Chapter 1 Measurements 1.7 Problem Solving

Initial and Final Units 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 meters (the unit of height). given unit = meters (m) The needed unit is inches (for the answer). needed unit = inches (in.)

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 = _______

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

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. 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

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

Setting up a Problem How many minutes are 2.5 h? Given unit = 2.5 h Needed unit = ? min Plan = h min Set up problem to cancel hours (h). Given Conversion Needed unit factor unit 2.5 h x 60 min = 150 min (2 SF) 1 h

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

Solution Given unit: 2.44 m Needed unit: cm Plan: m cm Equality: 1 m = 100 cm Factors: 1 m and 100 cm 100 cm 1 m Set up problem: 2.44 m x 100 cm = 244 cm (answer 2) 1 m

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 x factor 2 = Needed unit Unit 1 x Unit 2 x Unit 3 = Unit 3 Unit 1 Unit 2

Example: Problem Solving How many minutes are in 1.6 days? Given unit: days Needed unit: min Factor 1 Factor 2 Plan: days h min Set up problem: 1.6 days x 24 h x 60 min = 2300 = 2.3 x 103 min 1 day 1 h 2 SFs Exact Exact = 2 SFs

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. Example: What is wrong with the following setup? 1.4 day x 1 day x 1 h 24 h 60 min Units = day2/min, which is not the unit needed Units don’t cancel properly. Therefore, setup is wrong.

Guide to Problem Solving What is 165 lb in kg? STEP 1 Given: 165 lb Need: kg STEP 2 Plan: lb kg STEP 3 Equalities/conversion factors: 1 kg = 2.20 lb 2.20 lb and 1 kg 1 kg 2.20 lb STEP 4 Set up problem: 165 lb x 1 kg = 74.8 kg (3SF) 2.20 lb

Learning Check A bucket contains 4.65 L of water. How many gallons of water is that? Given: 4.65 L Need: gal Plan: L qt gal Equalities: 1.06 qt = 1 L 1 gal = 4 qt

Solution Given : 4.65 L Need: gal Plan: L qt gallon Equalities: 1.06 qt = 1 L; 1 gal = 4 qt Set up problem: 4.65 L x 1.06 qt x 1 gal = 1.23 gal 1 L 4 qt 3 SF 3 SF exact 3 SF

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

Solution Given: 3.0 ft Need: mm Plan: ft in. cm mm Equalities: 1 ft = 12 in. 2.54 cm = 1 in. 1 cm = 10 mm Set up problem: 3.0 ft x 12 in. x 2.54 cm x 10 mm = 910 (2SF, rounded) 1 ft 1 in. 1 cm Check initial unit: ft Check needed unit: mm Check factor setup: units cancel properly

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?

Solution Given: 7500 ft, 65 m/min Need: 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 (2SF)

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? 11% body fat, 11 kg fat 100 kg percent factor 86 kg x 11 kg fat = 9.5 kg of fat

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

Solution How many pounds (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 = 0.066 lb of sugar