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Our second foundational topic is Mad Science. This is our nickname for when we do things to a mathematical expression.
What is a mathematical expression?
Here are some examples:
47
20 − (2 × 5)
2x + 5
24 ÷ 8 + 10^{2} − 10 × ^{1}⁄_{5}
So a mathematical expression is one or more values (numbers or variables) that are linked with valid arithmetic operations.
We should notice three important things.
First, after we plug in any variables we can find the value of the entire expression. When we work the arithmetic correctly it will simplify into a single number.
Second, there are no equal signs in a mathematical expression. But the tools used to simplify an expression can be used to find y = something. This big topic will treat the expression 50 − (2 × 10) and the formula y = 50 − (2 × 10) as if there was no difference between them. Philopsophically they are distinct. But mathematically they are identical twins wearing slightly different clothes.
Third, invalid arithmetic operations do exist. You probably know at least one: the rule "never divide by zero". So an expression is like a recipe to follow, but we should not be asked to follow a gibberish recipe.
To summarize, we will explore more deeply how to use "mad science" to smoosh values together into a single, simple number. We use arithmetic: adding, subtracting, multiplying, dividing, and exponents. We will also study grouping structures and how they organize and order our arithmetic: parenthesis, fraction bars, square roots, and algorithms.
As we study this topic, work on making helpful and organized notes, so you have handy the comments, formulas, and example problems you need.
How do we do fraction multiplication? Let's use some pictures to get a sense of what happens.
Use the picture below to show multiplying onethird by onehalf. This picture starts with three vertical "cake cuts" and onethird shaded. Put the onehalf in the picture by making a horizontal "cut" across the middle, and shading either above or below our horizontal cut.
onesixth
How many pieces is the cake cut into?
How many pieces are shaded? Or after multiplying, how many pieces are overlapshaded?
To paraphrase, how many pieces do we bundle together into a group? In this problem there was no bundling. Each direction (horizontal or vertical) had only one shaded piece.
Use the picture below to show multiplying twothirds by threequarters. This time you must prepare the picture by drawing the twothirds yourself. Use different types of shading for two of the thirds and for three of the quarters.
sixtwelfths
The same thing: how many pieces is the cake cut into?
The same thing: how many pieces are shaded? Or after multiplying, how many pieces are overlapshaded?
To paraphrase, how many pieces do we bundle together into a group? In this problem there was bundling. When we cut thirds, two of those were bundled together. When we cut fourths, three of those were bundled together.
Definition
To multiply fractions, treat the situation as two independent multiplication problems. Multiply the numerators. Multiply the denominators.
The answer will be a fraction that may or may not need to be reduced.
We call this process the fraction multiplication algorithm.
The word algorithm is merely a fancy word for "recipe" or "process". It describes the steps to follow to do a task.
Here is an example where the answer does not need to be reduced.
In the numerators, 2 × 7 = 14. In the denominators, 3 × 5 = 15. So our answer is ^{14}⁄_{15}.
Here is an example where the answer does need to be reduced.
Using reducing: In the numerators, 5 × 1 = 5. In the denominators, 7 × 10 = 70. So our answer is ^{5}⁄_{70}, which can be reduced to ^{1}⁄_{14}.
Definition
During fraction multiplication, canceling is reducing early.
(This definition is not an algorithm. It defines a word. But it does not list the steps for a recipe or process.)
Canceling avoids big numbers when multiplying. Let's review it with an example.
Using canceling: We can divide by five in the first numerator and second denominator. Then ^{1}⁄_{7} × ^{1}⁄_{2} = ^{1}⁄_{14}.
We cannot escape dividing both numerator and denominator by 5. But can chose to do it before or after multiplying.
Your turn. Here are two more examples. Attempt them before we do them together. (If you have time, try doing this problem both ways: once with reducing and once with canceling.)
Using reducing: In the numerators, 3 × 1 = 3. In the denominators, 5 × 12 = 60. So our answer is ^{3}⁄_{60}, which can be reduced to ^{1}⁄_{20}.
Using canceling: We can divide by three in the first numerator and second denominator. Then ^{1}⁄_{5} × ^{1}⁄_{4} = ^{1}⁄_{20}.
Using reducing: In the numerators, 6 × 2 = 12. In the denominators, 11 × 3 = 33. So our answer is ^{12}⁄_{33}, which can be reduced to ^{4}⁄_{11}.
Using canceling: We can divide by three in the first numerator and second denominator. Then ^{2}⁄_{11} × ^{2}⁄_{1} = ^{4}⁄_{11}.
This example is quite important, because in shows how canceling can get a little complicated.
Using reducing: In the numerators, 10 × 6 = 60. In the denominators, 15 × 8 = 120. So our answer is ^{60}⁄_{120}, which can be reduced to ^{1}⁄_{2}.
Using canceling: We can divide a numerator and denominator by 5. (Yes, it is okay when "canceling" is merely reducing one of the fractions without touching the other.) Twice we can divide a numerator and denominator by 2. (Both times involve the denominator that starts out 8.) We can divide a denominator and numerator by 3. Then ^{1}⁄_{1} × ^{1}⁄_{2} = ^{1}⁄_{2}.
How do we multiply a fraction by a mixed number?
After we change 1 ^{4}⁄_{8} into ^{12}⁄_{8} we are solving ^{2}⁄_{3} × ^{12}⁄_{8}.
Then, in the numerators, 2 × 12 = 24. In the denominators, 3 × 8 = 24. So our answer is ^{24}⁄_{24}, which can be reduced to 1. (Or you could use canceling.)
Change 1 ^{2}⁄_{3} into ^{5}⁄_{3}, as well as changing 1 ^{4}⁄_{8} into ^{12}⁄_{8}.
Now we are solving ^{5}⁄_{3} × ^{12}⁄_{8}.
Then, in the numerators, 5 × 12 = 60. In the denominators, 3 × 8 = 24. So our answer is ^{60}⁄_{24}, which can be reduced to ^{5}⁄_{2}. (Or you could use canceling.)
Remember when we explained division as making piles of a fixed size? We did fraction division problems like these:
Review of Division as Making Piles of a Fixed Size
^{1}⁄_{2} ÷ ^{1}⁄_{4} =
Imagine having half a piece of paper. We rip it in half again, making quarters, while setting those quarters down as "piles". We get 2 piles.
So ^{1}⁄_{2} ÷ ^{1}⁄_{4} = 2
4 ÷ ^{2}⁄_{3} =
Imagine having 4 pieces of paper. We rip each in thirds while setting pairs of thirds down as "piles". We make 12 thirds by ripping. So we get 6 piles of twothirds.
So 4 ÷ ^{2}⁄_{3} = 6
Now what is happening to the numerators and denominators?
When we played with cake above, the denominators counted the number of pieces. The second numerator is also doing this.
For example, in 4 ÷ ^{2}⁄_{3} = 6 the 3 meant each of the four wholes was ripped into 3 pieces.
When we played with cake above, the numerators counted the number of shaded parts in a group. The second denominator is also saying how many pieces made a group.
For example, in 4 ÷ ^{2}⁄_{3} = 6 the 2 meant each of the piles bundled together 2 of the ripped thirds.
Definition
To divide fractions, flip the second fraction and multiply.
(Remember to reduce if needed.)
This definition is another algorithm. It describes the steps of a process. We can name it the fraction division algorithm.
Let's review fraction division with examples.
^{1}⁄_{2} ÷ ^{3}⁄_{4} = ^{1}⁄_{2} × ^{4}⁄_{3} = ^{2}⁄_{3}
^{3}⁄_{5} ÷ ^{1}⁄_{10} = ^{3}⁄_{5} × ^{10}⁄_{1} = 6
^{6}⁄_{9} ÷ ^{2}⁄_{3} = ^{6}⁄_{9} × ^{3}⁄_{2} = 1
^{10}⁄_{15} ÷ ^{6}⁄_{8} = ^{10}⁄_{15} × ^{8}⁄_{6} = ^{8}⁄_{9}
^{5}⁄_{3} ÷ ^{10}⁄_{8} = ^{5}⁄_{3} × ^{8}⁄_{10} = ^{4}⁄_{3}
Video example problems have three images to click on. You can see a video stepbystep answer, a written stepbystep answer, or only the answer. If you find yet more helpful videos, please let your instructor know so that this website can be updated and improved!
The Organic Chemistry Tutor
Multiplying Fractions  The Easy Way!
YouTube Problems
Simplify: ^{4}⁄_{3} × 24 Rewrite as ^{4}⁄_{3} × ^{24}⁄_{1} and use division by 3 to cancel with the 3 and 24. Then we see ^{4}⁄_{1} × ^{8}⁄_{1} = ^{32}⁄_{1} = 32 32
Changing these YouTube problems into a trio of links for video, steps, and only the answer is a lot of work. This is as far as I have gotten.
Are you using these? Does the textbook's videos for every example problem make this idea obsolete?
Chapter 2 Test, Problem 25: Simplify: 5 × ^{3}⁄_{10}
Chapter 2 Test, Problem 26: Simplify: ^{2}⁄_{3} × ^{15}⁄_{4}
Chapter 2 Test, Problem 27: Simplify: ^{22}⁄_{15} × ^{5}⁄_{33}
Chapter 2 Test, Problem 31: Simplify: ^{1}⁄_{5} ÷ ^{1}⁄_{8}
Chapter 2 Test, Problem 32: Simplify: 12 ÷ ^{2}⁄_{3}
Chapter 2 Test, Problem 33: Simplify: ^{24}⁄_{5} ÷ ^{28}⁄_{15}
Textbook Exercises for Fraction × and ÷
This list of recommended oddnumbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more oddnumbered problems. If your math foundation is strong, do fewer.
Please read the advice on doing homework in the study skills page.
Section 2.4 (Page 128) # 1, 3, 7, 11, 13, 21, 27, 29
You probably know that we need to have common denominators when adding fractions.
The denominator of a fraction acts like a word describing the type of thing we are considering. The fraction ^{3}⁄_{4} says "I am considering fourths, and care about three of them."
In this way, "threefourths" is very similar to the phrases "3 inches" or "3 centimeters".
Because of how denominators are conceptually like labels, trying to add fractions with unlike denominators works as badly as trying to add inches and centimeters. We need to change the fractions so their numerators are counting the same type of thing. We need to make their denominators match.
Consider the cake below. Amy gets to eat the big shaded half. Beatrice gets to eat the smaller shaded fourth. How much of the cake have they eaten?
We cannot combine the shaded half and fourth into a single thing without changing one of them to match the other.
Yes.
We could change both to halves. Then we would say they ate 1 ^{1}⁄_{2} halves.
We could change both to quarters. Then we would say they ate 3 quarters.
The answer "3 quarters" is a lot more friendly than the answer "1 ^{1}⁄_{2} halves".
Tangentially, how did fraction multiplication avoid the need for common denominators? Ponder that sometime, until you have a satisfying answer. Here is a clue.
Try using the picture below to show adding onethird and onehalf. This picture starts with all the "cake cuts" going in one direction, and onethird shaded. Put the onehalf in the picture by making a "cut" in the other direction. Then try to also shade the top or bottom half to add its amount without doublecounting any of the shaded pieces.
No. We get stuck because a portion of the half is already shaded! What should we do?
What should we do about a portion of the half already being shaded?
We need to "wrap around" as we shade in the half. This means we were forced to be aware that we are using sixths before looking for our final answer. The process itself requires sixths.
For the similar fraction multiplication problem earlier, we did not need to think about the new denominator until after the process is complete, when we are ready to write our answer. The process itself did not require sixths. Only naming our answer did.
When we ask with addition, "How much is onehalf more than onethird?" then overlapping the shaded pieces is not allowed.
When we ask with multiplication, "How much is onehalf of than onethird?" then the overlap is our answer.
See the difference?
When we change the denominators of two fractions so they match, there are three situations.
Sometimes one denominator is a multiple of the other.
Twelve is a multiple of six.
We only have to change the onesixth.
We unreduce ^{1}⁄_{6} × ^{2}⁄_{2} = ^{2}⁄_{12}
This example is like the cake pieces Amy and Beatrice ate. We should chop up the largersize piece so it matches the smallersize piece.
Sometimes the denominators have no common factors.
Fourteen and fifteen have no common factors.
We have to change both fractions.
We unreduce ^{1}⁄_{14} × ^{15}⁄_{15} = ^{15}⁄_{210}
We unreduce ^{1}⁄_{15} × ^{14}⁄_{14} = ^{14}⁄_{210}
We can always "brute force" a common denominator by multiplying the two denominator numbers.
In this example the "brute force" technique is the best we can do.
Definition
Two or more numbers are relatively prime if they have no common factors except for 1.
(Is this definition is an algorithm? Why or why not?)
For fractions with relatively prime denominators, the "brute force" technique will always be the best we can do.
Sometimes the denominators have factors in common, but neither is a multiple of the other.
Twelve and fifteen have a common factor of three.
We have to change both fractions.
We unreduce ^{1}⁄_{12} × ^{5}⁄_{5} = ^{5}⁄_{60}
We unreduce ^{1}⁄_{15} × ^{4}⁄_{4} = ^{4}⁄_{60}
In this example the "brute force" technique is not what we want. There is a common denominator less than 12 × 15 = 180.
We could use 180. But whenever we use a needlessly large common denominator we will eventually do more reducing than otherwise.
Definitions
When we look at a group of two or more fractions, the smallest number we can unreduce every denominator to match is the least common denominator.
How do we find it? It will be the least common multiple of the starting denominators.
(Are these definitions algorithms? Why or why not?)
If we use the least common denominator we might have to reduce our answer. But it is no longer a step that will always be needed.
Hm. We will need to find least common multiples. How do we do that?
The simplest method is to simply start making a list of multiples for each number. The smallest number the lists have in common is our answer.
Multiples of 10 are 10, 20, 30, 40, 50, 60, 70, ...
Multiples of 12 are 12, 24, 48, 60, 72, ...
Multiples of 15 are 15, 30, 45, 60, 75, ...
In that example we picked 60 for our answer because it was the smallest number on every list.
The quickest method is to write the prime factorization of each number (the long way, do not use exponents to condense your answer), merge factors shared in two or more prime factorizations, and then multiply the surviving factors.
The prime factorization of 10 is 2 × 5. The prime factorization of 12 is 2 × 2 × 3. The prime factorization of 15 is 3 × 5
The least common multiple is 2 × 2 × 3 × 5 = 60
In that example we merged the green 2s, the red 3s, the blue 5s. Every time a factor appears in more than one prime factorization we smoosh together one instance of it from each prime factorization that has it. Notice that the second 2 in the prime factorization of 12 was not merged. Neither of the other prime factorizations also had a 2 to smoosh it with.
Multiples of 24 are 24, 48, 72, 96, 120, 144, 168, ...
Multiples of 36 are 36, 72, 108, 144, 180, ...
Multiples of 48 are 48, 96, 144, 192, ...
The prime factorization of 24 is 2 × 2 × 2 × 3. The prime factorization of 36 is 3 × 2 × 2 × 3. The prime factorization of 48 is 2 × 2 × 2 × 2 × 3
We merge the red 2s because they appear in more than one prime factorization. And we merge the green 2 × 2 × 3 because it appears in more than one prime factorization. There is also an unmerged 2 and an an unmerged 3.
The least common multiple is 2 × 3 ×2 × 2 × 2 × 3 = 144
Those examples were pretty tricky because we found the least common multiple of three numbers. Most of the time when we add fractions we are only adding two fractions. Finding the least common multiple of two numbers will be a lot easier!
Let's do the same fraction addition problem in three different ways.
The first method we'll nickname the brute force method. We find a new denominator very quickly be multiplying each old denominator by the other. This always works, but forces us to deal with large numbers.
We unreduce ^{1}⁄_{30} × ^{42}⁄_{42} = ^{42}⁄_{1,260}
We unreduce ^{1}⁄_{42} × ^{30}⁄_{30} = ^{30}⁄_{1,260}
Then we add ^{42}⁄_{1,260} + ^{30}⁄_{1,260} = ^{72}⁄_{1,260} = ^{2}⁄_{35}
We started quickly! We could rush right into unreducing the two fractions. But then we had to do a lot of work reducing ^{72}⁄_{1,260} to get our final answer.
The second method is using the list of multiples method to find the least common denominator. This method is slow, but is favored by some students who prefer to work with multiples instead of factors.
Multiples of 30 are 30, 60, 90, 120, 150, 180, 210, 240, ... So we change ^{1}⁄_{30} into ^{7}⁄_{210} (Notice that 210 is the seventh multiple in the list.)
Multiples of 42 are 42, 84, 126, 168, 210, 252, ... So we change ^{1}⁄_{42} into ^{5}⁄_{210} (Notice that 210 is the fifth multiple in the list.)
Then we add ^{7}⁄_{210} + ^{5}⁄_{210} = ^{12}⁄_{210} = ^{2}⁄_{35}
We started slowly, finding lists of multipes. But we had a lot less reducing at the end!
The third method is using the prime factorization method to find the least common denominator. This method is quickest for students who can think quickly about factors.
The prime factorization of 30 is 2 × 3 × 5. The prime factorization of 42 is 2 × 3 × 7.
We merge the green 2 × 3. There is also an unmerged 5 and an an unmerged 7. So the least common multiple is 2 × 3 × 5 × 7 = 210.
Then we add ^{1}⁄_{30} + ^{1}⁄_{42} = ^{7}⁄_{210} + ^{5}⁄_{210} = ^{12}⁄_{210} = ^{2}⁄_{35}
Notice in that last example that we had hints about how to unreduce the numerators of our starting gractions: the unmerged factor(s) from the other prime factorization!
Time for a few more examples. You try these before seeing the instructor's work. Use any of the three methods.
^{1}⁄_{36} + ^{1}⁄_{6} = ^{1}⁄_{36} + ^{6}⁄_{6} = ^{7}⁄_{36}
^{3}⁄_{4} − ^{1}⁄_{12} = ^{9}⁄_{12} − ^{1}⁄_{12} = ^{8}⁄_{12} = ^{2}⁄_{3}
^{1}⁄_{5} + ^{1}⁄_{4} = ^{4}⁄_{20} + ^{5}⁄_{20} = ^{9}⁄_{20}
^{1}⁄_{12} + ^{2}⁄_{15} = ^{5}⁄_{60} + ^{8}⁄_{60} = ^{13}⁄_{60}
^{5}⁄_{42} − ^{1}⁄_{24} = ^{20}⁄_{168} − ^{7}⁄_{168} = ^{13}⁄_{168}
Note that the same techniques work when we have more than two fractions. The problem just takes longer because it is more work to find the prime factorization of more than two numbers.
^{1}⁄_{4} + ^{1}⁄_{6} − ^{1}⁄_{15} = ^{15}⁄_{60} + ^{10}⁄_{60} − ^{4}⁄_{60} = ^{21}⁄_{60} = ^{7}⁄_{20}
Bittinger Chapter Tests, 11th Edition
Chapter 3 Test, Problem 3: Simplify: ^{1}⁄_{2} + ^{5}⁄_{2}
Chapter 3 Test, Problem 4: Simplify: ^{7}⁄_{8} + ^{2}⁄_{3}
Chapter 3 Test, Problem 5: Simplify: ^{7}⁄_{10} + ^{19}⁄_{100} + ^{31}⁄_{1,000}
Chapter 3 Test, Problem 6: Simplify: ^{5}⁄_{6} − ^{3}⁄_{6}
Chapter 3 Test, Problem 7: Simplify: ^{5}⁄_{6} − ^{3}⁄_{4}
Chapter 3 Test, Problem 8: Simplify: ^{17}⁄_{24} − ^{1}⁄_{15}
Textbook Exercises for Fraction + and −
This list of recommended oddnumbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more oddnumbered problems. If your math foundation is strong, do fewer.
Please read the advice on doing homework in the study skills page.
Section 3.2 (Page 159) # 1, 5, 9, 13, 21, 23, 27, 29, 33, 35, 39, 47, 51, 55
There are two ways to think about subtracting with mixed numbers.
We could treat the fractions as a place value column and borrow from the 1's column if we need to do so.
Subract the whole numbers and fractions separately, then squish the answers together.
3 ^{1}⁄_{2} − 2 ^{1}⁄_{4} = 3 − 2 and then ^{1}⁄_{2} − ^{1}⁄_{4} = 1 and then ^{1}⁄_{4} = 1 ^{1}⁄_{4}
Subract the whole numbers and fractions separately, then squish the answers together.
5 ^{1}⁄_{4} − 3 ^{1}⁄_{2} = 5 ^{1}⁄_{4} − 3 ^{2}⁄_{4} = 5 − 3 and then ^{1}⁄_{4} − ^{2}⁄_{4}. We get stuck with not enough fourths to start that subtraction!
We borrow from the whole numbers by changing 5 into both 4 and ^{4}⁄_{4}.
Now we have plenty of fourths: the one we started with and four more (so five total)
So 4 − 3 = 1 and then ^{5}⁄_{4} − ^{2}⁄_{4} = ^{3}⁄_{4}, for a final answer of 1 ^{3}⁄_{4}
We could change both mixed numbers to improper fractions and then subtract.
3 ^{1}⁄_{2} − 2 ^{1}⁄_{4} = ^{7}⁄_{2} − ^{9}⁄_{4} = ^{14}⁄_{4} − ^{9}⁄_{4} = ^{5}⁄_{4}
5 ^{1}⁄_{4} − 3 ^{1}⁄_{2} = ^{21}⁄_{4} − ^{7}⁄_{2} = ^{21}⁄_{4} − ^{14}⁄_{2} = ^{7}⁄_{4}
It is helpful to be fluent with both these methods of subtracting mixed numbers. For some problems the first method will be easier. For other problems the second method will be easier.
Notice that treating the fractions as a place value column naturally resulted in a mixed number answer, whereas changing both mixed numbers to improper fractions naturally resulted in an improper fraction answer.
Bittinger Chapter Tests, 11th Edition
Chapter 3 Test, Problem 19: Simplify: 10 ^{1}⁄_{6} − 5 ^{7}⁄_{8}
Textbook Exercises for Mixed Number Subtraction
This list of recommended oddnumbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more oddnumbered problems. If your math foundation is strong, do fewer.
Please read the advice on doing homework in the study skills page.
Section 3.3 (Page 170) # 1, 9, 13, 21, 29, 49, 51, 53, 55
Many math students are taught to memorize the acronym PEMDAS to help solve resolve order of operations.
PEMDAS
P = Parenthesis
E = Exponents
M = Multiplication
D = Division
A = Addition
S = Subtraction
This acronym can be very helpful if it was taught well.
Unfortunately, PEMDAS can cause problems if it was taught poorly.
Which answer for 8 ÷ 4 × 2 is Correct?
Do the 4 × 2 first because M comes before D
8 ÷ 4 × 2 = 8 ÷ 4 × 2 = 8 ÷ 8 = 1
or
Do the 8 ÷ 4 first because we go left to right
8 ÷ 4 × 2 = 8 ÷ 4 × 2 = 2 × 2 = 4
The second is correct. Multiplication and division have equal priority, and should be done lefttoright.
To solve an order of operations problem we cannot simply do PEMDAS in order! This is because some operations have equal priority. These must be done lefttoright when only these are left to do.
Multiplication and division have equal priority.
Addition and subtraction have equal priority.
Stay alert!
Also note that PEMDAS does not tells us what to do with fractions. Fraction bars are actually another grouping symbol, just like parenthesis. We could rewrite fractions problems as parenthesis problems.
We could try to "fix" PEMDAS by rewriting it.
(PF)E(MD)(AS)
PF = Parenthesis and Fraction Bars
E = Exponents
MD = Multiplication and Division
AS = Addition and Subtraction
Parenthesis and fraction bars have first priority. Then exponents. Then multiplication and division. Then addition and subtraction. Work from left to right with items of equal priority.
Unfortunately, this is stupidly awkward. There must be something better!
The best way to deal with order of operations is to not replace the (AS) step of PEMDAS with a new kind of thinking.
Terms
A term is an portion of an expression separated from the equal sign only by addition or subtraction.
How many terms are in these expressions? (Hint: Try underlining them.)
(4 + 2) ÷ 3 + 8 × 2^{2} =
Two. They are separated by the only + symbol outside of parenthesis.
^{1}⁄_{2} × 4 × 5 − 3 × (9 − 1) ÷ 2^{2} + 5 =
Three. The first stops at the − symbol outside of parenthesis. The final 5 is its own term.
11 + 8 × 2 − 14 ÷ 7 + (19 − 4) ÷ 3 − 90 ÷ 3^{2} =
Five.
Why are terms the easiest way to think about order of operations?
First, each term is totally independent until as the very last step we combine them. It does not matter which term we simplify first.
Second, within a term we only have two priorities: first grouping structures (parenthesis, fraction bars, and exponents) and then going lefttoright with multiplication and division.
As a tangential comment, exponents are not actually grouping structures, but they demand using them. When we see an exponent we must know clearly how much it covers. So mathematicians include exponents when talking about grouping structures.
Consider the expression (3 + 2)^{2}. It is different from the expressions (3 + 2^{2}) and 3^{2} + 2^{2}. Wanting to square the sum of 3 + 2 demanded putting parenthesis around that sum.
Now we should actually simplify those three expressions.
(4 + 2) ÷ 3 + 8 × 2^{2} =
2 + 32 = 34
^{1}⁄_{2} × 4 × 5 − 3 × (9 − 1) ÷ 2^{2} + 5 =
10 − 6 + 5 = 9
11 + 8 × 2 − 14 ÷ 7 + (19 − 4) ÷ 3 − 90 ÷ 3^{2} =
11 + 16 − 2 + 5 − 10 = 20
As one last comment about terms, remember that we can cancel factors but we cannot cancel terms.
Bittinger Chapter Tests, 11th Edition
Chapter 1 Test, Problem 41: Simplify: 35 − 1 × 28 ÷ 4 + 3
Chapter 1 Test, Problem 42: Simplify: 10^{2} − 2^{2} ÷ 2
Chapter 1 Test, Problem 43: Simplify: (25 − 15) ÷ 5
Chapter 1 Test, Problem 44: Simplify: 2^{4} + 24 ÷ 12
Chapter 4 Test, Problem 50: Simplify: 256 ÷ 3.2 ÷ 2 − 1.56 + 78.325 × 0.02
Chapter 4 Test, Problem 51: Simplify: (1 − 0.08)^{2} + 6 × [5 × (12.1 − 8.7) + 10 × (14.3 − 9.6)]
Textbook Exercises for Order and Terms
This list of recommended oddnumbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more oddnumbered problems. If your math foundation is strong, do fewer.
Please read the advice on doing homework in the study skills page.
Section 1.7 (Page 84) as much of # 2773 odd as helps you, 77, 79, 81
Section 2.4 (Page 129) # 31, 43, 45, 47, 49, 51
Section 3.4 (Page 178) # 1, 3, 9, 13, 39, 41, 43, 45, 51
Understanding order of operations and terms allows us to smush together numbers using the basic arithmetic operations: +, −, ×, and ÷.
There is a very useful but less wellknown fifth arithmetic operation that looks like ∝. It also smushes numbers together. But it is used with rates instead of with plain numbers.
We will call it goeswith because it was invented by the famous yet ficticious mathematician and mad scientist, Lucille Hyperious Goeswith. Other math books and websites might call it the "proportional operator".
To use ∝ we follow a three step algorithm.
Let's look at an example.
First we list the rates. It helps to write them as fractions.
Second we write our goal. For this problem 1 day = _____ dollars. This tells us to start our chain by writing 1 day and keep going until we get to dollars.
Third we make our goeswith chain. The only rate that involves days is 7 hours per day.
1 day ∝ 7 hours ∝ ?
The other rate that involves hours is her average speed. We unreduce it by multiplying the top and bottom by 7 to get 420 miles per 7 hours. Then we can continue building our goeswith chain.
1 day ∝ 7 hours ∝ 420 miles ∝ ?
The other rate that involves miles is her old car's disappointing mileage. We unreduce it by multiplying the top and bottom by 21 to get 420 miles per 21 gallons. Then we can continue building our goeswith chain.
1 day ∝ 7 hours ∝ 420 miles ∝ 21 gallons ∝ ?
The other rate that involves gallons is the price of gasoline. We unreduce it by multiplying the top and bottom by 21 to get 63 dollars per 21 gallons. Then we can continue building our goeswith chain.
1 day ∝ 7 hours ∝ 420 miles ∝ 21 gallons ∝ 63 dollars
The goeswith chain is complete. We now have five items that go with 1 day, including the dollar amount we were looking for.
Let's look at more examples.
6 people ∝ 24 slices ∝ 2 pizzas ∝ $28
1 quart water ∝ 0.25 gallon water ∝ 2 teaspoons permethrin
120 miles ∝ 4 gallons gas ∝ ^{1}⁄_{3} tank (which is unfortunately more gas than you have left in the tank)
1 year ∝ 365 days ∝ 182.5 packs ∝ $1,095
Goeswith chains can be really useful! The process is pretty foolproof. It tells us when to multiply and when to divide. Yes, we have to write a lot. But if our intuition about when to multiply or divide is not accurate, it is worth writing more to avoid making a mistake.
We recently studied one step measurement conversions. Goeswith chains allow us to solve problems with more than one step.
Unfortunately, goeswith chains are terribly impractical. All of that fraction unreducing and reducing takes a lot of time. There must be a more efficient way to do this type of problem!
Fortunately, there is a better algorithm! We will study it soon. It is called unit analysis. However, unit analysis does nothing to help build an understanding for why this type of math problem smushes rates together two at a time to eventually form a chain that ends with the desired answer. So for the sake of building up a deeper understanding and intuition we look at goeswith chains now, as a temporary step between one step measurement conversions and unit analysis.
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Textbook Exercises for Goeswith Chains
This list of recommended oddnumbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more oddnumbered problems. If your math foundation is strong, do fewer.
Please read the advice on doing homework in the study skills page.
(Our textbook has no exercises for this topic.)
Try these ten exercises on scratch paper. Work in a study group if you can! Notice where your notes need improvement. After you are very happy with your answers, you can use this form to ask me to check your work. Can you get at least 8 out of 10 correct?
1. A cookie recipe calls for ^{2}⁄_{3} cup of flour. If you are making a double batch, how much flour will you use?
2. What is fourfifths times twenty?
3. What is fourfifths times twothirds?
4. What is twothirds divided by threehalves?
5. ^{22}⁄_{15} ÷ ^{33}⁄_{5} =
6. What is the least common multiple of 12 and 18?
7. What is fivesixths plus threefourths?
8. What is fivesixths minus threefourths?
9. A piece of fabric that is 1 ^{3}⁄_{4} yards long is cut into seven equal pieces. How long is each piece?
10. Grandma Jorgensen left ^{2}⁄_{3} of her ^{7}⁄_{8} pound silver bullion bar to her son Lloyd. Lloyd gave each of his four children a ^{1}⁄_{4} share. How much silver did each of Lloyd's children receive?
Try these exercises on scratch paper. Work in a study group if you can! Notice where your notes need improvement. Check your work when you are done.
The word of in a word problem tells us to multiply.
This makes sense. Making copies is what multiplication does. When visiting a candy store, when you tell the clerk "I would like ten of those" then you are saying "I would like ten copies of one of those".
$25 × 5 = $125
The phrase "5 of them" turned into × 5.
7 minutes × 30 = 210 minutes
The phrase "a class of 30" turned into × 30.
The way the English language works also makes it very natural to put the word "of" into a word problem with a fractional amount.
350 total × ^{3}⁄_{7} are boys = 150 boys
The phrase "threesevenths of the children" turned into × ^{3}⁄_{7}.
^{3}⁄_{8} remaining × ^{1}⁄_{2} of that = ^{3}⁄_{16} of a pizza
The phrase "onehalf of what is left" turned into × ^{1}⁄_{2}.
The third occasion when the English language naturally uses "of" to mean multiplication is with the phrase "percent of..."
Use RIP LOP to change the percentage into a decimal. Then multiply.
200 × 0.4 = 80
The phrase "40% of..." turned into × 0.4.
The "Percent Of..." Algorithm
When we are asked to find the percentage of an amount, we can use RIP LOP on the percentage and then multiply.
The amount could be a normal number.
200 × 0.6 = 120
The amount could be a labeled amount.
50 hours × 0.44= 22 hours
The amount could even be another percentage. Treat the other percertage symbol as a label.
50 percent × 0.1 = 5 percent = 5%
When comparing "percent of..." answers be wary of when they have the same or different initial amounts.
The initial amount in the first claim is "all Americans" which includes children. And the percentage measures how many have any health insurance.
The initial amount in the first claim is "adult Americans" which is a smaller group. And the percentage measures how many have one kind of health insurance.
The two claims are not compatible.
Here is a trick that allows you to solve more "percent of..." problems without needing to use a calculator. Impress your friends!
Remember that when multiplying two number the order of the numbers does not matter. Also remember that "percent" merely means to divide by 100 at the end. Let's put these two facts together.
Definition
The Switcheroo Trick says that when multiplying a number in percent format by a normal number we can switch which number is the percentage.
As an example, we can solve "What is 4% of 50?" by doing 4% × 50. However, finding 4% of a number might be tricky for you without a calculator. But finding 50% is easy—that is half of the number. So we switch which number is the percentage. 4 × 50% = 2.
We can switcheroo to ask, "What is 25% of 24?" Then with mental arithmetic we can more easily see that onequarter of 24 is 6.
We can switcheroo to ask, "What is 900% of 8?" Then with mental arithmetic we can more easily see that 9 times 8 is a $72 discount.
A common application for "percent of..." word problems is tipping at a restaurant. You can learn a few shortcuts to impress your friends.
Before continuing, make sure you are not confused. We know that being in percent format is equivalent to two decimal point scoots. Why does the shortcut for finding 10% only involve only one decial point scoot?
Have you figured that out?
When we find 10% we end up multipying by × 0.1
We know from our examination of decimal point scoots that × 0.1 is the same as one scoot to the left.
Let's practice using those shortcuts.
10% of $25 becomes $25 × 0.1 = $2.50
But we want 20% instead of that 10%, so to get twice as much we double the 10% result and get a $5 tip.
10% of $20 becomes $20 × 0.1 = $2
Next we see that 5% of $20 is half that, which is $1
But we want 15% so we combine the 10% result and the 5% result and get a $3 tip.
10% of $30 becomes $30 × 0.1 = $3.00
Next we see that 5% of $30 is half that, which is $1.50
But we want 15% so we combine the 10% result and the 5% result and get a $4.50 tip.
Notice that the word "of" did not appear in the previous problem. But we could rephrase the problems to ask "What is 15% of $30?"
In other words, merely having the option to rephrase a situation using the phrase "percent of..." is still an instruction to multiply.
Tipping is one instance when estimation is very useful. If my restaurant bill was $21.87, the tip will not change much if I estimate based on $22.00.
Estimate by using $40 for the meal cost.
10% of $40 becomes $40 × 0.1 = $4
But we want 15% so we use oneandahalf that result and get a $6 tip.
Estimate by using $24 for the meal cost.
10% of $24 becomes $24 × 0.1 = $2.40
But we want 20% so we double that result and get a $4.80 tip.
Another common application for "percent of..." word problems is a retail item that goes on sale with a percentage price reduction.
20% of $12 becomes $12 × 0.2 = $2.40 discount
35% of $60 becomes $60 × 0.35 = $21 reduction
Price reductions are another situation where estimation is very useful.
Estimate by using $80 for the jacket cost.
60% of $80 becomes $80 × 0.6 = $48 reduction
Notice that the word "of" did not appear in the previous two problems. But we could rephrase the problems to ask "What is 35% of $60?" and "Estimate 60% of $79.99."
Once again, merely having the option to rephrase a situation using the phrase "percent of..." is still an instruction to multiply.
All of these "percent of..." problems told us a percentage. We were multiplying that percentage by an amount.
It is possible to write an English sentence using the words "percent of..." that asks for the percentage instead. This is a different type of problem. We will learn about it later.
Guppies—Asking for the Percentage
In a tank of 10 fish, 8 are guppies. What percent of the fish are guppies?
Notice that this problem did not tell us a percentage. The "percent of..." algorithm does not work. There is no percentage to RIP LOP as the first step of that algorithm.
The Organic Chemistry Tutor
Textbook Exercises for Percent Of
This list of recommended oddnumbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more oddnumbered problems. If your math foundation is strong, do fewer.
Please read the advice on doing homework in the study skills page.
(Our textbook has no exercises for this topic.)
Many retail sale situations ask you to find the percentage.
Compared to the problems we just did, we are working backwards. So we do the opposite of multiplying, which is dividing. And instead of starting with RIP LOP we finish with RIP LOP.
Think of 15¢ and 75¢ as being a part and whole.
Perhaps imagine the fraction ^{15¢}⁄_{75¢}
Then 15¢ ÷ 75¢ = 0.2 = 20% decrease
Often the monetary amount of the decrease is hidden. We must subtract to find it.
The decrease is 80¢ − 60¢ = 20¢
Perhaps imagine the fraction ^{20¢}⁄_{80¢}
Then 20¢ ÷ 80¢ = 0.25 = 25% decrease
To avoid confusion, it is helpful to memorize a formula.
Percent Change Formula
percent change = change ÷ original
Then use RIP LOP to change the decimal answer into percent format.
This definition is an algorithm dressed up as a formula. The steps of the algorithm are "change ÷ original". We can write those steps as a formula by putting in front "percent change =". Notice that any formula of similar structure, which says "your answer is what happens when you correctly simplify this expression" is an algorithm dressed up as a formula.
This formula reminds us to find the change if the word problem is worded to only give us the old and new values. It also reminds us to divide by the original instead of dividing incorrectly by the new value.
The percent change formula works for increases as well as decreases.
We do change ÷ original = $15,000 ÷ $120,000 = 0.125 = 12.5% increase
So far so good. Unfortunately, the word problems get trickier.
We already saw one issue to keep us on our toes: problems can show or hide the numeric amount of the change.
The increase is $170,000 − $150,000 = $20,000.
Then change ÷ original = $20,000 ÷ $150,000 ≈ 0.13 = 13% increase
Also, some problems ask for the change, but others ask for the change combined with the original amount.
3% of $200 becomes $200 × 0.03 = $6 increase
3% of $200 becomes $200 × 0.03 = $6 increase
Then add $200 + $6 = $206 new value
Let's do a variety of problems to make sure you understand how the wording can ask for different things.
5% of 163,000 becomes 163,000 × 0.05 = 8,150 increase
Then add 163,000 + 8,150 = 171,150 people for the new value
The increase as a number was 163,000 − 145,000 = 18,000 people
So the percent change is change ÷ original = 18,000 ÷ 145,000 ≈ 0.12 = 12% increase
First we change the ounces to pounds. 40 ounces ÷ 16 ounces per pound = 2.5 pounds
Then we find the current price, which is 80% of $24. $24 × 0.8 = $19.20
Then we find the unit rate by dividing. $19.20 ÷ 2.5 = $7.68 per pound
12% of $100 becomes $100 × 0.12 = $12 increase
Then add $100 + $12 = $112 new value
Here is a diagram to help us remember both issues when doing "percent change..." problems.
Plettski Productions Math
Percent of Increase and Decrease
Mathispower4u
Determine a Percent of Change (Decrease)
Determine a Percent of Change (Increase)
Bittinger Chapter Tests, 11th Edition
Chapter 6 Test, Problem 9: The number of foreign children adopted by Americans declined from 20,679 in 2006 to 19,292 in 2007. Find the percent of the decrease.
Chapter 6 Test, Problem 13: The marked price of a DVD player is $200 and the item is on sale for 20% off. What are the discount (in dollars) and sale price?
Chapter 6 Test, Problem 19: A television that normally costs $349 is on sale for $299. What is the discount in dollars? What is the discount rate?
Textbook Exercises for Percent Change
This list of recommended oddnumbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more oddnumbered problems. If your math foundation is strong, do fewer.
Please read the advice on doing homework in the study skills page.
Section 6.5 (Page 347) # 1, 3, 9, 11, 13, 15, 17
Notice that in the previous problem, when we calculated $100 original value × 0.12 rate = $12 change, the initial amount of $100 was used but then disappeared. We had to add it back as a second step.
What if it did not disappear?
We know that multiplying any number by 1 does not change it. Let's use that.
The One Plus Trick
When we multiply by a percentage increase, we can first add 1 to that percentage to keep the original amount around.
So we can redo the previous problem as $100 original value × 1.12 rate = $112 new value
That is great! Sticking a 1 in front of the percentage is trivial. That is much nicer than our first method of solving the problem that involved two steps.
Remember, × 1.12 is a combination of doing × 1.0 to keep the initial amount around, and doing × 0.12 to find the change.
$5,000 original value × 1.08 rate = $5,400 new value
$200 original value × 1.04 rate = $208 new value
$140 original value × 1.18 rate = $165.20 new value
$750 original value × 1.05 rate = $787.50 new value
For a percent decrease problem we can use the same trick but it looks slightly different. We still add one. But the change we are combining with 1 is negative. So we need to subtract the percentage from 1.
The One Minus Trick
When we multiply by a percentage decrease, we can first subtract the percentage from 1 to keep the original amount around.
$200 × (1 − 0.15) = $200 × 0.85 = $170
$22,500 × (1 − 0.1) = $22,500 × 0.9 = $20,250
Try to explain in your own words the process for increasing a quantity by 10%.
Try to explain in your own words the process for increasing a quantity by 50%.
Is it possible to increase a quantity by 100%? Explain what that would mean.
None yet
Textbook Exercises for the One Plus Trick
This list of recommended oddnumbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more oddnumbered problems. If your math foundation is strong, do fewer.
Please read the advice on doing homework in the study skills page.
(Our textbook has no exercises for this topic.)
Try these ten exercises on scratch paper. Work in a study group if you can! Notice where your notes need improvement. After you are very happy with your answers, you can use this form to ask me to check your work. Can you get at least 8 out of 10 correct?
1. What is 25% of 228?
2. What is 40% of 750?
3. The price tag (before sales tax) on an item says $100. The sales tax rate is 5%. What is the total price (including tax)?
4. During a sale, a dress decreased in price from $90 to $72. What was the percent of decrease?
5. An investment increases in value from $200 to $216. What is the percent increase?
6. A pinball machine that normally sells for $3,999 is on sale for $3,150. What is the rate of discount?
7. An investment of $3,000 increases in value by 4%. What is the increase?
8. An investment of $4,000 increases in value by 8%. What is the new value?
9. An investment of $5,000 decreases in value by 2.4%. What is the new value?
10. An investment of $6,000 decreases in value by 2%. Then it decreases by another 5%. What is the final value?
Try these exercises on scratch paper. Work in a study group if you can! Notice where your notes need improvement. Check your work when you are done.
Remember that onestep conversions were the wrong tool for some measurement unit conversion problems. We needed a new technique if we did not know a single rate that translated from the old unit to the new unit.
Our new technique is a fivestep process called Unit Analysis (nicknamed "Canceling Words"). It is a simple and powerful tool. Here are the steps without context, but only for later reference. To first meet the new technique it is best to see it in action in an example.
Unit Analysis
The process of Unit Analysis uses five steps to convert measurement units.
 write the given measurement as a fraction
 write some "empty" rates without numbers
 fill in the numbers for the rates
 multiply
 simplify the fraction answer
Unit Analysis is an algorithm whose process is too complicated to be written as a formula.
Here is a simple example of Unit Analysis. We could do this problem much more efficiently as a onestep conversion. But we will use it now merely as a model of the five step process.
Detailed Example of Needless Unit Analysis
How many inches is 3.7 feet?
Our first step is to write the given measurement as a fraction. If it is not already a fraction we put the value over 1.
Our second step is to write some "empty" rates without numbers, multiplying. We do this until each unit we want to get rid of appears in a numeratoranddenominator pair so it will cancel.
Our third step is to fill in the numbers for the rates. To do this we need to memorize or look up the unit conversion rates.
Our fourth step is to multiply. This works just like any other fraction multiplication. Notice how units cancel. We do not normally reduce before multiplying.
Our fifth step is to simplify the fraction answer . Often the denominator is not 1, so we simplify by doing "numerator ÷ denominator" like old unit rate problems.
That was very methodical: if we follow the five steps we have almost no chance of making a mistake.
Now for a problem when we actually need to use Unit Analysis because the unit conversion has more than one step.
Detailed Example of Needing Unit Analysis
How many miles per hour is 200 inches per minute?
Our first step is to write the given measurement as a fraction. If it is not already a fraction we put the value over 1.
Our second step is to write some "empty" rates without numbers, multiplying. We do this until each unit we want to get rid of appears in a numeratoranddenominator pair so it will cancel.
(Be alert! We want distance on top and time on bottom. So we must start that way! Starting with the fraction upside down is our biggest potential danger—besides a careless calculator button error it is the only way to mess up.)
Our third step is to fill in the numbers for the rates. To do this we need to memorize or look up the unit conversion rates.
Our fourth step is to multiply. This works just like any other fraction multiplication. Notice how units cancel. We do not normally reduce before multiplying.
Our fifth step is to simplify the fraction answer . Often the denominator is not 1, so we simplify by doing "numerator ÷ denominator" like old unit rate problems.
To summrize, Unit Analysis is an important process because it works no matter how many rates are needed to translate from the old units to the new units. The process keeps track of when to multiply and when to divide, so we do not have to keep track in our heads. All we need to do is look up (or memorize) the unit conversion rates and line them up.
Now for more example problems from the textbook. We will always use Unit Analysis, even for the problems that are onestep unit conversions.
^{9.6 yards}⁄_{1} = ^{9.6 yards}⁄_{1} × ^{feet}⁄_{yard} = ^{9.6 yards}⁄_{1} × ^{3 feet}⁄_{1 yard} = ^{28.8 feet}⁄_{1} = 28.8 feet
^{3 yards}⁄_{1} = ^{3 yards}⁄_{1} × ^{feet}⁄_{yard} × ^{inches}⁄_{feet} = ^{3 yards}⁄_{1} × ^{3 feet}⁄_{1 yard} × ^{12 inches}⁄_{1 foot} = ^{108 inches}⁄_{1} = 108 inches
^{1 inch}⁄_{1} = ^{1 inch}⁄_{1} × ^{feet}⁄_{inches} = ^{1 inch}⁄_{1} × ^{1 foot}⁄_{12 inches} = ^{1}⁄_{12} foot (or if you prefer a decimal ≈ 0.83 foot)
First change 6 ^{1}⁄_{3} yard into ^{19}⁄_{3} yard.
^{19}⁄_{3} yard = ^{19 yard}⁄_{3} = ^{19 yard}⁄_{3} × ^{feet}⁄_{yard} × ^{inches}⁄_{feet} = ^{19 yards}⁄_{3} × ^{3 feet}⁄_{1 yard} × ^{12 inches}⁄_{1 foot} = ^{228 inches}⁄_{1} = 228 inches
^{11 gallons}⁄_{1} = ^{11 gallons}⁄_{1} × ^{quarts}⁄_{gallons} × ^{pints}⁄_{quarts} = ^{11 yards}⁄_{1} × ^{4 quarts}⁄_{1 gallon} × ^{2 pints}⁄_{1 quart} = ^{88 pints}⁄_{1} = 88 pints
Except for inches to centimeters, the unit conversion rates that move between Standard and SI are rounded. Because they are rounding that happens before the problem is finished they introduce some error. Let's do the same problem in two different ways to see this happen.
^{3.171 quarts}⁄_{1} = ^{3.171 quarts}⁄_{1} × ^{liters}⁄_{quarts} = ^{3.171 quarts}⁄_{1} × ^{1 liters}⁄_{1.057 quarts} = ^{3 liters}⁄_{1} = 3 liters
^{3.171 quarts}⁄_{1} = ^{3.171 quarts}⁄_{1} × ^{liters}⁄_{quarts} = ^{3.171 quarts}⁄_{1} × ^{0.946 liters}⁄_{1 quarts} = ^{2.999766 liters}⁄_{1} ≈ 3 liters
Is the answer slightly less than three liters or not? It is actually slightly more! But because our original number had quite a few decimal places the unit conversion rates we used did not have enough decimal places to produce answers with enough accuracy.
If your original numbers have many decimal places (thus asking your answers to also have many decimal places) you will need unit conversion rates with lots of decimal places.
Mathispower4u
Find the Number of Meters Traveled in 3 seconds Given Kilometers Per Hour
KetzBook
Unit Conversion the Easy Way (Dimensional Analysis)
Bittinger Chapter Tests, 11th Edition
Chapter 8 Test, Problem 1: How many inches is 4 feet?
Chapter 8 Test, Problem 2: How many feet is 4 inches?
Chapter 8 Test, Problem 5: How many meters is 200 yards?
Chapter 8 Test, Problem 6: How many miles is 2,400 kilometers?
Chapter 8 Test, Problem 11: How many ounces is 4 pounds?
Chapter 8 Test, Problem 12: How many pounds is 4.11 tons?
Chapter 8 Test, Problem 16: How many minutes is 5 hours?
Chapter 8 Test, Problem 17: How many hours is 15 days?
Chapter 8 Test, Problem 18: How many quarts is 64 pints?
Chapter 8 Test, Problem 19: How many ounces is 10 gallons?
Chapter 8 Test, Problem 20: How many ounces is 5 cups?
Textbook Exercises for Unit Analysis
This list of recommended oddnumbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more oddnumbered problems. If your math foundation is strong, do fewer.
Please read the advice on doing homework in the study skills page.
Section 7.1 (Page 371) # 13, 41, 43, 45, 47, 49, 51
Section 7.2 (Page 382) # 55, 59, 65
Section 7.3 (Page 388) # 35, 39, 41
There are two common ways to measure the size of a shape: perimeter and area.
Perimeter
The perimeter of a shape is the distance around its edges.
Area
The area of a shape is how much room the shape's surface takes up.
We'll start by talking about perimeter.
Let's explore the perimeter of polygons.
Polygon
Polygons are closed shapes whose edges are straight lines.
Here are a bunch of polygons: squares, rectangles, triangles, parallelograms, and trapezoids.
Perimeter is the distance around a shape. There is nothing more complicated to say, since all polygon perimeter problems are simply adding up the lengths of sides.
Finding a perimeter is easy. Just mark a corner (so you don't carelessly forget an edge) and begin adding.
For the first 3 inches + 4 inches + 5 inches = 14 inches
For the second it might help to label the sides that start without labels. 3 + 6.1 + 3 + 6.1 = 18.2 inches
There are only two possible ways to be tricky.
The first way to be tricky with perimeter is demonstrated with this problem:
We need to first change the sides labeled with 2.5 feet so they are labeled with 30 inches.
Then 30 inches + 20 inches + 30 inches + 20 inches = 100 inches
The trickiness comes from hiding a measurement unit conversion problem inside the perimeter problem.
The first way to be tricky is to ask a complicated question while being obtuse about providing all the required information, like this problem:
Oops. That is not a math problem. Let's try again.
This problem is a great excuse to introduce a sixstep problem solving process.
Step one is Determine what you are looking for. Read the problem two or three times so you understand it and notice all the details. Write down (or pretend to write) in an English sentence what you are looking for. Don't wander off track and forget what you are looking for.
Consider our example problem. We are looking for a number of miles. We need to add a bunch of numbers.
Step two is Draw pictures. Draw a picture or diagram of the situation. Then label things in the picture or diagram! Your visualization will not be much help without labels.
Consider our example problem. The diagram is already provided. Hooray!
But two sides are not labeled. We can use subtraction to find the missing lengths.
For the top we added ½ mile + 1 mile = 1 ½ mile
For the far left we subtracted ¾ mile − ½ mile = ¼ mile
Step three is Name Things. Write (or pretend to write) English sentences to give a one letter name to each quantity. Be aware of when two quantities are related to each other and can be expressed using the same letter. Check that each piece of information you are given is really relevant to the problem.
Consider our example problem. We are not told where on the route he starts (and stops). But that does not matter. We should just pick a corner as our "start" and label it. Let's use the top left corner.
Step four is Make equations. Express each relationship you know as an equation. Write (or pretend to write) an English sentence explaining each equation you write.
Consider our example problem. The sum is 1 ½ mile + ¾ mile + 1 mile + ½ mile + ½ mile + ¼ mile
Step five is Solve the equations.
Consider our example problem. The two ½'s add to 1. The ¼ and ¾ also add to 1. The total is 1 mile + 1 mile + 1 mile + 1 ½ miles = 4 ½ miles
Step six is Check your answer. Check that your answer is a reasonable amount. Make sure that your final answer is in units that make sense.
Consider our example problem. An answer of four ½ miles seems reasonable.
What can you do if you get stuck? Here are five ways to try to get unstuck.
For most people, the hardest part of a word problem is drawing the picture, naming the amounts, and especially setting up the equation.
Also remember which English words and phrases correspond with which arithmetic operations. Addition is usually "sum", "total", "increased by", or "more than". Subtraction is usually "difference", "how much more", "decreased by", or "less than". Multiplication is usually "of" or "times". Division usually lacks a phrase but is about finding equal portions.
Finding the area of a rectangle is easy. The formula Area = length × width is well known to most students.
Squares are just the same.
Take three paper rectangles to use while we figure out how other area formulas relate to the area for rectangles.
With a partner, take one of your paper rectangles and invent a way to use it to explain how to measure the area this triangle. Imagine that all you know is how to find the area of a rectangle.
One fold will make this kind of triangle out of a rectangle.
We can see the triangle uses up half of the rectangle. So its area is half of the rectangle's area.
The previous triangle had one side straight up and down. How about this other triangle with two slanted sides?
Two folds will make this kind of triangle out of a rectangle.
Which still is half of the rectangle, as we can see by moving the small piece over.
So for any triangles—whether they have a vertical side or not—we can see the triangle uses up half of the rectangle. Any triangle's area is half of the area of the rectangle that snugly contains it.
For historical reasons we normally do not write Area = ^{1}⁄_{2} × length × width for the area of a triangle. Instead of "length" and "width" we call those measuremets "base" and "height".
Why does Area = ^{1}⁄_{2} × base × height make people happier? No idea.
Students who prefer decimals to fractions will use the variation Area = base × height ÷ 2.
Notice that finding the area of a triangle requires its height, not diagonal edge lengths.
With a partner, take another of your paper rectangles and invent a way to use it to explain how to measure the area this parallelogram. As before, imagine that all you know is how to find the area of a rectangle.
We can make a rectangle out of a parallelogram by cutting off a triangular shape and sliding it over.
We could reverse this to make a paralellogram from a rectangle.
The area of a paralellogram is thus the same as the area of the rectangle.
Notice that finding the area of a parallelogram requires its height, not diagonal edge length.
One last time, with a partner, take another of your paper rectangles and invent a way to use it to explain how to measure the area this trapezoid. As before, imagine that all you know is how to find the area of a rectangle.
Sure, we could cut it apart into two triangles and a rectangle but that is too much work.
The area of a trapezoid is thus half of the area of a bigger parallelogram. This bigger parallelogram has a base whose length is equal to top + bottom for the original trapezoid.
We can write Area = (top + bottom) × height ÷ 2.
Notice that finding the area of a trapezoid requires both horizontal edges and its height, but not either diagonal edge length.
The formula for the area of a trapezoid can instead be written using an average. It looks like "average the horizontal sides, then multiply by the height" if we rewrite it as Area = (top + bottom) ÷ 2 × height.
As a special challenge, try proving that the area of a trapezoid equals its perimeter times onequarter its height. (If you need a hint, click here.)
We just invented a bunch of algorithms! These polygon area formulas should make sense to you now.
If you want to doublecheck your understanding before the final exam, you can use this summary page. What is the area of each shape? How could you prove it to a firstgrader using scissors and perhaps a crayon?
Mathispower4u
Bittinger Chapter Tests, 11th Edition
Chapter 9 Test, Problem 1: A rectangle has length 9.4 cm and width 7.01 cm. Find its perimeter and area.
Chapter 9 Test, Problem 2: A square has sides of length 4 ^{7}⁄_{8} inches. Find its perimeter and area.
Chapter 9 Test, Problem 3: A parallelogram has base 10 cm and height 2.5 cm. Find its area.
Chapter 9 Test, Problem 4: A triangle has base 8 meters and height 3 meters. Find its area.
Chapter 9 Test, Problem 5: A trapezoid has a bottom 8 feet long, top 4 feet long, and height of 3 feet. Find its area.
Chapter 9 Test, Problem 36: A rectangle has length 8 feet and width 3 inches. Find its area in square feet.
Chapter 9 Test, Problem 37: A triangle has base 5 yards and height 3 inches. Find its area in square feet.
Textbook Exercises for Polygon Perimeter and Area
This list of recommended oddnumbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more oddnumbered problems. If your math foundation is strong, do fewer.
Please read the advice on doing homework in the study skills page.
Section 1.2 (Page 23) # 85, 89, 91
Section 1.5 (Page 60) # 105, 107, 109, 113
Section 2.4 (Page 129) # 57, 59, 61
Section 8.1 (Page 422) # 1, 3, 5, 7, 9, 11, 13, 35, 37, 39
Section 8.2 (Page 430) # 1, 3, 5, 7, 17, 19, 21
For many students, finding the area of a rectangle as length × width makes a lot of sense when they see a drawing.
This makes a link in our brain. Multiplying two things can be pictured in our minds as a rectangle's area.
We can use drawings of a rectangle's area to help explain why different examples of multiplication make sense.
Let's start with the traditional multiplication algorithm.
First we use the traditional multiplication algorithm to find 31 × 25 without a calculator.
In our answer each of the four multiplication steps was given its own color. You can see why while looking at the rectangle version of this multiplication.
That rectangle diagram does not include the final answer.
We still need to add up the numbers in its four parts. 600 + 20 + 150 + 5 = 775.
Notice how the in the traditional multiplication algorithm we added those same four numbers but in a different way.
Now it can be your turn with another multiplication problem.
Play around with this more, on your own.
What happens if one number has three digits?
How special is the ghostly gray zero?
We saw earlier how to use place value to think about mixed number subtraction. Can you draw a rectangle diagram for 4 ¼ × 8 ½ = 36 ⅛?
When using a rectangle to help picture 31 × 25 we needed to realize that 31 = 30 + 1 to label one side as two lengths, and 25 = 20 + 5 to label the other side as another two lengths.
Word problems that similarly involve both addition and multiplication can sometimes be written as rectangle diagrams.
Explaining how rectangle diagrams work with algebra would involve more algebra than you might have already learned.
But just in case you have already seen enough algebra to understand, here is one final rectangle diagram.
Maybe that makes you happy?
If that type of algebra is new to you, but makes sense so far, you can search online for a video about the "distributive property".
Time to talk about about circles.
Consider your arm. How many "arm heights" does it take to go around your arm?
How about your head? How many "head widths" does it take to go around your head?
Our answers about measuring our body parts will vary, because arms and heads are not very circular.
If we were measuring actual circles, the answer to the question "How many circle widths does it take to go around the outside?" would be a number bigger than 3 but less than 4.
We need some more careful language.
Circle Width
The width of a circle, going through the center, is its diameter.
Circle Perimeter
The perimeter of a circle is called its circumference.
(Don't ask me why. "Perimeter" was already a perfectly usable word for this.)
So we can reword what we said. If we were measuring actual circles, the answer to the question "How many diameters does it take to go around the circumference?" would be a number bigger than 3 but less than 4.
If we were to write this as formulas, the answer is between C = 3 × d and C = 4 × d.
What number between 3 and 4 is the correct coefficient?
This answer for actual circles is roughly 3.14159 but its decimal digits keep on going forever without any repeating pattern. We call this number π (also written as "pi").
Thus we can look really fancy and professional and write a formula, using either words or just letters.
Circumference = π × diameter
C = π × d
But all this formula really says is "the distance around a circle is a bit more than three times its width".
This animation from Wikipedia shows what is happening nicely:
Consider this small square, of area A = r^{2}.
If we make four of them, the area is then A = 4 × r^{2}.
Now we ask a second question about circles: how large a circle can we fit into the big square of area 4 × r^{2}?
That little line going from the circle's center to its edge (half a diameter) is really useful. Let's give it a name.
Radius
Half the diameter is the radius.
It looks like the circle covers more than 3 of the little squares, but not all 4.
So the area of the circle is between A = 3 × r^{2} and A = 4 × r^{2}.
What number between 3 and 4 is the correct coefficient?
The answer is the same number as the answer to the circumference question: π!
Area = π × radius^{2}
A = π × r^{2}
Again, this looks fancy and professional but really only says "a circle covers a bit more than three squares with side length equal to the circle's radius".
Wonderful! If π was not the answer to both the circumference question and the area question then we would need two buttons on our calculator instead of merely one.
Are the circle formulas algorithms? They can be. Math problems can give you a radius or diameter and ask you to treat the formulas as a process to follow to calculate the circumference or area. But keep in mind that these formulas are also plain descriptions of how circles behave. A bit more than three diameters are required to cover the circumference. A circle covers a bit more than three squares whose side length is the radius.
Naturally, rounding π before the end of a problem will cause incorrect answers. To develop optimal math habits, you should always use the calculator key for π so that you do not round in the middle of the problem. However, some textbooks or websites use a rounded version of π (either 3.14 or ^{22}⁄_{7}). This means your answers will not quite match because yours are more accurate.
A final note of warning: some students learn the circumference formula as C = π × 2 × r. This makes it look a lot like the area formula: both have a π, an r, and a 2. Don't get mixed up! The easiest way to avoid a careless mistake is to use the circumference formula with a diameter.
Circumference = π × diameter = π × 4 inches ≈ 12.6 inches
Area = π × radius^{2} = π × 2^{2} square inches ≈ 12.6 square inches
Circumference = π × diameter = π × 2 inches ≈ 6.3 inches
Area = π × radius^{2} = π × 1^{2} square inches ≈ 3.1 square inches
Circumference = π × diameter = π × ½ inches ≈ 1.57 inches
Area = π × radius^{2} = π × ½^{2} square inches ≈ 0.78 square inches
First we break apart the shape into a halfcircle and a rectangle.
For the perimeter, add halfway around the circle with three sides of the rectangle.
For the area, add half of the circle's area with all of the rectangle.
The perimeter is about 14.6 inches.
The answer is about 16.9 inches.
Bittinger Chapter Tests, 11th Edition
Chapter 9 Test, Problem 6: A circle has a radius of oneeighth of an inch. What is its diameter?
Chapter 9 Test, Problem 7: A circle has a diameter of 18 centimeters. What is its radius?
Textbook Exercises for Circle Circumference and Area
This list of recommended oddnumbered textbook problems is designed for a hypothetical student with a "typical" math background. If your math foundation is weak, do even more oddnumbered problems. If your math foundation is strong, do fewer.
Please read the advice on doing homework in the study skills page.
Section 8.1 (Page 422) # 15, 17, 21, 23
Section 8.2 (Page 529) # 13, 15, 17, 23, 31, 37
Try these ten exercises on scratch paper. Work in a study group if you can! Notice where your notes need improvement. After you are very happy with your answers, you can use this form to ask me to check your work. Can you get at least 8 out of 10 correct?
1. Two years is how many minutes?
2. A football field is 4,844 cm wide. What is that width in yards? (1 meter ≈ 1.09361 yards)
3. A plane flying 6,500 feet per minute is going about how many miles per hour?
4. Including the end zones, a football field is 360 feet long and 160 feet wide. If you run three laps around a football field, about what percentage of a mile do you run?
5. You want to buy Sculpey III modeling clay online. At amazon.com it costs $7.29 for 8 ounces. At amazon.co.uk it costs £2.20 for 57 grams. (1 ounce ≈ 28.35 grams. If £1 ≈ $1.31), which is a better buy?
6. A triangle has base 4 m and height 3.5 m. What is the area?
7. Most billiard tables are twice as long as they are wide. What is the perimeter of a billiard table that measures 4.5 feet by 9 feet?
8. A parallelogram has base 2.3 cm and height 3.5 cm. What is the area?
9. A trapezoid has base 25 cm, top 16 cm, and height 35 cm. What is the area?
10. The standard backyard trampoline has a diameter of 14 feet. What is its area?
Try these exercises on scratch paper. Work in a study group if you can! Notice where your notes need improvement. Check your work when you are done.