1. 1 Chapter 2 Measurements 2.7 Problem Solving Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

2. 2 To solve a problem Identify the given unit Identify the needed unit. Problem: A person has a height of 2.0 meters. What is that height in inches? The given unit is the initial unit of height. given unit = meters (m) The needed unit is the unit for the answer. needed unit = inches (in.) Given and Needed Units

3. 3 Learning Check An injured person loses 0.30 pints of blood. How many milliliters of blood would that be? Identify the given and needed units in this problem. Given unit= _______ Needed unit = _______

4. 4 Solution An injured person loses 0.30 pints of blood. How many milliliters of blood would that be? Identify the given and needed units in this problem Given unit=pints Needed unit =milliliters

5. 5 STEP 1 State the given and needed units. STEP 2 Write a plan to convert the given unit to the needed unit. STEP 3 Write equalities/conversion factors that connect the units. STEP 4 Set up problem with factors to cancel units and calculate the answer. Unit 1 x Unit 2 = Unit 2 Unit 1 Given Conversion Needed unit factor unit Guide to Problem Solving (GPS)

6. 6 Setting up a Problem How many minutes are 2.5 hours? given unit= 2.5 hr needed unit= ? min plan= hr min Set up problem to cancel units (hr). given conversion needed unit factor unit 2.5 hr x 60 min = 150 min 1 hr (2 sig figs) Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

7. 7 A rattlesnake is 2.44 m long. How long is the snake in centimeters? 1) 2440 cm 2) 244 cm 3) 24.4 cm Learning Check

8. 8 A rattlesnake is 2.44 m long. How long is the snake in centimeters? 2)244 cm given conversion needed unit factor unit 2.44 m x 100 cm = 244 cm 1 m 3 SF exact 3 SF Solution

9. 9 Often, two or more conversion factors are required to obtain the unit needed for the answer. Unit 1 Unit 2 Unit 3 Additional conversion factors can be placed in the setup to cancel each preceding unit Given unit x factor 1 x factor 2 = needed unit Unit 1 x Unit 2 x Unit 3 = Unit 3 Unit 1 Unit 2 Using Two or More Factors

10. 10 How many minutes are in 1.4 days? Given unit: 1.4 days Needed unit: min Plan: days hr min Equalties: 1 day = 24 hr 1 hr = 60 min Set up problem: 1.4 days x 24 hr x 60 min = 2.0 x 10 3 min 1 day 1 hr 2 SF exact exact = 2 SF Example: Problem Solving

11. 11 Be sure to check your unit cancellation in the setup. The units in the conversion factors must cancel to give the correct unit for the answer. What is wrong with the following setup? 1.4 day x 1 day x 1 hr 24 hr 60 min Units = day 2 /min is Not the needed unit Units don’t cancel properly. Check the Unit Cancellation

12. 12 Using the GPS What is 165 lb in kg? STEP 1 Given 165 lb Need kg STEP 2 Plan STEP 3 Equalities/Factors 1 kg = 2.205 lb 2.205 lb and 1 kg 1 kg 2.205 lb STEP 4 Set Up Problem

13. 13 A bucket contains 4.65 L water. How many gallons of water is that? Given: 4.65 LNeed:gal Unit plan: L qt gallon Equalities:1 L = 1.057 qt 1 gal = 4 qt Set up Problem: Learning Check

14. 14 Given: 4.65 L Needed: gal Plan: L qt gal Equalities: 1 L = 1.057 qt1 gal = 4 qt Set Up Problem: 4.65 L x 1.057 qt x 1 gal = 1.23 gal 1 L 4 qt 3 SF 4 SF exact 3 SF Solution

15. 15 If a ski pole is 3.0 feet in length, how long is the ski pole in mm? Learning Check

16. 16 3.0 ft x 12 in x 2.54 cm x 10 mm = 1 ft 1 in. 1 cm Calculator answer: 914.4 mm Correct answer:910 mm (2 SF rounded) Check factor setup: Units cancel properly Check needed unit: mm Solution

17. 17 If your pace on a treadmill is 65 meters per minute, how many minutes will it take for you to walk a distance of 7500 feet? Learning Check

18. 18 Given: 7500 ft65 m/minNeed: min Plan: ft in. cm m min Equalities: 1 ft = 12 in. 1 in. = 2.54 cm 1 m = 100 cm 1 min = 65 m (walking pace) Set Up Problem: 7500 ft x 12 in. x 2.54 cm x 1m x 1 min 1 ft 1 in. 100 cm 65 m = 35 min final answer (2 SF) Solution

19. 19 Percent Factor in a Problem If the thickness of the skin fold at the waist indicates 11% body fat, how much fat is in a person with a mass of 86 kg? percent factor 86 kg mass x 11 kg fat 100 kg mass = 9.5 kg fat Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

20. 20 How many lb of sugar are in 120 g of candy if the candy is 25% (by mass) sugar? Learning Check

21. 21 How many lb of sugar are in 120 g of candy if the candy is 25%(by mass) sugar? % factor 120 g candy x 1 lb candy x 25 lb sugar 453.6 g candy 100 lb candy = 0.066 lb sugar Solution