• # Percent Word Problems

Posted by Mark Rinaldi on 3/9/2020

When looking at percent word problems, there are two main "outlines" we can use to set up our problem.  We can set it up using equivalent ratios or we can set up it as a number sentence.

Let's look at a sample problem:

Heather and her family are going to the grand opening of a new amusement park. There is a special price on tickets this weekend. Tickets cost \$56 each. This is 70% of the cost of a regular price ticket. What is the cost of the regular price ticket?

Now let's look at the two different ways we've been using in class to approach this type of problem.

The Number Sentence:

All percent problems can be boiled down to this simple number sentence:

part = % x whole

They will always give you two of the three parts somewhere in the word problem.  You can read the world problem and plug in the values that they give you to find the one that is missing.  In this problem, the percent is obvious - 70%.  Then we have the \$56.  We have to figure out if this is the part or the whole.  Since it says that \$56 is 70% of the cost of a regular ticket price, then \$56 is part of the price.

\$56 is 70% of the cost.

56 = .70 x ______

Since we are missing part of a multiplication problem, we can reverse it into division.

56 ÷ .70 = ____

If we solve that problem, we get \$80, so that is the cost of a regular ticket.

Equivalent Ratios:

We can also try to take the information and plug them into the same equivalent ratios that work for any percent problem: Again, looking at the problem, we have 70% and \$56 for the part.

70/100 = 56/?

We can solve this using equivalent rations by reducing the percent and then working our way back up.

70/100 = 7/10

7 x 8 = 56 so 10 x 8 = 80.

70/100 = 56/80 so the whole ticket price is \$80.

If students can remember one or both of those two basic setups, it will help make finding the solution to the problems that much easier.

You can check out this section on Khan Academy to see some sample problems explained.

• # Converting between Percent, Decimal, and Fraction

Posted by Mark Rinaldi on 3/5/2020

We have been working with converting between percent, decimal, and fraction.  There are six basic conversions students need to be able to do.

Decimal to Percent - Move the decimal point over two places to the right.

0.49 = 49%, 0.07 = 7%

Percent to Decimal - Move the decimal point over two places to the left.

84% = 0.84, 62.7% = 0.627, 129% = 1.29

Decimal to Fraction - Read the decimal and rewrite it as a fraction read the same way.  Then simplify if possible.  0.047 is read as forty-seven thousandths.  Rewrite it as 47/1000 so it is still read as forty-seven thousandths.

Fraction to Decimal - There are two options.  One is to try and make an equivalent fraction with the denominator being a decimal place value (10ths, 100ths, 100ths, etc.).  You can then turn it into its synonymous decimal, like in Decimal to Fraction above.  Also, you can remember that a fraction is just another way to write a division problem.  Turn it into a division problem and solve.  You may have to round it off if there is a repeating decimal.  5/8 = 5 ÷ 8 = 0.625

Fraction to Percent - One way is to set the fraction up equivalent to another fraction over 100.  If it can't easily be turned into something over 100, you can turn it into a decimal first and then turn the decimal into a percent.

Percent to Fraction - Change the percent into a fraction over 100.  73% = 73/100.  Simplify if possible.

This section of Khan Academy has some videos and practice problems you may find helpful.

• # Rates and Unit Rates

Posted by Mark Rinaldi on 2/3/2020

We are taking our work with ratios and applying it to rates.  A rate is a ratio where the two quantities being compared are measured in different units.  A simple example of a rate is speed.  Speed is measures as distance over time.  The most common rate of speed seen in everyday life is miles per hour.  Notice the two values are measured in different units: miles and hours.

An important type of rate we are working with is the unit rate.  A unit rate is a rate in which the second quantity in the comparison is always one unit.  For example, speed is almost always given as a unit rate.  We don't see signs on the highway saying the speed limit is 195 miles per 3 hours.  It is always something per 1 hour.

Many of the examples we are working with in class are related to finding unit rates.  For example, if you are told that one bottle of juice costs \$3.84 for a 16 ounce bottle and another bottle costs \$4.50 for 25 ounces, you would want to look at the unit price to see which is the better deal.

\$3.84 / 16 ounces = ? / 1 ounce

We would divide 16 by 16 to get to 1 so we divide \$3.84 by 16 to get to \$0.24 so the first bottle is \$0.24 per ounce.

\$4.50 25 ounce = ? / 1 ounce

We would divide 25 by 25 to get to 1 so we divide #4.50 by 25 to get to \$0.18 so the second bottle is \$0.18 per ounce.  The second bottle is a better value.

The Rates section has some videos about unit rates and some practice problems.

• # Ratio Word Problems

Posted by Mark Rinaldi on 1/28/2020

We have been working on solving word problems using ratios.  I've been trying to get them to set up their problems the same way every time.  Let's look at an example:

The ratio of boys to girls in the school is 8 to 11.  If there are 336 boys in the building, how many girls are there?

The first thing we do is write the ratio using the words of what we are comparing.  In this case, our starting ratio would be:

boys/girls

Then we write the starting ratio:

boys/girls = 8/11

Then we write the new information they give us to solve the problem:

boys/girls = 8/11 = 336/?

Now we have equivalent ratios and we can find the multiplicative relationship between the ratios.  We need to figure out 8 times what equals 336.  So we divide 336 by 8 and get 42.  The multiplicatve relationship between the ratios is x42.  So then we multiply 11 by 42 to find the number of girls.

11 x 42 = 462.

There are 462 girls in the building.

• # Ratios

Posted by Mark Rinaldi on 1/27/2020

We have begun our unit on Ratios, Rates and Proportions.  Our basic definition of a ratio is just a comparison of two quantities.

Let's look at this set of letters for example:

A D A B A A B C C A D A C

If I asked for the ratio of A's to B's, we would count the A's and count the B's.  Since there are 6 A's and 2 B's, we would say the ratio is 6 to 2.  The order is important as the numbers have to match the quantities we are looking for.

We also discussed that there are different ways to write ratios.  The answer in the above example could be written as 6 to 2, 6:2, or 6/2.

Introduction to Ratios with some sample problems.

• # Division of Decimals

Posted by Mark Rinaldi on 1/20/2020

In division, the students already know that a decimal in the dividend does not cause problems.  We just divide like normal and bring the decimal point up.  They just have to make sure they are keeping their columns lined up so they bring the decimal point up to the right place.  The part that we have to watch out for is if there is a decimal in the divisor.  Let's look at an example. In this math problem, we see there is a decimal in the divisor (the number outside the box).  To make this easier to solve, we are going to move the decimal point over as many places as needed to make it a whole number.  We would have to move the decimal two places to turn 0.91 into 91.  Since we moved that decimal two places, we have to keep it equivalent and move the other one two places so 50.96 becomes 5096. We then solve it as a normal division problem and get our answer. You can check out Khan Academy for review:

• # Multiplying Decimals

Posted by Mark Rinaldi on 1/13/2020

Students have seen that multiplying numbers with decimals points in them is the same as multiplying whole numbers except for the last step where they determine where the decimal point goes in their final answer.  Let's look at a simple example.

4.57 x 9.3 =

We can ignore the decimals and look at this problem as 457 x 93.  If we do out the problem we get an answer of 42501.  Then we go back and look at the original problem.  If we look at our two factors, the first factor has two (2) decimal places after the decimal point and our second factor has one (1) decimal place after the decimal point.  If we add that (2+1 = 3) we get three.  That means our answer needs to have three decimal places.  Therefore, the answer would be 42.501 since that gives us three decimal places.

Khan Academy practice: Multiplying Decimals practice problems

• # Unit Test Review

Posted by Mark Rinaldi on 12/16/2019

These are the topics you should be reviewing for the upcoming unit test:

• # Dividing Mixed Numbers

Posted by Mark Rinaldi on 12/4/2019

We are now looking at the division of fractions.  The trick to doing division of fractions is to not try and divide fractions.  Instead, we turn it into something that we do know how to do, which is multiplication of fractions.

As we have been saying all along, the first step to any division or multiplication of fractions/mixed numbers problem is to turn everything into a fraction.  In previous posts, we've gone over how to turn a mixed number into a fraction.  You multiply the denominator by the whole number, add that product to the numerator, and keep the denominator the same.  If it is just a whole number, put it over one (example: 9 = 9/1).  Once everything is in fraction form, we can continue.

There is only one new step for division compared to multiplication.  We change the division sign into its inverse operation which is multiplication.  After that, we "flip" the second fraction into its reciprocal or multiplicative inverse.

multiplicative inverse examples:

4/7 --> 7/4

8/13 --> 13/8

4 2/3 = 14/3 --> 13/4

6 = 6/1 --> 1/6

One of the most important things to remember is that only the fraction after the division sign gets "flipped".  Once you have done that, the problem has become a multiplication of fractions problem which is something we have been doing for a while.  We also discussed why the process works.

Example:

9 1/3 ÷ 5 5/6 =

becomes:

28/3 ÷ 35/6 =

becomes:

28/3 * 6/35 =

Now we solve that as we would a normal multiplication of fractions problem.  We can look for anything that can simplify before multiplying.  Look at the previous posts to see the methods for simplifying before multiplying.

28/3 * 6/35

becomes

4/1 * 2/5 = (4*2)/(1*5) = 8/5 = 1 3/5

This video and the one that follows it are good refreshers

• # Multiplying Mixed Numbers

Posted by Mark Rinaldi on 11/19/2019 9:00:00 AM

Since we have been working on multiplying fractions, today we decided to throw some mixed numbers into the problem to see what happens.  We discussed the fact that the easiest way to multiply mixed numbers is to not multiply mixed numbers.  Instead, we turn them into fraction form and just multiply the fractions as we have already been practicing.

Step 1.
Turn any whole numbers or mixed numbers into fractions. To turn a whole number into a fraction, just make it over 1.

Example: For mixed numbers, you multiply the whole number by the denominator and add it to the numerator. The denominator stays the same.

Example: 3 * 2 = 6 + 1 = 7

Step 2. Once they are in fraction form, we can see if there is anything that can simplify.  If you look at the previous post, we discussed a couple ways to simplify before multiplying.

Example: becomes 27/4 and 32/15 when we turn them into fractions. As we already looked at, we can simplify before multiplying at this point.  We can either look for common factors or we can break them down into prime factors.  Look at the previous blog post to see an example of how we did that.

Step 3. Multiply across and it becomes 72/5

Step 4. If the answer is greater than a whole, you use division to make it back into a mixed number. with a remainder of 2

So the final answer is 