How to Calculate Percent Increase and Decrease

In 9.2 you took a percent of a fixed number. Here the number itself changes and you want the new amount. A $30\%$ rise means the new amount is $130\%$ of the original, one multiply by $1.30$, the multiplier. A drop of $20\%$ leaves $80\%$, a single multiply by $0.80$.

Problem

A garden trail of $240$ m was extended by $35\%$. Find $35\%$ of $240$, then add it to $240$. How many meters of trail are there now?

Show a hint

The extra trail is $35\%$ of the ORIGINAL $240$ meters, not of some new total. Use part $=$ percent $\times$ whole, so the added length is $0.35 \times 240$.
Once you know the added piece is $84$ meters, the new trail is just the old length plus that piece, $240 + 84$. Adding the extra back on is what makes this an increase.

Show the full solution

$35\%$ of $240$ is $0.35 \times 240 = 84$ meters of new trail, and $240 + 84 = 324$. The trail is now $\boxed{324}$ meters. Faster route, keeping the original and adding $35\%$ more is $135\%$, so $240 \times 1.35 = 324$ in one step.

Problem

A streaming channel had $480$ subscribers and grew by $35\%$. The new count is $135\%$ of $480$. How many subscribers does the channel have now?

Show a hint

The original count is $100\%$ of itself, and an increase piles more on top of that. A $35\%$ increase means the new amount is $100\% + 35\% = 135\%$ of the original.
Turn $135\%$ into the multiplier $1.35$, then compute $480 \times 1.35$ in one step instead of adding the part separately.

Show the full solution

The new count is $100\% + 35\% = 135\%$ of $480$, a multiplier of $1.35$. $$480 \times 1.35 = 648$$ The channel now has $\boxed{648}$ subscribers. Adding the part separately gives the same thing, since $0.35 \times 480 = 168$ and $480 + 168 = 648$.

Problem

A coastal marsh had $350$ wading birds at dawn. By midday, $12\%$ had flown off. The survivors are $88\%$ of $350$. How many birds remain at midday?

Show a hint

A $12\%$ decrease does not leave $12\%$. It leaves the rest. Find the leftover percent by taking $100\% - 12\% = 88\%$, and that fraction of the flock is what you want.
Turn $88\%$ into the multiplier $0.88$ and multiply the original count by it, $350 \times 0.88$. If you prefer the long way, take $12\%$ of $350$ and subtract it from $350$. Both roads land on the same number.

Show the full solution

A $12\%$ drop leaves $100\% - 12\% = 88\%$, so the multiplier is $0.88$. $$350 \times 0.88 = 308$$ At midday there are $\boxed{308}$ birds. The long way agrees, since $0.12 \times 350 = 42$ birds flew off and $350 - 42 = 308$.

Problem

A kayak rental costs $\$90$ a day. A $40\%$ discount removes $40\%$ and leaves $60\%$. What is the discounted price in dollars?

Show a hint

A $40\%$ discount removes $40\%$ of the price, so what is left is $100\% - 40\% = 60\%$ of it. That surviving $60\%$ is your multiplier.
The multiplier for a $40\%$ decrease is $1 - 0.40 = 0.60$. Multiply the original price by it, $90 \times 0.60$.

Show the full solution

A $40\%$ discount leaves $100\% - 40\% = 60\%$ of the price, so the multiplier is $0.60$. $$90 \times 0.60 = 54$$ The discounted price is $\boxed{54}$ dollars. The multiplier folds the subtracting into one step, since $0.40 \times 90 = 36$ and $90 - 36 = 54$.

Both strips start from the same original, the full $100\%$ bar. Adding $30\%$ stacks a new block on the right, so the total is $100\% + 30\% = 130\%$, which is the multiplier $\times 1.30$. Cutting $20\%$ peels a slice off the right, leaving $100\% - 20\% = 80\%$, the multiplier $\times 0.80$. An increase always lands above $1$ and a decrease always lands below $1$, and either way you are scaling the original $100\%$.

Problem

A podcast went from $250$ plays to $320$. The change is $70$ plays. Divide by the original $250$, not the new $320$. What percent increase was this? Give the number only.

Show a hint

First find the change in plays, the new amount minus the old amount. Then remember a percent change is measured against the ORIGINAL, so the comparison you want is that change divided by $250$, not divided by $320$.
You have $70$ more plays out of the starting $250$. Compute $\tfrac{70}{250}$ and turn that fraction into a percent by multiplying by $100$.

Show the full solution

The change is $320 - 250 = 70$ plays, and a percent change divides by the original. $$\frac{70}{250} \times 100 = 28$$ The increase is $\boxed{28}$. Forward check, a $28\%$ rise scales by $1.28$, and $250 \times 1.28 = 320$. Dividing by the new $320$ answers a different question.

Problem

A library had $640$ visitors on a sunny Saturday and $480$ on a rainy one, a drop of $160$. Divide by the original $640$. What percent decrease was this? Give the number only.

Show a hint

First find the drop, then ask what fraction of the ORIGINAL it is. The drop is $640 - 480 = 160$, and the original is $640$, so the percent change is $\tfrac{160}{640}$ written as a percent.
Simplify $\tfrac{160}{640}$. Both top and bottom share a factor of $160$, so it becomes $\tfrac{1}{4}$, and $\tfrac{1}{4} = 25\%$. Resist the urge to divide by $480$. The original count is the whole here.

Show the full solution

The drop is $640 - 480 = 160$ visitors, and a percent change divides by the original $640$. $$\frac{640 - 480}{640} = \frac{160}{640} = \frac{1}{4} = 0.25 = 25\%$$ So the answer is $\boxed{25}$. Dividing by $480$ instead gives about $33\%$, which answers how much the count would have to grow to climb back to $640$.

Problem

A bamboo shoot grew from $14$ cm to $35$ cm, a change of $21$ cm. Since $21 > 14$, the percent will sail past $100\%$. What percent increase was this? Give the number only.

Show a hint

The change is $35 - 14 = 21$ centimeters. For a percent change you always divide by the ORIGINAL amount, the $14$ centimeters it started at, not the $35$ it ended at.
Compute $\tfrac{21}{14}$ and turn it into a percent. Since $21$ is more than $14$, this fraction is bigger than $1$, so the percent lands above $100\%$.

Show the full solution

The change is $35 - 14 = 21$ centimeters, and a percent change divides by the original $14$. $$\frac{35 - 14}{14} = \frac{21}{14} = \frac{3}{2} = 1.5$$ That is $150\%$, so the increase is $\boxed{150}$. A plain doubling to $28$ would be a $100\%$ increase, and this went past double, so the percent lands above $100$.

A podcast climbs from $250$ plays to $320$. The gold slice is the change, $320 - 250 = 70$, and the percent change is that change measured against where we started, $\tfrac{70}{250} = 0.28 = 28\%$. Dividing by the new amount, $\tfrac{70}{320}$, is the trap, since the change is always compared to the original, never to where you landed.

Problem

After a $20\%$ increase, a shelf holds 738 books. To recover last year's count, divide by $1.20$, not subtract $20\%$ of 738. How many books were there before?

Show a hint

The tempting wrong path is $738 - 20\%$ of $738 = 590.4$, which is not even a whole number of books. The $20\%$ growth was a slice of last year's count, not of this year's. So the right question is what number, when grown by $20\%$, lands on $738$.
A $20\%$ increase multiplies the original by $1.20$, so $1.20 \times \text{original} = 738$. To undo a multiplication, divide. Compute $738 \div 1.20$.

Show the full solution

A $20\%$ increase multiplies by $1.20$, so $1.20 \times \text{original} = 738$. Undo the multiplication by dividing, $738 \div 1.20 = 615$. Check it forward, $1.20 \times 615 = 738$. Last year the shelf held $\boxed{615}$ books. Subtracting $20\%$ of $738$ gives $590.4$, too low, because the $20\%$ was a slice of last year's smaller count.

Problem

A ferry pass now costs $\$510$ after a $15\%$ fare cut. A $15\%$ decrease multiplied the old price by $0.85$. What was the original season pass price in dollars?

Show a hint

A $15\%$ decrease keeps $100\% - 15\% = 85\%$ of the original, so it multiplies the old price by $0.85$ to give $510$. The $510$ is the new amount, not the original.
Since $\text{original} \times 0.85 = 510$, undo the multiplier by dividing. Compute $510 \div 0.85$. Adding $15\%$ of $510$ back on is the wrong move, because that $15\%$ would be a slice of the smaller current price, not of the original.

Show the full solution

A $15\%$ cut leaves $100\% - 15\% = 85\%$, so $\text{original} \times 0.85 = 510$. Divide by the multiplier to recover the old price. $$\text{original} = \frac{510}{0.85} = 600$$ The season pass was $\boxed{600}$ dollars. Adding $15\%$ of $510$ back gives $586.50$, short of the mark, since $15\%$ of the smaller current price is smaller.

Problem

A bee colony of $200$ grew by $30\%$ one week, then shrank by $30\%$ the next. Many guess it returns to $200$. It does not. How many workers are in the colony at the end?

Show a hint

Do the two weeks in order, one change at a time. A $30\%$ increase multiplies by $1.30$, so the boosted colony is $200 \times 1.30$. Whatever that comes to is the new starting size for week two.
For week two, a $30\%$ decrease multiplies by $0.70$. The catch is that the $30\%$ is taken off the bigger boosted number, not off the original $200$, so it removes more bees than the increase added.

Show the full solution

A $30\%$ rise multiplies by $1.30$ and a $30\%$ fall by $0.70$. Week one gives $200 \times 1.30 = 260$ workers, and week two gives $260 \times 0.70 = 182$. The colony ends at $\boxed{182}$ workers. Chained, $1.30 \times 0.70 = 0.91$, under $1$, because the drop came off the larger $260$ and removed more than the rise added.

Going up $20\%$ then down $20\%$ does not undo itself. The rise multiplies by $1.20$ and the cut multiplies by $0.80$, and $1.20 \times 0.80 = 0.96$, so you land at $100 \times 0.96 = 96$, a hair below the dashed start line, not back at $100$. The reason is that the $20\%$ cut came off the taller $120$, which is $24$, while the $20\%$ rise had added only $20$. Same percent, different size to act on, so the trip down outweighs the trip up.

Every move in this lesson is one multiplier, above $1$ for an increase and below $1$ for a decrease. A chain of changes is just those multipliers multiplied together in order, no guessing whether they cancel.

Problem

A recycling program collects $5{,}000$ kg in Q1 and expects each later quarter to grow by $10\%$. Each quarter multiplies by $1.10$. How many kilograms are expected in Q4 (three quarters later)?

Show a hint

A $10\%$ increase multiplies the amount by $1.10$. One quarter later it is $5{,}000 \times 1.10$. Each new quarter multiplies the previous quarter's amount by $1.10$ again.
Three quarters later means three multipliers stacked, so compute $5{,}000 \times 1.10 \times 1.10 \times 1.10$. Take it one step at a time, $5{,}000$ to $5{,}500$ to $6{,}050$, then once more.

Show the full solution

Each quarter multiplies by $1 + \tfrac{10}{100} = 1.10$, and three quarters later means three of those multipliers. $$5{,}000 \times 1.10^3 = 5{,}000 \times 1.331 = 6{,}655$$ Q4 is expected to bring $\boxed{6655}$ kilograms. Step by step that is $5{,}500$, then $6{,}050$, then $6{,}655$, with the jumps growing because each $10\%$ comes off a bigger amount.

Problem

An art class had $400$ students. Enrollment rose $50\%$ for summer, then fell $40\%$ from that summer peak. A $50\%$ rise and a $40\%$ fall do not cancel. What is the fall enrollment?

Show a hint

Do it in two steps, one change at a time. A $50\%$ increase multiplies by $1.50$, so the summer peak is $400 \times 1.50$. Whatever that comes to, the fall drop of $40\%$ is measured off that peak, not off the original $400$.
A $40\%$ decrease multiplies by $1 - 0.40 = 0.60$, so multiply the summer peak by $0.60$. If you like, chain the multipliers first, $1.50 \times 0.60 = 0.90$, which already tells you the fall figure lands at $90\%$ of the original, not back at $100\%$.

Show the full solution

A $50\%$ rise multiplies by $1.50$, so the summer peak is $400 \times 1.50 = 600$. A $40\%$ fall multiplies by $0.60$, so the fall count is $600 \times 0.60 = 360$. Enrollment is $\boxed{360}$. The two changes do not cancel, since $1.50 \times 0.60 = 0.90$. The rise was figured on $400$ but the fall on the larger $600$.

Practice these ideas

Practice

A bakery sold $80$ loaves on Tuesday and $15\%$ more on Wednesday. How many loaves did the bakery sell on Wednesday?

Show the solution

A $15\%$ increase multiplies by $1 + \tfrac{15}{100} = 1.15$, so Wednesday's total is $$80 \times 1.15 = 92.$$ The bakery sold $\boxed{92}$ loaves. The long way matches, since $0.15 \times 80 = 12$ and $80 + 12 = 92$.

Practice

A winter jacket priced at $\$250$ is marked $30\%$ off. What is the sale price in dollars?

Show the solution

A $30\%$ decrease keeps $70\%$ of the price, so the multiplier is $1 - 0.30 = 0.70$. Apply it to the original price. $$250 \times 0.70 = 175$$ The sale price is $\boxed{175}$ dollars.

Practice

A stadium that held $600$ fans expanded its seating by $45\%$. How many seats are there now?

Show the solution

A $45\%$ increase makes the new seating $100\% + 45\% = 145\%$ of the old, a multiplier of $1.45$. $$600 \times 1.45 = 870$$ The stadium now holds $\boxed{870}$ seats.

Practice

A bamboo shoot rose from $40$ cm to $54$ cm, a change of $14$ cm. Divide by the original. What percent increase is this? Give the number only.

Show the solution

The change is $54 - 40 = 14$ centimeters, and a percent change divides by the original $40$. That gives $\tfrac{14}{40} = 0.35$, or $35\%$, so the increase is $\boxed{35}$. Forward check, $40 \times 1.35 = 54$.

Practice

A reservoir dropped from $900$ to $720$ megalitres, a fall of $180$. What percent decrease is this? Give the number only.

Show the solution

The drop is $900 - 720 = 180$ megaliters, and a percent change divides by the original $900$. $$\frac{180}{900} = \frac{1}{5} = 0.20$$ That is a $20\%$ decrease, so the answer is $\boxed{20}$. The multiplier agrees, since $\tfrac{720}{900} = 0.80$ means the reservoir kept $80\%$ of its water.

Practice

A startup grew from $25$ customers to $75$, a jump of $50$. Since $50 > 25$, the percent lands above $100\%$. What percent increase is this? Give the number only.

Show the solution

The change is $75 - 25 = 50$ customers, and a percent increase divides by the original $25$. That gives $\tfrac{50}{25} = 2$, which as a percent is $200\%$, so the answer is $\boxed{200}$. The new count is $3$ times the old, and a multiplier of $3$ is $1 + 2$, the original $100\%$ plus a $200\%$ increase on top.

Practice

After a $25\%$ increase, a road bike sells for $\$500$. The $25\%$ was on the OLD price. What was the original price in dollars?

Show the solution

A $25\%$ increase multiplies by $1 + \tfrac{25}{100} = 1.25$, so $1.25 \times \text{original} = 500$. Divide to undo it. $$\text{original} = \frac{500}{1.25} = 400$$ The original price was $\boxed{400}$ dollars. Subtracting $25\%$ of $500$ would give $375$, which is wrong because the $25\%$ was taken on the smaller original.

Practice

After a $20\%$ decrease, a sofa sells for $\$480$. The multiplier was $0.80$. What was the original price in dollars?

Show the solution

A $20\%$ decrease keeps $100\% - 20\% = 80\%$, so $\text{original} \times 0.80 = 480$. Divide by the multiplier instead of adding $20\%$ back. $$\text{original} = \frac{480}{0.80} = 600$$ The original price was $\boxed{600}$ dollars. Forward check, $600 \times 0.80 = 480$.

Practice

A vlogger started with $200$ subscribers, gained $40\%$, then lost $25\%$ of that new total. How many subscribers does the channel have after both months?

Show the solution

A $40\%$ gain multiplies by $1.40$ and a $25\%$ loss by $0.75$. Month one gives $200 \times 1.40 = 280$ subscribers, and month two gives $280 \times 0.75 = 210$. The final count is $\boxed{210}$. It does not land back at $200$, because the $25\%$ drop came off the larger $280$.

Practice

A koi pond holds $2{,}000$ liters and gains $20\%$ each day. How many liters does the pond hold after three full days of growth?

Show the solution

A $20\%$ increase multiplies by $1.20$, and three days means three of those multipliers. $$2{,}000 \times 1.20^3 = 3{,}456$$ The pond holds $\boxed{3456}$ liters. Day by day that is $2{,}400$, then $2{,}880$, then $3{,}456$.

Practice

A factory used $1{,}500$ kWh last month and cut usage by $8\%$ this month. How many kWh did the factory use this month?

Show the solution

An $8\%$ decrease leaves $100\% - 8\% = 92\%$ of the original, so the multiplier is $0.92$. Multiply the original amount by it. $$1{,}500 \times 0.92 = 1{,}380$$ So this month the factory used $\boxed{1380}$ kilowatt-hours.

Practice

A scout troop grew from $120$ to $156$ members, a change of $36$. Divide by the original. What percent increase is this? Give the number only.

Show the solution

The change is $156 - 120 = 36$ members, and a percent change divides by the original $120$. $$\frac{36}{120} = 0.30$$ That is a $30\%$ increase, so the answer is $\boxed{30}$. Forward check, $120 \times 1.30 = 156$.

Practice

A camping tent is marked down $40\%$ and now sells for $\$90$. The sale price is $60\%$ of the original. What was the original price in dollars?

Show the solution

A $40\%$ markdown leaves $1 - 0.40 = 0.60$ of the price, so $0.60 \times \text{original} = 90$. Divide to undo the multiplier, $90 \div 0.60 = 150$. The original price was $\boxed{150}$ dollars. Check it forward, $150 - 0.40 \times 150 = 150 - 60 = 90$.

Practice

A club had $\$500$. A fundraiser raised it by $20\%$, then supply costs cut it by $20\%$. How many dollars are left in the fund?

Show the solution

The fundraiser multiplies by $1 + 0.20 = 1.20$ and the supply costs by $1 - 0.20 = 0.80$. $$500 \times 1.20 \times 0.80 = 600 \times 0.80 = 480$$ The fund has $\boxed{480}$ dollars left. It lands below $500$ because the $20\%$ drop was taken from the larger $600$, so it removed more than was added.