CEMC's Open Courseware - Lesson 10: Revisiting Percentages

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1. $12.5\% = \dfrac{12.5}{100} = \dfrac{125}{1000} = 0.125$
2. $0.3\%=\dfrac{0.3}{100}=\dfrac{3}{1000} = 0.003$
3. $130\% = \dfrac{130}{100} = \dfrac{13}{10} = 1.3$
4. $2000\% = \dfrac{2000}{100} = 20$
1. $\dfrac{19}{40} = 47.5 \%$
2. $1.15 = 115\%$
3. $10.55 = 1055\%$
4. $\dfrac{15}{4}=375\%$
5. $0.004 = 0.4\%$
6. $4\dfrac{1}{5} = 420\% $
There are many diagrams that can represent $32.5\%$. One possibility is
1. $220\% = \dfrac{220}{100} = \dfrac{22}{10} = 2.2$
2. We are told that $100\%$ of the daily intake is $60$ mg. One glass provides $220\%$ and\[220\% = 100\% + 100\% + 20\%\]Since $20\%$ of $60$ is $0.2 \times 60 = 12$, there are\[60\text{ mg} + 60\text{ mg} + 12\text{ mg} = 132\text{ mg}\] in one cup of orange juice.
  We can find the number $132$ directly by calculating $220\%$ of $60$ or $2.2 \times 60 = 132$.
1. Yes. Rainfall can be more than $100\%$ of the average. For example, if the average rainfall is $100$ millimetres, then a year with $120$ millimetres would represent $120\%$ of the average.
2. No. You cannot eat more than $100\%$ of the available bananas.
3. Yes. This is a small profit increase, but it is definitely possible. If the profit in one year is $$10~000$ and the profit in the next year is $$10~500$ then the annual profit increased by $0.5\%$.
4. Yes. A charity can receive more than $100\%$ of their goal for donations. This means that the charity received twice as many donations as they were hoping to receive.
1. Since $10\%$ of $75$ is $7.5$, this means that $110\%$ of $75$ is $75+7.5 = 82.5$. So we estimate the total cost to be around $$ 82.50$.
  Since $113\%$ is larger than $110\%$, our estimate is too low.
  Since $1\%$ of $75$ is $0.75$, our estimate of $82.5$ we will be less than $$3$ from the correct total.
2. Since $113\%$ of $75$ is $1.13 \times 75 = 84.75$, the actual total cost is $$84.75$. As we predicted in part a), this is higher than our estimate, but within $$3$ of our estimate.
1. Company A: $36-15=21$ more shirts in Week 2 than in Week 1 and $108-36 = 72$ more shirts in Week 3 than in Week 2.
  Company B: $360 - 240 = 120$ more shirts in Week 2 than in Week 1 and $648 - 360=288$ more shirts in Week 3 than in Week 2.
2. Company A: Since $\dfrac{21}{15} =1.4 = 140\%$, sales increased by $140\%$ from Week 1 to Week 2.
  Since $\dfrac{72}{36} = 2 = 200\%$, sales increased by $200\%$ from Week 2 to Week 3.
  Company B: Since $\dfrac{120}{240} = 0.5 = 50\%$, sales increased by $50\%$ from Week 1 to Week 2.
  Since $\dfrac{288}{360}=0.8 = 80\%$, sales increased by $80\%$ from Week 2 to Week 3.
3. Both companies are selling more shirts each week so production is growing for each company. From part a), Company B has larger growth in terms of total number of shirts. From part b), Company A has larger growth when measured in terms of percentages. Often percentage increase is considered a more fair way to measure growth than total number of items.
1. The decrease is price is $$15.00-$12.75=$2.25$. Then, since $\dfrac{2.25}{15} = \dfrac{15}{100}$, the percentage decrease is $15\%$.
2. The discount is $15\%$ of $$15.00$ which is $$2.25$. Therefore, the sale price of the game is $$12.75$.
3. Answers will vary.
Using the equation for increased value, we have

$\begin{align*} \text{Jan Population} &= \text{Oct Population}+(0.38\%~\text{of Oct Population}) \\ &= 998~832 + \left( \dfrac{0.38}{100} \times 998~832 \right) \\ &= 998~832 + \left( 0.0038 \times 998~832 \right) \\ &= 998~832 + 3795.5616 \\ &= 1~002~627.5616 \end{align*}$

Since this number represents a population, we round up and determine that the population of Nova Scotia on January 1, 2022 is approximately $1~002~628$.
First, we determine that $0.1\%$ of $1000$ is $0.001 \times 1000 = 1$.
Next, we determine that $0.9\%$ of $1000$ is $0.009 \times 1000 = 9$.
Then, two consective whole numbers between $1$ and $9$ could be chosen from the following list:
- $2$ and $3$,
- $3$ and $4$,
- $4$ and $5$,
- $5$ and $6$,
- $6$ and $7$, or
- $7$ and $8$.
1. Since $\dfrac{24-8}{8} = \dfrac{16}{8} = 2$, your travel time increased by $200\%$ from Monday to Tuesday.
2. Your biking speed is $\dfrac{2 \text{ km}}{8 \text{ min}}$, which is equivalent to $\dfrac{1}{4}$ km/min or $15$ km/h.
  Your walking speed is $\dfrac{2 \text{ km}}{24 \text{ min}}$, which is equivalent to $\dfrac{1}{12}$km/min or $5$ km/h.
3. Using the unit of km/h, since $\dfrac{15-5}{15} = \dfrac{10}{15} = 0.\bar{6}$, then your travel speed decreased by $66\dfrac{2}{3}\%$ from Monday to Tuesday.
  Using the unit of km/min, the decrease in speed would be $\dfrac{1}{4} - \dfrac{1}{12} = \dfrac{1}{6}$, and so the percentage decrease would be\[\begin{align*} \left(\dfrac{1}{4}-\dfrac{1}{12}\right)\div \dfrac{1}{4} &= \left(\dfrac{1}{6}\right) \div \dfrac{1}{4} \\ &= \dfrac{1}{6} \times \dfrac{4}{1} \\ &= \dfrac{4}{6} \\ &= 0.\bar{6} \\ &= 66\dfrac{2}{3}\% \end{align*}\]This is the same as before.
  The percentage decrease does not depend on the particular choice of units, but you must use the same units for both rates.