For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka

Applications of Percents

Calculator for Percents

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

For JAMB and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

Solve all questions.
Use at least two methods (two or more methods) whenever applicable.
Show all work.

(1.) Benedict and Benedicta went to dine at La Casa restaurant.
They left a $12\%$ tip of their bill of $\$34.99$.
How much was their tip?

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

(2.) The cost of a Blackberry Z10 phone at Godwin's Phones is $\$220.98$.
This cost excludes a $7\%$ sales tax.
How much is the sales tax?

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

(3.) The cost of a Blackberry Z10 phone at Emmanuel's Electronics is $\$220.98$.
This cost includes a $7\%$ sales tax.
How much is the sales tax?

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 = $\$14.46$

(4.) A gaming laptop at Godwin's Electronics is on sale for $9\%$ off.
The initial price of the laptop is $\$960$
Calculate the discount and the sale price.

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 $

(5.) Expecting new arrivals, Emmanuel's Electronics laptops were marked down for $12\%$ off the initial price.
A customer purchases a laptop for $\$550.00$.
Determine the initial price of the laptop.

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 $

(6.) Benedict and Benedicta went to dine at La Casa restaurant.
They left a $12\%$ tip of their bill of $\$34.99$.
How much was the total cost?

$ 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 $\$39.19$

(7.) Expecting new arrivals, Emmanuel's Electronics laptops were marked down for $20\%$ off the initial price.
A customer purchases a laptop for $\$588.00$.
This cost excludes a $5\%$ sales tax.
Determine the initial price of the laptop.

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 = $\$735.00$

(8.) Expecting new arrivals, Godwin's Electronics laptops were marked down for $20\%$ off the initial price.
A customer purchases a laptop for $\$588.00$.
This cost includes a $5\%$ sales tax.
Determine the initial price of the laptop.

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 $

(9.) ACT A calculator has a regular price of $\$59.95$ before taxes.
It goes on sale at $20\%$ below the regular price.
Before taxes are added, what is the sale price of the calculator?

$ A.\:\: \$11.99 \\[3ex] B.\:\: \$29.98 \\[3ex] C.\:\: \$39.95 \\[3ex] D.\:\: \$47.96 \\[3ex] E.\:\: \$54.95 \\[3ex] $

$ 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 $

(10.) ACT A coat originally priced at $\$80$ is discounted to $\$60$.
What is the percent of discount on this coat?

$ A.\:\: 13\% \\[3ex] B.\:\: 20\% \\[3ex] C.\:\: 25\% \\[3ex] D.\:\: 30\% \\[3ex] E.\:\: 33\dfrac{1}{3}\% \\[5ex] $

$ 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\% $

(11.) ACT A shirt has a sale price of $\$30.40$, which is $20\%$ off the original price.
How much less than the original price is the sale price?

$ A.\:\: \$0.38 \\[3ex] B.\:\: \$1.52 \\[3ex] C.\:\: \$6.08 \\[3ex] D.\:\: \$7.60 \\[3ex] E.\:\: \$10.40 \\[3ex] $

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 $

(12.) WASSCE A trader made a profit of $15\%$ by selling an article for $Le\:345.00$.
Calculate the actual profit.

$ A.\:\: Le\:300.00 \\[3ex] B.\:\: Le\:117.00 \\[3ex] C.\:\: Le\:51.75 \\[3ex] D.\:\: Le\:45.00 \\[3ex] $

$ \%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 $

(13.) JAMB A trader bought goats for $₦4,000$ each.
He sold them for $₦180,000$ at a loss of $25\%$.
How many goats did he buy?

$ A.\:\: 50 \\[3ex] B.\:\: 60 \\[3ex] C.\:\: 36 \\[3ex] D.\:\: 45 \\[3ex] $

$ 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 $

(14.) ACT Of the $900$ students enrolled at Sierra Elementary School, $45\%$ live south of Highway $R$.
Of the students who live south of Highway $R$, $20\%$ do NOT ride the bus to school.
How many students who live south of Highway $R$ ride the bus to school?

$ A.\:\: 81 \\[3ex] B.\:\: 180 \\[3ex] C.\:\: 324 \\[3ex] D.\:\: 585 \\[3ex] E.\:\: 720 \\[3ex] $

$ 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 $

(15.) ACT Widely considered one of the greatest film directors, Alfred Hitchcock directed over $60$ films.
The table below gives some information about Hitchcock's last $12$ films.

Title	Year of release	Length(minutes)
The Trouble with Harry	$1955$	$99$
The Man Who Knew Too Much	$1956$	$120$
The Wrong Man	$1956$	$105$
Vertigo	$1958$	$128$
North by Northwest	$1959$	$136$
Psycho	$1960$	$109$
The Birds	$1963$	$119$
Marnie	$1964$	$130$
Torn Curtin	$1966$	$128$
Topaz	$1969$	$143$
Frenzy	$1972$	$?$
Family Plot	$1976$	$?$

Recently, a director made a new version of Vertigo.
The new version is $20\%$ shorter in length than Hitchcock's version.
Which of the following values is closest to the length, in minutes, of the new version?

$ F.\:\: 64 \\[3ex] G.\:\: 102 \\[3ex] H.\:\: 105 \\[3ex] J.\:\: 108 \\[3ex] K.\:\: 125 \\[3ex] $

$ 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.

(16.) JAMB $P$ sold his bicycle to $Q$ at a profit of $10\%$.
$Q$ sold it to $R$ for $₦209$ at a loss of $5\%$.
How much did the bicycle cost $P$?

$ A.\:\: ₦200 \\[3ex] B.\:\: ₦196 \\[3ex] C.\:\: ₦180 \\[3ex] D.\:\: ₦205 \\[3ex] E.\:\: ₦150 \\[3ex] $

$ \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 $₦200.00$

(17.) ACT The changes in a city's population from one decade to the next decade for $3$ consecutive decades were a $20\%$ increase, a $30\%$ increase, and a $20\%$ decrease.
About what percent was the increase in the city's population over the $3$ decades?

$ A.\:\: 10\% \\[3ex] B.\:\: 20\% \\[3ex] C.\:\: 25\% \\[3ex] D.\:\: 30\% \\[3ex] E.\:\: 70\% \\[3ex] $

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\% $

(18.) JAMB $22\dfrac{1}{2}\%$ of the Nigerian Naira, $₦$ is equal to $17\dfrac{1}{10}\%$ of a foreign curreny $M$.
What is the conversion rate of the $M$ to the Naira?

$ A.\:\: 1M = \dfrac{15}{57}₦ \\[5ex] B.\:\: 1M = 2\dfrac{11}{57}₦ \\[5ex] C.\:\: 1M = 1\dfrac{18}{57}₦ \\[5ex] D.\:\: 1M = 38\dfrac{1}{4}₦ \\[5ex] E.\:\: 1M = 384\dfrac{3}{4}₦ \\[5ex] $

$ 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} $

(19.) JAMB If the length of a square is increased by $20\%$ while its width is decreased by $20\%$ to form a rectangle, what is the ratio of the area of the rectangle to the area of the square?

$ A.\:\: 6.5 \\[3ex] B.\:\: 25.24 \\[3ex] C.\:\: 5.6 \\[3ex] D.\:\: 24.25 \\[3ex] $

$ 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 $

(20.) JAMB The length of a notebook $15\:cm$ was measured as $16.8\:cm$
Calculate the percentage error to $2$ significant figures.

$ A.\:\: 12.00\% \\[3ex] B.\:\: 11.00\% \\[3ex] C.\:\: 10.71\% \\[3ex] D.\:\: 0.12\% \\[3ex] $

$ \% 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\% $

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(21.) Sarah loves ice cream so much that after returning from Abraham's Market with some ice cream, she eats $25\%$ of the ice cream she just bought.
The next day, she eats $25\%$ of the remaining ice cream and continues to eat another $25\%$ each day.
What percent of the ice cream will Sarah have left after five days?
Round your answer to the nearest whole percent.

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\% $

(22.) JAMB A taxpayer is allowed $\dfrac{1}{8}th$ of his income tax free, and pays $20\%$ on the remainder.
If he pays $₦490.00$ tax, what is his income?

$ A.\:\: ₦560.00 \\[3ex] B.\:\: ₦2,450.00 \\[3ex] C.\:\: ₦2,800.00 \\[3ex] D.\:\: ₦2,920.00 \\[3ex] $

$ 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 $

(23.) Love Electronics reduced the price of a Smart Television from $\$1000$ to $\$900$.
What is the percent of decrease?

$ 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\% $

(24.) If the first chapter of a book is $15$ pages long and make up $5\%$ of the book, how many pages does the entire book have?

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 $

(25.) Jacob increased the wholesale price of an item by $30\%$.
He offered a $15\%$ discount off the sticker price.
The wholesale price of the item is $\$900$
Calculate:
(a.) the sticker price (the price including the mark-up).
(b.) the savings (the discount removed from the sticker price)
(c.) the amount paid for the item by the customer
(d.) the profit made by Jacob
(e.) the percent profit made by Jacob

$ (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\% $

(26.) JAMB A cinema hall contains a certain number of people.
If $22\dfrac{1}{2}\%$ are children, $47\dfrac{1}{2}\%$ are men, and $84$ are women, find the number of men in the hall.

$ A.\:\: 113 \\[3ex] B.\:\: 133 \\[3ex] C.\:\: 84 \\[3ex] D.\:\: 63 \\[3ex] $

$ 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 $

(27.) CSEC John bought a car for $\$65000$.
If the value of the car depreciates by $8\%$ each year, how much will the car be worth at the end of $2$ years?

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 $

(28.) Martha's weekly paycheck is thirty percent less than Mary's weekly paycheck.
The sum of the two paychecks is six hundred and sixty five dollars.
Calculate the amount of each paycheck.

$ 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 $\$273.82$

(29.) A group of adults were asked how many children they have in their families.
The histogram below shows the number of adults who indicated each number of children.

(a.) How many adults were questioned?
(b.) What percentage of the adults questioned had a child?

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\% $

(30.) Judith categorized her spending for this month into four categories: Rent, Food, Fun, and Other.
The percents she spent in each category are shown below.

If Judith spent a total of $\$2500$ this month, how much did she spend on Fun?

$ 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.

ACT Use the following information to answer questions $31 - 33$
In $2012$, pollsters for the gallup Organization asked a random sample of $1,014$ adults.
"On the average, about how much does your family spend on food each week?"
The table below lists the percent of the sample that gave each response.
For example, approximately $21\%$ of adults in the sample responded that, on average, they spend no less than $\$200$ but no more than $\$299$ on food each week.

Average amount spent	Percent of sample
Less than $\$50$	$8\%$
$\$50$ to $\$99$	$17\%$
$\$100$ to $\$124$	$22\%$
$\$125$ to $\$149$	$4\%$
$\$150$ to $\$199$	$15\%$
$\$200$ to $\$299$	$21\%$
$\$300$ or more	$10\%$
Did not give an amount	$3\%$

(31.) ACT Which of the following expressions is equal to the approximate number of adults from the sample that said they spend an average of less than $\$100$ each week on food?

$ A.\:\: 1,014(22) \\[3ex] B.\:\: 1,014(25) \\[3ex] C.\:\: 1,014(47) \\[3ex] D.\:\: 1,014(0.22) \\[3ex] E.\:\: 1,014(0.25) \\[3ex] $

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\% $

(32.) ACT What percent of adults in the sample responded that they spend, on average, at least $\$150$ each week on food?

$ F.\:\: 15\% \\[3ex] G.\:\: 46\% \\[3ex] H.\:\: 49\% \\[3ex] J.\:\: 51\% \\[3ex] K.\:\: 66\% \\[3ex] $

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

(33.) ACT A pollster will create a circle graph using the information in the table.
One sector of the circle graph will represent the percent of adults in the sample who said they spend an average of $\$300$ or more on food each week.
What will be the measure of the central angle for that sector?

$ A.\:\: 10^\circ \\[3ex] B.\:\: 13^\circ \\[3ex] C.\:\: 36^\circ \\[3ex] D.\:\: 45^\circ \\[3ex] E.\:\: 47^\circ \\[3ex] $

$ 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 $

(34.) JAMB By selling $20$ oranges for $₦1.35$, a trader makes a profit of $8\%$.
What is his percentage gain or loss if he sells the same $20$ oranges for $₦1.10$?

$ A.\:\: 8\% \\[3ex] B.\:\: 10\% \\[3ex] C.\:\: 12\% \\[3ex] D.\:\: 15\% \\[3ex] $

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 $12\%$

(35.) CSEC The table below shows the results obtained by a student in her CSEC Mathematics examination.
The maximum mark for each paper is given in the third column of the table.

Paper	Percentage Obtained	Maximum Mark for Paper
$01$	$55$	$30$
$02$	$60$	$50$
$03$	$80$	$20$
Total		$\boldsymbol{100}$

Determine, as a percentage, the student's final mark for the Mathematics examination.

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\% $

(36.) ACT Douglas wants to draw a circle graph showing the favorite colors of his friends.
When he polled his friends asking each their favorite color, $25\%$ of his friends said red; $30\%$ of his friends said blue; $20\%$ of his friends said green; $10\%$ of his friends said purple; and the remaining friends said colors other than red, blue, green, and purple.
The colors other than red, blue, green, and purple will be grouped together in an Other sector.
What will be the degree measure of the Other sector?

$ A.\:\: 108^\circ \\[3ex] B.\:\: 54^\circ \\[3ex] C.\:\: 27^\circ \\[3ex] D.\:\: 15^\circ \\[3ex] E.\:\: 10^\circ \\[3ex] $

$ 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 $54^\circ$

(37.) WASSCE A man bought $250$ oranges for $D1,000.00$
He kept $20\%$ of the oranges for himself, sold $115$ at $D6.50$ each and the rest at $D5.00$ each.
Calculate his percentage profit.

$ 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\% $

(38.) Lucy left a $20\%$ tip of her bill of $\$32$ at a restaurant.
How much would she have to pay at check-out?

$ 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 $\$38.40$

(39.) WASSCE The population of a village increases by $20\%$ every year.
The District Assembly grants the village $GH¢15.00$ per head at the beginning of every year.
If the population of the village was $3,000$ in the year $2003$, calculate the Assembly's total grant to the village from $2003$ to $2007$

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] $

(40.) The United States Weather Bureau has a station in Mauna Loa in Hawaii that has measured carbon$IV$oxide levels since $1959$.
At that time, there were $304$ parts per million of carbon dioxide in the atmosphere.
In $2005$, the figure was $379$ parts per million.
Determine the increase and the percent of increase in carbon$IV$oxide levels to the nearest tenth of a percent.

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 $24.7\%$

Top

(41.) WASSCE A trader purchased $10$ dozen eggs at $₦300.00$ per dozen.
One getting to his shop, he found that $20$ eggs were broken.
How much did he sell the remaining eggs if he made a profit of $10\%$

$ 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 $10\%$

(42.) Hannah's current salary is $\$68,804.00$
If she received a $3\%$ raise last year, what was her salary last year before the raise?

$ 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 $\$66,800.00$

(43.) WASSCE A publisher prints $30,000$ copies of a book at $GH¢ 2.00$ each and sold them for $GH¢ 2.76$ each.

The publisher agrees to pay the author $10\%$ of the selling price for the first $6,000$ copies sold and $12\dfrac{1}{2}\%$ of the selling price for all copies sold in excess of $6,000$.

If $25,380$ copies of the book were sold,

(a.) calculate, correct to the nearest Ghana Cedi, the:
(i) total amount received by the author;
(ii) net profit the publisher makes after he has paid the author.

(b) find, correct to one decimal place, the publisher's net profit as a percentage of the author's total receipt.

$ \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 $20.5\%$

(44.) ACT The oxygen saturation level of a river is found by dividing the amount of dissolved oxygen the river water currently has per liter by the dissolved oxygen capacity per liter of the water and then converting to a percent.
If the river currently has $7.3$ milligrams of dissolved oxygen per liter of water and the dissolved oxygen capacity is $9.8$ milligrams per liter, what is the oxygen saturation level, to the nearest percent?

$ A.\:\: 34\% \\[3ex] B.\:\: 70\% \\[3ex] C.\:\: 73\% \\[3ex] D.\:\: 74\% \\[3ex] E.\:\: 98\% \\[3ex] $

$ 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 $74\%$.

(45.) ACT The Harrisburg Recreation Center recently changed its hours to open $1$ hour later and close $3$ hours later than it had previously.
Residents of Harrisburg aged $16$ or older were given a survey, and $560$ residents replied.
The survey asked each resident his or her student status (high school, college, or nonstudent) and what he or she thought about the change in hours (approve, disapprove, or no opinion).
The results are summarized in the table below.

Student status	Approve	Disapprove	No opinion
High school College Nonstudent	$30$ $14$ $85$	$4$ $10$ $353$	$11$ $6$ $47$
Total	$129$	$367$	$64$

After constructing the table, it was discovered that the student status of $15$ residents who replied that they approved had been incorrectly classified as nonstudents.
After correcting the errors, exactly $60\%$ of the college students has replied that they approved.
To the nearest $1\%$, what percent of high school students replied that they approved?

$ F.\:\: 60\% \\[3ex] G.\:\: 67\% \\[3ex] H.\:\: 70\% \\[3ex] J.\:\: 75\% \\[3ex] K.\:\: 82\% \\[3ex] $

$ \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\% $

(46.) ACT Mario plays basketball on a town league team. The table below gives Mario's scoring statistics for last season.
How many points did Mario score playing basketball last season?

Type of shot	Number attempted	Percent successful
$1-point$ free throw $2-point$ field goal $3-point$ field goal	$80$ $60$ $60$	$75\%$ $90\%$ $25\%$

$ A.\:\: 129 \\[3ex] B.\:\: 190 \\[3ex] C.\:\: 213 \\[3ex] D.\:\: 330 \\[3ex] E.\:\: 380 \\[3ex] $

$ \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 $

(47.) CSEC A jeweller paid $\$6800$ for $165$ bracelets in China.
The Customs Department in his country charged him $\$1360$ in duty.

(i) Calculate the TOTAL cost of the $165$ bracelets inclusive of duty.

(ii) The jeweller sold the $165$ bracelets at a selling price of $\$68.85$ EACH.
a) Calculate the TOTAL profit he made on the sale of the $165$ bracelets.
b) Calculate the profit as a percentage of the cost price, giving your answer to the nearest whole number.

$ 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\% $

(48.) JAMB

The bar chart above shows the distribution of marks scored by $60$ pupils in a test in which the maximum score was $10$.
If the pass mark was $5$, what percentage of the pupils failed the test?

$ A.\:\: 59.4\% \\[3ex] B.\:\: 50.0\% \\[3ex] C.\:\: 41.7\% \\[3ex] D.\:\: 25.0\% \\[3ex] $

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.

(49.) JAMB A car dealer bought a second-hand car for $₦250,000.00$ and spent $₦70,000.00$ refurbishing it.
He then sold the car for $₦400,000.00$
What is the percentage gain?

$ A.\:\: 60\% \\[3ex] B.\:\: 32\% \\[3ex] C.\:\: 25\% \\[3ex] D.\:\: 20\% \\[3ex] $

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\% $

(50.) ACT If the length of a rectangle is increased by $25\%$ and the width is decreased by $10\%$, the area of the resulting rectangle is larger than the area of the original rectangle by what percent?

$ A.\:\: 2.5\% \\[3ex] B.\:\: 12.5\% \\[3ex] C.\:\: 15\% \\[3ex] D.\:\: 22.5\% \\[3ex] E.\:\: 35\% \\[3ex] $

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?

(51.) ACT Keanu bought a new laptop computer and paid a discount price that was $20\%$ less than the $\$1,000$ list price.
He also paid tax on the laptop equal to $6\%$ of the discount price.
What is the total amount Keanu paid for the laptop computer?

$ A.\:\: \$752 \\[3ex] B.\:\: \$806 \\[3ex] C.\:\: \$848 \\[3ex] D.\:\: \$860 \\[3ex] E.\:\: \$986 \\[3ex] $

$ 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 $

(52.) ACT While shopping, Alston decided to purchase a shirt on sale that was marked down $40\%$ from its original price of $\$10.00$
At the checkout, he used an in-store coupon for $10\%$ off the sale price.
What was the final price of the shirt before any tax was added?

$ F.\:\: \$3.60 \\[3ex] G.\:\: \$4.00 \\[3ex] H.\:\: \$4.40 \\[3ex] J.\:\: \$5.00 \\[3ex] K.\:\: \$5.40 \\[3ex] $

$ 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 $

(53.) ACT Mikea, an intern with the Parks and Recreation Department, is developing a proposal for the new trapezoidal Springdale Park.
The figure below shows her scale drawing of the proposed park with $3$ side lengths and the radius of the merry-go-round given in inches.
In Mikea's scale drawing, $1\:inch$ represents $1.5\:feet$.

The length of the south side of the park is what percent of the length of the north side?

$ F.\:\: 112\% \\[3ex] G.\:\: 124\% \\[3ex] H.\:\: 142\dfrac{6}{7}\% \\[5ex] J.\:\: 175\% \\[3ex] K.\:\: 250\% \\[3ex] $

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.

(54.) ACT A $500-square-mile$ national park in Kenya has large and small protected animals.
The number of large protected animals at the beginning of $2014$ is given in the table below.

Large animal	Number
Elephant Rhinoceros Lion Leopard Zebra Giraffe	$600$ $100$ $200$ $300$ $400$ $800$
Total	$2,400$

At the beginning of $2014$, the number of all protected animals in the park was $10,000$.
Zoologists predict that for each year from $2015$ to $2019$, the total number of protected animals in the park at the beginning of the year will be $2\%$ more than the number of protected animals in the park at the beginning of the previous year.
At the beginning of $2014$, the number of lions in the park was $p$ percent of the total number of large animals.
Which of the following is closest to the value of $p?$

$ F.\:\: 2 \\[3ex] G.\:\: 8 \\[3ex] H.\:\: 9 \\[3ex] J.\:\: 11 \\[3ex] K.\:\: 12 \\[3ex] $

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.

(55.) JAMB A father decided to give $20\%$ of his monthly income to his three children as their monthly allowance.
The eldest child got $45\%$ of the allowance and the youngest got $25\%$.
How much was the father's monthly income if the second child got $₦3000?$

$ A.\:\: ₦33,000 \\[3ex] B.\:\: ₦45,000 \\[3ex] C.\:\: ₦50,000 \\[3ex] D.\:\: ₦60,000 \\[3ex] $

$ 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 $₦50,000$

(56.) ACT The table below gives the price per gallon of unleaded gasoline at Gus's Gas Station on January $1$ for $5$ consecutive years in the $1990s$.
At Gus's, a customer can purchase a car wash for $\$4.00$.

Year	Price
$1$ $2$ $3$ $4$ $5$	$\$1.34$ $\$1.41$ $\$1.41$ $\$1.25$ $\$1.36$

The price for gas on January $1$ of Year $6$ was $3\%$ higher than the price on January $1$ of Year $5$.
To the nearest $\$0.01$, how much was the price per gallon on January $1$ of Year $6?$

$ A.\:\: \$1.39 \\[3ex] B.\:\: \$1.40 \\[3ex] C.\:\: \$1.66 \\[3ex] D.\:\: \$1.77 \\[3ex] E.\:\: \$2.39 \\[3ex] $

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 $5$.

(57.) Philemon has an eighty four percent average before the final exam for the Statistics course.
That score includes all grades besides the final exam, which counts for twenty percent of the course final grade.

(a.) Determine the best course final grade Philemon can earn.

(b.) Philemon wants to have an $A$ in the course.
If he cannot make an $A$ in the course, he does not want to settle for anything less than a $B$
(i) Can he make an $A$? If he can make an $A$, what minimum score should he make on the final exam in order to earn an $A$?
(i) If he cannot make an $A$, can he make a $B$? If he can make a $B$, what minimum score should he make on the final exam in order to earn a $B$?
(NOTE: Assume $90 - 100$ for an $A$ and $80 - 89$ for a $B$ using the United States Grading Scale)

For more practice with questions like this, please review Questions $25$ and $27$ of this website

$ 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\% $