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# Percentages, simple and compound interest; time, distance and speed

Percentages, simple and compound interest; time, distance and speed

Percentages, simple and compound interest; time, distance and speed

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### Percentages, simple and compound interest; time, distance and speed

1. 1. PERCENTAGES, SIMPLE AND COMPOUND INTEREST, TIME, DISTANCE AND SPEED NUMBERS
2. 2. PERCENTAGES Introduction on how to calculate a percentage. https://www.youtube.com/watch?v=rR95Cbcjz us&t=237s
3. 3. FOR EXAMPLE Type1: What is the 23% of \$500? So we write 23% as a fraction: 23 100 Then we multiply times \$500: 23 100 × \$500 = \$115 Type 2: If 17% of the total price is \$1445, what is the total price? \$1445 is 17% of some number So we call X = 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑡ℎ𝑎𝑡 1445 𝑖𝑠 17% 𝑜𝑓 17% × X = 0.17 × X = 1445  now we solve for X 𝑋 = \$1445 0.17 = \$8500
4. 4. FOR EXAMPLE Type 2: (continued) 18% of the students at a school play tennis. If the school has 216 tennis players, then how many students are there? 216 is 18% of some number So we call: 𝑋 = 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑡ℎ𝑎𝑡 216 𝑖𝑠 18% 𝑜𝑓 18% × 𝑋 = 0.18 × 𝑋 = 216  Solve for X 𝑋 = 216 0.18 = 1200 students Type 3: What percentage is 85 of 250? Now we will use the following formula: Percentage = 𝑎𝑚𝑜𝑢𝑛𝑡 𝑡𝑜𝑡𝑎𝑙 × 100% So we have: 𝑎𝑚𝑜𝑢𝑛𝑡=85 𝑡𝑜𝑡𝑎𝑙=250 × 100% = 34%
5. 5. PERCENTAGE CHANGE – EXAMPLES In a sale the Price of a shirt was reduced from \$40 to \$32. Find the percentage decrease. This is a similar Type 3 question, so we will be using the following formula: 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑐ℎ𝑎𝑛𝑔𝑒 = 𝐶ℎ𝑎𝑛𝑔𝑒 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 × 100% In this case, the original is \$40 and the change in the Price is \$40-\$32=\$8 So 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑐ℎ𝑎𝑛𝑔𝑒 = 8 32 × 100% = 20% Type 4: A car worth \$5000 loses 15% of its value in a year. What is it worth after one year? If it loses 15% of its value after a year, it means that after one year the car is going to cost 85% of the value that cost at the beginning of the year. It means that it is going to cost 85% of \$5000: 85% × \$5000 = 0.85 × \$5000 = \$4250
6. 6. PERCENTAGE CHANGE - EXAMPLES Type 4: (continued) A footbal club’s average attendance was 41,200. The following year the attendance rose by 4%. What was its new average attendance? If the attendance rose by 4% it means that 4% of 41,200 will be added to the following year: 4% of 41,200 = 0.04x41200 = 1648 So the new average attendance will be: 41200+1648=42848 Another way: since it increased by 4%, the new average attendance will be 104% of 41200  1.04x41200 = 42848 The value of a house in 2007 was \$300000. What was its value in 2008 after a fall of 8%? Since there was a fall of 8%, it means that the house now costs 92% of the cost in 2007 92% of \$300000 = 0.92x\$300000 = \$276000 Another way: since there was a fall of 8%, 0.08x\$300000 = \$24000  so the new cost will be: \$300000 - \$24000 = \$276000
7. 7. SIMPLE AND COMPOUND INTEREST • When you invest money in a bank offering simple interest you only get interest on the original principal. • I = PRT (I = interest earned, P is the investment, R is the percentage rate and T is the time) Simple interest • When you invest money in a bank offering compound interest, the interest you get each year is added to your principal and the next year’s interest is paid on the increased amount in your account. Compound interest If you sabe money in a savings scheme (for example, with a bank), the initial amount you invest is called the PRINCIPAL and you recieve interest on your money.
8. 8. FOR EXAMPLE – SIMPLE INTEREST Joseph invests \$500 at 4% per annum simple interest. How much will he have at the end of 5 years? (per annum=per year) Interest = 4% of \$500 = 0.04 x \$500 = \$20 per year Total interest in 5 years = \$20 x 5 = \$100 Joseph will have: \$500 + \$100 = \$600 Aruna invests \$480 at 5% simple interest per annum. What is the investment worth after: a) 7 months A whole year’s interest = 5% of \$480 = 0.05 x \$480 = \$24 7 months interest = \$24 x 7 12 = \$14 After 7 months, Aruna’s investment will be worth: \$480 + \$14 = \$494 b) 3 years 3 years interest = 3 x \$24 = \$72 After 3 years Aruna’s investment will be worth: \$480 + \$72 = \$552
9. 9. FOR EXAMPLE – COMPOUND INTEREST Ali invests \$200 at 3% compound interest. What amount will he have after 3 years? Year 1: 3% of \$200 = 0.03 x \$200 = \$6 \$200 + \$6 = \$206  at the start of year 2 Year 2: 3% of \$206 = 0.03 x \$206 = \$6.18 \$206 + \$6.18 = \$212.18  at the start of year 3 Year 3: 3% of \$212.18 = \$6.37 (to 2 decimal places) \$212.18 + \$6.37 = \$218.55  at the end of year 3 Rose borrows \$800 at 7% compound interest for 2 years. How much does she owe at the end of the 2 years? 7% of \$800 = 0.07 x \$800 = \$56  after 1 year, Rose owes \$856 7% of \$856 = 0.07 x \$856 = \$59.92  after 2 years Rose owes \$915.92
10. 10. FOR EXAMPLE – COMPOUND INTEREST (CONTINUED) A car loses 30% of its value each year. If it cost \$18000 then, how much is it worth after 3 years? One way: to calculate the value of the car after each year \$18000 x 0.3 = \$5400  the value after year 1: \$18000 - \$5400 = \$12600 \$12600 x 0.3 = \$3780  the value after year 2: \$12600 - \$3780 = \$8820 \$8820 x 0.3 = \$2646  the value after year 3: \$8820 - \$2646 = \$6174 Another way: NOTE: At the end of each year the car is worth 70% (0.7) of its value at the start of that year. \$18000 x 0.7 = \$12600  the value after year 1 \$12600 x 0.7 = \$8820  the value after year 2 \$18000 × 0.73 = \$6174 \$8820 x 0.7 = \$6174  the value after year 3
11. 11. WORKING WITH TIME Units used for measuring time: 1 year = 365 days = 8,760 hours = 525,600 minutes = 31,536,000 seconds Since 1 day has 24 hours & 1 hour has 60 minute & 1 minute has 60 seconds. This numbers are the ones you have to use to go from one unit to another. There are two ways to express time: 24-hr or 12-hr. (The 12-hr has AM and PM) Remember: when doing the calculations on your calculator, the decimal isn’t the minute!!! For example: 5.3 hours ≠ 5 hours and 3 minutes or 5 hours and 30 minutes, it means 5 hours and 18 minutes. You can try with www.springfrog.com/converter/decimal-time.htm
12. 12. FOR EXAMPLE A girl arrived at a rehearsal at 1935 and left 2129. For how long was she at the rehearsal? 1935 to 2135 is 2 hours, but she stayed till 2129 (6 minutes less than 2 hours)  she stayed for 1 hour and 54 minutes A plane leaves Shanghai at 1055 local time and arrives at London the same day at 1550 local time. The website says that the flight takes 12h 55min. How many hours is the time in London behind the time in Shanghai? First: I want to know at what time from Shanghai did the plane arrive to London (A que hora de Shanghai llego el avión a Londres) 1055 + 12h 55min = 2350  so the plane arrives at 2350 (Shanghai time). This is 1550 London time, so 2350 – 1550 = 8hrs  so London is 8 hours behind Shanghai.
13. 13. FOR EXAMPLE (CONTINUED) Derek wanted to get from Twickenham to Wembley. A website gave him the information shown in the table: (a) How long did the journey take in total? (b) At which station did Derek wait for 8 minutes? (a) He departed at 1158 and arrived at 1302. 1158 + 1hr = 1258 + 4 minutes = 1302  So the journey took 1 hour and 4 minutes (b) 1158 – 1203 = 5min; 1211 – 1203 = 8min; 1230 – 1211 = 9min; 1235 – 1230 = 5min; 1242 – 1235 = 7min; 1248 – 1242 = 6min; 1302 – 1248 =4min.  Derek waited for 8 minutes at the Richmond Rail Station. Depart Twickenham Rail Station 1158 Arrive Richmond Rail Station 1203 Depart Richmond Rail Station 1211 Arrive Willesden Junction Underground Station 1230 Depart Willesden Junction Underground Station 1235 Arrive Wembley Central Station 1242 Depart Wembley Central Station 1248 Arrive Wembley Stadium 1302
14. 14. TIME, DISTANCE AND SPEED 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑠𝑝𝑒𝑒𝑑(𝑆) = 𝑡𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑒𝑑 (𝐷) 𝑡𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 (𝑇) Example: A car travels at 189km at an average speed of 60km/hr, how long does the journey take? S = D/T  T = D/S  𝑇 = 189𝑘𝑚 60 𝑘𝑚 ℎ𝑟 = 3.15ℎ𝑟𝑠 = 3ℎ𝑟 9𝑚𝑖𝑛