Math 7 · 7.RP.A.3 · F.IF.C.8B

Two percent changes one after the other

A drawing is enlarged by thirty percent and the copy is then reduced by thirty percent. Why is it not back to its old size?

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The multiplier for a percent changeMath 7 A fraction of a fraction of an amountGrade 5

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Discounts, sales tax, tips and commissionMath 7 Simple interestMath 7

Where you are 4 in Percent change

Fractions, ratio and percent3

Percent8

Percent change4

Percent increase and decrease Percent error The multiplier for a percent change

Two percent changes one after the other

When you reach for it

A quantity changes by a percent, and then the amount it has become changes by a percent again. A drawing goes through a photocopier twice. A collectible loses value one year and loses more the next. An insect population grows season after season. Each time the question is the same: what is the single change from the very start to the very end?

The tempting answer is to add the percents, and it is wrong. Percent increase and decrease already showed the first sign of this: a rise and a fall of the same size do not land back where they began. The reason is that every percent is measured against the amount just before it, and after the first change that amount is no longer the original.

This lesson stands on the multiplier for a percent change: an increase of 30\% is one multiplication by 1.3, and a decrease of 30\% is one multiplication by 0.7. So a chain of percent changes is a chain of multiplications, which collapses into one.

The same move answers a question that looks different: what is 40\% of the 25\% of a group? That is a part of a part, the percent version of a fraction of a fraction of an amount.

The lesson runs over two days. Part 1 combines two changes, and takes a percent of a percent. Part 2 repeats one change many times, undoes a change, and finds a change that is missing.

The procedure

Part 1: two changes, one multiplier

A drawing is 20 cm wide. A photocopier set to 130\% makes a copy 30\% wider. That copy goes back into the copier set to 70\%, which makes a copy 30\% narrower.

20 \times 1.3 = 26 \qquad 26 \times 0.7 = 18.2

The last copy is 18.2 cm wide, not 20 cm. It is 1.8 cm narrower than the drawing, and \displaystyle \frac{1.8}{20} = 0.09, so it is 9\% narrower.

Figure 1The enlargement adds 30\% of 20, which is 6. The reduction removes 30\% of 26, which is 7.8. The same percent of a bigger amount is a bigger piece.

The picture shows where the 1.8 cm went. The enlargement added 30\% of 20 cm, which is 6 cm. The reduction took away 30\% of 26 cm, which is 7.8 cm. The two percents are equal, but they are percents of different amounts, and the second amount is bigger.

Multiply the multipliers. Write the two copies as one line:

(20 \times 1.3) \times 0.7$$ $$= 20 \times (1.3 \times 0.7) = 20 \times 0.91

Moving the brackets is allowed by the associative property of multiplication, one of the properties of multiplication: when three numbers are multiplied, it does not matter which two are multiplied first. So the two changes together are one change with multiplier 1.3 \times 0.7 = 0.91. Nothing about these numbers mattered. For any start and any two changes with multipliers m_1 and m_2,

\text{end} = \text{start} \times m_1 \times m_2

and m_1 \times m_2 is the overall multiplier. The same property regroups any number of factors, so three or more changes in a row are also one change, whose multiplier is the product of all of theirs.

Reading the overall change. An overall multiplier above 1 is an increase, and the part above 1 is the percent: 1.32 is an increase of 32\%. A multiplier below 1 is a decrease, and the gap below 1 is the percent: 0.91 is 1 - 0.91 = 0.09 below 1, a decrease of 9\%.

The order does not matter. m_1 \times m_2 = m_2 \times m_1 by the commutative property of multiplication. Reduce the drawing first and then enlarge it, and 20 \times 0.7 = 14, then 14 \times 1.3 = 18.2: the same 18.2 cm.

Why the percents do not add. Split the overall multiplier into pieces. Write 1.3 as 1 + 0.3, and the distributive property gives

1.3 \times 0.7 = 1 \times 0.7 + 0.3 \times 0.7

The first piece, 0.7, is the reduction acting on the drawing itself; the second is the reduction acting on the 0.3 the enlargement added. Split that 0.7 as 1 - 0.3 and distribute again:

0.3 \times 0.7 = 0.3 - 0.3 \times 0.3 = 0.3 - 0.09

Altogether 1.3 \times 0.7 = 0.7 + 0.3 - 0.09 = 1 - 0.09. Adding the percents, +30\% and -30\%, gives the 0.7 + 0.3 = 1 and nothing else. The extra -0.09 is the reduction taking 30\% of the added 30\%: in centimeters, 30\% of the 6 cm the enlargement added, which is the missing 1.8 cm.

The same two splits work for any percent p, written as a decimal, going up and then down:

(1 + p)(1 - p) = 1 - p + p - p^2 = 1 - p^2

which is always below 1 when p is not 0. Whichever comes first, a rise and a fall by the same percent end in a loss of p^2: a loss of 9\% for p = 0.3, and of 25\% for p = 0.5.

Part 1: a percent of a percent

Over one summer, a nature reserve catches 480 birds, fits each with a small leg ring so it can be recognized later, and lets it go. Of these birds, 25\% are swallows, and 40\% of the swallows are young birds hatched that year. What percent of all the birds are young swallows?

Count first. 25\% of 480 is 0.25 \times 480 = 120 swallows, and 40\% of those is 0.4 \times 120 = 48 young swallows. As a share of all the birds, \displaystyle \frac{48}{480} = 0.1, which is 10\%.

The count was not needed. Taking a percent of an amount is multiplying it by the percent written as a decimal, so the two steps were 480 \times 0.25 \times 0.4, and by the associative property that is 480 \times (0.25 \times 0.4). The part of a part is

0.4 \times 0.25 = 0.1 = 10\%

of the whole, for any number of birds. It is the same picture as a fraction of a fraction: cut off a quarter, then cut 40\% of that quarter.

Figure 2The column, the swallows, is 25\% of the square. The shaded part, the young swallows, is 40\% of the column, which is 10\% of the square, because 0.4 \times 0.25 = 0.1.

Only the multipliers differ from the two copies. A change keeps the whole and adds or removes a piece, so its multiplier is 1 plus or minus something. A part keeps only the piece, so its multiplier is the percent itself, 0.25 or 0.4.

Examples 1 to 3 go with Part 1.

Part 2: the same change many times

An insect population grows by 10\% each season. One season multiplies it by 1.1. Two seasons multiply it by 1.1 \times 1.1 = 1.21, and three by 1.1 \times 1.1 \times 1.1 = 1.331. A repeated multiplication is written with an exponent:

\text{after } n \text{ seasons:} \quad \text{start} \times 1.1^n

Two seasons of 10\% growth make 21\%, not 20\%. Three make 33.1\%, not 30\%. Each season’s 10\% is of a population already bigger than last time, so the gap from adding widens every season. In general, a change of r (as a decimal, negative for a decrease) repeated n times has overall multiplier

(1 + r)^n

A repeated decrease runs the other way. Three falls of 10\% give 0.9^3 = 0.729, a decrease of 27.1\% rather than 30\%, because each 10\% is taken from a smaller amount than the last. However many times it is repeated, 0.9^n is never 0: each step keeps 90\% of what there was, and 90\% of a positive amount is still positive.

Undoing a change. To get back to the start, the second multiplier has to make the overall multiplier 1:

m_1 \times m_2 = 1 \qquad\text{so}\qquad m_2 = \frac{1}{m_1}

A copy made at 50\% has multiplier 0.5, so the way back is \displaystyle \frac{1}{0.5} = 2, a setting of 200\%: a decrease of 50\% is undone by an increase of 100\%. A rise of 30\% is undone by \displaystyle \frac{1}{1.3} \approx 0.769, a fall of only about 23.1\%. Undoing a fall always takes a bigger percent rise, and undoing a rise takes a smaller percent fall, because the way back is measured against the changed amount.

Finding a missing change. If the overall multiplier M and the first multiplier m_1 are known, then m_1 \times m_2 = M, and dividing both sides by m_1 gives

m_2 = \frac{M}{m_1}

Examples 4 to 7 go with Part 2.

Worked examples

Example 1

A trading card over two years

Wiremu buys a signed trading card for 350 dollars. It loses 20\% of its value in the first year, and 10\% of its new value in the second year. What is it worth after two years, and what is the overall percent change?

Solution

A loss of 20\% leaves 80\%, multiplier 0.8. A loss of 10\% leaves 90\%, multiplier 0.9. The overall multiplier is their product:

0.8 \times 0.9 = 0.72

So the card keeps 72\% of its value, and 350 \times 0.72 = 252. It is worth 252 dollars, a decrease of 1 - 0.72 = 0.28, or 28\%, not the 30\% that adding gives.

Check year by year: 350 \times 0.8 = 280, then 280 \times 0.9 = 252. The second year’s 10\% was 28 dollars, 10\% of 280, not 35 dollars, 10\% of the price paid.

Example 2

The rest of the swallows

At the nature reserve from Part 1, 25\% of the 480 birds were swallows and 40\% of the swallows were young. What percent of all the birds were adult swallows, and how many birds is that?

Solution

If 40\% of the swallows were young, the other 60\% of the swallows were adults. That is a percent of a percent again:

0.6 \times 0.25 = 0.15

So 15\% of all the birds were adult swallows, and 0.15 \times 480 = 72 birds. Check against the counts: there were 120 swallows and 48 were young, which leaves 120 - 48 = 72. The two shares, 10\% and 15\%, add to 25\%, all the swallows, as they must.

Example 3

The area of an enlarged copy

Giacomo copies a drawing 20 cm wide and 15 cm tall on a photocopier set to 120\%. By what percent does the area of the drawing grow?

Solution

A copier multiplies every length by the same multiplier, so the width and the height both grow by 20\%. The copy is 20 \times 1.2 = 24 cm wide and 15 \times 1.2 = 18 cm tall. Its area is 24 \times 18 = 432 square centimeters, against 20 \times 15 = 300 for the drawing, and

\frac{432}{300} = 1.44

The area grows by 44\%, not 20\%. Area is width times height, and each is multiplied by 1.2, so the area is multiplied by 1.2 \times 1.2 = 1.44: two percent changes in a row, one for each direction. This is the idea behind scale factors and area.

Example 4

A population over three seasons

A colony of 5{,}000 ladybirds in an orchard grows by 20\% each season. How many are there after 3 seasons, and what is the overall percent change?

Solution

Each season multiplies the colony by 1.2, so three seasons multiply it by

1.2^3 = 1.2 \times 1.2 \times 1.2 = 1.728

After 3 seasons there are 5{,}000 \times 1.728 = 8{,}640 ladybirds, an increase of 72.8\%.

Season	Ladybirds
start	5{,}000
1	6{,}000
2	7{,}200
3	8{,}640

The table shows why it is more than 3 \times 20\% = 60\%. The first season adds 1{,}000, the second adds 1{,}200, the third adds 1{,}440: each 20\% is taken of a bigger colony.

Example 5

Getting a copy back to its original size

Imogen copies an 18 cm drawing at 60\%. At what setting must she copy the copy to get back to 18 cm?

Solution

The first copy has multiplier 0.6, so it is 18 \times 0.6 = 10.8 cm wide. The way back needs a multiplier that makes the overall one 1:

m_2 = \frac{1}{0.6} = \frac{10}{6} = \frac{5}{3}

As a percent, \displaystyle \frac{5}{3} = 166\tfrac{2}{3}\%, an increase of 66\tfrac{2}{3}\%. Check: \displaystyle 10.8 \times \frac{5}{3} = \frac{54}{3} = 18.

The way back is measured against 10.8 cm, so it takes far more than 40\%.

Example 6

Finding the second year’s change

Lorcan’s vinyl record rose in value by 60\% in its first year. Over the two years together its value rose by 20\%. What happened in the second year?

Solution

The overall multiplier is 1.2 and the first is 1.6. The second multiplier m_2 must satisfy 1.6 \times m_2 = 1.2, so

m_2 = \frac{1.2}{1.6} = \frac{12}{16} = 0.75

A multiplier of 0.75 is a decrease of 25\%. Check with a price: a record bought for 50 dollars rises to 50 \times 1.6 = 80 dollars, then falls to 80 \times 0.75 = 60 dollars, and 60 is 50 \times 1.2.

Subtracting the percents would have said 60 - 20 = 40, a fall of 40\%. That is wrong because the second year’s fall is a percent of 80 dollars, not of 50.

Example 7

How many years to double

A population of cranes on a wetland reserve grows by 15\% each year. After how many whole years is it more than double what it was at the start?

Solution

No starting number is needed: the population has more than doubled exactly when 1.15^n is more than 2. Build the powers a year at a time, multiplying by 1.15:

Years n	1.15^n
1	1.15
2	1.3225
3	1.520875
4	about 1.749
5	about 2.011

After 4 years the multiplier is still below 2; after 5 it is above. So the population first passes double after 5 years.

Adding the percents would say that doubling, a 100\% increase, takes \displaystyle \frac{100}{15} \approx 6.7 years, so 7. Growth by a fixed percent gets there sooner, because each year’s growth also grows. This steady multiplying is the start of exponential growth.

Boundaries

Each percent must be of the amount just before it. That is what makes the multipliers chain. If every change is a percent of the original instead, the changes are fixed amounts and they do add. A card that loses 20\% of its purchase price every year goes 350 \to 280 \to 210, a loss of 40\% after two years, whereas losing 20\% of its current value each year gives 0.8^2 = 0.64, a loss of 36\%. A fixed amount each year is how simple interest works.
A percent of a percent needs the second percent to be of the first group. “40\% of the swallows were young” multiplies with the 25\%. “40\% of all the birds were young” does not: it says nothing about how many of the young birds are swallows.
A decrease of 100\% cannot be undone. Its multiplier is 0, everything becomes 0, and no multiplier turns 0 back into the start. Any smaller decrease has multiplier above 0 and is undone by \displaystyle \frac{1}{m}.
\text{start} \times (1 + r)^n assumes exactly the same percent every step. It says nothing about steps that break the pattern, and a count of animals worked out this way has to be rounded to a whole number.
Percentage points add. When the quantity is itself a percent and it changes by percentage points, from 8\% to 10\% and then to 13\%, the changes are amounts of the same kind and they add: 5 points in all. Multipliers are for changes measured as a percent of the amount before.

Where this goes next

Repeating the same multiplier, \text{start} \times (1 + r)^n, is the first form of exponential growth and decay, where n becomes the input of a function and the graph curves upward or flattens toward 0. Money that earns interest on the interest it has already earned grows the same way, which is compound interest; it is set against interest on the original amount only in simple interest.

Closer to hand, a price with a discount taken off and a tax then added on is two multipliers in a row, worked in discounts, sales tax, tips and commission.

Exercises

Part 1: one multiplier for two changes

For each pair of changes, one after the other, find the single multiplier and the overall percent change.

An increase of 10\%, then an increase of 20\%.
A decrease of 20\%, then a decrease of 25\%.
An increase of 60\%, then a decrease of 50\%.
A decrease of 10\%, then an increase of 10\%.
An increase of 5\%, then another increase of 5\%.

Part 1: in context

A drawing 40 cm wide is copied at 150\%, and the copy is copied again at 80\%. How wide is the final copy, and what is the overall percent change?
Hollis owns a vintage camera worth 400 dollars. Its value rises by 15\% in one year and falls by 20\% the next. What is it worth now, and what is the overall percent change?
A bee colony has 20{,}000 bees. It grows by 40\% over the spring and then shrinks by 40\% over the autumn. How many bees are there at the end, and what is the overall percent change?
At a craft fair with 80 stalls, 30\% of the stalls sell pottery, and 50\% of the pottery stalls also sell glass. What percent of all the stalls sell both, and how many stalls is that?
In a choir, 60\% of the singers take part in the winter concert, and 15\% of those sing a solo. What percent of the whole choir sings a solo in the concert?
A drawing 12 cm wide is copied at 120\%, that copy at 125\%, and that copy at 50\%. How wide is the last copy, and what is the overall percent change?

Part 2: the same change repeated

A population grows by 30\% each season for 2 seasons. Find the single multiplier and the overall percent change.
A value falls by 20\% each year for 3 years. Find the single multiplier and the overall percent change.
There are 1{,}000 aphids on a rose bush, and their number grows by 50\% each week. How many are there after 3 weeks, and what is the overall percent change?
Keziah’s model train was worth 250 dollars, and its value falls by 10\% each year. What is it worth after 2 years, and what is the overall percent change?

Part 2: undoing a change, and finding a missing one

A drawing is copied at 50\%. At what setting must the copy be copied to get back to the original size, and what percent increase is that?
A population falls by 60\%. What percent increase would bring it back to where it started?
A collectible’s value rises by 60\%. What percent decrease would bring it back to where it started?
A signed soccer lost 30\% of its value in its first year. After two years it was worth 56\% of what it cost. By what percent did its value fall in the second year?
A bird population rose by 20\% during the breeding season, and over the whole year it rose by only 8\%. What percent change happened during the rest of the year?

Reasoning

Imogen says that a 50\% increase followed by a 50\% decrease leaves an amount where it started. Is she right? Explain, and give the true overall change.
Explain why it does not matter which of two percent changes happens first.
Show that an increase by p followed by a decrease by p, with p a decimal between 0 and 1, always ends below the start, and say by how much.
Lorcan says that two increases of 10\% in a row are the same as one increase of 21\%. Keziah says they are the same as one increase of 20\%. Who is right, and why?
A population falls by 10\% every season. Hollis says that after 10 seasons nothing will be left. Explain why this is wrong, and say roughly what percent of the population remains.

Answers

1.32, an increase of 32\%. 1.1 \times 1.2 = 1.32.
0.6, a decrease of 40\%. 0.8 \times 0.75 = 0.6, and 1 - 0.6 = 0.4.
0.8, a decrease of 20\%. 1.6 \times 0.5 = 0.8. A rise followed by a fall of a smaller percent can still end lower.
0.99, a decrease of 1\%. 0.9 \times 1.1 = 0.99, which is 1 - 0.1^2.
1.1025, an increase of 10.25\%. 1.05 \times 1.05 = 1.1025: the second 5\% is also taken of the first 5\%, which adds the extra 0.25\%.
48 cm, an increase of 20\%. 1.5 \times 0.8 = 1.2, and 40 \times 1.2 = 48.
368 dollars, a decrease of 8\%. 1.15 \times 0.8 = 0.92, and 400 \times 0.92 = 368.
16{,}800 bees, a decrease of 16\%. 20{,}000 \times 1.4 = 28{,}000, then 28{,}000 \times 0.6 = 16{,}800. The single multiplier is 1.4 \times 0.6 = 0.84, which is 1 - 0.4^2.
15\% of the stalls, which is 12 stalls. 0.5 \times 0.3 = 0.15, and 0.15 \times 80 = 12.
9\% of the choir. 0.15 \times 0.6 = 0.09.
9 cm, a decrease of 25\%. 1.2 \times 1.25 \times 0.5 = 0.75, and 12 \times 0.75 = 9.
1.69, an increase of 69\%. 1.3^2 = 1.3 \times 1.3 = 1.69.
0.512, a decrease of 48.8\%. 0.8^3 = 0.512, and 1 - 0.512 = 0.488.
3{,}375 aphids, an increase of 237.5\%. 1.5^3 = 3.375, and 1{,}000 \times 3.375 = 3{,}375.
202.50 dollars, a decrease of 19\%. 0.9^2 = 0.81, and 250 \times 0.81 = 202.5.
A setting of 200\%, an increase of 100\%. The first multiplier is 0.5, and \displaystyle \frac{1}{0.5} = 2.
An increase of 150\%. The multiplier was 0.4, and \displaystyle \frac{1}{0.4} = 2.5, which is 1 plus 1.5.
A decrease of 37.5\%. \displaystyle \frac{1}{1.6} = 0.625, and 1 - 0.625 = 0.375.
A fall of 20\%. 0.7 \times m_2 = 0.56, so \displaystyle m_2 = \frac{0.56}{0.7} = 0.8.
A decrease of 10\%. 1.2 \times m_2 = 1.08, so \displaystyle m_2 = \frac{1.08}{1.2} = 0.9.
No. The overall multiplier is 1.5 \times 0.5 = 0.75, a decrease of 25\%. The 50\% decrease is taken of the raised amount, which is bigger than the original, so it removes more than the increase added. In the form 1 - p^2, the loss is 0.5^2 = 0.25.
Each change multiplies the amount by its multiplier, so two changes multiply it by m_1 \times m_2 in one order and m_2 \times m_1 in the other. These are equal by the commutative property of multiplication, so the end amount is the same. The amount in between is different, but the end is not.
The overall multiplier is (1 + p)(1 - p). By the distributive property this is 1 - p + p - p^2 = 1 - p^2. Since p is not 0, p^2 is positive, so 1 - p^2 < 1 and the end is below the start. It is below by p^2 of the start: for a rise and fall of 30\%, by 0.09, which is 9\%.
Lorcan. The first increase multiplies by 1.1 and so does the second, so together they multiply by 1.1 \times 1.1 = 1.21, an increase of 21\%. The second 10\% is 10\% of an amount that is already 110\% of the start, so it adds 11\% of the start, not 10\%.
Each season keeps 90\% of the population, and 90\% of a positive amount is still positive, so the overall multiplier 0.9^{10} is never 0. In fact 0.9^{10} \approx 0.349, so about 34.9\% remains after 10 seasons. Adding ten falls of 10\% to get 100\% treats every fall as 10\% of the original, but each one is 10\% of what was left.