Use the correct reference whole for percent change and percent error

Learning goal: Compute percent increase, decrease and error using the appropriate reference amount. Explain why reversing a change or applying successive equal percentages does not generally undo the original change.

Percent means per hundred of a specified reference quantity. If an amount grows from 40 to 50, the change is ten, but the percent change compares that ten with the original 40. The ratio 10/40 = 1/4 = 25/100 shows a 25% increase. The original quantity represents 100%, and the new quantity represents 125% of that original. A change amount and a final amount answer different questions.

For an original amount a and new amount b, the signed relative change is (b − a)/a when a is nonzero. Multiplying by 100% expresses it as a percentage. A positive result indicates increase and a negative result indicates decrease for the positive reference amounts used here. You can instead report a positive magnitude with the word decrease. If the original amount is zero, ordinary percent change using this division is undefined; an absolute increase may still be described.

Reverse the first change: an amount falls from 50 to 40. The decrease is again ten, but the new reference is the original 50 for this second comparison. Ten divided by 50 is 20%, so this is a 20% decrease. The equal absolute differences do not imply equal percentages because the reference wholes differ. Label the original quantity before computing a ratio.

The diagram shows 40 and 50 on a common length scale. The ten-unit extension is one quarter of the 40-unit bar but one fifth of the 50-unit bar. The visible whole is the entire outlined amount, not just the changed piece. If a model highlights the extra ten, the unhighlighted 40 remains part of the new total. Count all parts when interpreting the new amount.

A percent multiplier can simplify calculation. Increasing a positive amount by r of itself gives a + ra = (1 + r)a. A 15% increase therefore multiplies by 1.15. Decreasing by 15% multiplies by 0.85. The decimal r represents the percent as a fraction of one, so 15% is 0.15, not 15. Use an estimate to catch a hundredfold error.

Successive percentages apply to changing reference amounts unless the problem says otherwise. Start with 100 model units, increase by 20% to get 120, then decrease the new amount by 20% to get 96. The second decrease is 24, not 20, because it is based on 120. Multipliers explain the result: 1.20 × 0.80 = 0.96. Equal percentage increase and decrease do not cancel because the second whole has changed.

Percent error compares the size of a measurement or estimate's difference with an accepted reference value. For positive nonzero reference R and estimate E, percent error is |E − R|/R × 100%. If a model measurement is 19 centimeters and the accepted comparison is 20 centimeters, the absolute error is one centimeter and the percent error is 1/20 × 100% = 5%. The absolute value makes this measure of error size nonnegative.

An estimate of 21 centimeters against the same 20-centimeter reference also has 5% error, although one estimate is low and the other high. Preserve that direction separately if it matters. Do not divide by the estimate just because it appears first in the sentence; the accepted reference is the denominator in this definition. In real measurement, an accepted value may itself have uncertainty, so state what reference the problem uses.

Percent change is different from a change in percentage points. If a survey share rises from 40% to 50%, the difference is ten percentage points. Relative to the original 40%, it is a 25% increase in the share. These two descriptions use different comparison units and are both meaningful when labeled correctly. Saying only “up ten percent” would leave the intended relationship ambiguous.

Worked increase: A modeled quantity rises from 80 to 92. The increase is 12, and 12/80 = 0.15, so the increase is 15%. Check with the multiplier: 80 × 1.15 = 92. Dividing 12 by 92 would answer a different comparison with the final amount.

Worked decrease and reversal: A value decreases from 80 to 60, a drop of 20/80 = 25%. Returning from 60 to 80 requires an increase of 20/60 = 1/3, or 33 1/3%. The absolute change is the same twenty, but the reference for the return is sixty.

Worked error: A student estimates a length as 31 centimeters; the supplied reference is 30 centimeters. Percent error is |31 − 30|/30 × 100% = 3 1/3%. The estimate is high by one centimeter. Reporting the direction alongside the error magnitude preserves information that the absolute value alone removes.

Reference-whole practice: Make a two-row record for a change and its reversal. In each row label original, new, absolute difference, comparison denominator and percent. Explain why the denominators change. Use an accurately partitioned bar to show a 25% increase from 40 to 50 and a 20% decrease back.

Independent performance: Create one increase, one decrease, one successive-change and one percent-error problem using positive reference amounts. Solve them with labeled ratios and check at least two with multipliers. Include a case starting at zero and explain why an ordinary percent-change answer cannot be calculated. Exchange work with a partner who checks the reference whole before checking arithmetic. Finish by distinguishing ten percentage points from a ten-percent relative increase in a survey example.