Percent Increase and Decrease: Free Response

The amount of change is the new figure minus the earlier one:

$88 - 64 = 24.$

The number of places rose, so this is an increase. Percent change measures that rise against the figure the programme started the year on, so the $24$ goes over the earlier $64$, and the ratio is rescaled to a denominator of one hundred:

$\frac{24}{64} \times 100\% = 0.375 \times 100\% = 37.5\%.$

The swim school rose by $37.5\%$, a gain of $24$ weekly places. The $64$ is the right baseline because the change grew out of it: the rise is being reported as a part of what was already there. Dividing by this spring's $88$ would answer a question nobody asked, namely what share of this year's places the gain accounts for.

Subtract to find the change:

$189 - 216 = -27.$

The programme lost $27$ weekly places, so this is a decrease, and its size is compared with last spring's figure of $216$:

$\frac{27}{216} \times 100\% = 0.125 \times 100\% = 12.5\%.$

The lane sessions fell by $12.5\%$. Now set the two programmes side by side, and they disagree about which move was bigger. The lane sessions lost $27$ places while the swim school gained $24$, so counting a rise and a fall alike the lane sessions moved by more places. But the swim school moved by $37.5\%$ against the lane sessions' $12.5\%$, so the swim school moved by far the larger percent.

Both statements are true at once, and that is the whole reason the centre reports percents. The lane sessions began the year on $216$ places against the swim school's $64$, more than three times as many, so each place lost from the lane sessions is a much smaller share of its programme than each place gained by the swim school.

The duty manager's arithmetic is sound; only the denominator is wrong. Dividing the loss by the figure the programme ended the year on gives

$\frac{27}{189} \times 100\% \approx 14.3\%,$

which says the places lost are about a seventh of the places still running. That is a genuine fact about this spring's timetable, but it is not a measure of how big the move was.

Percent change compares a change with the amount that was there before it, because that is the amount the change happened to. The programme ran $216$ weekly places and lost $27$ of them, so the fair measure is

$\frac{27}{216} \times 100\% = 12.5\%.$

The two figures differ because they sit over different denominators, and with the same change on top, the smaller denominator always produces the larger percent. Publishing $14.3\%$ would make the lane sessions look like a steeper decline than they suffered, which matters most in a report whose purpose is to set one programme against another.

Take the two-step route first. A rise of $22\%$ means finding $22\%$ of the current price and adding it on, and $22\% = 0.22$:

$0.22 \times 85 = 18.70, \qquad 85 + 18.70 = 103.70.$

Now the one-step route. The poster keeps the whole of its old price and gains $22\%$ more, so it ends on $122\%$ of what it was, and the number to store is

$1 + \frac{22}{100} = 1.22, \qquad 1.22 \times 85 = 103.70.$

Both routes land on $103.70$ dollars, because the multiplier is the two-step calculation packed into a single number.

A stored number below $1$ shrinks whatever it multiplies, so this is a decrease. Read off how much of the price survives. Since $0.78 = \frac{78}{100}$, the frame keeps $78\%$ of what it cost, and what was taken off is the rest:

$100\% - 78\% = 22\%.$

So the record describes a $22\%$ decrease. Applying it,

$0.78 \times 245 = 191.10.$

The frame's new price is $191.10$ dollars. Checking by the two-step route, $22\%$ of $245$ is $0.22 \times 245 = 53.90$ dollars off, and $245 - 53.90 = 191.10$, the same price.

Write the two-step calculation on one line, before simplifying anything:

$85 + 0.22 \times 85.$

The old price appears twice. Once it appears whole, because the shop keeps every dollar it was already charging, and once it appears multiplied by $0.22$, because that is the part being added. Anything that multiplies the same amount in two places can be collected:

$85 + 0.22 \times 85 = (1 + 0.22) \times 85 = 1.22 \times 85.$

That is what the stored number is made of. The $1$ is the price you keep, which is $100\%$ of itself, and the $0.22$ is the extra $22\%$ laid on top, so a single multiplication carries the old price and the increase together. It is not a new rule, only the two-step calculation written without the intermediate stop.

Multiplying by $0.22$ alone throws away the first of those two pieces. It returns $18.70$ dollars, which is the increase by itself rather than the price after the increase, and it is why a shopper who reaches for the bare percent gets an answer far too small to be a price.

A discount is a percent decrease, so what matters is the share of the list price that survives it. Taking $28\%$ off leaves

$100\% - 28\% = 72\%,$

so the multiplier is $0.72$ and the trade price is

$0.72 \times 275 = 198.$

The router costs $198$ dollars on the trade account. As a check by the two-step route, $28\%$ of $275$ is $0.28 \times 275 = 77$ dollars off, and $275 - 77 = 198$, the same price.

Tax is an increase, and it is charged on the money actually changing hands, which is the trade price rather than the list price. Adding $9\%$ leaves the bill at $109\%$ of that amount, a multiplier of $1.09$:

$1.09 \times 198 = 215.82.$

The total is $215.82$ dollars. Step by step this is $0.09 \times 198 = 17.82$ dollars of tax, and $198 + 17.82 = 215.82$, which agrees.

Charging the tax on the list price instead would have added $0.09 \times 275 = 24.75$ dollars, so the cabinetmaker would have paid tax on $77$ dollars that were never spent.

Give each route its single multiplier and the comparison settles itself.

The shortcut takes $19\%$ off the list price and stops there, so it multiplies by

$1 - 0.19 = 0.81.$

The real transaction multiplies by $0.72$ and then by $1.09$, and two multiplications in a row can be gathered into one number:

$0.72 \times 1.09 = 0.7848.$

Since $0.7848$ is smaller than $0.81$, the real route charges less than the shortcut on any list price at all. On this one,

$0.81 \times 275 = 222.75, \qquad 0.7848 \times 275 = 215.82,$

a difference of $6.93$ dollars in the cabinetmaker's favour.

The shortcut fails because it adds and subtracts percents that are not taken of the same amount. The $28\%$ comes off the list price of $275$ dollars, where it is worth $77$ dollars; the $9\%$ is charged on the discounted $198$ dollars, where it is worth only $17.82$ dollars. Subtracting $9$ from $28$ would be legitimate only if both percents sat on the same base, and the entire effect of the discount is that they do not.

A reduction of $12\%$ leaves the rest of the price standing, so the shelf label is

$100\% - 12\% = 88\%$

of the price the helmet carried before. That makes $154$ dollars a known part of the whole we are after, and a whole is recovered from a known part by dividing:

$\frac{154}{0.88} = 175.$

The helmet was $175$ dollars. Check it forward: $12\%$ of $175$ is $0.12 \times 175 = 21$ dollars off, and $175 - 21 = 154$, the figure on the label. No equation was needed anywhere, only the observation that the label is a known share of the amount we want.

A surcharge is an increase, so the invoice sits above the fee that was quoted. Adding $16\%$ leaves the total at

$100\% + 16\% = 116\%$

of the quoted fee, which is a multiplier of $1.16$. Dividing the invoice by that multiplier undoes the surcharge:

$\frac{377}{1.16} = 325.$

The fee quoted was $325$ dollars. Check it forward: $16\%$ of $325$ is $0.16 \times 325 = 52$ dollars, and $325 + 52 = 377$, the invoice total.

Test his figure the way any answer is tested, by putting it back through the change it claims to have undone. A $12\%$ reduction keeps $88\%$ of whatever it acts on, so

$0.88 \times 172.48 \approx 151.78.$

That is not the $154$ dollars on the label, so $172.48$ dollars cannot be the price the helmet started from.

The reason is the base. The shop's $12\%$ was taken of the earlier, larger price, where it was worth $0.12 \times 175 = 21$ dollars. The assistant takes $12\%$ of the label instead, where the same percent is worth only $0.12 \times 154 = 18.48$ dollars, so he puts back $2.52$ dollars less than was removed and his answer lands exactly $2.52$ dollars short.

Adding a percent back on can never undo taking that percent off, because the two percents are charged on different amounts, and the smaller amount always yields the smaller adjustment. What does undo it is division. The label is a fixed share of the price we want, namely $88\%$ of it, so dividing the label by $0.88$ recovers the whole from the part, every time and without any guesswork.

Take the years one at a time, and apply each change to the population that year actually begins with.

The first year is a rise of $15\%$, a multiplier of $1.15$:

$1.15 \times 6{,}400 = 7{,}360.$

The second year is a fall of $15\%$, which leaves $85\%$ standing, and it acts on the $7{,}360$ residents the town now has:

$0.85 \times 7{,}360 = 6{,}256.$

The town ends on $6{,}256$ residents, which is $6{,}400 - 6{,}256 = 144$ fewer than it began with. Measured against the starting population,

$\frac{144}{6{,}400} \times 100\% = 2.25\%,$

so the two years together come to a decrease of $2.25\%$ rather than a return to the start.

Apply each raise to the salary in force when it is awarded. The first multiplies by $1.05$:

$1.05 \times 32{,}000 = 33{,}600.$

The second raise is $4\%$ of the new salary, a multiplier of $1.04$:

$1.04 \times 33{,}600 = 34{,}944.$

A single raise of $9\%$ on the original salary would instead have paid

$1.09 \times 32{,}000 = 34{,}880.$

The two raises therefore pay $34{,}944 - 34{,}880 = 64$ dollars more than the single raise, so they are worth slightly more than $9\%$, not exactly $9\%$.

Neither rule survives, and in both cases the reason is the amount the second percent is taken of.

For the town, the first $15\%$ was added to $6{,}400$ residents while the second was taken off $7{,}360$:

$0.15 \times 6{,}400 = 960, \qquad 0.15 \times 7{,}360 = 1{,}104.$

The same percent is worth more when it is taken of a larger amount, so the fall removes $144$ more than the rise had added, which is exactly the shortfall found in part A. A rise and a fall of the same percent always land below the start for this reason, and the shortfall has a tidy description: it is that percent of itself, since $15\%$ of $15\%$ is $2.25\%$.

For the salary, the first raise is worth $0.05 \times 32{,}000 = 1{,}600$ dollars. A single $9\%$ raise charges its whole $9\%$ against the original salary. The second raise instead charges $4\%$ against a salary that already contains that $1{,}600$, so it collects $4\%$ of the raise as well:

$0.04 \times 1{,}600 = 64.$

That single line is the whole of the difference, and it is why two raises in a row come to $1.05 \times 1.04 = 1.092$, a rise of $9.2\%$ rather than $9\%$.

The condition is the one both examples break: percents may be added only when both are taken of the same amount. A tax and a tip each figured on the same menu price add without any trouble, because there is one base and it never moves. A second raise figured on an already raised salary does not, because by the time it is applied the base has changed underneath it.