COLLEGE OF SOUTHERN NEVADA MATH 120 - FUNDAMENTALS OF COLLEGE MATHEMATICS LATEST 2023/2024 LECTURE NOTES TO ACCOMPANY THE BOOK.

COLLEGE OF SOUTHERN NEVADA MATH 120 - FUNDAMENTALS OF COLLEGE MATHEMATICS LATEST 2023/2024 LECTURE NOTES TO ACCOMPANY THE BOOK. Chapter 1 1.2 Percents • A percent is a fraction whose denominator is 100. Example: 5%  5 . 100 • To convert a percent to a decimal, remove the % sign and move the decimal point two places to the left. Example: 5%  0.05 . • To convert a decimal to a percent, move the decimal point two places to the right and attach the % sign to the number. Example: 0.275  27.5% • To convert a fraction to a percent, divide the numerator by the denominator to get a decimal, then convert the decimal to a percent. Example: 5 16  0.3125  31.25% . • To convert a percent to a fraction, change the % sign to 100 in the denominator, then reduce or simplify. Example: 12.5%  12.5  125  1 . 100 1000 8 • The phrase “percent of” signifies a multiplication. Examples: (a) 8% of 30 = 8% 30  0.08  30  2.4 . (b) 6% of what number is 24? If we let x be the unknown number, then we need to solve the equation 6%x  24 , i.e., 0.06 x  24 . Divide both sides by 0.06 we find the answer x  24  400 . 0.06 • Percent change is defined as %  new  old 100% , you can also write the formula as old %   new   old 1100%. It also has an equivalent form new  old  old  %. Since 100% equals  1, the 100% in the formula is simply a reminder that you need to write the final answer as a percent. Percent change can be positive or negative. If it is negative, the new number is less than the old number. Examples: (a) Last week a dozen eggs cost $2.00, this week they cost $2.30, the 2.30  2.00 percent change is 2.00 100%  0.15 100%  15% . (b) A city had a population of 30,000 last year. The population has since declined 15%. What is the city’s current population? Solution: Since the population has declined, we know the percent change is negative, i.e., %  15% . So the current population is new  old  old %  30000 30000 (15%)  30000 4500  25500. 1.3 Simple and Compound Interests • Simple interest formula I  PRT , where I is the amount of interest, P the principal, R the interest rate, and T the amount of time. Even though the formula is very simple (no pun intended), you need to remember that the units for R and T must match. Normally R is given as annual rate, then T must be given as number of years. But if R is given as a daily rate, then T must be given as the number of days. Examples: (a) What is the amount of interest if the principal is $500, the annual rate is 18%, and the time is 3 years? Solution: Follow the formula, I  PRT  500  0.18  3  270 . (b) If the interest amount is $30 on a loan of $1,000 after one month, what is the annual rate of the loan? Solution: Here we are to find the annual rate, but time is given in months, so we first need to convert one month to years. Since there are 12 months in a year, one month = 1/12 of a year. So I  PRT becomes 30  1000  R  1 12 . Multiply both sides of the equation by 12 we get R  0.36 . So is the answer 0.36%? No, we need to move the decimal point two places to the right to convert a decimal to a percent, the rate is 36%: 0.36 = 36%. nt • Compound interest formula A    r  , where A = balance (interest and principal combined),    n  P = principal, r = annual rate, t = time measured in years, and n = number of times interest is compounded per year. Examples: (1) Invest $500 at annual rate 6% with interest compounded monthly. What is the balance after 3 years? Solution: Here