Applications of Percents

$ Tip = 12\% \:\:of\:\:\$34.99 \\[3ex] = \dfrac{12}{100} * 34.99 \\[5ex] = 4.1988 \approx 4.20 \\[3ex] $ The tip is $\$4.20$

$ Sales\:\: tax\:\: = 7\% \:\:of\:\: 220.98 \\[3ex] = \dfrac{7}{100} * 220.98 \\[5ex] = 15.4686 \approx 15.47 \\[3ex] $ Sales tax = $\$15.47$

You may begin with a preview to ensure students understand it.
Preview is in black color.
Assume the cost of a phone is $\$200.00$
Include a $5\%$ sales tax
How much is the sales tax?

$ 5\% = \dfrac{5}{100} = 0.05 \\[5ex] 5\%\:\:sales\:\:tax = 0.05(200) = \$10 \\[3ex] Checkout\:\:price = \$200 + \$10 = \$210 \\[3ex] $ Ask students this question:
Given: $\$210$, with $5\%$ sales tax (included), how can we get $\$10$?
In other words, Given the checkout price which includes the sales tax, how do we figure out the sales tax?
Introduce and discuss the concept of Algebra (using variables).
First, we need to find the cost of the phone without the sales tax.
This is also known as the initial cost of the phone.

$ Let\:\:the\:\:initial\:\:cost\:\:of\:\:the\:\:phone = x \\[3ex] 5\%\:\:sales\:\:tax = 0.05(x) = 0.05x \\[3ex] Initial\:\:cost + Sales\:\:tax = Checkout\:\:price \\[3ex] x + 0.05x = 210 \\[3ex] 1.05x = 210 \\[3ex] x = \dfrac{210}{1.05} \\[5ex] x = 200 \\[3ex] Sales\:\:tax = 0.05x = 0.05(200) = \$10 \\[3ex] OR \\[3ex] Sales\:\:tax = Checkout\:\:price - Initial\:\:cost \\[3ex] Sales\:\:tax = 210 - 200 = \$10 $

Let the initial cost of the phone = $p$

$ 7\%\:\: sales\:\: tax\:\: of\:\: p = 7\% * p \\[3ex] = \dfrac{7}{100} * p \\[5ex] = 0.07 * p \\[3ex] = 0.07p \\[3ex] $ The initial price of the phone and the tax equals $\$220.98$

$ p + 0.07p = 220.98 \\[3ex] 1.07p = 220.98 \\[3ex] p = \dfrac{220.98}{1.07} \\[5ex] p = 206.5233645 \\[3ex] 7\%\:\: sales\:\: tax = 0.07p \\[3ex] = 0.07 * 206.5233645 \\[3ex] = 14.45663551 \\[3ex] $ The initial cost of the phone = $\$206.52$
The sales tax = $\$14.46$

The initial price of the laptop = $960$

$ Discount = 9\%\:\:of\:\: 960 \\[3ex] = 9\% * 960 \\[3ex] = 0.09 * 960 \\[3ex] = 86.4 \\[3ex] Discount = \$86.40 \\[3ex] Sale\:\: price\:\:means\:\: 9\%\:\:off\:\: 960 \\[3ex] Sale\:\:price = Initial\:\:price - Discount \\[3ex] = 960 - 86.4 \\[3ex] = 873.60 \\[3ex] Sale\:\:price = \$873.60 $

You may begin with a preview to ensure students understand it.
Preview is in black color.
Assume the initial cost of a phone is $\$200.00$
Assume a $5\%$ discount (Assume it is on sale for $5\%$ off)
How much is the sale price?

$ 5\% = \dfrac{5}{100} = 0.05 \\[5ex] Discount = 5\%\:\:of\:\:200 = 0.05(200) = \$10 \\[3ex] Sale\:\:price = 5\%\:\:off\:\:200 = \$200 - \$10 = \$190 \\[3ex] Sale\:\:price = \$190 \\[3ex] $ Ask students this question:
Given: $\$190$, with $5\%$ discount (included), how can we get $\$200$?
In other words, Given the sale price which includes the discount, how do we figure out the actual price (initial price)?
Introduce and discuss the concept of Algebra (using variables).

$ Let\:\:the\:\:initial\:\:cost\:\:of\:\:the\:\:phone = x \\[3ex] Discount = 5\%\:\:of\:\:x = 0.05(x) = 0.05x \\[3ex] Sale\:\:price = 5\%\:\:off\:\:x = x - 0.05x = 0.95x \\[3ex] Sale\:\:price = 190 \\[3ex] \rightarrow 0.95x = 190 \\[3ex] x = \dfrac{190}{0.95} \\[5ex] x = 200 \\[3ex] Initial\:\:price = \$200 $

$ Let\:\:the\:\:initial\:\:price = p \\[3ex] 12\% = \dfrac{12}{100} = 0.12 \\[5ex] Sale\:\:price = \$550 \\[3ex] Discount = 12\%\:\:of\:\:p = 0.12p \\[3ex] Sale\:\:price = 12\%\:\:off\:\:p = p - 0.12p = 0.88p \\[3ex] \rightarrow 0.88p = 550 \\[3ex] p = \dfrac{550}{0.88} \\[5ex] p = 625 \\[3ex] $ The initial price of the laptop is $\$625.00$

$ \underline{Check} \\[3ex] Initial\:\:price = \$625.00 \\[3ex] 12\%\:\:discount\:\:of\:\:625 \\[3ex] = \dfrac{12}{100} * 625 \\[5ex] = 0.12 * 625 \\[3ex] = 75 \\[3ex] Sale\:\:price = 625 - 75 = \$550 $

$ Tip = 12\% \:\:of\:\:\$34.99 \\[3ex] = \dfrac{12}{100} * 34.99 \\[5ex] = 4.1988 \\[3ex] Total\:\:cost = Cost\:\:of\:\:meal + Tip \\[3ex] Total\:\:cost = 34.99 + 4.1988 \\[3ex] Total\:\:cost = 39.1888 \approx 39.19 \\[3ex] $ The total cost is $\$39.19$

We can do this question in two ways.
Use any method you prefer.

$ Let\:\:the\:\:initial\:\:price = p \\[3ex] \underline{First\:\:Method:\:\:Quantitative\:\:Literacy} \\[3ex] Marked\:\:down = 20\%\:\:off \\[3ex] Remaining = 100\% - 20\% = 80\% \\[3ex] Exclude\:\:tax \\[3ex] 80\% \:\:of\:\:what\:\:is\:\: 588 \\[3ex] \dfrac{is}{of} = \dfrac{\%}{100}...Percent-Proportion \\[5ex] \dfrac{588}{p} = \dfrac{80}{100} \\[5ex] Cross\:\:Multiply \\[3ex] p * 80 = 588 * 100 \\[3ex] p = \dfrac{588 * 100}{80} \\[5ex] p = \dfrac{5880}{8} \\[5ex] p = 735 \\[3ex] $ The initial price = $\$735.00$

$ \underline{Second\:\:Method:\:\:Algebraic\:\:Thinking} \\[3ex] Selling\:\:price = \$588.00 \\[3ex] Discount = 20\%\:\:of\:\: p \\[3ex] = 20\% * p \\[3ex] = \dfrac{20}{100} * p \\[5ex] = 0.2 * p \\[3ex] = 0.2p \\[3ex] Sale\:\:price = Initial\:\:price - Discount \\[3ex] Sale\:\: price = p - 0.2p \\[3ex] = 0.8p \\[3ex] Exclude\:\:tax \\[3ex] Sale\:\:price = Selling\:\:price \\[3ex] \therefore 0.8p = 588 \\[3ex] p = \dfrac{588}{0.8} \\[5ex] p = 735 \\[3ex] $ The initial price = $\$735.00$

You may begin with a preview to ensure students understand it.
Preview is in black color.

Assume the initial cost of a phone is $\$200.00$
Assume a $5\%$ discount (Assume it is on sale for $5\%$ off)
Include a $10\%$ sales tax
How much is the checkout price?

$ 5\% = \dfrac{5}{100} = 0.05 \\[5ex] Discount = 5\%\:\:of\:\:200 = 0.05(200) = \$10 \\[3ex] Sale\:\:price = 5\%\:\:off\:\:200 = \$200 - \$10 = \$190 \\[3ex] Sale\:\:price = \$190 \\[3ex] 10\% = \dfrac{10}{100} = 0.1 \\[5ex] Tax = 10\%\:\:of\:\:190 = 0.1(190) = \$19 \\[3ex] Checkout\:\:price = Sale\:\:price + Tax \\[3ex] Checkout\:\:price = 190 + 19 = \$209 \\[3ex] $ Ask students this question:
Given: $\$209$, with $5\%$ discount (included) and $10\%$ tax (included), how can we get $\$200$?
In other words, Given the checkout price which includes the discount and the tax, how do we figure out the actual price (initial price)?
Introduce and discuss the concept of Algebra (using variables).

$ Let\:\:the\:\:initial\:\:cost\:\:of\:\:the\:\:phone = x \\[3ex] Discount = 5\%\:\:of\:\:x = 0.05(x) = 0.05x \\[3ex] Sale\:\:price = 5\%\:\:off\:\:x = x - 0.05x = 0.95x \\[3ex] Tax = 10\%\:\:of\:\:0.95x = 0.1(0.95x) = 0.095x \\[3ex] Checkout\:\:price = Sale\:\:price + Tax \\[3ex] \rightarrow 209 = 0.95x + 0.095x \\[3ex] 209 = 1.045x \\[3ex] 1.045x = 209 \\[3ex] x = \dfrac{209}{1.045} \\[5ex] x = 200 \\[3ex] Initial\:\:price = \$200 $

$ Let\:\:the\:\:initial\:\:price = p \\[3ex] Selling\:\:price = \$588.00 \\[3ex] Discount = 20\%\:\:of\:\: p \\[3ex] = 20\% * p \\[3ex] = \dfrac{20}{100} * p \\[5ex] = 0.2 * p \\[3ex] = 0.2p \\[3ex] Sale\:\:price = Initial\:\:price - Discount \\[3ex] Sale\:\: price = p - 0.2p \\[3ex] = 0.8p \\[3ex] Include\:\:tax \\[3ex] 5\%\:\: sales\:\: tax\:\: of\:\: 0.8p \\[3ex] = 0.05 * 0.8p \\[3ex] = 0.04p \\[3ex] Sale\:\:price + Tax = Selling\:\:price \\[3ex] \therefore 0.8p + 0.04p = 588 \\[3ex] 0.84p = 588 \\[3ex] p = \dfrac{588}{0.84} \\[5ex] p = 700.00 \\[3ex] $ The initial price = $$700.00$

$ \underline{Check} \\[3ex] Initial\:\:price = \$700.00 \\[3ex] 20\%\:\:discount\:\:of\:\:700 \\[3ex] = \dfrac{20}{100} * 700 \\[5ex] = 20 * 7 \\[3ex] = 140 \\[3ex] Sale\:\:price = 700 - 140 = 560 \\[3ex] 5\%\:\:sales\:\:tax\:\:of\:\:560 \\[3ex] = \dfrac{5}{100} * 560 \\[5ex] = \dfrac{56}{2} \\[5ex] = 28 \\[3ex] Selling\:price = 560 + 28 = 588 \\[3ex] Selling\:\:price = \$588.00 $

$ Initial\:\:Price = 59.95 \\[3ex] 20\%\:\:Discount = \dfrac{20}{100} * 59.95 = 0.2 * 59.95 = 11.99 \\[3ex] Sale\:\:Price = Initial\:\:Price - Discount \\[3ex] Sale\:\:Price = 59.95 - 11.99 \\[3ex] Sale\:\:Price = \$47.96 $

$ Initial\:\:Price = 80 \\[3ex] Sale\:\:Price = 60 \\[3ex] Discount = Initial\:\:Price - Sale\:\:Price \\[3ex] Discount = 80 - 60 = 20 \\[3ex] \%Discount = \dfrac{Discount}{Initial\:\:Price} * 100 \\[5ex] \%Discount = \dfrac{20}{80} * 100 \\[5ex] \%Discount = 25\% $

We can do a basic part of this question in two ways.
Use any method you prefer.

$ Let\:\:the\:\:original\:\:price = x \\[3ex] \underline{First\:\:Method:\:\:Quantitative\:\:Literacy} \\[3ex] Discount = 20\%\:\:off \\[3ex] Remaining = 100\% - 20\% = 80\% \\[3ex] Exclude\:\:tax \\[3ex] 80\% \:\:of\:\:what\:\:is\:\: 30.40 \\[3ex] \dfrac{is}{of} = \dfrac{\%}{100}...Percent-Proportion \\[5ex] \dfrac{30.4}{x} = \dfrac{80}{100} \\[5ex] Cross\:\:Multiply \\[3ex] x * 80 = 30.4 * 100 \\[3ex] x = \dfrac{30.4 * 100}{80} \\[5ex] x = \dfrac{304}{8} \\[5ex] x = 38 \\[3ex] $ The original price = $\$38.00$

$ \underline{Second\:\:Method:\:\:Algebraic\:\:Thinking} \\[3ex] 20\%\:\:Discount = \dfrac{20}{100} * x \\[5ex] = 0.2 * x = 0.2x \\[3ex] Sale\:\:Price = Original\:\:Price - Discount \\[3ex] = x - 0.2x \\[3ex] = 0.8x \\[3ex] Sale\:\:Price = \$30.40 \\[3ex] \rightarrow 0.8x = 30.40 \\[3ex] x = \dfrac{30.40}{0.8} \\[5ex] x = 38 \\[3ex] x = \$38.00 \\[3ex] $ How much less than $\$38.00$ is $\$30.40$

$ Difference = 38.00 - 30.40 \\[3ex] Difference = \$7.60 $

$ \%Profit = \dfrac{Profit}{Cost\:\:Price} * 100 \\[5ex] \rightarrow \%Profit = \dfrac{Selling\:\:Price - Cost\:\:Price}{Cost\:\:Price} * 100 \\[5ex] Let\:\:Cost\:\:Price = C \\[3ex] Selling\:\:Price = 345.00 = 345 \\[3ex] \%Profit = 15\% = \dfrac{15}{100} \\[5ex] \rightarrow \dfrac{15}{100} = \dfrac{345 - C}{C} \\[5ex] Cross\:\:Multiply \\[3ex] 15C = 100(345 - C) \\[3ex] 15C = 34500 - 100C \\[3ex] 15C + 100C = 34500 \\[3ex] 115C = 34500 \\[3ex] C = \dfrac{34500}{115} \\[5ex] C = 300 \\[3ex] Profit = Selling\:\:Price - Cost\:\:Price \\[3ex] Profit = 345 - 300 \\[3ex] Profit = Le\:45.00 $

$ Let\:\:the\:\:number\:\:of\:\:goats = g \\[3ex] Cost\:\:Price = 4000 * g = 4000g \\[3ex] Selling\:\:Price = 180000 \\[3ex] \%Loss = 25\% = \dfrac{25}{100} = \dfrac{1}{4} \\[5ex] \%Loss = \dfrac{Cost\:\:Price - Selling\:\:Price}{Cost\:\:Price} \\[5ex] \rightarrow \dfrac{1}{4} = \dfrac{4000g - 180000}{4000g} \\[5ex] Cross\:\:Multiply \\[3ex] 4000g = 4(4000g - 180000) \\[3ex] 4000g = 16000g - 720000 \\[3ex] 720000 = 16000g - 4000g \\[3ex] 720000 = 12000g \\[3ex] 12000g = 720000 \\[3ex] g = \dfrac{720000}{12000} \\[5ex] g = 60 \\[3ex] $ He bought $60$ goats.

$ \underline{Check} \\[3ex] Cost\:\:price\:\:of\:\:one\:\:goats = ₦4000 \\[3ex] Cost\:\:price\:\:of\:\:all\:\:goats = 60 * 4000 = ₦240000 \\[3ex] 25\%\:\:of\:\:240000 \\[3ex] = \dfrac{25}{100} * 240000 \\[5ex] = 25 * 2400 \\[3ex] = 60000 \\[3ex] 25\%\:\:Loss \\[3ex] = 240000 - 60000 \\[3ex] = 180000 \\[3ex] Selling\:\:price = ₦180,000.00 $

$ Live\:\:south\:\:of\:\:Highway\:\:R = 45\%\:\:of\:\:900 \\[3ex] = \dfrac{45}{100} * 900 \\[5ex] = 45 * 9 \\[3ex] = 405 \\[3ex] Do\:\:NOT\:\:ride\:\:the\:\:bus\:\:to\:\:school = 20\%\:\:of\:\:405 \\[3ex] = \dfrac{20}{100} * 405 \\[5ex] = \dfrac{405}{5} \\[5ex] = 81 \\[3ex] Rides\:\:the\:\:bus\:\:to\:\:school \\[3ex] \underline{First\:\:Method} \\[3ex] 405 - 81 = 324 \\[3ex] \underline{Second\:\:Method} \\[3ex] Rides\:\:the\:\:bus\:\:to\:\:school = 80\%\:\:of\:\:405 \\[3ex] = \dfrac{80}{100} * 405 \\[5ex] = \dfrac{4}{5} * 405 \\[5ex] = 4 * 81 \\[3ex] = 324 $

$ Hitchcock's\:\:Vertigo = 128\:\:minutes \\[3ex] 20\%\:\:of\:\:128 = \dfrac{20}{100} * 128 = 0.2 * 128 = 25.6 \\[5ex] 20\%shorter = 128 - 25.6 = 102.4 \\[3ex] $ The new version of Vertigo is about $102$ minutes in length.

$ \underline{Q\:\:to\:\:R} \\[3ex] \%Loss = 5\% = \dfrac{5}{100} \\[5ex] Selling\:\:Price = 209 \\[3ex] Cost\:\:Price = C \\[3ex] \%Loss = \dfrac{Cost\:\:Price - Selling\:\:Price}{Cost\:\:Price} \\[5ex] \dfrac{5}{100} = \dfrac{C - 209}{C} \\[5ex] Cross\:\:Multiply \\[3ex] 5C = 100(C - 209) \\[3ex] 5C = 100C - 20900 \\[3ex] 20900 = 100C - 5C \\[3ex] 20900 = 95C \\[3ex] 95C = 20900 \\[3ex] C = \dfrac{20900}{95} \\[5ex] C = 220 \\[3ex] $ $Q$ bought the bicycle at the cost price of $₦220.00$
This cost price is the price that $P$ sold it to $Q$
This means that:
The cost price that $Q$ bought the bicycle is the selling price that $P$ sold the bicycle

$ \underline{P\:\:to\:\:Q} \\[3ex] \%Profit = 10\% = \dfrac{10}{100} \\[5ex] Selling\:\:Price = 209 \\[3ex] Cost\:\:Price = C \\[3ex] \%Profit = \dfrac{Selling\:\:Price - Cost\:\:Price}{Cost\:\:Price} \\[5ex] \dfrac{10}{100} = \dfrac{220 - C}{C} \\[5ex] 10C = 100(220 - C) \\[3ex] 10C = 22000 - 100C \\[3ex] 10C + 100C = 22000 \\[3ex] 110C = 22000 \\[3ex] C = \dfrac{22000}{110} \\[5ex] C = 200 \\[3ex] $ $P$ bought the bicycle at the cost price of $₦200.00$

Please be careful of this common mistake!

$ 20\% + 30\% - 20\% = 30\% \\[3ex] $ This is one of the common misuses of Percents.
Please take note!
If you are unsure of how to begin, try Arithmetic (with a number).
It is okay to try with a number in this case because the initial population was not given.
So, you can assume a number.
Then, you can do it with Algebra (with a variable)
You are encouraged to do with a variable so you can get used to Algebra (working with variables)

$ \underline{Arithmetic} \\[3ex] Assume\:\:Initial\:\:Population = 100 \\[3ex] 20\%\:\:of\:\:100 = \dfrac{20}{100} * 100 = 0.2 * 100 = 20 \\[3ex] 20\%\:Increase = 100 + 20 = 120 \\[3ex] 30\%\:\:of\:\:120 = \dfrac{30}{100} * 120 = 0.3 * 120 = 36 \\[3ex] 30\%\:Increase = 120 + 36 = 156 \\[3ex] 20\%\:\:of\:\:156 = \dfrac{20}{100} * 156 = 0.2 * 156 = 31.2 \\[3ex] 20\%\:Decrease = 156 - 31.2 = 124.8 \\[3ex] New\:\:Population = 124.8 \\[3ex] Change = New - Initial \\[3ex] Change = 124.8 - 100 = 24.8 \\[3ex] Change\:\:is\:\:an\:\:increase \\[3ex] \%Increase = \dfrac{Increase}{Initial} * 100 \\[5ex] \%Increase = \dfrac{24.8}{100} * 100 \\[5ex] \%Increase = 24.8\% \\[3ex] \%Increase \approx 25\% \\[3ex] \underline{Algebra} \\[3ex] Let\:\:Initial\:\:Population = x \\[3ex] 20\%\:\:of\:\:x = \dfrac{20}{100} * x = 0.2 * x = 0.2x \\[3ex] 20\%\:Increase = x + 0.2x = 1.2x \\[3ex] 30\%\:\:of\:\:1.2x = \dfrac{30}{100} * 1.2x = 0.3 * 1.2x = 0.36x \\[3ex] 30\%\:Increase = 1.2x + 0.36x = 1.56x \\[3ex] 20\%\:\:of\:\:1.56x = \dfrac{20}{100} * 1.56x = 0.2 * 1.56x = 0.312x \\[3ex] 30\%\:Decrease = 1.56x - 0.312x = 1.248x \\[3ex] New\:\:Population = 1.248x \\[3ex] Change = New - Initial \\[3ex] Change = 1.248x - x = 0.248x \\[3ex] Change\:\:is\:\:an\:\:increase \\[3ex] \%Increase = \dfrac{Increase}{Initial} * 100 \\[5ex] \%Increase = \dfrac{0.248x}{x} * 100 \\[5ex] \%Increase = 24.8\% \\[3ex] \%Increase \approx 25\% $

$ 22\dfrac{1}{2} = \dfrac{2 * 22 + 1}{2} = \dfrac{44 + 1}{2} = \dfrac{45}{2} \\[5ex] 22\dfrac{1}{2}\% = \dfrac{22\dfrac{1}{2}\%}{100} \\[5ex] = \dfrac{\dfrac{45}{2}}{100} \\[5ex] = \dfrac{45}{2} \div \dfrac{100}{1} \\[5ex] = \dfrac{45}{2} * \dfrac{1}{100} \\[5ex] = \dfrac{9}{2} * \dfrac{1}{20} \\[5ex] = \dfrac{9 * 1}{2 * 20} \\[5ex] = \dfrac{9}{40} \\[5ex] 17\dfrac{1}{10} = \dfrac{10 * 17 + 1}{10} = \dfrac{170 + 1}{10} = \dfrac{171}{10} \\[5ex] 17\dfrac{1}{10}\% = \dfrac{17\dfrac{1}{10}\%}{100} \\[5ex] = \dfrac{\dfrac{171}{10}}{100} \\[5ex] = \dfrac{171}{10} \div \dfrac{100}{1} \\[5ex] = \dfrac{171}{10} * \dfrac{1}{100} \\[5ex] = \dfrac{171 * 1}{10 * 100} \\[5ex] = \dfrac{171}{1000} \\[5ex] \underline{Proportional\:\:Reasoning\:\:Method} \\[3ex] $

$M$	$₦$
$\dfrac{171}{1000}$	$\dfrac{9}{40}$
$1$	$p$

$ \dfrac{\dfrac{171}{1000}}{1} = \dfrac{\dfrac{9}{40}}{p} \\[7ex] Cross\:\:Multiply \\[3ex] \dfrac{171p}{1000} = 1\left(\dfrac{9}{40}\right) \\[5ex] \dfrac{171p}{1000} = \dfrac{9}{40} \\[5ex] \dfrac{1000}{171} * \dfrac{171p}{1000} = \dfrac{1000}{171} * \dfrac{9}{40} \\[5ex] p = \dfrac{1000}{171} * \dfrac{9}{40} \\[5ex] p = \dfrac{75}{57} \\[5ex] p = \dfrac{25}{19} \\[5ex] p = 1\dfrac{6}{19} \\[5ex] \therefore 1M = 1\dfrac{6}{19}₦ \\[5ex] Option\:\:C \\[3ex] 1\dfrac{18}{57} = 1\dfrac{6}{19} $

$ Let\:\:L = length\:\:of\:\:the\:\:Square \\[3ex] Let\:\:W = width\:\:of\:\:the\:\:Square \\[3ex] Area\:\:of\:\:the\:\:Square = L * W = LW \\[3ex] 20\% = \dfrac{20}{100} = 0.2 \\[3ex] 20\%\:\:increase\:\:in\:\:length = L + 0.2L = 1.2L \\[3ex] 20\%\:\:decrease\:\:in\:\:width = W - 0.2W = 0.8W \\[3ex] Area\:\:of\:\:the\:\:Rectangle = 1.2L * 0.8W = 0.96LW \\[3ex] Ratio = \dfrac{Area\:\:of\:\:the\:\:Rectangle}{Area\:\:of\:\:the\:\:Square} \\[5ex] Ratio = \dfrac{0.96LW}{LW} \\[5ex] = \dfrac{0.96}{1} \\[5ex] = 0.96 $

$ \% Error = \dfrac{Measured - Actual}{Actual} * 100 \\[5ex] \% Error = \dfrac{16.8 - 15}{15} * 100 \\[5ex] = \dfrac{1.8}{15} * 100 \\[5ex] = \dfrac{180}{15} \\[5ex] = \dfrac{60}{5} \\[5ex] = 12\% $

We can solve this question in two ways
Use any method you like

$ Let\:\:the\:\:amount\:\:of\:\:initial\:\:ice\:\:cream = C \\[3ex] \underline{First\:\:Method:\:\:Quantitative\:\:Reasoning - More\:\:Work} \\[3ex] 25\% = \dfrac{25}{100} = 0.25 \\[5ex] First\:\:Day:\:\:Eats\:\:25\%\:\:of\:\:C = 0.25 * C = 0.25C \\[3ex] Remaining:\:\: C - 0.25C = 0.75C \\[3ex] Second\:\:Day:\:\:Eats\:\:25\%\:\:of\:\:0.75C = 0.25 * 0.75C = 0.1875C \\[3ex] Remaining:\:\: 0.75C - 0.1875C = 0.5625C \\[3ex] Third\:\:Day:\:\:Eats\:\:25\%\:\:of\:\:0.5625C = 0.25 * 0.5625C = 0.140625C \\[3ex] Remaining:\:\: 0.5625C - 0.140625C = 0.421875C \\[3ex] Fourth\:\:Day:\:\:Eats\:\:25\%\:\:of\:\:0.421875C = 0.25 * 0.421875C = 0.10546875C \\[3ex] Remaining:\:\: 0.421875C - 0.10546875C = 0.31640625C \\[3ex] Fifth\:\:Day:\:\:Eats\:\:25\%\:\:of\:\:0.31640625C = 0.25 * 0.31640625C = 0.0791015625C \\[3ex] Remaining:\:\: 0.31640625C - 0.0791015625C = 0.237304688C \\[3ex] Remaining\:\:in\:\:\% = 0.237304688C * 100 = 23.7304688\%C \\[3ex] \approx 24\%C \\[4ex] \underline{Second\:\:Method:\:\:Declining\:\:Balance\:\:Method:\:\:Formula - Less\:\:Work} \\[3ex] V = N(1 - r)^t \\[4ex] N = 1 \\[3ex] r = 25\% = \dfrac{25}{100} = 0.25 \\[5ex] t = 5\:days \\[3ex] V = 1(1 - 0.25)^5 \\[4ex] V = 1(0.75)^5 \\[4ex] V = 0.237304688 \\[3ex] to\:\:\%: V = 0.237304688 * 100 \\[3ex] V = 23.7304688\% \\[3ex] V \approx 24\% $

$ Let\:\:his\:\:income = x \\[3ex] Tax-free = \dfrac{1}{8}th\:\:of\:\:x \\[5ex] = \dfrac{1}{8} * x \\[5ex] = \dfrac{1}{8}x \\[5ex] = \dfrac{1 * x}{8} \\[5ex] = \dfrac{x}{8} \\[5ex] Remainder = x - \dfrac{x}{8} \\[5ex] = \dfrac{8x}{8} - \dfrac{x}{8} \\[5ex] = \dfrac{8x - x}{8} \\[5ex] = \dfrac{7x}{8} \\[5ex] 20\%\:\:tax\:\:on\:\:Remainder \\[5ex] = \dfrac{20}{100} * \dfrac{7x}{8} \\[5ex] = \dfrac{1}{5} * \dfrac{7x}{8} \\[5ex] = \dfrac{1 * 7x}{5 * 8} \\[5ex] = \dfrac{7x}{40} \\[5ex] This\:\:is\:\:equal\:\:to\:\; 490 \\[3ex] \rightarrow \dfrac{7x}{40} = 490 \\[5ex] \dfrac{7x}{40} = \dfrac{490}{1} \\[5ex] Cross\:\:Multiply \\[3ex] 7x(1) = 40(490) \\[3ex] 7x = 40 * 490 \\[3ex] x = \dfrac{40 * 490}{7} \\[5ex] x = 40 * 70 \\[3ex] x = 2800 \\[3ex] x = ₦2,800.00 $

$ Initial\:\:Price = 1000 \\[3ex] New\:\:Price = 900 \\[3ex] Decrease = 900 - 1000 = - 100 \\[3ex] \%Decrease = \dfrac{Decrease}{Initial\:\:Price} * 100 \\[5ex] = \dfrac{100}{1000} * 100 \\[5ex] = 10\% $

We can reword this question as:
$5\%$ of what (the entire book) is $15$?
We can solve this in two ways.
Use any method you prefer.

$ Let\:\:the\:\:pages\:\:of\:\:the\:\:book = x \\[3ex] \underline{First\:\:Method:\:\:Percent-Proportion} \\[3ex] \dfrac{is}{of} = \dfrac{\%}{100} \\[5ex] \dfrac{15}{x} = \dfrac{5}{100} \\[5ex] Cross\:\:Multiply \\[3ex] x * 5 = 15 * 100 \\[3ex] x = \dfrac{15 * 100}{5} \\[5ex] x = 3 * 100 \\[3ex] x = 300\:\:pages \\[3ex] \underline{First\:\:Method:\:\:Percent-Equation} \\[3ex] 5\%\:\:of\:\:what\:\:is\:\:15? \\[3ex] \dfrac{5}{100} * x = 15 \\[5ex] \dfrac{100}{5} * \dfrac{5}{100} * x = \dfrac{100}{5} * 15 \\[5ex] x = 100 * 3 \\[3ex] x = 300\:\:pages $

$ (a.) \\[3ex] Sticker\:\:Price = 900 \\[3ex] 30\%\:\:of\:\:Sticker\:\:Price = \dfrac{30}{100} * 900 = 30 * 9 = 270 \\[5ex] 30\%\:\:Increase = 900 + 270 = 1170 \\[3ex] New\:\:Sticker\:\:Price = \$1,170.00 \\[3ex] (b.) \\[3ex] 15\%\:\:of\:\:Sticker\:\:Price = \dfrac{15}{100} * 1170 = 175.5 \\[5ex] Discount = \$175.50 \\[3ex] (c.) \\[3ex] 15\%\:\:off = 1170 - 175.5 = 994.5 \\[3ex] New\:\:Sticker\:\:Price = \$994.50 \\[3ex] (d.) \\[3ex] Profit = Selling\:\:Price - Cost\:\:Price \\[3ex] = 994.50 - 900 \\[3ex] = 94.5 \\[3ex] Profit = \$94.50 \\[3ex] (e.) \\[3ex] \% Profit = \dfrac{Profit}{Cost\:\:Price} * 100 \\[5ex] = \dfrac{94.5}{900} * 100 \\[5ex] = \dfrac{94.5}{9} \\[5ex] = 10.5\% \\[3ex] Profit = 10.5\% $

$ Let\:\:the\:\:\%\:\:number\:\:of\:\:women = w \\[3ex] 22\dfrac{1}{2} + 47\dfrac{1}{2} + w = 100...Total\:\:Percent \\[5ex] 22 + \dfrac{1}{2} + 47 + \dfrac{1}{2} + w = 100 \\[5ex] 69 + 1 + w = 100 \\[3ex] 70 + w = 100 \\[3ex] w = 100 - 70 \\[3ex] w = 30\% \\[3ex] Let\:\:the\:\:number\:\:of\:\:people\:\:in\:\:the\:\:hall = p \\[3ex] \rightarrow 30\%\:\:of\:\:x = 84...number\:\:of\:\:women \\[3ex] \dfrac{30}{100} * x = 84 \\[5ex] \dfrac{100}{30} * \dfrac{30}{100} * x = \dfrac{100}{30} * 84 \\[5ex] x = \dfrac{10 * 84}{3} \\[5ex] x = 10 * 28 \\[3ex] x = 280\:\:people \\[3ex] Number\:\:of\:\:men = 47\dfrac{1}{2}\% \:\:of\:\:280 \\[5ex] = 47\dfrac{1}{2} \div 100 * 280 \\[5ex] = \dfrac{95}{2} \div \dfrac{100}{1} * 280 \\[5ex] = \dfrac{95}{2} * \dfrac{1}{100} * 280 \\[5ex] = \dfrac{95 * 1 * 280}{2 * 100} \\[5ex] = 19 * 7 \\[3ex] = 133\:\:men $

We shall use two different depreciation methods to solve this question

The Declining Balance Method and the Straight-Line Depreciation Method

$ \underline{Declining\:\:Balance\:\:Method:\:\:Manually} \\[3ex] 8\% = \dfrac{8}{100} \\[5ex] At\:\:the\:\:end\:\:of\:\:1st\:\:year \\[3ex] Depreciation = 8\%\:\:of\:\:65000 = \dfrac{8}{100} * 65000 = 8 * 650 = 5200 \\[5ex] Value = 65000 - 5200 = 59800 \\[3ex] At\:\:the\:\:end\:\:of\:\:2nd\:\:year \\[3ex] Depreciation = 8\%\:\:of\:\:59800 = \dfrac{8}{100} * 59800 = 8 * 598 = 4784 \\[5ex] Value = 59800 - 4784 = 55016 \\[3ex] Value = \$55,016.00 \\[3ex] \underline{Declining\:\:Balance\:\:Method:\:\:Formula} \\[3ex] V = N(1 - r)^t \\[4ex] N = \$65000 \\[3ex] r = 8\% = \dfrac{8}{100} = 0.08 \\[5ex] t = 2\:years \\[3ex] V = 65000(1 - 0.08)^2 \\[4ex] V = 65000(0.92)^2 \\[4ex] V = 65000(0.8464) \\[3ex] V = 55016 \\[3ex] Value = \$55,016.00 \\[5ex] \underline{Straight-Line\:\:Depreciation\:\:Method:\:\:Formula} \\[3ex] V = N(1 - rt) \\[4ex] N = \$65000 \\[3ex] r = 8\% = \dfrac{8}{100} = 0.08 \\[5ex] t = 2\:years \\[3ex] V = 65000(1 - 0.08(2)) \\[3ex] V = 65000(1 - 0.16) \\[3ex] V = 65000(0.84) \\[3ex] V = 54600 \\[3ex] Value = \$54,600.00 $

$ Let\:\:Mary's\:\:paycheck = p \\[3ex] Let\:\:Martha's\:\:paycheck = k \\[3ex] 30\% = \dfrac{30}{100} = 0.3 \\[5ex] 30\%\:\:of\:\:p = 0.3 * p = 0.3p \\[3ex] Remaining:\:\: p - 0.3p \\[3ex] \rightarrow k = p - 0.3p \\[3ex] k = 0.7p \\[3ex] k + p = 665 \\[3ex] \rightarrow 0.7p + p = 665 \\[3ex] 1.7p = 665 \\[3ex] p = \dfrac{665}{1.7} \\[5ex] p = 391.176471 \\[3ex] p \approx 391.18 \\[3ex] k = 0.7p = 0.7(391.176471) \\[3ex] k = 273.82353 \\[3ex] k \approx 273.82 \\[3ex] $ Mary's weekly paycheck is $\$391.18$ while Martha's weekly paycheck is $\$273.82$

Based on the histogram:

$2$ adults had no child

$8$ adults had one child

$5$ adults had $2$ children

$2$ adults had $3$ children

No adult had $4$ children

$2$ adults had $5$ children

$ (a.) \\[3ex] Total\:\:Number\:\:of\:\:adults = 2 + 8 + 5 + 2 + 0 + 2 = 19\:\:adults \\[3ex] (b.) \\[3ex] \%\:\:of\:\:adults\:\:with\:\:one\:\:child \\[3ex] = \dfrac{Number\:\:of\:\:adults\:\:with\:\:one\:\:child}{Total\:\:number\:\:of\:\:adults} * 100 \\[5ex] = \dfrac{8}{19} * 100 \\[5ex] = \dfrac{800}{19} \\[5ex] = 42.1052630\% $

$ Amount\:\:spent\:\:on\:\:Fun \\[3ex] = 16\%\:\:of\:\:2500 \\[3ex] = \dfrac{16}{100} * 2500 \\[5ex] = 16 * 25 \\[3ex] = 400 \\[3ex] $ Judith spent $\$400.00$ on Fun activities.

Less than $\$100$ "on average" includes the percent of adults who spend less than $\$50$ "on average" as well as the percent of adults who spend no less than $\$50$ and no more than $\$99$ "on average"

$ Less\:\:than\:\:\$100\:\:on\:\:average \\[3ex] = 8\% + 17\% \\[3ex] = 25\% \\[3ex] = \dfrac{25}{100} \\[5ex] = 0.25...Option\:\:E \\[3ex] $ NOTE:

$ 25 \ne 25\% \\[3ex] 0.25 = 25\% $

$ At\:\:least\:\:\$150\:\:means \ge \$150 \\[3ex] \%\:\:of\:\:adults\:\:that\:\:spend\:\:\ge \$150 \\[3ex] = 15\% + 21\% + 10\% \\[3ex] = 46\% $

$ Frequency\:\:for\:\:\$300\:\:or\:\:more = 10\% \\[3ex] Total\:\:Frequency, \Sigma F = 100\% \\[3ex] Sectorial\:\angle \:\:for\:\:\$300\:\:or\:\:more \\[3ex] = \dfrac{Frequency\:\:for\:\:\$300\:\:or\:\:more}{Total\:\:Frequency} * 360^\circ \\[5ex] = \dfrac{10}{100} * 360 \\[5ex] = 1 * 36 \\[3ex] = 36\^circ $

This is a somewhat "tricky" question.
We cannot just assume that the trader will make a loss by selling the $20$ oranges for $₦1.10$
This is because we do not know how much he bought those $20$ oranges
So, we need to find the cost of those $20$ oranges first.
We find the cost price based on the first sentence that was given...based on what we know...that he made an $8\%$ profit by selling the $20$ oranges for $₦1.35$
Then, we determine if that cost is greater than or less than $₦1.10$ to find if he made a profit or loss.
Then, we calculate the percentage gain or loss.

$ \underline{First:\:\:Determine\:\:the\:\:Cost\:\:Price} \\[3ex] Let\:\:C = cost\:\:price \\[3ex] Let\:\:S = selling\:\:price \\[3ex] \%Profit = \dfrac{S - C}{C} \\[5ex] S = ₦1.35 \\[3ex] \%Profit = 8\% \\[3ex] 8\% = \dfrac{1.35 - C}{C}...\%\:\:is\:\:given \\[5ex] \dfrac{8}{100} = \dfrac{1.35 - C}{C} \\[5ex] Cross\:\:Multiply \\[3ex] 8 * C = 100(1.35 - C) \\[3ex] 8C = 135 - 100C \\[3ex] 8C + 100C = 135 \\[3ex] 108C = 135 \\[3ex] C = \dfrac{135}{108} = \dfrac{15}{12} = \dfrac{5}{4} \\[5ex] C = ₦1.25 \\[3ex] S = ₦1.10 \\[3ex] \underline{Second:\:\:Profit\:\:or\:\:Loss} \\[3ex] ₦1.25 \gt ₦1.10...he\:\:made\:\:a\:\:loss \\[3ex] He\:\:bought\:\:it\:\:for\:\:more\:\:than\:\:what\:\:he\:\:sold\:\:it \\[3ex] \underline{Third:\:\:Calculate\:\:the\:\:\%\:\:Loss} \\[3ex] \%Loss = \dfrac{C - S}{C} * 100...\%\:\:needs\:\:be\:\:found \\[5ex] \%Loss = \dfrac{1.25 - 1.10}{1.25} * 100 \\[5ex] = \dfrac{0.15}{1.25} * 100 \\[5ex] = \dfrac{15}{125} * 100 \\[5ex] = \dfrac{3}{25} * 100 \\[5ex] = 3 * 4 \\[3ex] = 12\% \\[3ex] $ Be selling the $20$ oranges for $₦1.10$, he made a loss of $12\%$

We can do this question in two ways.
Choose any method you prefer.

$ \underline{First\:\:Method:\:\:Quantitative\:\:Reasoning} \\[3ex] Mark\:\:obtained\:\:for\:\:01\:\:Paper \\[3ex] = 55\%\:\:of\:\:30 \\[3ex] = \dfrac{55}{100} * 30 \\[5ex] = \dfrac{11 * 3}{2} \\[5ex] = 16.5 \\[3ex] Mark\:\:obtained\:\:for\:\:02\:\:Paper \\[3ex] = 60\%\:\:of\:\:50 \\[3ex] = \dfrac{60}{100} * 50 \\[5ex] = 6 * 5 \\[3ex] = 30 \\[3ex] Mark\:\:obtained\:\:for\:\:03\:\:Paper \\[3ex] = 80\%\:\:of\:\:20 \\[3ex] = \dfrac{80}{100} * 20 \\[5ex] = 8 * 2 \\[3ex] = 16 \\[3ex] Final\:\:Mark = 16.5 + 30 + 16 = 62.5\:\:out\:\:of\:\:100 \\[3ex] \dfrac{62.5}{100} = 62.5\% \\[3ex] Final\:\:Mark = 62.5\% \\[3ex] \underline{Second\:\:Method:\:\:Weighted\:\:Average\:\:Method} \\[3ex] $ This is similar to the Grading Method Mr. C used for your class.

Assessment	Weight	Your Score	Weighted Score
$01$	$30$	$55$	$30 * 55 = 1650$
$02$	$50$	$60$	$50 * 60 = 3000$
$03$	$20$	$80$	$20 * 80 = 1600$

$ \Sigma Weighted\:\:Scores = 1650 + 3000 + 1600 = 6250 \\[3ex] \Sigma Weights = 30 + 50 + 20 = 100 \\[3ex] Final\:\:Grade = \dfrac{\Sigma Weighted\:\:Scores}{\Sigma Weights} \\[5ex] = \dfrac{6250}{100} \\[5ex] = 62.5\% $

$ Let\:\:R = red \\[3ex] Let\:\:B = blue \\[3ex] Let\:\:G = green \\[3ex] Let\:\:P = purple \\[3ex] Let\:\: K = Other\:\:sector \\[3ex] S = sample\:\:space \\[3ex] n(S) = 100\% \\[3ex] n(R) = 25\% \\[3ex] n(B) = 30\% \\[3ex] n(G) = 20\% \\[3ex] n(P) = 10\% \\[3ex] n(K) = 100 - (25 + 30 + 20 + 10) \\[3ex] n(K) = 100 - 85 = 15\% \\[3ex] Sectorial\:\angle \:\:for\:\:K \\[3ex] \dfrac{n(K)}{n(S)} * 360 \\[5ex] = \dfrac{15}{100} * 360 \\[5ex] = \dfrac{3}{2} * 36 \\[5ex] = 3 * 18 \\[3ex] = 54^\circ \\[3ex] $ The sectorial angle for the Other sector is $54^\circ$

$ Cost\:\:price\:\:for\:\:250\:\:oranges = D1000 \\[3ex] Cost\:\:price\:\:for\:\:1\:\:orange = \dfrac{1000}{250} = D4 \\[5ex] Kept\:\:for\:\:himself:\:\: 20\% \:\:of\:\: 250 \\[3ex] = \dfrac{20}{100} * 250 \\[5ex] = \dfrac{250}{5} \\[5ex] = 50\:\:oranges \\[3ex] Remaining:\:\: 250 - 50 = 200\:\:oranges \\[3ex] Sold:\:\: 115\:\:oranges\:\:@\:\:D6.50\:\:each = 115(6.5) = D747.5 \\[3ex] Remaining:\:\: 200 - 115 = 85\:\:oranges \\[3ex] Sold:\:\: 85\:\:oranges\:\:@\:\:D5.00\:\:each = 85(5) = D425 \\[3ex] \underline{For\:\:the\:\:oranges\:\:sold} \\[3ex] 200\:\:oranges \\[3ex] Cost\:\:price\:\:for\:\:200\:\:oranges\:\:@\:\:D5.00\:\:each = 200(4) = D800 \\[3ex] Selling\:\:price = D747.5 + D425 = D1172.5 \\[3ex] Profit = Selling\:\:price - Cost\:\:price \\[3ex] = D1172.5 - D800 = D372.5 \\[3ex] \%\:\:Profit = \dfrac{Profit}{Cost\:\:price} * 100 \\[5ex] = \dfrac{372.5}{800} * 100 \\[5ex] = \dfrac{372.5}{8} \\[5ex] = 46.5625\% $

$ Bill = \$32 \\[3ex] Tip\:\:amount \\[3ex] = 20\%\:\:of\:\:32 \\[3ex] = \dfrac{20}{100} * 32 \\[5ex] = \dfrac{32}{5} \\[5ex] = 6.4 \\[3ex] Check-out\:\:bill = 32 + 6.4 = 38.4 \\[3ex] Check-out\:\:bill = \$38.40 \\[3ex] $ Lucy has to pay $\$38.40$

We can do this question in two ways.
Use any method you prefer.

$ \underline{First\:\:Method:\:\:Arithmetically - Long\:\:Way} \\[3ex] \underline{2003} \\[3ex] Population = 3000 \\[3ex] \underline{Beginning\:\:of\:\:2003} \\[3ex] GH¢15.00\:\:per\:\:head = 3000(15) = GH¢45000 \\[3ex] 20\%\:\:increase\:\:during\:\:2003 \\[3ex] = 20\%\:\:of\:\:3000 \\[3ex] = \dfrac{20}{100} * 3000 \\[5ex] = 20(30) \\[3ex] = 600 \\[3ex] Population = 3000 + 600 = 3600 \\[3ex] \underline{Beginning\:\:of\:\:2004} \\[3ex] GH¢15.00\:\:per\:\:head = 3600(15) = GH¢54000 \\[3ex] 20\%\:\:increase\:\:during\:\:2004 \\[3ex] = 20\%\:\:of\:\:3600 \\[3ex] = \dfrac{20}{100} * 3600 \\[5ex] = 20(36) \\[3ex] = 720 \\[3ex] Population = 3600 + 720 = 4320 \\[3ex] \underline{Beginning\:\:of\:\:2005} \\[3ex] GH¢15.00\:\:per\:\:head = 4320(15) = GH¢64800 \\[3ex] 20\%\:\:increase\:\:during\:\:2005 \\[3ex] = 20\%\:\:of\:\:4320 \\[3ex] = \dfrac{20}{100} * 4320 \\[5ex] = 2(432) \\[3ex] = 864 \\[3ex] Population = 4320 + 864 = 5184 \\[3ex] \underline{Beginning\:\:of\:\:2006} \\[3ex] GH¢15.00\:\:per\:\:head = 5184(15) = GH¢77760 \\[3ex] 20\%\:\:increase\:\:during\:\:2006 \\[3ex] = 20\%\:\:of\:\:5184 \\[3ex] = \dfrac{20}{100} * 5184 \\[5ex] = 0.2(5184) \\[3ex] = 1036.8 \\[3ex] Population = 5184 + 1036.8 = 6220.8 \approx 6221...no\:\:decimal\:\:human\:\:being \\[3ex] \underline{Beginning\:\:of\:\:2007} \\[3ex] GH¢15.00\:\:per\:\:head = 6221(15) = GH¢93315 \\[3ex] Total\:\:Grant\:\:from\:\:2003\:\:to\:\:2007 \\[3ex] = 45000 + 54000 + 64800 + 77760 + 93315 \\[3ex] = GH¢334,875 \\[3ex] $

Let the levels of carbon IV oxide in parts per million be $C$
Let parts per million = $ppm$

$ \underline{1959} \\[3ex] Initial\:\:C = 304ppm \\[3ex] \underline{2005} \\[3ex] New\:\:C = 379ppm \\[3ex] Increase = New\:\:C - Initial\:\:C \\[3ex] = 379 - 304 \\[3ex] = 75ppm \\[3ex] \%Increase = \dfrac{Increase}{Initial\:\:C} * 100 \\[5ex] = \dfrac{75}{304} * 100 \\[5ex] = \dfrac{7500}{304} \\[5ex] = 24.6710526\% \\[3ex] \approx 24.7\% \\[3ex] $ The increase is $75$ parts per million
The percent increase is about $24.7\%$

$ Cost\:\:Price\:\:for\:\:10\:\:dozen\:\:eggs\:\:@\:\:₦300\:\:per\:\:dozen \\[3ex] = 10(300) \\[3ex] = ₦3000 \\[3ex] Number\:\:for\:\:10\:\:dozen\:\:eggs\:\:@\:\:12\:\:eggs\:\:per\:\:dozen \\[3ex] = 10(12) \\[3ex] = 120\:\:eggs \\[3ex] But:\:\: 20\:\:eggs\:\:were\:\:broken \\[3ex] Remaining:\:\: 120 - 20 = 100\:\:eggs \\[3ex] Profit = 10\% = \dfrac{10}{100}...already\:\:given \\[5ex] \%Profit = \dfrac{Selling\:\:Price - Cost\:\:Price}{Cost\:\:Price} \\[5ex] \dfrac{10}{100} = \dfrac{Selling\:\:Price - 3000}{3000} \\[5ex] Cross\:\:Multiply \\[3ex] 100(Selling\:\:Price - 3000) = 10(3000) \\[5ex] Selling\:\:Price - 3000 = \dfrac{10 * 3000}{100} \\[5ex] Selling\:\:Price - 3000 = 10 * 30 \\[3ex] Selling\:\:Price - 3000 = 300 \\[3ex] Selling\:\:Price = 300 + 3000 \\[3ex] Selling\:\:Price = 3300 \\[3ex] $ He sold the remaining $100$ eggs ($8$ dozen eggs and $4$ eggs) for $₦3,300$ in order to make a profit of $10\%$

$ Let\:\:her\:\:salary\:\:last\:\:year = p \\[3ex] 3\%\:\:raise\:\:of\:\:p \\[5ex] = \dfrac{3}{100} * p \\[5ex] = \dfrac{3p}{100} = 0.03p \\[5ex] Current\:\:salary = p + 0.03p = 1.03p \\[3ex] \rightarrow 1.03p = 68804 \\[3ex] p = \dfrac{68804}{1.03} \\[5ex] p = 66800 \\[3ex] $ Hannah's salary last year was $\$66,800.00$

$ \underline{Publisher} \\[3ex] Number\:\:of\:\:copies\:\:printed = 30000 \\[3ex] Cost\:\:price = 30000\:\:@\:\:GH¢2.00\:\:each = 30000(2) = GH¢60000 \\[3ex] But\:\:25,380\:\:copies\:\:were\:\:sold\:\:@\:\:GH¢2.76\:\:each \\[3ex] Selling\:\:price = 25380\:\:@\:\:GH¢2.76\:\:each = 25380(2.76) = GH¢70048.8 \\[3ex] \underline{Paid\:\:to\:\:Author} \\[3ex] First\:\:Payment:\:\:10\%\:\:of\:\:selling\:\:price\:\:of\:\:first\:\:6000\:\:copies \\[3ex] Selling\:\:price\:\:for\:\:first\:\:6000\:\:copies = 6000\:\:@\:\:GH¢2.76\:\:each = 6000(2.76) = GH¢16560 \\[3ex] First\:\:Payment = 10\%\:\:of\:\:16560 \\[3ex] = \dfrac{10}{100} * 16560 \\[5ex] = 1 * 1656 \\[3ex] = GH¢1656 \\[3ex] Second\:\:Payment:\:\:12\dfrac{1}{2}\%\:\:of\:\:selling\:\:price\:\:of\:\:all\:\:copies\:\:in\:\:excess\:\:of\:\:6000\:\:copies \\[3ex] All\:\:copies\:\:in\:\:excess\:\:of\:\:6000 = Remaining\:\:copies \\[3ex] Remaining\:\:copies = 25380 - 6000 = 19380 \\[3ex] Selling\:\:price\:\:for\:\:remaining\:\:19380\:\:copies = 19380\:\:@\:\:GH¢2.76\:\:each = 19380(2.76) = GH¢53488.8 \\[3ex] Second\:\:Payment = 12\dfrac{1}{2}\%\:\:of\:\:53488.8 \\[5ex] = 12.5\%\:\:of\:\:53488.8 \\[3ex] = \dfrac{12.5}{100} * 53488.8 \\[5ex] = 0.125 * 53488.8 \\[3ex] = GH¢6,686.1 \\[3ex] (i) \\[3ex] Total\:\:amount\:\:received\:\:by\:\:Author = 1656 + 6686.1 \\[3ex] = GH¢8342.10 \\[3ex] (ii) \\[3ex] Net\:\:profit\:\:made\:\:by\:\:Publisher = Selling\:\:price - (Cost\:\:price + Total\:\:amount\:\:paid\:\:to\:\:Author) \\[3ex] = 70048.8 - (60000 + 8342.1) \\[3ex] = 70048.8 - 68342.1 \\[3ex] = GH¢1,706.70 \\[3ex] (b.) \\[3ex] What\:\:\%\:\:of\:\:the\:\:Author's\:\:total\:\:receipt\:\:is\:\:the\:\:Publisher's\:\:net\:\:profit? \\[3ex] Let\:\:the\:\:\% = p \\[3ex] \underline{Percent-Proportion\:\:Method} \\[3ex] \dfrac{is}{of} = \dfrac{p}{100} \\[5ex] \dfrac{1706.7}{8342.1} = \dfrac{p}{100} \\[5ex] Cross\:\:multiply \\[3ex] 8342.1 * p = 1706.7 * 100 \\[3ex] p = \dfrac{1706.7 * 100}{8342.1} \\[5ex] p = \dfrac{170670}{8342.1} \\[5ex] p = 20.4588773\% \\[3ex] p \approx 20.5\% \\[3ex] $ The publisher's net profit as a percentage of the author's total receipt is about $20.5\%$

$ Oxygen\:\:saturation\:\:level \\[3ex] = \dfrac{7.3}{9.8} * 100 \\[5ex] = \dfrac{7.3 * 100}{9.8} \\[5ex] = \dfrac{730}{9.8} \\[5ex] = 74.4897959 \\[3ex] \approx 74\% \\[3ex] $ The oxygen saturation level is approximately $74\%$.

$ \underline{Nonstudents} \\[3ex] Initially\:\:approved = 85 \\[3ex] Error = 15 \\[3ex] Correctly\:\:approved = 85 - 15 = 70 \\[3ex] \underline{Approved} \\[3ex] Total = 129 \\[3ex] Nonstudents = 70 \\[3ex] High\:\:school\:\:and\:\:College\:\:students = 129 - 70 = 59 \\[3ex] \underline{College\:\:students} \\[3ex] Disapprove = 10 \\[3ex] No\:\:opinion = 6 \\[3ex] Approved = ???\:\:(changed\:\:because\:\:of\:\:the\:\:error\:\:with\:\:misidentification\:\:of\:\:Nonstudents) \\[3ex] Total = ???\:\:changed\:\:as\:\:well \\[3ex] Let\:\:the\:\:total\:\:number\:\:of\:\:College\:\:students = p \\[3ex] Exactly\:\:60\%\:\:of\:\:p\:\:Approved \\[3ex] This\:\:implies\:\:that\:\:(100\% - 60\%)\:\:did\:\:not\:\:Approve \\[3ex] Exactly\:\:40\%\:\:of\:\:p\:\:Disapprove\:\:and\:\:No\:\:opinion \\[3ex] Disapprove\:\:and\:\:No\:\:opinion = 10 + 6 = 16 \\[3ex] 40\%\:\:of\:\:p = 16 \\[3ex] \dfrac{40}{100} * p = 16 \\[3ex] 0.4 * p = 16 \\[3ex] p = \dfrac{16}{0.4} \\[5ex] p = 40\:\:students \\[3ex] Total = 40\:\:students \\[3ex] Approved = 40 - 16 = 24\:\:students \\[3ex] \underline{High\:\:school\:\:students} \\[3ex] High\:\:school\:\:and\:\:College\:\:students\:\:who\:\:Approved = 59 \\[3ex] College\:\:students\:\:who\:\:Approved = 24 \\[3ex] \therefore High\:\:school\:\:students\:\:who\:\:Approved = 59 - 24 = 35 \\[3ex] Disapprove = 4 \\[3ex] No\:\:opinion = 11 \\[3ex] Total = 35 + 4 + 11 = 50 \\[3ex] \%\:\:who\:\:Approved \\[3ex] = \dfrac{Approved}{Total} * 100 \\[5ex] = \dfrac{35}{50} * 100 \\[5ex] = 35(2) \\[3ex] = 70\% $

$ \underline{1-point\:\:free\:\:throw} \\[3ex] Number\:\:of\:\:points \\[3ex] = 75\%\:\:of\:\:80 \\[3ex] = \dfrac{75}{100} * 80 \\[5ex] = 0.75 * 80 \\[3ex] = 60\:points \\[3ex] \underline{2-point\:\:free\:\:throw} \\[3ex] Number\:\:of\:\:points \\[3ex] = 60\%\:\:of\:\:90 \\[3ex] = \dfrac{60}{100} * 90 \\[5ex] = 0.6 * 90 \\[3ex] = 54\:\:(2-points) \\[3ex] = 54(2) = 108\:points \\[3ex] \underline{3-point\:\:free\:\:throw} \\[3ex] Number\:\:of\:\:points \\[3ex] = 60\%\:\:of\:\:25 \\[3ex] = \dfrac{60}{100} * 25 \\[5ex] = 0.6 * 25 \\[3ex] = 15\:\:(3-points) \\[3ex] = 15(3) = 45\:points \\[3ex] Total\:\:Number\:\:of\:\:points = 60 + 108 + 45 = 213\:\:points $

$ Number\:\:of\:\:bracelets = 165 \\[3ex] Cost\:\:price\:\:of\:\:all\:\:bracelets = \$6800 \\[3ex] Customs\:\:duty = \$1360 \\[3ex] (i) \\[3ex] Total\:\:cost = \$6800 + \$1360 = \$8160 \\[3ex] (ii) \\[3ex] 165\:\:bracelets\:\:@\:\:\$68.85\:\:per\:\:bracelet \\[3ex] Selling\:\:price\:\:of\:\:all\:\:bracelets = 165(68.85) = \$11360.25 \\[3ex] (a.) \\[3ex] Total\:\:profit = \$11360.25 - \$8160 = \$3200.25 \\[3ex] (b.) \\[3ex] \%Profit\:\:based\:\:on\:\:cost\:\:price \\[3ex] \%Profit = \dfrac{Profit}{Cost\:\:price} * 100 \\[5ex] = \dfrac{3200.25}{6800} * 100 \\[5ex] = \dfrac{320025}{6800} \\[5ex] = 47.0625\% \\[3ex] \approx 47\% $

From the Bar Graph,
$1$ student earned $0$
$3$ students earned $1$
$4$ students earned $2$
$7$ student earned $3$
$10$ students earned $4$
$8$ students earned $5$
$7$ student earned $6$
$9$ students earned $7$
$8$ students earned $8$
$2$ students earned $9$
$1$ students earned $10$

$ Number\:\:of\:\:Students = 60 \\[3ex] Verify:\:\: 1 + 3 + 4 + 7 + 10 + 8 + 7 + 9 + 8 + 2 + 1 = 60 \\[3ex] \underline{Below\:\:Pass\:\:Mark} \\[3ex] Marks = 4, 3, 2, 1, 0 \\[3ex] Number\:\:of\:\:students = 10 + 7 + 4 + 3 + 1 = 25 \\[3ex] \%\:\:of\:\:students \\[3ex] = \dfrac{25}{60} * 100 \\[5ex] = \dfrac{25}{3} * 5 \\[5ex] = \dfrac{25 * 5}{3} \\[5ex] = \dfrac{125}{3} \\[5ex] = 41.6666667\% \\[3ex] \approx 41.7\% \\[3ex] $ About $41.7\%$ of students failed the test.

The refurbishing cost should be included in the cost price

Cost price = $250000 + 70000 = 320000$

Selling Price = $400000$

Gain = Selling Price - Cost Price

Gain = $400000 - 320000 = 80000$

$ \% Gain = \dfrac{Gain}{Cost\:\:Price} * 100 \\[5ex] \% Gain = \dfrac{80000}{320000} * 100 \\[5ex] \% Gain = \dfrac{100}{4} \\[5ex] \% Gain = 25\% $

Area of a rectangle = Length * Width

Let the length of the original rectangle = $L$

Let the width of the original rectangle = $W$

Area of the original rectangle = $L * W$

Length of the new rectangle/resulting rectangle = $L + 25\% = L + 0.25L = 1.25L$

Width of the new rectangle/resulting rectangle = $W - 10\% = W - 0.10W = 0.9W$

Area of the new rectangle/resulting rectangle = $(1.25L)(0.9W) = 1.125LW$

Area of resulting rectangle - Area of original rectangle = $1.125LW - LW = 0.125LW$

$0.125LW = 0.12LW * 100 = 12.5\%LW$

Area of resulting rectangle is larger than the Area of original rectangle by $12.5\%$

Student: Why did you subtract?
Teacher: Because of what the question asked: By how much is the area of the resulting rectangle larger than the area of the original rectangle?

$ Discount\:\:Price = Sale\:\:Price \\[3ex] List\:\:Price = 1000 \\[3ex] Discount = 20\%\:\:of\:\:1000 \\[3ex] = \dfrac{20}{100} * 1000 \\[5ex] = 20(10) = 200 \\[3ex] Sale\:\:Price = 20\%\:\:off\:\:1000 \\[3ex] = 1000 - 200 = 800 \\[3ex] Tax = 6\%\:\:of\:\:800 \\[3ex] = \dfrac{6}{100} * 800 \\[5ex] = 6(8) = 48 \\[3ex] Total\:\:Price = 800 + 48 = \$848 $

$ First\:\:Discount \\[5ex] = 40\%\:\:of\:\: 10 \\[5ex] = \dfrac{40}{100} * 10 \\[5ex] = 0.4(10) = 4 \\[3ex] First\:\:Sale\:\:Price = 40\%\:\:off\:\:10 \\[3ex] = 10 - 4 = 6 \\[3ex] Second\:\:Discount = Coupon \\[3ex] = 10\%\:\:of\:\:6 \\[3ex] = \dfrac{10}{100} * 6 \\[5ex] = 0.1(6) = 0.6 \\[3ex] Second\:\:Sale\:\:Price = 10\%\:\:off\:\:6 \\[3ex] = 6 - 0.6 = 5.4 \\[3ex] Final\:\:Sale\:\:Price = \$5.40 $

We can solve this question in at least two ways.
Use any method you prefer

$ Length\:\:of\:\:south\:\:side = 40\:inches \\[3ex] Length\:\:of\:\:north\:\:side = 28\:inches \\[3ex] 40\:inches\:\:is\:\:what\:\:percent\:\:of\:\:28\:inches? \\[3ex] Let\:\:the\:\:percent = p \\[3ex] \underline{First\:\:Method:\:\:Percent-Proportion} \\[3ex] \dfrac{is}{of} = \dfrac{\%}{100} \\[5ex] \dfrac{40}{28} = \dfrac{p}{100} \\[5ex] Cross\:\:Multiply \\[3ex] 28 * p = 40 * 100 \\[3ex] p = \dfrac{40 * 100}{28} \\[5ex] p = \dfrac{10 * 100}{7} \\[5ex] p = \dfrac{1000}{7} \\[5ex] p = 142\dfrac{6}{7} \\[5ex] p = 142\dfrac{6}{7}\% \\[5ex] \underline{Second\:\:Method:\:\:Percent-Equation} \\[3ex] 40\:inches\:\:is\:\:what\:\:percent\:\:of\:\:28\:inches? \\[3ex] Let\:\:the\:\:percent = p \\[3ex] 40 = p\% * 28 \\[3ex] 40 = \dfrac{p}{100} * 28 \\[5ex] Swap \\[3ex] \dfrac{p}{100} * 28 = 40 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\: \dfrac{100}{28} \\[5ex] \dfrac{100}{28} * \dfrac{p}{100} * 28 = \dfrac{100}{28} * 40 \\[5ex] p = \dfrac{100 * 40}{28} \\[5ex] p = 142\dfrac{6}{7} \\[5ex] p = 142\dfrac{6}{7}\% \\[5ex] $ The length of the south side of the park is $142\dfrac{6}{7}\%$ of the length of the north side.

We can solve this question in at least two ways.
Use any method you prefer

$ At\:\:the\:\:beginning\:\:of\:\:2014: \\[3ex] Number\:\:of\:\:lions = 200 \\[3ex] Total\:\:number\:\:of\:\:large\:\:animals = 2400 \\[3ex] 200\:\:was\:\:p\:\:percent\:\:of\:\:2400? \\[3ex] \underline{First\:\:Method:\:\:Percent-Proportion} \\[3ex] \dfrac{is}{of} = \dfrac{\%}{100} \\[5ex] \dfrac{200}{2400} = \dfrac{p}{100} \\[5ex] Cross\:\:Multiply \\[3ex] 2400 * p = 200 * 100 \\[3ex] p = \dfrac{200 * 100}{2400} \\[5ex] p = \dfrac{100 * 2}{24} \\[5ex] p = \dfrac{200}{24} \\[5ex] p = 8.33333333 \\[3ex] p \approx 8\% \\[3ex] \underline{Second\:\:Method:\:\:Percent-Equation} \\[3ex] 200\:\:was\:\:p\:\:percent\:\:of\:\:2400? \\[3ex] 200 = p\% * 2400 \\[3ex] 200 = \dfrac{p}{100} * 2400 \\[5ex] Swap \\[3ex] \dfrac{p}{100} * 2400 = 200 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\: \dfrac{100}{2400} \\[5ex] \dfrac{100}{2400} * \dfrac{p}{100} * 2400 = \dfrac{100}{2400} * 200 \\[5ex] p = \dfrac{100 * 2}{24} \\[5ex] p = \dfrac{200}{24} \\[5ex] p = 8.33333333 \\[3ex] p \approx 8\% \\[3ex] $ The number of lions in the park was approximately $8\%$ percent of the total number of large animals.

$ Let\:\:the\:\:father's\:\:income = p \\[3ex] 20\% = \dfrac{20}{100} = \dfrac{1}{5} \\[5ex] Monthly\:\:allowance\:\:given\:\:to\:\:his\:\:three\:\:children = 20\%\:\:of\:\:p \\[3ex] = \dfrac{1}{5} * p \\[5ex] = \dfrac{1}{5}p \\[5ex] Eldest\:\:child's\:\:share = 45\%\:\:of\:\:\dfrac{1}{5}p \\[5ex] = \dfrac{45}{100} * \dfrac{1}{5}p \\[5ex] = \dfrac{9}{100}p \\[5ex] Youngest\:\:child's\:\:share = 25\%\:\:of\:\:\dfrac{1}{5}p \\[5ex] = \dfrac{25}{100} * \dfrac{1}{5}p \\[5ex] = \dfrac{5}{100}p \\[5ex] = \dfrac{1}{20}p \\[5ex] Remaining = Second\:\:child's\:\:share \\[3ex] = 100\% - (45\% + 25\%) \\[3ex] = 100\% - 70\% \\[3ex] = 30\% \\[3ex] Second\:\:child's\:\:share = 30\%\:\:of\:\:\dfrac{1}{5}p \\[5ex] = \dfrac{30}{100} * \dfrac{1}{5}p \\[5ex] = \dfrac{6}{100}p \\[5ex] = \dfrac{3}{50}p \\[5ex] Second\:\:child's\:\:share = ₦3,000 \\[3ex] \implies \dfrac{3}{50}p = 3000 \\[5ex] Multiply\:\:both\:\:sides\:\:by\:\:\dfrac{50}{3} \\[5ex] \dfrac{50}{3} * \dfrac{3}{50}p = \dfrac{50}{3} * 3000 \\[5ex] p = 50 * 1000 \\[3ex] p = ₦50,000 \\[3ex] $ The father's monthly income is $₦50,000$

Price for gas on January $1$ of Year $6$ was $3\%$ higher than the price on January $1$ of Year $5$.
This means that price for gas on January $1$ of Year $6$ is $3\%$ higher than $\$1.36$

$ 3\%\:\:of\:\:1.36 \\[3ex] = \dfrac{3}{100} * 1.36 \\[5ex] = 0.03(1.36) \\[3ex] = \$0.0408 \\[3ex] Price\:\:for\:\:gas\:\:on\:\:January\:1\:\:of\:\:Year\:6 \\[3ex] = 1.36 + 0.0408 \\[3ex] = 1.4008 \approx \$1.40 \\[3ex] $ The price for gas on January $1$ of Year $6$ was approximately $\$1.40$ higher than the price on January $1$ of Year $5$.

$ Final\:\:exam\:\:is\:\:worth\:\:20\%\:\:of\:\:the\:\:final\:\:grade \\[3ex] This\:\:implies\:\:that\:\:all\:\:other\:\:assessments\:\:besides\:\:the\:\:final\:\:exam\:\:are\:\:worth\:\:80\%\:\:(100\% - 20\%) \\[3ex] Philemon\:\:made\:\:an\:\:average\:\:of\:\:84\%\:\:on\:\:all\:\:those\:\:assessments \\[3ex] (a.) \\[3ex] \underline{All\:\:assessments\:\:besides\:\:the\:\:final\:\:exam} \\[3ex] \bar{x} = mean\:\:or\:\:average \\[3ex] \bar{x} = 84\% \\[3ex] weights = 80\% \\[3ex] \bar{x} = \dfrac{\Sigma weighted\:\:scores}{weights} \\[5ex] \Sigma weighted\:\:scores = \bar{x} * weights \\[3ex] \Sigma weighted\:\:scores = 84 * 80 = 6720 \\[3ex] \underline{Final\:\:exam} \\[3ex] weight = 20\% \\[3ex] To\:\:find\:\:the\:\:best\:\:course\:\:final\:\:grade,\:\:we\:\:should\:\:assume\:\:he\:\:makes\:\:a\:\:100\%\:\:on\:\:the\:\:final\:\:exam \\[3ex] score = 100\% \\[3ex] weighted\:\:score = 20 * 100 = 2000 \\[3ex] \underline{Course\:\:final\:\:grade:\:\:All\:\:assessments\:\:including\:\:the\:\:final\:\:exam} \\[3ex] Final\:\:grade = \dfrac{\Sigma weighted\:\:scores}{\Sigma weights} \\[5ex] Final\:\:grade = \dfrac{6720 + 2000}{80 + 20} \\[5ex] Final\:\:grade = \dfrac{8720}{100} \\[5ex] Final\:\:grade = 87.20\% \\[3ex] The\:\:best\:\:final\:\:score\:\:he\:\:can\:\:earn\:\:is\:\:87.20\% \\[3ex] The\:\:best\:\:final\:\:grade\:\:he\:\:can\:\:earn\:\:is\:\:a\:\:B \\[3ex] (b.) \\[3ex] (i) \\[3ex] He\:\:cannot\:\:make\:\:an\:\:A \\[3ex] (ii) \\[3ex] He\:\:can\:\:make\:\:a\:\:B \\[3ex] \underline{To\:\:determine\:\:the\:\:minimum\:\:score\:\:on\:\:the\:\:final\:\:exam\:\:in\:\:order\:\:to\:\:make\:\:a\:\:B} \\[3ex] Let\:\:the\:\:minimum\:\:score = p \\[3ex] weight = 20\% \\[3ex] score = p\% \\[3ex] weighted\:\:score = 20 * p = 20p \\[3ex] For\:\:a\:\:B, \\[3ex] Final\:\:grade = 80\% \\[3ex] Final\:\:grade = \dfrac{\Sigma weighted\:\:scores}{\Sigma weights} \\[5ex] 80 = \dfrac{6720 + 20p}{80 + 20} \\[5ex] 80 = \dfrac{6720 + 20p}{100} \\[5ex] 6720 + 20p = 80(100) \\[3ex] 6720 + 20p = 8000 \\[3ex] 20p = 8000 - 6720 \\[3ex] 20p = 1280 \\[3ex] p = \dfrac{1280}{20} \\[5ex] p = 64\% \\[3ex] The\:\:minimum\:\:score\:\:on\:\:the\:\:final\:\:exam\:\:needed\:\:for\:\:Philemon\:\:to\:\:make\:\:a\:\:B\:\:in\:\:the\:\:course\:\:is\:\:64\% $

$ \underline{Grade\;\;4} \\[3ex] regular\;\;enrollment\;\;fee = \$400 \\[3ex] 20\%\;\;of\;\;400 \\[3ex] = \dfrac{20}{100} * 400 \\[5ex] = \$80 \\[3ex] 20\%\;\;off = 400 - 80 = \$320 \\[3ex] \underline{Grade\;\;6} \\[3ex] regular\;\;enrollment\;\;fee = \$400 \\[3ex] 20\%\;\;of\;\;500 \\[3ex] = \dfrac{20}{100} * 500 \\[5ex] = \$100 \\[3ex] 20\%\;\;off = 500 - 100 = \$400 \\[3ex] Total\;\;amount \\[3ex] = 320 + 400 \\[3ex] = \$720 $

$ 8/24:\;\; initial = 8600 \\[3ex] 9/30:\;\; new = 7,630 \\[3ex] change = new - initial \\[3ex] change = 7630 - 8600 = -970 \\[3ex] The\;\;change\;\;is\;\;a\;\;decrease \\[3ex] \%\;decrease = \dfrac{change}{initial} * 100 \\[5ex] \%\;decrease \\[3ex] = \dfrac{970}{8600} * 100 \\[5ex] = 11.27906977\% \\[3ex] \approx 11.3\% $

In other words, which of the options best explains why the decline from the August 24 closing value to the September 30 closing value was relatively large (-970)?
All of the options are correct. However, which option is the most correct?
As observed and also written, there are more declines than advances.
But, the main reason for the large decline is because the average of the declines is much greater than the average of the advances.

Ask students to verify that statement...the last option...Option K. by calculating the:
(i.) average of the declines
(ii.) average of the advances

$ 9/13:\;\; 7945 \\[3ex] 9/14:\;\; 8020 \\[3ex] 9/15:\;\; 8090 \\[3ex] 9/16:\;\; 7870 \\[3ex] 9/17:\;\; 7895 \\[3ex] Average\;\;closing\;\;value\;\;for\;\;the\;\;5-day\;\;period \\[3ex] = \dfrac{7945 + 8020 + 8090 + 7870 + 7895}{5} \\[5ex] = \dfrac{39820}{5} \\[5ex] = 7964 $