1

Chapter 2 Measurements

2.7Problem Solving

Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings

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 thatheight 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. Howmany milliliters of blood would that be?

Identify the given and needed units in thisproblem.

Given unit = _______

Needed unit = _______

4

Solution

An injured person loses 0.30 pints of blood. Howmany milliliters of blood would that be?

Identify the given and needed units in thisproblem

Given unit = pints Needed unit = milliliters

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 2Unit 1 Given Conversion Needed

unit factor unit

Guide to Problem Solving (GPS)

6

Setting up a ProblemHow many minutes are 2.5 hours?given unit = 2.5 hrneeded unit = ? minplan = 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

A rattlesnake is 2.44 m long. How long is the snakein centimeters?1) 2440 cm2) 244 cm3) 24.4 cm

Learning Check

8

A rattlesnake is 2.44 m long. How long is thesnake 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

• 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 unitGiven unit x factor 1 x factor 2 = needed unitUnit 1 x Unit 2 x Unit 3 = Unit 3

Unit 1 Unit 2

Using Two or More Factors

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 103 min

1 day 1 hr 2 SF exact exact = 2 SF

Example: Problem Solving

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 = day2/min is Not the needed unitUnits don’t cancel properly.

Check the Unit Cancellation

12

Using the GPS What is 165 lb in kg?STEP 1 Given 165 lb Need kgSTEP 2 PlanSTEP 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

A bucket contains 4.65 L water. How many gallons of water is that?Given: 4.65 L Need: galUnit plan: L qt gallon

Equalities: 1 L = 1.057 qt 1 gal = 4 qt

Set up Problem:

Learning Check

14

Given: 4.65 L Needed: gal

Plan: L qt gal

Equalities: 1 L = 1.057 qt 1 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

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

Learning Check

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 properlyCheck needed unit: mm

Solution

17

If your pace on a treadmill is 65 meters perminute, how many minutes will it take for you towalk a distance of 7500 feet?

Learning Check

18

Given: 7500 ft 65 m/min Need: minPlan: ft in. cm m minEqualities: 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

Percent Factor in a ProblemIf the thickness of the skin fold atthe waist indicates 11% body fat,how much fat is in a person with amass of 86 kg?

percent factor86 kg mass x 11 kg fat

100 kg mass

= 9.5 kg fat

Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings

20

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

Learning Check

21

How many lb of sugar are in 120 g of candy if thecandy 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