1. Chapter 1 Measurements
1.7 Problem Solving

2. 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.)

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. 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 x Unit = Unit 2
Unit 1
Given x Conversion = Needed
unit factor unit

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

7. 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

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

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

10. Using Two or More Factors
Often, two or more conversion factors are required to obtain the unit needed for the answer.
Unit 1 Unit 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 x Unit x Unit 3 = Unit 3
Unit Unit 2

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

12. 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 h
24 h min
Units = day2/min, which is not the unit needed
Units don’t cancel properly. Therefore, setup is wrong.

13. 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 lb and kg
1 kg lb
STEP 4 Set up problem:
165 lb x kg = kg (3SF)
2.20 lb

14. 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

15. 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 qt x 1 gal = gal
1 L qt
3 SF SF exact SF

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

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

18. 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?

19. Solution Given: 7500 ft, 65 m/min Need: min Plan: ft in. cm m min
Equalities: 1 ft = 12 in in. = 2.54 cm 1 m = 100 cm
1 min = 65 m (walking pace)
Set up problem:
7500 ft x 12 in. x cm x 1m x 1 min
1 ft in. 100 cm m
= 35 min (2SF)

20. 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, kg fat
100 kg
percent factor
86 kg x kg fat = 9.5 kg of fat

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

22. 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 lb candy
= lb of sugar