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

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 Students

Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For NSC Students

__For the Questions:__

Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from
behind.

Any comma included in a number indicates a decimal point.

__For the Solutions:__

Decimals are used appropriately rather than commas

Commas are used to separate digits appropriately.

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 $\$4.20$

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 = $\$15.47$

This cost

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?

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 = $\$14.46$

This cost

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] $

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 $

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.

$ 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 $

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] $

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 $\$39.19$

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 = $\$735.00$

A customer purchases a laptop for $588.00.

This cost

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.

$ 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 $

A customer purchases a laptop for $588.00.

This cost

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] $

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 $

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\% $

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 $

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

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 $

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 $

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 $

Of the students who live south of Highway

How many students who live south of Highway

$ 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.

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.

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

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

(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 $₦200.00$

How much did the bicycle cost

$ 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\% $

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] $

$ \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} $

What is the conversion rate of the

$ 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 $

$ 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\% $

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\% $

(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\% $

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 $

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\% $

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 $

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\% $

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 $

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 $

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 $\$273.82$

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\% $

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.

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\% $

$ 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\% $

$ 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 $

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 $12\%$

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.

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.

$ \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\% $

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

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 $54^\circ$

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 $D$1,000.00

He kept 20% of the oranges for himself, sold 115 at $D$6.50 each and the rest at $D$5.00 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\% $

He kept 20% of the oranges for himself, sold 115 at $D$6.50 each and the rest at $D$5.00 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 $\$38.40$

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] $

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 $24.7\%$

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 $24.7\%$

(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 $10\%$

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 $\$66,800.00$

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 $20.5\%$

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

$ \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 $74\%$.

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.

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\% $

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?

$ 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 $

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\% $

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.

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\% $

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?

$ 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 $12.5\%$

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 $

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 $

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.

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.

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.

The number of

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

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

Which of the following is closest to the value of

$ 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

(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 $₦50,000$

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.

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 $5$.

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)

*
To practice more of these types of questions, please review Questions $25$ and $27$ on the Descriptive Statistics 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\% $

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

(i) Can he make an

(i) If he cannot make an

(NOTE: Assume 90 - 100 for an

$ 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\% $

(58.) **ACT** A weeklong summer camp is held in June for children in Grades 3 - 6

Parents and guardians who enrolled their children for camp by May 15 received a 20% discount off the regular enrollment fee for each child enrolled.

For each grade, the table below gives the number of children enrolled by May 15 as well as the*regular*
enrollment fee per child.

The grade of any child is that child's grade in school as of May 15.

By May 15, Mr. Ramirez had enrolled his 2 children for camp.

One child was in Grade 4, and the other was in Grade 6.

What was the total amount Mr. Ramirez paid to enroll his 2 children?

$ A.\;\; \$720 \\[3ex] B.\;\; \$800 \\[3ex] C.\;\; \$820 \\[3ex] D.\;\; \$860 \\[3ex] E.\;\; \$880 \\[3ex] $

$ \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 $

Parents and guardians who enrolled their children for camp by May 15 received a 20% discount off the regular enrollment fee for each child enrolled.

For each grade, the table below gives the number of children enrolled by May 15 as well as the

The grade of any child is that child's grade in school as of May 15.

Grade | Enrollment by May 15 | Regular enrollment fee |

3 4 5 6 |
20 15 28 18 |
$350 $400 $450 $500 |

By May 15, Mr. Ramirez had enrolled his 2 children for camp.

One child was in Grade 4, and the other was in Grade 6.

What was the total amount Mr. Ramirez paid to enroll his 2 children?

$ A.\;\; \$720 \\[3ex] B.\;\; \$800 \\[3ex] C.\;\; \$820 \\[3ex] D.\;\; \$860 \\[3ex] E.\;\; \$880 \\[3ex] $

$ \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 $

(59.)

(60.)

**ACT** Use the following information to answer questions 61 - 63

The Dow Jones Industrial Average (DJIA) is an index of stock values.

The chart below gives the DJIA closing values from August 24 through September 30 of a certain year and the
change in
the closing value from the previous day.

A minus sign indicates a *decline* (a closing value less than the previous day's closing value).

A plus sign indicates an *advance* (a closing value greater than the previous day's closing value).

Date | Closing value | Change | Date | Closing value | Change |

8/24 8/25 8/26 8/27 8/30 8/31 9/01 9/02 9/03 9/07 9/08 9/09 9/10 |
8,600 8,515 8,160 8,050 7,540 7,825 7,780 7,680 7,640 8,020 7,860 8,045 7,795 |
-85 -355 -110 -510 +285 -45 -100 -40 +380 -160 +185 -250 |
9/13 9/14 9/15 9/16 9/17 9/20 9/21 9/22 9/23 9/24 9/27 9/28 9/29 9/30 |
7,945 8,020 8,090 7,870 7,895 7,930 7,900 8,150 8,000 8,025 8,110 8,080 7,845 7,630 |
+150 +75 +70 -220 +25 +35 -30 +250 -150 +25 +85 -30 -235 -215 |

(61.) Which of the following is closest to the percent of decrease from the August 24 closing value to the September 30
closing value?

$ A.\;\; 7.9\% \\[3ex] B.\;\; 8.9\% \\[3ex] C.\;\; 11.3\% \\[3ex] D.\;\; 12.7\% \\[3ex] E.\;\; 88.7\% \\[3ex] $

$ 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\% $

$ A.\;\; 7.9\% \\[3ex] B.\;\; 8.9\% \\[3ex] C.\;\; 11.3\% \\[3ex] D.\;\; 12.7\% \\[3ex] E.\;\; 88.7\% \\[3ex] $

$ 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\% $

(62.) The chart shows 4 more declines than advances.

All of the following statements are true.

Which one best explains why the decline from the August 24 closing value to the September 30 closing value was relatively large?

**F.** The greatest change in the chart was a decline.

**G.** The least change in the chart was an advance.

**H.** The greatest number of consecutive declines was greater than the greatest number of consecutive advances.

**J.** The first change was a decline.

**K.** The average of the declines was much greater than the average of the advances.

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

All of the following statements are true.

Which one best explains why the decline from the August 24 closing value to the September 30 closing value was relatively large?

In other words, which of the options

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.

(i.) average of the declines

(ii.) average of the advances

(63.) What is the average closing value for the 5-day period from September 13 through September 17?

$ A.\;\; 7,895 \\[3ex] B.\;\; 7,920 \\[3ex] C.\;\; 7,964 \\[3ex] D.\;\; 7,980 \\[3ex] E.\;\; 8,090 \\[3ex] $

$ 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 $

$ A.\;\; 7,895 \\[3ex] B.\;\; 7,920 \\[3ex] C.\;\; 7,964 \\[3ex] D.\;\; 7,980 \\[3ex] E.\;\; 8,090 \\[3ex] $

$ 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 $

(64.)

(65.)

(66.)