Problem Solving
The Ralph Way :)
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The following is a power point that has been designed to help
you work through dimensional analysis and alternative method
to solving story problems.
We will work with dimensional analysis first.
Then we will tackle Story problems.
(As a help to those in Ralph’s and Nancy’s Online Class after
each problem we will show you what you would enter into the
test box to show your work)
Dimensional Analysis
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Dimensional analysis is the scientist fancy
word for converting from one unit to another.
It can be scary at the beginning but we hope
to easy you into it.
To start let’s work with something we are all
familiar with, Money.
If I ask how many quarters are in two dollars?
If I ask how many quarters are in two dollars?
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You of course knew the answer was 8.
Correct?
But let us look at this as a scientist.
Two dollars = $2.00
There are 4 quarters in a dollar or
$1.00 = 4 quarters
So $2.00 X 4 quarters =>
$1.00
Cancel the dollar signs so;
$2.00 x 4 quarters = 8 quarters
$1.00
For online test $2.00 x (4qtr/$1.00)= 8 quarters
Try another money problem
How many twenty dollar bills does it take to
have $140?
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Again you know the answer is 7.
But using math $140/$20 = 7 bills
In dimensional analysis
We know 1 bill = $20
So $140 x 1 bill = 7 bills
$20
For online test $140 x (1bill / $20) = 7 bills
We can use Dimensional Analysis to convert
between any two similar units. Just like the
money.
We have broken it down to a set of
questions/steps to help you.
 What unit are you canceling?
 Place that unit on the opposite side of the
line.
 Place the other unit.
 Determine which unit is bigger
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Place a 1 in front of that unit
Place conversion number in front of other
unit.
Another money problem. (do not cheat, Try it
through dimensional analysis)
How many nickels are in 65 cents?
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So we are starting out with 65 cents, write 65
cents on your paper. 65 cents X
=
And we know 1 nickel = 5 cents
What unit are you canceling?
Cents, correct?
Place that unit on the opposite side of the
line.
So 65 cents X
=
cents
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How many nickels are in 65 cents?
Place the other unit.
65 cents X
nickels =
cents
Determine which unit is bigger
Place a 1 in front of that unit
So a nickel is bigger because 5 cents = 1 nickel so it
gets the one.
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65 cents X
1 nickels =
cents
Place conversion number in front of other
unit.
65 cents X 1 nickels =
5 cents
How many nickels are in 65 cents?
We then cancel the units
> 65 cents X
1 nickels =
5 cents
We want our answer in nickels so I circle it.
> 65 cents X
1 nickels =
nickels
5 cents
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So our units are now nickels and we can do the math
by multiplying across the top and dividing across the
bottom.
65 X 1 nickels / 5 = 13
And our units are Nickels so
13 nickels.
For online test 65cent x (1 nickel / 5 cent) = 13 nickels
We know you are thinking why do we need to
do money the math way? Isn’t that making it
harder?
Yes and no, We are trying to teach you how to use
Dimensional analysis. What we just used for money
works for all conversions. Let’s look at a different
conversion problem and apply the same dimensional
analysis techniques.
Try
How many kilograms are 2056 g?
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So we are starting out with 2056 g, write 2056
g on your paper. 2056 g X
=
What unit are you canceling?
grams, correct?
How many kilograms are 2056 g?
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Place that unit on the opposite side of the line.
So 2056 g X
=
g
Place the other unit.
2056 g X
kg =
g
Determine which unit is bigger
 Place a 1 in front of that unit
So 1 kilogram is 1000g
2056 g X
1 kg =
g
Place conversion number in front of other unit.
2056 g X
1 kg =
1000 g
How many kilograms are 2056 g?
We then cancel the units
> 2056 g X
1 kg =
1000 g
We want our answer in kg so I circle it.
> 2056 g X
1 kg =
1000 g
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So our units are now kg and we can do the math by
multiplying across the top and dividing across the
bottom.
2056 X 1 kg / 1000 = 2.056
And our units are kg so
2.056 kg.
For online test 2056g x (1kg /1000g) = 2.056 kg
Now that you have a feel for
Dimensional Analysis, we will move
on to solving story problems which
includes the use of Dimensional
Analysis.
Story Problem Solving
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We have condensed the steps down to 7 step. The next few
pages are step up to show you the steps then to talk you
through each of the steps similar to what would be presented in
a lecture. The steps will then be used to help you solve several
problems to help ingrain the process.
Problem Solving
The Ralph Way :)
1. Read the whole problem
 2. Read the problem carefully for facts.
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a. What is known
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b. What are you trying to find, What
units is it in?
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c. What could be known.
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d. Other info.
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 3.
Select which number to start
with
 4. Put units in formula
 5. Cancel units, top L -> R, Bot L ->
R, Using Dimensional Analysis
 6. use Calculator
 7 Write down answer (with units),
Determine Sig. Figs., Recheck
Calculations
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1. Read the whole problem
 You want to read the problem without worrying
about numbers. You are doing this to work on the
flow of the question and story. It is often useful to
just skip the numbers or say number.
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2. Read the problem carefully for facts.
 This time you are reading the problem to look at
the numbers and the information you will need to
complete the problem.

a. What is known
 What are the numbers given in the problem. Write
them down with their units.
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2 b. What are you trying to find, What
units is it in?
 This is where you are looking at what number you are
solving for. I.E. Why are you even doing this problem. Try
to figure out what type of number you are looking for. Make
sure to include units.
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c. What could be known.
 What conversions could you look up, or do you
know? I try to think about what conversions I may
need to solve the problem.
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d. Other info.
 This is if there is some other piece of information
you want to note like temperature or date or etc.
 3.
Select which number to start
with
 4. Put units in formula
 5. Cancel units, top L -> R, Bot L ->
R, Using Dimensional Analysis
 6. use Calculator
 7 Write down answer (with units),
Determine Sig. Figs., Recheck
Calculations
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3. Select which number to start with
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This is often the hardest part. Often times you will be presented with
many numbers in a problem to pick from. So which number do you
pick? First ask yourself, “is there any conversions in the problem?”
Like 5 dogs equal a cat, this is a ratio or a conversion. So write it as
such 5 dogs , if it is not a ratio write it normal 5 mL.
1 cat
The easiest number to start with is one that contains similar units to
what you are solving for. If your answer needs a volume for
instance pick a number that has volume in it. Even if the units do
not match, you can always convert units.
So if you are solving for cats you would want to start with the 5
dogs/1 cat instead of the 5ml.
You should then place this number at the start of the problem.
Be mindful of the units and the numbers. If you want cats on top for the
answer you will need cats on top for the problem also so 1/5
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4. Put units in formula
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Now you need to include units with your number. Always write out the
units as well as the numbers they can be a great cross check of your
answer later so 1 cat
5 dogs
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5. Cancel units, top L -> R, Bot L -> R,
Using Dimensional Analysis
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See the next slide for the dimensional analysis steps
I always like to work very systematically when doing my canceling of units. I will
work across the top row from left to right of the problem canceling units as I go.
When I get an unit that I need for the answer I often circle it to tell me that I do
not need to cancel it. After I have canceled all the top row units I then work from
left to right across the bottom row.
Some times in the solving of a problem you place a unit from your problem that
does not have a conversion on the other side of the equation. You can place a
one on the other side of the problem if that helps you feel better about the
problem. This is because one (1) times or divided into anything does not
change the numbers.
Here is an example of a complex problem, (Don’t Panic this is just an example)
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Dimensional Analysis
which we covered earlier.
What unit are you canceling?
 Place that unit on the opposite side of
the line.
 Place the other unit.
 Determine which unit is bigger
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 Place
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a 1 in front of that unit
Place conversion number in front of
other unit.
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6. use Calculator
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Now that all the units are canceled you are ready to use the calculator.
The easiest way to do this without brackets and to much calculator error is to
use the Times (x) Key for everything above the horizontal line and use the
divide key for everything below the line.
You do not want to press the equals key till the very end! This is because
calculators introduce rounding errors when the equal key is pressed. Another
common error is when students try to times the top, write it down, then times the
bottom, right it down then try to put them back in the calculator to do the divide.
You have added to many steps and to many errors to get a good anwser.
Using the old Example back two pages
45 x 4.5 / 1000 / 4.184 = 0.0483987
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7 Write down answer (with units),
Determine Sig. Figs., Recheck
Calculations
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This is where you finish the problem off. Write down the answer, add the units, figure out
the significant figures and then redo the calculator one time to make sure the answer is right
Example 0.048/39 = 0.048 kg/J
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Now let us Try a few together
If a Recipe calls for 600. g of chocolate chips, how
many pounds of chips do you need?
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1. Read problem
2. read for facts
 a. what is known?
600.g
 b. what are we finding?
? in pounds
 c. What do we know?
 (from book) 1 lbs= 454g
 d. other info? none
3. select which number to
start with? You only have 600.g
4. put units formula
600 g
1
If a Recipe calls for 600. g of chocolate chips, how
many pounds of chips do you need?
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5. Cancel units, top L-> R, Bot L-> R
 600 g
 1
1 lbs
454 g
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6. use calculator
600 /454 =
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7. Write down answer,
 1.3215859 lbs
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determine sig. figs
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“3”
1.32 lbs
recheck calculations
For online test 600 g x (1lbs/454g) = 1.32158 = 1.32
lbs
You want to change oil on your car, it need 4.0L, how
many quarts is that?
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1. Read problem
2. read for facts
 a. what is known?
4.0 L
 b. what are we finding?
? in Quarts
 c. What do we know?
1 qts= 0.9463529 L
 d. other info? none
3. Select which number to
start with
4.0 L
4. Put units in formula
4.0 L
1
You want to change oil on your car, it need 4.0L, how
many quarts is that?
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5. Cancel units, top L->R, Bot L->R
4.0 L
1 qts
1
0.9463529 L
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6. use calculator
4.0 / 0.9463529 =
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7. Write down answer
 4.2267530 qts.
 determine sig. figs
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“2”
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4.2 qts.
 recheck calculations
For online test 4.0L x (1 qts/0.9463529L) = 4.2267 = 4.2 qts
A Board is 27.0in long, It weights 1.2kg. Can it be used
to connect a 0.56m opening?
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1. Read problem
2. Read for facts
 a. what is known?
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27.0in B
1.2kg B
Opening= 0.56m
b. what are we finding?
? m, will it fit?
or ? in is it long enough?
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c. What do we know?
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1 in= 2.54 cm
1 m= 100cm
d. other info?
3. Select which number to start with
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27.0 in
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or
0.56 m
4. put units in formula
27.0 in
1
or
0.56 m
1
A Board is 27.0in long, It weights 1.2kg. Can it be used
to connect a 0.56m opening?
5. Cancel units, top L-> R, Bot. L->R
27.0 in 2.54 cm
or 0.56 m 100cm
1
1 in
1
1m
then
27.0 in 2.54 cm
1m
or 0.56 m 100 cm 1 in
1
1 in 100 cm
1
1 m 2.54
cm
 6. use calculator
 7. Write down answer
 0.6858 m
22.047in
 determine sig. figs
 “3”
“2”
 0.686 m
22 in
 recheck calculations
 Yes it Could
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For Online test 27.0 in x (2.54cm/1in)(1m/100cm)=0.6858=0.686 m Yes it could
Or 0.56m x (100cm/1m)(1in/2.54cm) = 22.047 = 22in Yes it could
How many minutes does it take a car traveling
85 km/hr to go 42 miles? How many seconds is that?
First “Don’t Panic” this is a harder problem to show you how this
story problem system still works on a harder problem. In Fact
you may find that this system makes what looks hard easy.
 1. Read problem
 2. Read for facts
 a. what is known?
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85 km/hr
42 miles
b. what are we finding?
? Time, in minutes
And ? time in seconds
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c. What do we know?
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60 minutes in a hour
60 seconds in a minute
1.609 km = 1 mile
d. other info?
How many minutes does it take a car traveling
85 km/hr to go 42 miles? How many seconds is that?
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3. Select which number to start with
85 km/hour has time in it so it may be a good starting point.
Ask yourself then where do you want time in the final answer
on the top or the bottom?
Top correct.
So we want
hrs
km
Does the 85 go with hours or km?
km right
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4. put units in formula
1 hrs
85 km
How many minutes does it take a car traveling
85 km/hr to go 42 miles? How many seconds is that?
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5. Cancel units, top L-> R, Bot. L->R
So we would start with hrs. where would it go top or bottom?
1 hrs
85 km
hrs
What are we going to?
Minutes, so
1 hrs
min
85 km
hrs
Which is bigger it gets a 1 and how many of the smaller in the
bigger? And cancel
1 hrs
60 min
85 km
1 hrs
We want minutes so circle it
1 hrs
60 min
85 km
1 hrs
ARE WE DONE?
No, so next page
How many minutes does it take a car traveling
85 km/hr to go 42 miles? How many seconds is that?
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5. Cancel units, top L-> R, Bot. L->R continued
The next unit we need to get rid of is?
kilometers or km where does it go top or bottom?
1 hrs
60 min
km
85 km
1 hrs
What are we converting it to?
Miles, SO
1 hrs
60 min
km
85 km
1 hrs
miles
Place the numbers in with Miles getting the 1 and km getting
the conversion
1 hrs
85 km
60 min
1 hrs
1.609 km
1 miles
Cancel the units. Are we done?
1 hrs
85 km
60 min
1 hrs
1.609 km
1 miles
How many minutes does it take a car traveling
85 km/hr to go 42 miles? How many seconds is that?
5. Cancel units, top L-> R, Bot. L->R continued 2
We still need to get rid of Miles Correct?
What do we know about miles?
There are 42 miles.
So where are we going to put miles top or bottom?
Top right?
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1 hrs
85 km
60 min
1 hrs
1.609 km
miles
1 miles
What goes under miles?
Nothing, because we are getting rid of miles so put a one
there. Place the 42 in front of miles and cancel.
1 hrs
85 km
60 min
1 hrs
1.609 km 42 miles
1 miles
1
We are now ready for step 6 on the next page.
How many minutes does it take a car traveling
85 km/hr to go 42 miles? How many seconds is that?
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6. use calculator
1 hrs
60 min 1.609 km 42 miles
85 km
1 hrs
1 miles
1
Typed into the calculator times across top, divide across bottom
1 x 60 x 1.609 x 42 /85 /1 /1 /1 =
(p.s. you can leave the ones out if you want)
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7. Write down answer
 47.70211765
 determine sig. figs
 “2”
 48 minutes
 recheck calculations
For Online test
(1hrs/85km)(60min/1hrs)(1.609km/1mile)(42 miles/1)= 47.702=48 minutes
How many minutes does it take a car traveling
85 km/hr to go 42 miles? How many seconds is that?
Are you done with this problem?
 No we still need to find seconds
So un-circle the minutes and add one more step
What are you getting rid of ? Minutes
Where does it go? What goes on other side?

1 hrs
60 min 1.609 km 42 miles 60 sec
85 km
1 hrs
1 miles
1
1 min
Typed into the calculator times across top, divide across bottom
60 x 1.609 x 42x60 /85 =
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7. Write down answer
 2862.13
 determine sig. figs
 “2”
 2.9 x 103 sec
 recheck calculations
(1hrs/85km)(60min/1hrs)(1.609km/1mile)(42 miles/1)(60sec/1min)= 2862.13=2.9e3 sec
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It is now time to see what you know?
To aid you and give you more practice at
solving story problems, on the next few
pages, there are many examples of story
problems. We would like you to try each
problem on your own. Then you can click the
screen and the complete formula with
answers will show up so that you can check
your work.
Also check the Assignments for the week
because there is additional worksheets to
work through.
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1. If you have a 3498mL of a substance how
many L is that? How many ML is it?
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3498 mL 1L
= 3.498 L
1
1000mL
and 3498 mL 1ML
= 3.498e-9 ML
1
1,000,000,000 (or 1e9)mL
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2. What is the cost of a Kg of sugar if it cost
$1.37 per 5 lb bag?
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$1.37
5 Lbs
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3. You have a 10.12lb container which
contains 50 apples. Each apple weights 120.
g. How much will the apples weight?
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50 apples
1
2.2 Lbs
1Kg
120. g
1 apple
= 0.6028 => $0.60/Kg
= 6000 => 6.00e3 g
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4. If you have a car traveling at 30m/sec how
many Km/hr would that be?
30 m 1 km
60 sec 60 min = 108 => 1e2 km/hr
1 sec 1000 m 1 min 1 hr
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5. A New drug has a dose of 2.3mL/kg body
weight. If you have a patient who weights
198lbs. How much drug should you give them?
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2.3 mL
1 kg
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6. If you have a 12cm square box. And you
have 200.mL of sand weighting 1000g, can you
fit the sand in the box in one trip?
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12 x 12 x 12 = 1728 cm3 => 1.7e3 cm3 = ml So 200ml will fit
1 kg
198 lbs = 207 => 2.1e2 mL
2.2 lbs
1
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7. If the temperature outside is 25oC how hot is
that in oF? In K?
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(1.8 x 25oC)+32= 77oF
25oC + 273.15 = 298.15=> 298K
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8. If you have 200.g of a substance and it takes
up 45ml what is the density of the substance?
Will it float on water?
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200.g / 45ml = 4.44444=> 4.4g/ml
Water has a density of 1.0g/ml so the substance is more
dense than water so it will sink.
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9. If you have a compound with the density of
4.5g/ml and a mass of 75g what volume does
it take up?
1ml
4.5g
75g = 16.66666=> 17ml
1
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10. If you have a compound that is 6.03 ml
and has a density of 0.74g/ml what is its
mass?
_
0.74 g
ml
6.03ml = 4.4622 => 4.5 g
1