinction explains why repeated decreases cannot generally be combined by adding their percentage numbers.

A single year of depreciation

Problem
A TV valued at ₹21,000 depreciates by 5% in one year. Find its new value.

  1. 1.Reduction = 5% of ₹21,000 = 0.05 × 21,000 = ₹1050.
  2. 2.New value = ₹21,000 − ₹1050 = ₹19,950.
  3. 3.Alternatively, retain 95%: 0.95 × 21,000 = ₹19,950.
  4. 4.Check: the reduction and the remaining value add back to ₹21,000.
Two years at the same decline rate

Problem
Continue the TV model for a second year of 5% depreciation on current value. Find the value and total percentage decrease from purchase.

  1. 1.The second year begins at ₹19,950.
  2. 2.After the second decrease, value = 19,950 × 0.95 = ₹18,952.50.
  3. 3.Total decrease = ₹21,000 − ₹18,952.50 = ₹2047.50.
  4. 4.Overall decrease percentage = (2047.50/21,000) × 100 = 9.75%.
  5. 5.Two successive 5% decreases do not give 10% overall, because the second uses a smaller base.
Repeated declineLaTeX
P is the initial quantity, r the decimal decrease rate per period and t the number of equal periods. V is the final quantity in the same unit as P. This model uses 0 ≤ r ≤ 1 and no intervening additions.

The exponent counts repetitions, just as in compound growth. A 10% decrease each decade leaves a multiplier of 0.9 each decade. Three decades produce 0.9 × 0.9 × 0.9 = 0.729. Thus 72.9% remains and the overall decrease is 27.1%.

The rate and period must match. Three decades means three applications of a per-decade rate, not thirty applications. Read the time label before choosing the exponent.

A declining model population

Problem
A village has a current population of 1250, and a model predicts a decrease of 10% each decade. Estimate the population after three decades.

  1. 1.Each decade keeps 90% of the previous population, giving the multiplier 0.9.
  2. 2.After one decade: 1250 × 0.9 = 1125.
  3. 3.After two decades: 1125 × 0.9 = 1012.5.
  4. 4.After three decades: 1012.5 × 0.9 = 911.25.
  5. 5.The model estimate is approximately 911 people to the nearest whole person, or about 910 to the nearest ten.
  6. 6.Keep decimal intermediate values in the model and round at the end. The model is an estimate; it does not claim that a fraction of a person exists.
Initially1250After one decade1125After two decades1012.5After three decades911.25Each new bar is 90% of the preceding bar.
A fixed percentage decline gives shrinking amounts

Rounding at every stage changes the later bases. If you round 1012.5 to 1013 before the final step, you obtain 911.7 and might report 912. Keeping the unrounded model value produces 911.25 and rounds to 911. Both displays may look close, but they follow different rounding procedures.

Unless a question explicitly requires whole-number rounding after each stage, preserve exact or sufficiently precise intermediate values. State the final precision so that “911”, “911.25” and “about 910” are not presented as conflicting exact answers.

Recovering an original value after repeated decline

Problem
A model machine value becomes ₹7290 after three successive decreases of 10%. Find its initial value.

  1. 1.The combined multiplier is 0.9³ = 0.729.
  2. 2.If P is the initial value, 0.729P = 7290.
  3. 3.Divide by the combined multiplier: P = 7290 ÷ 0.729 = ₹10,000.
  4. 4.Check the sequence: ₹10,000 → ₹9000 → ₹8100 → ₹7290. The result matches the stated final value.

For changing rates, write a separate factor for each period. A 5% increase gives 1.05. A 2% decrease gives 0.98. A later 3% decrease gives 0.97. Multiplying the starting amount by all three factors follows the evolving base automatically.

Adding +5 − 2 − 3 gives zero, but that addition hides the different amounts used as bases. The second percentage applies after the first change, and the third applies after the second.

A rise followed by two falls

Problem
A model population p changes by +5%, then −2%, then −3% in successive months. Is it back at p?

  1. 1.After the first month: 1.05p.
  2. 2.After the second: 1.05 × 0.98p = 1.029p.
  3. 3.After the third: 1.029 × 0.97p = 0.99813p.
  4. 4.The final population is 99.813% of the initial population.
  5. 5.The overall decrease is 100% − 99.813% = 0.187%. It is slightly below p, not equal to p.
Two fare increases

Problem
Bus fares increase by 3% one year and 4% the next. Find the overall percentage increase.

  1. 1.Let the initial fare be F. The two factors are 1.03 and 1.04.
  2. 2.Final fare = F × 1.03 × 1.04 = 1.0712F.
  3. 3.The final fare is 107.12% of the original, so the overall increase is 7.12%.
  4. 4.For an initial fare of ₹100, the successive fares are ₹103 and ₹107.12. The second 4% increase is ₹4.12, not ₹4.

Equal percentage increases and decreases usually do not cancel. Starting with 100, a 10% increase produces 110. A 10% decrease of 110 removes 11, leaving 99. The larger second base makes the reduction larger than the original increase.

Reversing a percentage change requires the reciprocal multiplier. After a 25% increase, the multiplier is 1.25. Returning to the original needs multiplication by 1/1.25 = 0.8, which is a 20% decrease. This extends the reverse-percentage reasoning from a single change.

Keeping a rectangle’s area unchanged

Problem
A rectangle’s length increases by 10%. Its area must stay unchanged. By what exact percentage must its breadth decrease?

  1. 1.Let original length be L and breadth be B, so the original area is LB.
  2. 2.The new length is 1.1L. To keep area LB, the new breadth must be B/1.1 = (10/11)B.
  3. 3.The removed fraction of breadth is 1 − 10/11 = 1/11.
  4. 4.Therefore the decrease percentage is (1/11) × 100 = 100/11% = 9 1/11%.
  5. 5.Check: 1.1 × 10/11 = 1. A 10% breadth reduction would instead give 1.1 × 0.9 = 0.99 of the original area.
Activity: Test whether changes cancel

Begin with 200. Increase it by 20%, then decrease the result by 20%. You obtain 240 and then 192. Reverse the order to obtain 160 and then 192. In this pure percentage model the products match, but neither sequence restores 200. Explain why the percentages must be applied to each updated amount.

Quiz

Quick check

What multiplier describes a 12% decrease?

Quick check

Two successive decreases of 10% leave what percentage of the original?

Quick check

Why should intermediate model values usually be kept unrounded?

Quick check

What is the combined multiplier for a 5% rise followed by a 2% fall?

Quick check

A quantity rises by 10% and then falls by 10%. What happens overall?

Practice Problems

Practice Problems
  1. An item worth ₹8000 depreciates by 15% in one period. Find its new value. Solution: It retains 85%, so new value = 0.85 × 8000 = ₹6800. The reduction is ₹1200.
  2. A quantity of 2000 decreases by 5% per period for two periods. Find its final value and overall decrease percentage. Solution: Final = 2000 × 0.95² = 1805. Decrease = 195, so overall decrease percentage = 195/2000 × 100 = 9.75%.
  3. A value is ₹5120 after three successive 20% decreases. Find the original. Solution: The combined factor is 0.8³ = 0.512. Original = 5120 ÷ 0.512 = ₹10,000. Check the sequence: 10,000, 8000, 6400, 5120.
  4. A quantity increases by 20%, then decreases by 10%. Find the overall percentage change. Solution: The combined multiplier is 1.2 × 0.9 = 1.08. Therefore the final value is 108% of the original, an 8% increase.
  5. A rectangle’s length increases by 25% while its area stays constant. Find the required percentage decrease in breadth. Solution: The length factor is 1.25 = 5/4. Breadth must be multiplied by the reciprocal 4/5 = 0.8. Thus breadth decreases by 20%. Check: 1.25 × 0.8 = 1, preserving area.

Key Takeaways

Key Takeaways

• A repeated percentage decline applies to the current amount each time. • A decrease of p% retains the factor 1 − p/100. • Repeated constant decline gives P(1 − r)ᵗ with decimal rate r. • Multiply factors to combine successive changes; do not simply add the rates. • Keep precise intermediate values and state final rounding clearly. • An inverse multiplier reverses a change; the same opposite percentage usually does not.