Mathematics Study Guide for the TABE Test

Measurement, Data, and Probability: Data and Measurement—Part 2

This section covers particular types of measurement and how to use the results. Some of this will be reviewing concepts we’ve already covered, but with more complex ideas and example problems.

Measuring Time

Time can be thought of in two different ways: as a specific moment (e.g., \(10\text{:}27\) a.m.) or as the length of time that has passed from one moment to another (e.g., six hours and five minutes), called a time interval.

On the TABE, you may see problems that ask you one of two time-related questions: “What time did \(x\) happen?” and “How long did \(x\) take?” Both questions require you to be comfortable with measuring and calculating time.

Basic Time Measurement

We measure time in a variety of units, from the nearly instantaneous millisecond to the much longer millennia or even eons. In daily life, we generally track time with our phones or on a watch or even with a wall clock. When we’re timing something that is quick, we can use a stopwatch.

These are the basic conversions between common units of time:

• \(60\) seconds \(= 1\) minute
• \(60\) minutes \(= 1\) hour
• \(24\) hours \(= 1\) day
• \(7\) days \(= 1\) week
• \(365\) days \(= 1\) year

Note: Even though one year is considered to equal \(52\) weeks, there is actually one extra day. This is how we end up with \(365\) days.

Reading Time

For the standard \(12\)-hour time notation (as opposed to the \(24\)-hour military time notation), times of the day are written with hours first followed by a colon, then the number of minutes. If the time on a clock is \(20\) minutes past \(10\), for example, it is written as \(10\text{:}20\).

On a clock, the day is split into two \(12\)-hour periods, from \(12\text{:}00\) a.m. (midnight) to \(12\text{:}00\) p.m. (noon), and then from \(12\text{:}00\) p.m. to \(12\text{:}00\) a.m. Between \(12\) to \(12\), the hour goes from \(1\) to \(11\), while the minutes of each hour go from \(0\) to \(59\). The “a.m.” and “p.m.” indicate the different halves of the day. If it is seven in the morning, it is \(7\text{:}00\) a.m., while seven at night is \(7\text{:}00\) p.m.

Another use of this colon notation is in writing how much time has gone by. If you drove straight through from Boston to San Diego, it might take you something like \(26\) hours and \(30\) minutes, which would be written as \(26\text{:}30\).

Let’s do some practice problems of varying difficulty that involve time intervals.

Finding Elapsed Time

One of the more common time problem types asks you to determine how much time has passed between a starting time and an ending time. This amount of time is called the elapsed time or time interval. A useful strategy is to move forward in stages. First, move from the starting time to the next hour. Then count any remaining whole hours. Finally, add any remaining minutes. With time problems, it’s important to remember that an hour is \(60\) minutes, so when subtracting minutes, \(60\) is your starting point, not \(100\).

We’ll do a simple example problem to show you how this works.

A flight from Detroit leaves at \(8\text{:}40\) a.m. and arrives in Chicago at \(9\text{:}30\) a.m. How long did the flight take?

Solution

An easy way to visualize this problem is to put the times on a number line:

The figure shows that there are \(20\) minutes from \(8\text{:}40\) to \(9\text{:}00\) and \(30\) minutes from \(9\text{:}00\) to \(9\text{:}30\). That adds up to \(50\) minutes. Let’s do an example problem that is a bit more complicated.

A movie begins at \(2\text{:}45\) p.m. and ends at \(5\text{:}20\) p.m. How long is the movie?

Solution

The start is at \(2\text{:}45\) p.m. From \(2\text{:}45\) p.m. to \(3\text{:}00\) p.m. is:

\[15 \text{ min}\]

From \(3\text{:}00\) p.m. to \(5\text{:}00\) p.m. is:

\[2 \text{ hr}\]

From \(5\text{:}00\) p.m. to \(5\text{:}20\) p.m. is:

\[20 \text{ min}\]

Adding everything together, we get:

\[2 \text{ hr} + 15 \text{ min} + 20 \text{ min}\] \[= 2 \text{ hr }35 \text{ min}\]

Thus, the movie lasted two hours and \(35\) minutes.

Finding End Time

Sometimes you will be given a starting time and a time interval and you must determine when an event ends. This type of problem is similar to the elapsed time problem except for the information you know up front.

To solve these problems, add the hours first and then add the minutes. If the minutes total more than \(60\), convert the extra minutes into hours.

Let’s look at an example.

A basketball tournament begins at \(8\text{:}40\) a.m. and lasts five hours and \(45\) minutes. What time does it end?

Solution

Add the hours first:

\[8\text{:}40 + 5 \text{ hr}=1\text{:}40 \text{ p.m.}\]

Now, add the minutes:

\[1\text{:}40 + 45 \text{ min}=2\text{:}25 \text{ p.m.}\]

Thus, the tournament ends at \(2\text{:}25\) p.m.

Note: When adding minutes causes the total to exceed \(60\), convert every \(60\) minutes into one additional hour. For example:

\[4\text{:}45 + 30 \text{ min}=5\text{:}15\]

Since hours are \(60\) minutes, \(45+30=75\) minutes and \(75\) minutes equals \(1\) hour and \(15\) minutes.

Measuring Perimeter

As you’ve learned, the perimeter is the distance around the boundary of a figure. If you know the lengths of the sides, you just add them up to get the perimeter. Sometimes a perimeter problem has a little twist. You will be given the perimeter and all but one of the sides and asked for the length of the missing side.

For example, if a six-sided polygon has a perimeter of \(38\) and five of its sides have lengths of \(3, \,7,\, 7,\, 6,\) and \(9\), how long is the sixth side?

If we add the five sides we know, and the missing side, we will get \(38\), so we can write this:

\[3+7+7+6+9+x=38\] \[32 +x= 38\]

The missing side must be \(6\).

Note: Notice how we used the variable \(x\) in this problem. You will often see problems that blend different types of math, like algebra and geometry in this example.

Here’s a different kind of problem.

If you have \(16\) feet of fencing to make a small rectangular turtle pen, will every length and width give the same area inside?

Solution

Try a \(4 \times 4\) pen and a \(2 \times 6\) pen. They both have the same perimeter.

\[4+4+4+4=16\ \ \ \ \ \ \ 2+2+6+6 = 16\]

The \(4 \times 4\) has an area of \(4 \times 4 = 16\). The \(2 \times 6\) pen has an area of \(2 \times 6 = 12\).

So the answer is no, they do not have the same area inside. Length and width make a difference. In fact, a \(4 \times 4\) square gives you the most area possible for \(16\) feet of fencing.

Measuring Area

As you learned in the geometry section, area is the measure of a surface enclosed by a boundary. A common sheet of paper, for example, is a rectangle enclosed by the paper’s edges. A US quarter is a circle enclosed by its round edge. Fairly simple shapes, like rectangles, triangles, circles, and parallelograms, have formulas that can be used to calculate their areas. Oddly shaped things, like golf greens, don’t have simple ways to find their areas, but their areas can still be found.

Simple Area Formulas

You’ve learned these before, but here is a refresher on some of the basic area formulas:

Figure	Formula	Variables
Rectangle	\(A=l\cdot w\)	l = length w = width
Square	\(A = s^2\)	s = side length
Triangle	\(A=\frac{1}{2}b\cdot h\)	b = base length h = height
Circle	\(A = \pi r^2\)	\(\pi \approx\) 3.14 r = radius

Having these basic formulas memorized will save you time on the TABE and help you do more complex problems.

Measuring Area with Unit Squares

When working with polygons, you can think of measuring area as finding the number of one-by-one squares that fit in a figure. The squares are the units, which could be inches, feet, or miles. When you’re doing the basic math, though, the units don’t matter. For instance, how many squares are shown fitting into the rectangle below?

We could just count them, but that will take time. Instead, we’ll reason our way to the answer by noticing that there are \(12\) squares in each row, and there are seven rows. Rather than add all the squares, we can use simple multiplication to get the answer. Seven rows of \(12\) squares gives us:

\[7\times12 = 84\]

Notice how that aligns with the formula \(A = lw\).

Measuring Volume

As you’ve learned, we define volume as the measure of space enclosed by a surface, as in a balloon, or multiple surfaces, as in a cube. To review, these are a few some common formulas for volume:

Figure	Formula	Variables
Rectangular Solid	\(V=l w h\)	l = length w = width h = height
Cube	\(V = s^3\)	s = edge length
Cylinder	\(V=\pi r^2 h\)	r = radius h = height
Sphere	\(V = \frac{4}{3}\pi r^3\)	\(\pi\) = 3.14 r = radius

Just as we use square meters and square feet when measuring area, some volume units have the word “cubic” in them, as with cubic inches, cubic centimeters, cubic feet, and so on. There are, however, also other volume units, including gallons, liters, milliliters, quarts, and more. Here are some basic volume comparisons:

• \(1\) liter = \(1\text{,}000\) milliliters
• \(1\) milliliter = \(1\) cubic centimeter

Liquid Volume

Liquid volumes are often measured with a container that has a scale printed on it, such as a measuring cup in the kitchen or a beaker or graduated cylinder in the science lab, as illustrated below:

Retrieved from: https://publicdomainvectors.org/en/free-clipart/Beaker-with-water/52141.html

You should be able to do problems that combine or divide different volumes. Here are a couple example problems.

If a student has a beaker holding \(520\) milliliters of water and pours as much as possible into a \(150\)-milliliter beaker before pouring the rest into a \(500\)-milliliter beaker, how much water will be in the \(500\)-milliliter beaker?

Solution

The inclusion of beakers and milliliters shouldn’t obscure the fact that this is a simple subtraction problem. We need to subtract the amount poured from the initial beaker:

\[520-150=370\]

While the problem tells you there is another step, pouring the remaining water into a \(500\)-milliliter beaker, you already have your answer. There will be \(370\) milliliters in the beaker.

The same student again starts with \(520\) milliliters of water and uses it to fill \(26\) test tubes, with no water left in the original beaker. How many milliliters did each test tube hold?

Solution

This is another problem that involves a basic math operation, this time division. Since we know no water was left in the first beaker, we are dividing the full original amount by the number of test tubes:

\[520 \div 26 = 20\]

Therefore, each test tube is holding \(20\) milliliters of water.

Solid Volume

A cube that has dimensions \(1 \times 1 \times 1\) is called a unit cube. The one below happens to be a centimeter unit cube. It represents the volume of a cubic centimeter. There can be unit cubes of any length unit you can think of: a cubic inch, a cubic foot, a cubic yard, a cubic meter, etc. If you were measuring the volume of something really big, like the ocean, you might even want to use cubic miles.

One cubic centimeter:

To help visualize cubic units, sketching is a useful tool. The proportions don’t have to be exact for it to represent the number of units you need. You could draw something like this:

Retrieved from: https://publicdomainvectors.org/en/free-clipart/Dice-building/46859.html

When you’re thinking about measuring the volume of an object, you’re looking to see how many unit cubes will fit into the object. For example, if you stacked \(10\) one-centimeter unit cubes on top of each other, you will have made a little tower that has a volume of \(10\) cubic centimeters. Like any other form of measuring, you have a standard unit and you’re trying to see how many of that unit will fit into the object you’re measuring.

Combination 3-D Shapes

Many real-world objects are not made from a single solid shape. Instead, they are formed by combining two or more simpler solids. When a shape is made by joining multiple right prisms, we can find its volume by breaking it into smaller parts, finding the volume of each part, and then adding the results together. Recall that a right prism is a solid shape with two parallel, congruent bases and sides that meet the bases at right angles. Rectangular prisms are the most common type of right prism.

This strategy is especially useful when measuring the volume of buildings, storage containers, steps, platforms, and other structures made from rectangular sections.

To find the volume of a composite solid:

1. Divide the solid into simpler right prisms.
2. Find the volume of each prism using the formula \(V=l \cdot w \cdot h\).
3. Add the individual volumes together.

Note: You should recognize this as a similar process to finding the area of composite two-dimensional shapes.

Let’s do an example problem.

A storage platform is shaped like the composite solid shown below. The width of prism B is half of that of prism A. What is the total volume of the storage platform?

Solution

The length, width, and height of prism A is \(10\), \(4\), and \(3\), respectively. So, its volume is:

\[V_A = 10 \times 4 \times 3 = 120\]

Note: Since we’re finding volume but aren’t given the specific unit type, we say this is \(120\) cubic units.

The length of prism B is \(4\) and the height is \(2\). The width is half that of prism A, so the width will be half of \(4\), or \(2\). Thus, the volume of prism B is:

\[V_B = 4 \times 2 \times 2 = 16\]

Adding the two volumes, we get:

\[120 + 16 = 136\]

Thus, the volume of the whole figure is \(136\) cubic units.

Unit Analysis

Earlier in this guide, you saw how units are used in ratios. This introduced the concept of unit rates, such as \(30\) miles per hour, which you can think of as \(\frac{30\text{ mi}}{1\text{ hr}}\). When you have problems that involve units, you will often start with one unit and end with another. If you’re not careful, you can get the wrong unit and get a completely wrong answer. After all, there’s a big difference between \(20\) inches and \(20\) feet.

Here, you will learn the process of unit analysis, or dimensional analysis, which will make sure you’ve arrived at the correct answer with the correct unit.

Multi-Step Problems

When you’re doing problems with multiple steps, you need to pay attention to the units. They can tell you what to do and what kind of operations to perform. Furthermore, in many questions you will need to convert from one unit to another, and you may even need to use an intermediary unit. Make sure to read the question carefully and pay attention to what unit the answer is supposed to have. All other units must be canceled out.

Here are the steps for doing a multi-step unit problem:

1. Write down the given quantity that needs to be converted. As an example, say it is \(5\) ft.

2. Create a unit rate that has the same unit as the given quantity. We’re using feet as our unit, so this will be our unit rate:

\[\frac{ 1\text{ ft}}{12\text{ in}}\]

1. Set up the unit rate so it will cancel out the given unit. In our example, we would have:

\[5 \text { ft} \times \frac{ 12\text{ in}}{ 1\text{ ft}}\]

1. Multiply the numbers and cancel out the units. Continuing with our example, we have:

\[\require{cancel}\] \[5 \cancel{\text { ft}} \times \frac{ 12\text{ in}}{ 1 \cancel {\text{ ft}}}=60 \text{ in}\]

If you needed to convert inches to something else, like centimeters, you would just take \(60\) inches as the given quantity and multiply by \(\frac{2.54\text{ cm}}{1 \text{ in}}\).

Note: As we stated in the beginning of this guide, you will be provided with the conversions you need for whichever level of the test you are taking.

Unit Analysis with Formulas

Consider these three sets of expressions:

\[lw \ \ \ \ \ \ \ \frac{1}{2}bh \ \ \ \ \ \ \ \ \pi r^2\]

The variables are \(l\) for length, \(w\) for width, \(b\) for base, \(h\) for height, and \(r\) for radius. Let’s say all the variables are in meters. Based on that information, you should know the following:

• The expression \(lw\) will give us meters \(\times\) meters, which results in square meters (also written m\(^2\))).

• The expression \(\frac{1}{2} \times bh\) will give us \(\frac{1}{2} \times\) meters \(\times\) meters, which again results in square meters (m\(^2\)). We can ignore the \(\frac{1}{2}\) at the moment because it has no units.

• The expression \(\pi r^2\) will give us \(\pi \times\) meters \(\times\) meters, which yet again results in square meters. \(pi\) has no units.

You may have recognized the three expressions we started with. Normally they would be part of a formula:

\[A = lw \ \ \ \ \ \ \ \ A=\frac{1}{2}bh\ \ \ \ \ \ \ \ A= \pi r^2\]

The point of this is to show that any time you have two length units multiplied together, you will have squared units in your answer, which is an area unit. To turn this around, if you are calculating area and you carry the units along in the problem, you should get squared units of some kind for the answer. If not, recheck your work.

We could go through the same kind of discussion for volume units. They are always formed when three length units are multiplied, resulting in cubed units. Notice that each volume formula below has three length units multiplied together. Remember, \(\frac{4}{3}\) and \(\pi\) have no units so they don’t count.

\[V = lwh \ \ \ \ \ \ \ \ V = s^3\ \ \ \ \ \ \ \ V = \frac{4}{3} \pi r^3\] \[V = lwh \ \ \ \ \ \ \ \ V = s \times s \times s\ \ \ \ \ \ \ \ V = \frac{4}{3} \times \pi \times r \times r \times r\]

Measuring Mass

Although mass and weight are closely related, and the terms are often used interchangeably, they are not exactly the same. Weight measures the force of gravity pulling down on an object. Weight changes from place to place. Take a scale to the Moon and you will see that you weigh less there than on Earth.

Mass, by contrast, measures how hard it is to start an object moving or to stop an object that’s moving. Mass isn’t dependent on gravity, so it doesn’t change from place to place.

Imagine you’re going to kick a bowling ball with your bare foot. Broken toe and ER trip, right? Is that because of the ball’s weight or mass? Before you answer, imagine that you could travel to some asteroid in space where you were nearly weightless. Could your toe survive a barefoot kick to a bowling ball there? Could you throw a bowling ball forward like you would a baseball?

The answer to both questions is “no” because of the mass of the bowling ball. The bowling ball will still fight you if you try to speed it up (or slow it down), even though it’s essentially weightless. That’s the effect of mass for you.

Standard Units to Measure Mass

The standard metric units of mass are the gram (g) and the kilogram (kg). For estimating, think of a kilogram as a bit more than two pounds (lb). A liter (L) of water is a bit more than a quart and has a mass of one kilogram. A paper clip has a mass of about one gram, and an average banana has a mass of about \(125\) grams. Approximations like these can help you to estimate the masses of common objects. For a more accurate measure of masses, it’s common to use a balance or a scale.

These are some useful unit comparisons to be familiar with, even if you don’t memorize them:

• \(1\) ton = \(2\text{,}000\) pounds
• \(1\) kilogram = \(2.2\) pounds

Solving Problems Using Mass

On the TABE, you will be asked to solve one-step mass problems using basic math operations, such as addition, subtraction, multiplication, and division. Even if the concept and terms are different, the process for solving these mass problems will be the same as any other problem using those operations. Sometimes, drawing a picture of the information you have helps you determine what to do.

Let’s try two example problems.

Betty Baker wants to bake two dozen cookies. She knows that it takes two ounces of cookie dough to make three cookies. How many ounces of cookie dough will she need?

Solution

The drawing below is one way to help visualize the info in this question:

Once you know that Betty needs to make eight groups of three cookies, you can simply multiply \(8\) by \(2\). Thus, she will need \(16\) ounces of cookie dough.

A science student has a container holding \(2.5\) kilograms of sand. She pours \(850\) grams into Container A and \(650\) grams into Container B. How much sand remains in the original container?

Solution

The measurements are given in different units, so we must first convert them to the same unit.

Since one kilogram is equal to \(1{,}000\text{ grams}\) the original mass is:

\[2.5\text{ kg}=2{,}500\text{ g}\]

Next, we find the total mass removed:

\[850+650=1{,}500\text{ g}\]

Now, we’ll subtract the amount removed from the original amount:

\[2{,}500-1{,}500=1{,}000\text{ g}\]

So, \(1{,}000\) grams of sand remain in the original container, or one kilogram.

Density

Density is a word that loosely means how tightly packed something is. In an elementary school, teachers may be concerned with how many students are packed into each classroom. While \(20\) students per classroom might be fairly low density in some school districts, \(40\) would be very high almost anywhere. City dwellers live where there is a high population density, measured in people per square mile. Gold has a very high weight density, which could be measured in pounds per cubic foot.

The point is that there are different kinds of densities. What they have in common is that they all involve some ratio where the bottom number is one, and they use the word per in their description. In math, density is measured by dividing mass by volume, as represented by this formula:

\[\rho = \frac{m}{v}\]

You should be able to calculate a density given those two values. Let’s try an example problem.

If three cubic feet of maple wood weighs \(120\) pounds, what is its density in pounds per cubic foot?

Solution

We will be using the density formula to solve this. But which number goes on top? As you’ve learned, volume is measured in cubic feet and mass is measured in pounds. Even if you didn’t remember that, though, you could just know that whatever unit comes before the word per goes on top and the other unit goes on the bottom.

Here we have:

\[\frac{120\ \text{lb}}{3\ \text{cu ft}}\]

When we simplify that, it comes to \(40\) pounds per cubic feet.

Measuring Angles

Angles are a vital part of geometry, and being able to measure them is an essential skill. In the figure below, two rays, \(\overrightarrow{AF}\) and \(\overrightarrow{AN}\), create an angle by sharing point \(A\), the vertex of the angle:

In this section, you will learn different techniques for measuring an angle.

Using a Protractor

When measuring angles in the real world, you can use what is known as a protractor. The figure below shows a protractor lined up with a \(120^\circ\) angle. See how the vertex of the angle lines up with the center mark, and one ray of the angle lines up with the zero mark. The number of degrees of the angle then lines up with the second ray of the angle.

Adding Angles

Sometimes, you can’t physically measure an angle. In such cases, you may still be able to deduce the size of the angle if you have enough information. For instance, two angles can be combined to form a third larger angle, as shown below, if they have the same vertex (\(A\), in this case) and share a common side (\(\overrightarrow{AN}\)) between them. In such a situation, their measures can be added together to get the measure of the larger angle they make:

\[\angle NAB + \angle MAN = \angle MAB\] \[25^\circ + 65^\circ = 90^\circ\]

Recall that angles that share a vertex and common side are adjacent angles. Furthermore, \(\angle MAN\) and \(\angle NAB\) are complementary angles because they add up to \(90^\circ\). However, two adjacent angles can create a third angle of any size.

Solving for Angles

Just like solving word problems, solving problems with angles is determining what you know and what you don’t know. Everything else is just basic operations. Let’s look at a couple sample problems.

In the graphic below, if \(\angle MOB = 120^\circ\), what is \(\angle MOJ\)?

Solution

We know two things: \(\angle JOB\) is equal to \(25^\circ\), and when that angle is added to \(\angle MOJ\) we get \(120^\circ\), which is the measure of \(\angle MOB\). If we let \(\angle MOJ = x\), we can write this equation:

\[25^\circ + x = 120^\circ\]

To get \(x\), we subtract \(25^\circ\) from both sides of the equation:

\[25^\circ + x -25^\circ= 120^\circ - 25^\circ\] \[x = 120^\circ - 25^\circ\] \[x = 95^\circ\]

Therefore, \(\angle MOJ\) is \(95^\circ\).

Let’s do one more example.

Lucy baked a round apple pie to share with her friends. She cut a slice that was \(50^\circ\) for Jim. Marnie asked for a sliver, so Lucy cut her a slice that was \(25^\circ\). After baking all day, Lucy was very hungry, so she cut a large slice for herself. If the amount of pie that was uneaten measures \(195^\circ\), how big was the slice that Lucy ate?

Solution

The secret to answering this question is remembering that a circle is \(360^\circ\). All we have to do is add up the numbers we know and subtract them from that total. Let \(x\) be the size (in degrees) of the slice that Lucy cut for herself. Then, we know:

\[50^\circ + 25^\circ + 195^\circ + x =360^\circ\] \[270^\circ + x =360^\circ\] \[x = 360^\circ - 270^\circ = 90^\circ\]

Therefore, the slice that Lucy ate was \(90^\circ\).