1 Chapter 1 Measurements 1.7 Problem Solving Copyright © 2009 by Pearson Education, Inc.

2 To solve a problem, identify the given unit. identify the needed unit. Example: 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 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 given in this problem. Given unit= _______ Needed unit = _______

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 given in this problem. Given unit= pints Needed unit = milliliters

5 Write the given and needed units. Write a plan to convert the given unit to the needed unit. Write equalities and conversion factors that connect the units. Use conversion factors to cancel the given unit and provide the needed unit. Unit 1 x Unit 2 = Unit 2 Unit 1 Given x Conversion= Needed unit factor unit Problem Setup

6 Study Tip: Problem Solving Using GPS The steps in the Guide to Problem Solving (GPS) are useful in setting up a problem with conversion factors.

7 How many minutes are in 2.5 hours? Given unit= 2.5 h Needed unit=min Plan=h min Set Up Problem Given Conversion Needed unit unitfactor 2.5 h x 60 min = 150 min ( 2 SF ) 1 h Setting Up a Problem Copyright © 2009 by Pearson Education, Inc.

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

9 A rattlesnake is 2.44 m long. How many cm long is the snake? 2)244 cm Given Conversion Needed unit factor unit 2.44 m x 100 cm = 244 cm 1 m Solution

10 Often, two or more conversion factors are required to obtain the unit needed for the answer. Unit 1 Unit 2Unit 3 Additional conversion factors are placed in the setup problem 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

11 How many minutes are in 1.4 days? Given unit: 1.4 days Factor 1 Factor 2 Plan: days h min Set Up Problem: 1.4 days x 24 h x 60 min = 2.0 x 10 3 min 1 day 1 h (rounded) 2 SF Exact Exact = 2 SF Example: Problem Solving

12 Be sure to check the 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 h 24 h 60 min = day 2 /min is not the unit needed Units don’t cancel properly. Study Tip: Check Unit Cancellation

13 A bucket contains 4.65 L of water. Write the setup for the problem and calculate the gallons of water in the bucket. Plan: L qt gallon Equalities:1.06 qt = 1 L 1 gal = 4 qt Set Up Problem: 4.65 L x x 1.06 qt x 1 gal = 1.23 gal 1 L 4 qt Learning Check

14 Given: 4.65 L Needed: gallons Plan: L qt gallon Equalities: 1.06 qt = 1 L; 1 gal = 4 qt Set Up Problem: 4.65 L x x 1.06 qt x 1 gal = 1.23 gal 1 L 4 qt 3 SF 3 SF exact 3 SF Solution

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

16 Equalities: 1 ft = 12 in. 1 in. = 2.54 cm 1 cm = 10 mm Set Up Problem: 3.0 ft x 12 in. x 2.54 cm x 10 mm = 1 ft 1 in. 1 cm Calculator answer = mm Final answer= 910 mm (2 SF rounded) Check Factors in Setup: Units cancel properly Check Final Unit: mm Solution

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 Solution 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) Given: 7500 ft65 m/minNeed: min Plan: ft in. cm m 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)

19 Percent Factor in a Problem If the thickness of the skin fold at the waist indicates an 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 of fat Copyright © 2009 by Pearson Education, Inc.

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

21 Solution How many lb of sugar are in 120 g of candy if the candy is 25%(by mass) sugar? percent factor 120 g candy x 1 lb candy x 25 lb sugar 454 g candy 100 lb candy = lb of sugar