Chapter 1 Measurements 1 6 Problem Solving Copyright

Given and Needed Units 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. ) 2

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 = _______ 3

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 Needed unit = pints = milliliters 4

Problem Setup • Write the given and needed units. • Write a unit 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 Given unit x Unit 2 = Unit 2 Unit 1 Conversion = Needed factor unit 5

Guide to Problem Solving The steps in the Guide to Problem Solving are useful in setting up a problem with conversion factors. 6

Setting up a Problem How many minutes are 2. 5 hours? Given unit = 2. 5 hr Needed unit = min Unit Plan = hr min Setup problem to cancel hours (hr). Given Conversion Needed unit factor unit 2. 5 hr x 60 min = 150 min (2 SF) 1 hr Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings 7

Learning Check A rattlesnake is 2. 44 m long. How many centimeters long is the snake? 1) 2440 cm 2) 244 cm 3) 24. 4 cm 8

Solution A rattlesnake is 2. 44 m long. How many centimeters long is the snake? 2) 244 cm Given Conversion Needed unit factor unit 2. 44 m x 100 cm = 244 cm 1 m 9

Using Two or More Factors • 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 are placed in the setup to cancel each preceding unit Given unit x factor 1 Unit 1 x Unit 2 Unit 1 x factor 2 x Unit 3 Unit 2 = needed unit = Unit 3 10

Example: Problem Solving How many minutes are in 1. 4 days? Given unit: 1. 4 days Factor 1 Plan: days Factor 2 hr Set up problem: 1. 4 days x 24 hr x 60 min 1 day 1 hr 2 SF Exact min = 2. 0 x 103 min = 2 SF 11

Check the Unit Cancellation • 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 Units = x 1 day x 1 hr 24 hr 60 min day 2/min is not the unit needed Units don’t cancel properly. 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. 20 lb and 1 kg 2. 20 lb STEP 4 Set Up Problem 165 lb x 1 kg = 75. 0 kg 2. 20 lb 13

Learning Check A bucket contains 4. 65 L of water. How many gallons of water is that? Unit plan: L qt gallon Equalities: 1. 06 qt = 1 L 1 gal = 4 qt Set up Problem: 14

Solution Given: 4. 65 L Needed: gallons Plan: qt L gallon Equalities: 1. 06 qt = 1 L; 1 gal = 4 qt Set Up Problem: 4. 65 L x x 1. 06 qt 1 L 3 SF x 1 gal 4 qt exact = 1. 23 gal 3 SF 15

Learning Check If a ski pole is 3. 0 feet in length, how long is the ski pole in mm? 16

Solution 3. 0 ft x 12 in x 2. 54 cm x 10 mm = 1 ft 1 in. 1 cm Calculator answer: 914. 4 mm Needed answer: 910 mm (2 SF rounded) Check factor setup: Check needed unit: Units cancel properly mm 17

Learning Check 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? 18

Solution Given: 7500 ft Plan: ft 65 m/min in. Equalities: 1 ft = 12 in. Need: min cm m 1 in. = 2. 54 cm min 1 m = 100 cm 1 min = 65 m (walking pace) Set Up Problem: 7500 ft x 12 in. x 1 ft 2. 54 cm 1 in. x 1 m x 1 min 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 fat Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings 20

Learning Check How many lb of sugar are in 120 g of candy if the candy is 25% (by mass) sugar? 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 454 g candy x 25 lb sugar 100 lb candy = 0. 066 lb sugar 22