Percentages (I)

1. Percentages (I) Form 1 Mathematics Chapter 3

2. 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)

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

4. 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!

5. Simple Percentage Problems • P.118 Q14. 1100 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 = 165+35 = 200 • Number of students = 1100+700 = 1800 • Number of students passed = 1800-200 = 1600

6. Simple Percentage Problems • P.118 Q14. 1100 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 = 935-665 = 1600

7. Time for Practice • Page 117 of Textbook 1A • Questions 3 - 13 • Pages 54 - 55 of WB 1A • Questions 4 - 9

8. 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

9. 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

10. 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

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

12. Time for Practice • Page 127 of Textbook 1A • Class Practice • Page 129 of Textbook 1A • Questions 6 – 14 • Pages 57 - 58 of WB 1A • Questions 3 - 9

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

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

15. 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

16. 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

17. Time for Practice • Pages 134 - 135 of Textbook 1A • Class Practice • Page 137 of Textbook 1A • Questions 7 – 19 • Pages 61 - 62 of WB 1A • Questions 7 - 10

18. Discount (折扣,p.138) • Western style vs. Chinese style • 10% discount 9折 • 20% discount 8折 • 70% discount 3折 • 12% discount 88折 • 25% discount 75折 • 5% discount 95折

19. Discount • Marked Price > Selling Price

20. 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.

21. 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.

22. 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.

23. Time for Practice • Page 141 of Textbook 1A • Class Practice • Page 142 of Textbook 1A • Questions 3 – 10 • Pages 65 - 66 of WB 1A • Questions 7 - 10

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

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

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

27. Simple Interest • P.151 Q14. Judy borrows $1000000 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 $50000. • She pays $50000 less in Bank B from in Bank A.

28. Time for Practice • Page 149 of Textbook 1A • Class Practice • Page 150 of Textbook 1A • Questions 1 – 13 • Pages 69 - 70 of WB 1A • Questions 4 - 7

29. 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 • 12-12-12 (Wed) • 8:30 am – 9:30 am (1 hour) • Room 104

30. Enjoy the world of Mathematics!