Percents Review 1. Simple Problems Involving Percentages 2. Percentage Change 3. Profit and Loss 4. Discount 5. Simple Interest

1 Simple Problems Involving Percentages Using Percentage to Find a Number B Finding the Percentage C Finding the Original Number from a Given Percentage

2 Percentage Change Percentage Increase A Percentage Decrease B

3 Profit and Loss A Profit B Loss

y% of a number A = A × y% Using Percentage to Find a Number A) 1 Simple Problems Involving Percentages Example Using Percentage to Find a Number A) y% of a number A = A × y%

Find the value of each of the following. 1 Simple Problems Involving Percentages Find the value of each of the following. (a) 25% of 850 (b) 10% of 12.8 (a) 850 × 25% = 850 × 0.25 = 212.5 (b) 12.8 × 10% = 12.8 × 0.1 = 1.28

Number of staff who are university graduates = 1 400 × 88% 1 Simple Problems Involving Percentages There are 1 400 staff in a company. It is known that 88% of the staff in the company are university graduates. How many staff in that company are university graduates? Number of staff who are university graduates = 1 400 × 88% = 1 400 × Fulfill Exercise Objective Use percentage to find a number. = 1 232

Number of people aged under 15 = 6.8 × 16% million 1 Simple Problems Involving Percentages In 2002, the population of Hong Kong was 6.8 million and 16% of them were aged under 15. How many people were 15 or above? Number of people aged under 15 = 6.8 × 16% million = 6.8 × 0.16 million = 1.088 million ∴ Number of people aged 15 or above Fulfill Exercise Objective Use percentage to find a number. = (6.8 – 1.088) million = 5.712 million

b a × 100% a b × 100% Finding the Percentage B) 1 Simple Problems Involving Percentages Example Finding the Percentage B) To find out what percentage of a is b, we write . b a × 100% 2. To find out what percentage of b is a, we write . a b × 100%

(a) The required percentage = 1 Simple Problems Involving Percentages What percentage of 45 is 36? What percentage of 36 is 45? (a) The required percentage = = 80% (b) The required percentage = = 125%

(a) What percentage of the votes has gone to candidate A? 1 Simple Problems Involving Percentages Among the 800 people in the election team, 256 of the team members vote for candidate A and the rest vote for candidate B. (a) What percentage of the votes has gone to candidate A? (b) What percentage of the number of votes for A is that for B?

(a) The required percentage = 1 Simple Problems Involving Percentages Back to Question (a) The required percentage = = 32% (b) The number of votes that candidate B gets = 800 – 256 = 544 The required percentage = = 212.5% Fulfill Exercise Objective Find the required percentage.

y A = x% Finding the Original Number from a Given Percentage C) 1 Simple Problems Involving Percentages Example Finding the Original Number from a Given Percentage C) Given : x% of the original number A = y Then y x% A =

Find the unknown in each of the following. 1 Simple Problems Involving Percentages Find the unknown in each of the following. (a) $24 is 60% of $m. (b) 180 kg is 125% of n kg. (a) $m × 60% = $24 (b) n kg × 125% = 180 kg m × 0.6 = 24 m = 24 ÷ 0.6 = 40 n × 1.25 = 180 n = 180 ÷ 1.25 = 144

Let n be the total number of games the team played. 1 Simple Problems Involving Percentages A football team won 85% of its games last year. If they won 34 games altogether, how many games did the team play? Let n be the total number of games the team played. Then 85% of n is 34. i.e. 85% × n = 34 n × 0.85 = 34 n = 34 ÷ 0.85 = 40 Fulfill Exercise Objective Find the original number. ∴ The total number of games the team played was 40.

Percentage of female students 1 Simple Problems Involving Percentages In the academic year 1999 – 2000, 55% of the students in the University of Hong Kong were male. If the number of female students in that year was 6 300, then what was the total number of students? Percentage of female students = 100% – 55% = 45%

Let n be the total number of students. Then 45% of n is 6 300. 1 Simple Problems Involving Percentages Back to Question Let n be the total number of students. Then 45% of n is 6 300. i.e. 45% × n = 6 300 n × 0.45 = 6 300 n = 6 300 ÷ 0.45 = 14 000 ∴ The total number of students was 14 000. Fulfill Exercise Objective Find the original number.

1. Increase = New value – Original value 2 Percentage Change Example Percentage Increase A) 1. Increase = New value – Original value 2. Percentage increase = Increase Original value × 100% 3. Increase = Original value × Percentage increase 4. New value = Original value × (1 + Percentage increase)

Find the percentage increase in each of the following. 2 Percentage Change Find the percentage increase in each of the following. An increase from 120 to 156. An increase of 10.5 mL from 25 mL. (a) Increase = 156 – 120 = 36 (b) Increase = 10.5 mL ∴ Percentage increase ∴ Percentage increase = = = 30% = 42%

Increase each of the following quantities by the given percentage: 2 Percentage Change Increase each of the following quantities by the given percentage: (a) 82 by 15% (b) $144 by 35% (a) New value = 82 × (1 + 15%) = 82 × 1.15 = 94.3 (b) New value = $144 × (1 + 35%) = $144 × 1.35 = $194.4

2 Percentage Change The price increase of a movie recorded on DVD and on video CD is both $5. If the original prices of a DVD and a video CD are $125 and $40 respectively, find the percentage increase in the price of (a) a DVD, (b) a video CD.

Percentage increase in the price of a DVD 2 Percentage Change Back to Question Percentage increase in the price of a DVD = = 4% (b) Percentage increase in the price of a video CD = = 12.5% Fulfill Exercise Objective Find the percentage increase.

2 Percentage Change To celebrate the New Year, a certain brand of chocolate beans added 10% to the weight of each pack for free. If the original weight of each pack of the chocolate beans was 50 g, what was the weight of each pack after the increase? The new weight = 50 × (1 + 10%) g = 50 × 1.1 g = 55 g Fulfill Exercise Objective Find the new value.

1. Decrease = Original value – New value 2 Percentage Change Example Percentage Decrease B) 1. Decrease = Original value – New value 2. Percentage decrease = Decrease Original value × 100% 3. Decrease = Original value × Percentage decrease 4. New value = Original value × (1 – Percentage decrease)

Find the percentage decrease in each of the following. 2 Percentage Change Find the percentage decrease in each of the following. An decrease from 300 to 48. An decrease of 3 g from 12 g. (a) Decrease = 300 – 48 = 252 (b) Decrease = 3 g ∴ Percentage decrease ∴ Percentage decrease = = = 25% = 84%

Decrease each of the following quantities by the given percentage: 2 Percentage Change Decrease each of the following quantities by the given percentage: (a) 150 by 27% (b) 855 cm by 60% (a) New value = 150 × (1 – 27%) = 150 × 0.73 = 109.5 (b) New value = 855 × (1 – 60%) cm = 855 × 0.4 cm = 342 cm

The percentage decrease = 2 Percentage Change The worldwide population of the black rhinoceroes has dropped from 65 000 in the early 1970s to 2 405 in the late 1990s. What is the percentage decrease in the number of black rhinoceroes? Fulfill Exercise Objective Find the percentage decrease. The decrease = 65 000 – 2 405 = 62 595 The percentage decrease = = 96.3%

2 Percentage Change Kitty scored 80 marks in Mathematics last term. If her score is decreased by 5% this term, what is her score in this term? The score of this term = 80 × (1 – 5%) = 80 × 0.95 = 76 Fulfill Exercise Objective Find the new value.

+ – Percentage Change C) New value – Original value × 100% Example Percentage Change C) ‧ Percentage change = New value – Original value Original value × 100% New value > Original value New value < Original value Change Sign of percentage change Increase + Decrease –

(a) If 100 is changed to 110, then 2 Percentage Change (a) If 100 is changed to 110, then = percentage change = +10% (b) If 50 is changed to 40, then = percentage change = –20%

2 Percentage Change The price of a digital camera was $4 000 last year. This year the price becomes $3 400. Find the percentage change in price. = Percentage change = Fulfill Exercise Objective Find the percentage change. = –15%

If selling price > cost price, the merchant will make a profit. 3 Profit and Loss Profit A) When a merchant pays to buy goods, the amount he pays is called the cost price. When the merchant sells goods at a price, the amount he receives is called the selling price. If selling price > cost price, the merchant will make a profit. If we compare the profit with the cost price of the goods and express the result as a percentage, the percentage is called the profit per cent (written as profit %).

5. If selling price > cost price, 3 Profit and Loss Example Profit A) 5. If selling price > cost price, i. Profit = Selling price – Cost price ii. Profit % = Profit Cost price × 100% iii. Profit = Cost price × Profit % iv. Selling price = Cost price × (1 + Profit %)

The selling price of a toy car is $225, and the cost price is $180. 3 Profit and Loss The selling price of a toy car is $225, and the cost price is $180. Find the profit of the toy car. Find the profit per cent of the toy car. (a) Profit = $(225 – 180) = $45 = (b) Profit % = 25%

3 Profit and Loss Mr Wong bought a pair of jeans for $80 and sold them for $130. Mr Cheung bought a T-shirt for $40 and sold it for $90. Who made a greater profit %? Profit made by Mr Wong = $(130 – 80) = $50 = ∴ Profit % = 62.5%

Profit made by Mr Cheung = $(90 – 40) 3 Profit and Loss Back to Question Profit made by Mr Cheung = $(90 – 40) = $50 = ∴ Profit % = 125% ∴ Mr Cheung made a greater profit %. Fulfill Exercise Objective Find the profit %.

3 Profit and Loss A car was bought for $350 000 and was sold at a profit of 20%. How much was it sold for? Selling price = $350 000 × (1 + 20%) = $350 000 × 1.2 = $420 000 Fulfill Exercise Objective Find the selling price.

1. If selling price < cost price, there will be a loss. 3 Profit and Loss Loss B) 1. If selling price < cost price, there will be a loss. We can express loss as a percentage of the cost price. The result is called the loss per cent (written as loss %).

3. If selling price < cost price, 3 Profit and Loss Example Loss B) 3. If selling price < cost price, i. Loss = Cost price – Selling price ii. Loss % = Loss Cost price × 100% iii. Loss = Cost price × Loss % iv. Selling price = Cost price × (1 – Loss %)

If a bicycle is bought at $1 000 and sold for $800, then 3 Profit and Loss If a bicycle is bought at $1 000 and sold for $800, then loss = $(1 000 – 800) = $200 = and loss % = 20%

3 Profit and Loss A box of 100 coloured pencils costs $250. In a sale, each pencil is sold at $2 a piece.What is the loss per cent after all of the pencils have been sold? Total cost price = $250 Total selling price = $2 × 100 = $200 Total loss = $(250 – 200) = $50 Fulfill Exercise Objective Find the loss %. = ∴ Loss % = 20%

4 Discount Discount The difference between the marked price and the selling price is called the discount. We can express the discount as a percentage of the marked price, which is called the discount per cent (written as discount %).

3. i. Discount = Marked price – Selling price Example Discount 3. i. Discount = Marked price – Selling price ii. Discount % = Discount Marked price × 100% iii. Discount = Marked price × Discount % iv. Selling price = Marked price × (1 – Discount %)

4 Discount A sofa is marked at $2 800 and sold at $2 100 in a sale. What is the discount and discount %? = $(2 800 – 2 100) = $700 Discount = Discount % = 25%

Which pack of Battery is sold at a larger discount per cent? A pack of E-Power Battery marked at $24 is now sold for $21 only. A pack of D-cell Battery, on the other hand, marked at $20 is now sold for $17.9. Which pack of Battery is sold at a larger discount per cent?

∴ E-Power Battery is sold at a larger discount per cent. Back to Question For E-Power Battery, discount = $(24 – 21) = $3 = discount % = 12.5% For D-cell Battery, discount = $(20 – 17.9) = $2.1 Fulfill Exercise Objective Find the discount %. = discount % = 10.5% ∴ E-Power Battery is sold at a larger discount per cent.

Let $P be the original marked price of the radio. 4 Discount After a 44% discount, a radio is now sold for $112. What was the original marked price of the radio? 44% off Now $112 only Let $P be the original marked price of the radio. Then 112 = P × (1 – 44%) 112 = P × 0.56 P = 112 ÷ 0.56 = 200 Fulfill Exercise Objective Find the marked price. ∴ The original marked price was $200.

A dictionary marked at $260 is sold at a discount of 25%. 25% off A dictionary marked at $260 is sold at a discount of 25%. (a) Find the selling price. (b) If the cost price of the dictionary is $200, find the loss %. (a) The selling price = $260 × (1 – 25%) = $260 × 75% = = $195

= cost price – selling price = $(200 – 195) = $5 4 Discount Back to Question (b) Loss = cost price – selling price = $(200 – 195) = $5 = Loss % = 2.5% Fulfill Exercise Objective Miscellaneous problems.

5 Simple Interest Simple Interest If the interest in each period is calculated on the same principal, then the interest obtained is called simple interest. The calculation of interest is based on a percentage of the principal and that percentage is called the interest rate.

5 Simple Interest Simple Interest Let $I stand for the simple interest, $P stand for the principal, R% stand for the interest rate per annum, T years stand for the period of time, $A stand for the amount, then Example i. ii. Example

5 Simple Interest If $1 000 is deposited in a bank at an interest rate of 6% p.a., find the interest received a year later. = $1 000 × 6% = $60 Interest received

Simple interest received Deposit $6 000 in a bank at an interest rate of 8% p.a. for 2 years. What is the simple interest received? Simple interest received = = $960

Simple interest given by Bank A Simon wants to deposit $30 000 in a bank for 2.5 years. Bank A offers an interest rate of 8% p.a. and Bank B offers 7.5% p.a. How much more simple interest will Bank A give to Simon than Bank B? = Simple interest given by Bank A = $6 000

Simple interest given by Bank B Back to Question = Simple interest given by Bank B = $5 625 The difference between the simple interests = $(6 000 – 5 625) = $375 ∴ Bank A will give $375 more simple interest to Simon than Bank B. Fulfill Exercise Objective Find the simple interest.

Find the amount of money she borrows 5 Simple Interest Kimmy borrows some money from a bank to start her own business. The interest rate is 5% p.a. Find the amount of money she borrows from the bank if she has to pay a total of $30 000 as simple interest after 2 years.

Let $P be the amount of money Kimmy borrows, then 5 Simple Interest Back to Question Let $P be the amount of money Kimmy borrows, then 30 000 = 30 000 = P = 30 000 × 10 = 300 000 ∴ The amount of money Kimmy borrows is $300 000. Fulfill Exercise Objective Find the principal.

Amount received after 3 years 5 Simple Interest A bank pays simple interest for money deposited at an interest rate of 10% p.a. If $5 000 is deposited in the bank, find the amount received after 3 years. Amount received after 3 years = = $6 500

the amount received after 10 months, correct to the nearest $100, 5 Simple Interest A bank pays simple interest at an interest rate of 8% p.a. If $25 000 is deposited in the bank, find the amount received after 10 months, correct to the nearest $100, the time required to get the amount $27 500. (a) Amount received after 10 months = 10 12 = $26 700 , cor. to the nearest 100

(b) Let T years be the time required, then 5 Simple Interest Back to Question (b) Let T years be the time required, then 27 500 = = = 0.1 Fulfill Exercise Objective Find the amount. Find the period of time. T = = 1.25 ∴ The time required is 1.25 years, i.e. 1 year and 3 months.

(i) Find the amount received by John after 20 years. 5 Simple Interest 1A_Ch3(61) John deposits $1 000 000 in a bank where simple interest is calculated at a rate of R% p.a. If the amount received by John doubles the principal after 10 years, find the value of R. Suppose R = 8. (i) Find the amount received by John after 20 years. (ii) Peter borrows a sum of money from John and the interest rate is 8% p.a. If the amount that Peter has to pay back John after 2.5 years is $120 000, how much does Peter borrow? Soln Soln

The amount received by John after 10 years 5 Simple Interest Back to Question (a) The principal is $1 000 000. The amount received by John after 10 years = $1 000 000 × 2 = $2 000 000 Hence 2 000 000 = 2 = 1 + 0.1R 1 = 0.1R ∴ R = 10

(b) (i) Amount received after 20 years = 5 Simple Interest Back to Question (b) (i) Amount received after 20 years = = $2 600 000 (ii) If $P is the sum of money that Peter borrows from John, then 120 000 = 120 000 = P(1 + 0.2) Fulfill Exercise Objective Find the interest rate. Find the principal. P = = 100 000 ∴ Peter borrowed $100 000 from John.