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Please open your laptops, log in to the MyMathLab course web site, and open Quiz 2.5.

You will have access to the online calculator on your laptop during this quiz. No other calculator may be used . Please open your laptops, log in to the MyMathLab course web site, and open Quiz 2.5.

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Please open your laptops, log in to the MyMathLab course web site, and open Quiz 2.5.

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  1. You will have access to the online calculator on your laptop during this quiz. No other calculator may be used. Please open your laptops, log in to the MyMathLab course web site, and open Quiz 2.5. • IMPORTANT NOTE: If you have time left after you finish the problems on this quiz, use it to check your answers before you submit the quiz! • Remember to turn in your answer sheetto the TA when the quiz time is up.

  2. Please CLOSE YOUR LAPTOPS, and turn off and put away your cell phones, and get out your note-taking materials.

  3. Sections 2.6 and 2.7 More Problem Solving

  4. Solving Percent Problems Note: “Per cent” means “per 100”. For example, 17% = 17/100 = 0.17 . A percent problem has three different parts: amount = percent x base Any one of the three quantities may be unknown. 1. When we do not know the amount: n= 10% · 500 2. When we do not know the base: 50 = 10% ·n 3. When we do not know the percent: 50 = n· 500

  5. Solving a Percent Problem: Amount Unknown Example: What is 9% of 65? 5.85 is 9% of 65

  6. Solving a Percent Problem: Base Unknown 36 is 6% of what? Example: 36 is 6% of 600

  7. Solving a Percent Problem: Percent Unknown Example: 24 is what percent of 144?

  8. Percent Problem Tip: If you’re having trouble doing percent problems that give you a new value after a certain percent increase or decrease from an old value (such as sales tax problems), try thinking about it this way: Think about when you go shopping to buy, say, a TV. Usually you know how much the TV costs, for example $400, and the percent tax rate, for example 5.5%. Normally what you do (or the salesclerk’s computer does) is calculate the TOTAL COST by taking 5.5% of $400, then adding that amount back onto the $400 price of the TV to get the total cost to you.

  9. The working equation is PRICE + TAX = TOTAL COST. In words, here’s what you did (after writing the 5.5% as a decimal, 0.055): PRICE + .055 times PRICE = TOTAL COST Plugging in the numbers, we get 400 + .055 x 400 = 400 + 22 = 422. Notice that you’ve multiplied the OLD VALUE (the price before tax) by the .055.

  10. The same basic format applies to anything with a percent increase or decrease from an original amount: Old amount +/- % ofold amount = new amount (Remember to write the percent as a decimal.) This equation works for raises in pay, population increases or decreases, and many other percent change problems, especially where you’re given the new amount and the percent change and you need to work backwards to find out the old amount.

  11. Example: After a 6% pay raise, Nora’s 2013 salary is $39,703. What was her salary in 2012? (Round to the nearest dollar). Solution: Recall the equation: Old amount + % ofold amount = new amount The “old amount” is her 2012 salary, which is unknown, so we’ll call it X. This gives us the equation X + 0.06X = 39703

  12. Example (cont.) After a 6% pay raise, Nora’s 2013 salary is $39,703. What was her salary in 2012? (Round to the nearest dollar). 1X + 0.06X = 39703 This simplifies to 1.06X = 39703 Divide both sides X = 39703 by 1.06 to get X. 1.06 Answer: Her 2012 salary was $37,456

  13. Now checkyour answer: 37456 + .06 x 37456 = 37456 + 2247 = 39703 NOTE that this DOES NOT give you the same answer as if you subtracted 6% of the new salary (39703) from the new salary. Try it and you’ll see that it doesn’t work. (It’s not real far off, but enough to give you the wrong answer, and the bigger the percentage, the farther off you’ll be.) On these kinds of problems, we won’t give partial credit for those “close” answers on tests.

  14. Sample problem from today’s homework:

  15. Solving Mixture Problems Example: The owner of a candy store is mixing candy worth $6 per pound with candy worth $8 per pound. She wants to obtain 144 pounds of candy worth $7.50 per pound. How much of each type of candy should she use in the mixture? • Solution: • Let n = the number of pounds of candy costing $6 per pound. • Let 144 – n = candy costing $8 per pound.

  16. # of pounds of $7.50 candy # of pounds of $6 candy # of pounds of $8 candy Use a table to summarize the information. 6n + 8(144  n) = 144(7.5)

  17. (144  n) = 144  36 = 108 6n + 8(144  n) = 144(7.5) 6n + 1152  8n = 1080 1152  2n = 1080 2n = 72 n = 36 She should use 36 pounds of the $6 per pound candy. She should use 108 pounds of the $8 per pound candy.

  18. ? ? 6(36) + 8(108) = 144(7.5) 216 + 864 = 1080 ? 1080 = 1080 Check:Will using 36 pounds of the $6 per pound candy and 108 pounds of the $8 per pound candy yield 144 pounds of candy costing $7.50 per pound? 

  19. Distance Problems: distance = rate · time or d = r · t When the amount in the formula is distance, we refer to the formula as the distance formula.

  20. The assignment on this material (HW 2.6/7) is due at the start of the next class session, and there will be a short quiz on it as usual, either at the start or end of the class session.. Lab hours: Mondays through Thursdays 8:00 a.m. to 6:30 p.m.

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