Percentages (I) Form 1 Mathematics Chapter 3

Reminder Standard Homework (III), (IV) and (V) ▫30 Nov (Fri) Extra Tutorial ▫Tuesday’s last CTP ▫MUST: 1, 10, 11, 12, 14, 16, 26, 30, 36 ▫Mon & Tue after-school – Consultation on G/F Open Book Quiz ▫30 Nov (Fri) Close Book Quiz ▫5 Dec (Wed)

What is Percentage? (p.110) “%” is the symbol of percentage. Per cent means “per one hundred”. e.g.

Simple Percentage Problems P.116 Q2(c): 125% of $365.2 P.117 Q10(c): 132 g is 88% of t g Remark: same units!

Simple Percentage Problems P.118 Q boys and 700 girls took an examination. 15% of boys and 5% of girls failed. Find the total number of students passed. Number of boys failed = 1100  15% = 165 Number of girls failed = 700  5% = 35 Number of students failed = = 200 Number of students = = 1800 Number of students passed = = 1600

Simple Percentage Problems P.118 Q boys and 700 girls took an examination. 15% of boys and 5% of girls failed. Find the total number of students passed. Number of boys passed = 1100  (1 – 15%) = 1100  (85%) = 935 Number of girls passed = 700  (1 – 5%) = 700  (95%) = 665 Number of students passed = = 1600

Time for Practice Page 117 of Textbook 1A ▫Questions Pages of WB 1A ▫Questions 4 - 9

Percentage Increase (p.119) New value > Original value Last year, Andy was 150cm tall. His height is increased by 20% this year. His height now is

Percentage Decrease (p.122) New value < Original value The price of a car was $160,000 last month. If it is reduced by 15% this month, the new price

Percentage Change (p.126) We can summarize the two formulae into one and call percentage change. New value > Original value  + (increase) New value < Original value  – (decrease) The temperature was dropped from 30  C to 27  C in the evening. The percentage change is

Percentage Change P.130 Q19. James had $5000 in his savings account last year. This year, he has $ Change % James then takes out 40% of his savings. The amount now is = $10000  (1 – 40%) = $6000 New change %

Time for Practice Page 127 of Textbook 1A ▫Class Practice Page 129 of Textbook 1A ▫Questions 6 – 14 Pages of WB 1A ▫Questions 3 - 9

Profit ( 盈利, p.131) Selling Price ( 售價 ) > Cost Price ( 成本價 )

Loss ( 虧蝕, p.134) Selling Price ( 售價 ) < Cost Price ( 成本價 )

Profit and Loss A shop sold a car at the profit % of 20%. If the profit was $32000, what was the cost of the car? What was the selling price? Profit or Loss? Let n be the Cost Price. Then, So, the Selling Price is

Profit and Loss P.138 Q22. A merchant bought a chair for $500 and a desk for $750. He sold the chair at a loss of 30% and the desk at a profit of 24%. Selling price of chair Selling price of desk Total selling prices = $350 + $930 = $1280 Total cost prices = $500 + $750 = $1250 So, the profit % is

Time for Practice Pages of Textbook 1A ▫Class Practice Page 137 of Textbook 1A ▫Questions 7 – 19 Pages of WB 1A ▫Questions

Discount ( 折扣, p.138) Western style vs. Chinese style 10% discount9 折 20% discount8 折 70% discount3 折 12% discount88 折 25% discount75 折 5% discount95 折

Discount Marked Price > Selling Price

Discount A pair of scorer shoes is sold at 25% discount at a marked price of $650. Can Vincent buy the shoes if he has $500? Marked price = $650 Selling price of the shoes Since Vincent got $500, he can buy the shoes.

Discount A golden ring is sold at 30% discount with a selling price of $441.What is the marked price? Selling price = $441. Let n be Marked price. Then, So, the Marked price is $630.

Discount P.142 Q10. In shop A, the marked price of a MD is $1920 and the selling price is $1248. In shop B, the prices are $1760 and $1320 respectively. A’s discount % B’s discount % So, B’s discount is smaller.

Time for Practice Page 141 of Textbook 1A ▫Class Practice Page 142 of Textbook 1A ▫Questions 3 – 10 Pages of WB 1A ▫Questions

Simple Interest ( 單利息, p.144) Amount(A) = Principal(P) + Interest(I) Interest(I) = Principal(P)  Interest rate(r%)  Time(T) Formula:

Simple Interest If $40000 is deposited in a bank for 5 years at 3%p.a., find the simple interest and the amount. P=$40000r=3T=5 (years) Simple interest (I) is So, the amount (A) is

Simple Interest How long will $20000 amount to $30000 at 5%p.a. simple interest? P=$20000r=5A=30000 I = $30000-$20000=$10000 Let T be the time. Then, The time required is 10 years.

Simple Interest P.151 Q14. Judy borrows $ for 10 years. Bank A’s interest is 5% p.a. and B is 4.5% p.a. A’s interest B’s interest The difference is $ She pays $50000 less in Bank B from in Bank A.

Time for Practice Page 149 of Textbook 1A ▫Class Practice Page 150 of Textbook 1A ▫Questions 1 – 13 Pages of WB 1A ▫Questions 4 - 7

Reminder Term 1 Examination Syllabus (p.1-185) ▫Chapter 0: Basic Mathematics ▫Chapter 1: Directed Numbers ▫Chapter 2: Using Algebra to Solve Problems (I) ▫Chapter 3: Percentages (I) ▫Chapter 4: Using Algebra to Solve Problems (II) Date of Term 1 Examination ▫ (Wed) ▫8:30 am – 9:30 am (1 hour) ▫Room 104

Enjoy the world of Mathematics!