Mathematics Study Guide for the HESI Exam
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Military Time
Military time is a way to tell the time by looking at the hours in the day as \(\bf{24}\) hours rather than the standard two \(12\)-hour periods of a.m. (morning) and p.m. (evening).
Time Equivalents
Military time starts at midnight with \(0000\) (zero hundred hours) and continues up to \(2400\) (midnight again). This allows for less confusion if someone tells you to meet at, say \(8\) o’clock. In the \(12\)-hour clock system, or civilian time, this could be in the morning or the evening. In military time, \(0800\) hours can only be the morning, whereas \(8\) p.m. is \(2000\) (twenty hundred hours).
The last two digits of the military time are reserved for the minutes, just like in civilian time, and can show from \(00\) to \(59\).
Converting Times
To find the conversion between military and civilian time, we subtract twelve or count up from twelve. Here is a conversion chart between the two systems:
| Civilian Time | Military Time |
| 12:00 a.m. | 0000 hours |
| 1:00 a.m. | 0100 hours |
| 12:00 p.m. | 1200 hours |
| 1:00 p.m. | 1300 hours |
| 2:00 a.m. | 0200 hours |
| 2:00 p.m. | 1400 hours |
| 3:00 a.m. | 0300 hours |
| 3:00 p.m. | 1500 hours |
| 4:00 a.m. | 0400 hours |
| 4:00 p.m. | 1600 hours |
| 5:00 a.m. | 0500 hours |
| 5:00 p.m. | 1700 hours |
| 6:00 a.m. | 0600 hours |
| 6:00 p.m. | 1800 hours |
| 7:00 a.m. | 0700 hours |
| 7:00 p.m. | 1900 hours |
| 8:00 a.m. | 0800 hours |
| 8:00 p.m. | 2000 hours |
| 9:00 a.m. | 0900 hours |
| 9:00 p.m. | 2100 hours |
| 10:00 a.m. | 1000 hours |
| 10:00 p.m. | 2200 hours |
| 11:00 a.m. | 1100 hours |
| 11:00 p.m. | 2300 hours |
Let’s look at some example problems using military time.
Tiana has to report for work at \(1945\) hours. What time is this in civilian time?
Solution
The first two digits, \(19\), refer to the hour. We know it is in the second half of the day (p.m.) because the number is above \(12\). Now, we subtract \(\bf{12}\) from \(19\) and we’re left with \(7\). The minutes work the same way in both systems. So, Tiana has to work at \(7\text{:}45\) p.m.
Let’s do another example.
Iliana is planning her day. In her journal, she marks down “Workout―\(0545\) to \(0630\).” In civilian time, when is Iliana working out?
Solution
Since the hours are below \(1200\), we know this is in the morning, and the hours and minutes are the same as in civilian time. Iliana is working out from \(5\text{:}45\) a.m. to \(6text{:}30\) a.m.
Algebra
Algebra is the language of mathematics used to solve problems with unknowns. In nursing and healthcare, you’ll often need algebraic thinking when calculating dosages, interpreting formulas, or adjusting IV rates. In this section, we’ll break down the key terms, operations, and rules you’ll need to know.
Algebraic Terms
Understanding basic algebraic terms helps you read, set up, and solve problems. Each part of an equation or expression has a role, and knowing what they are makes complex problems much more manageable. These are the essential algebraic terms:
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variable—A variable is a symbol, usually a letter, that stands in for a number you don’t know yet. For example, in the equation \(x + 5 = 12\), the variable is \(x\). In healthcare, you might use \(d\) for an unknown dosage amount.
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constant—A constant is a number that does not change. It is a known value. For example, in \(y = 3x + 4\), the number \(4\) is a constant. If a medication always comes in \(500\)-milligram tablets, the number \(500\) could be a constant in a dosage-related formula.
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coefficient—A coefficient is any value multiplied by a variable. Similar to a constant, the coefficient is a known value. For example, in \(y = 3x + 4\), the number \(3\) is a coefficient. It means you have a multiple of a specific unknown quantity. For example, if a person needs to take two tablets every hour for an unknown number of hours, \(x\), the person will need a total of \(2x\) tablets to have enough medication.
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expression—An expression is a mathematical phrase that can contain numbers, variables, and operations, but it does not have an equal sign. For example, \(3x + 7\) is an expression. For instance, \(21d - 5\) could represent “21 doses minus five.”
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equation—An equation is a mathematical statement that shows two expressions are equal. It uses an equal sign (\(=\)). For example, \(5x - 2 = 13\) is an equation. If you know the total dosage and need to find a single dose, you might set up an equation.
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exponent—An exponent shows how many times a number is multiplied by itself. For example, \(3^2 = 3 \times 3 = 9\) and \(10^3 = 10 \times 10 \times 10 = 1\text{,}000\). In the first equation, \(2\) is the exponent, while in the next equation, \(3\) is the exponent. In healthcare, exponents are often seen in measurements like cubic centimeters (\(\text{cm}^3\)).
Evaluating Expressions
Evaluating an expression involves substituting specific values for the variables and simplifying the resulting expression. This is important because it allows you to calculate values in algebraic equations. Let’s look at an example.
Evaluate the expression \(3x + 2\) when \(x = 4\).
Solution
There is one variable in this expression, \(x\). So, we will start by substituting \(4\) for \(x\):
\[3(4) + 2\]Now, we’ll simplify to get the result:
\[12 + 2 = 14\]So, the evaluated expression is \(14\). If the expression had more than one \(x\), we would have substituted \(4\) for each \(x\).
Solving Equations
Solving an equation means finding the value of an unknown variable that makes the equation true. An equation often involves one or more variables, and solving it requires isolating a variable on one side of the equation. We can do this by performing the same action on both sides of the equation. Let’s look at a simple example.
Solve for \(x\) in the equation \(2x + 3 = 7\).
Solution
Our goal is to isolate the variable, \(x\). So, first, we subtract \(3\) from both sides:
\[2x + 3 - 3 = 7 - 3\] \[2x = 4\]Now, we divide both sides by \(2\) to get the answer:
\[\frac{2x}{2} = \frac{4}{2}\] \[x = 2\]Algebraic Formulas
In addition to the properties and order of operations discussed near the beginning of this study guide, here are some algebra formulas you should know and an explanation of them.
Squaring a Binomial
A binomial is any expression that includes two unlike terms. For instance, \(x+y\) and \(8x+4\) are both binomials. In certain situations, you will need to square a binomial (multiply the binomial by itself). This formula shows you how squaring a binomial expands it:
\[(a + b)^2 = a^2 + 2ab + b^2\]In other words, to expand \((a + b)^2\), you square the first term, then double the product of the two terms, and end by squaring the second term. Let’s try this with an example problem.
Expand: \((x + 3)^2\)
Solution
\[(x + 3)^2 = x^2 + 2(x)(3) + 3^2 = x^2 + 6x + 9\]So, \(x^2 + 6x + 9\) is your answer.
Squaring a Binomial Difference
If the binomial expression involves subtraction, the process is essentially the same, but the formula is slightly different:
\[(a - b)^2 = a^2 - 2ab + b^2\]As you can see, there is a negative sign in the second term. Let’s try this with another example.
Expand: \((x - 4)^2\)
Solution
\[(x - 4)^2 = x^2 - 2(x)(4) + 4^2 = x^2 - 8x + 16\]Factoring the Sum of Squares
The following formula shows that the sum of squares can be expressed as the difference between the square of the sum and twice the product of the two numbers:
\[a^2 + b^2 = (a + b)^2 - 2ab\]Using this formula with \(a = 3\) and \(b=5\), we have:
\[a^2 + b^2 = 3^2 + 5^2 = (3 + 5)^2 - 2(3)(5) = 8^2 - 30 = 64 - 30 = 34\]Thus, with the given values, \(a^2 + b^2 = 34\).
Finding the Difference of Squares
This formula shows that the difference between two squares can be factored into the product of a sum and a difference:
\[a^2 - b^2 = (a + b)(a - b)\]Using this formula with \(a = 7\) and \(b = 4\), we have:
\[a^2 - b^2 = 7^2 - 4^2 = (7 + 4)(7 - 4) = 11(3) = 33\]Thus, with the given values, \(a^2 - b^2 = 33\).
Cubing a Binomial
Just as you can square a binomial, you can cube a binomial (multiply it by itself three times). To expand \((a + b)^3\), you apply the powers and coefficients of \(a\) and \(b\) as shown in this formula:
\[(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\]Let’s try this with an example.
Expand: \((x + 2)^3\)
Solution
\[(x + 2)^3 = x^3 + 3x^2(2) + 3x(2)^2 + 2^3 = x^3 + 6x^2 + 12x + 8\]Cubing a Binomial Difference
Cubing a binomial that involves subtraction looks very similar to the previous formula, but with two important differences:
\[(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3\]As you can see, there are negative signs in the second and fourth terms. Here is an example.
Expand: \((x - 2)^3\)
Solution
\[(x - 2)^3 = x^3 - 3x^2(2) + 3x(2)^2 - 2^3 = x^3 - 6x^2 + 12x - 8\]Factoring the Sum of Cubes
The following formula shows that the sum of two cubes can be factored into the product of a binomial and a trinomial:
\[a^3 + b^3 = (a + b)(a^2 - ab + b^2)\]Using the formula for \(a = 2\) and \(b = 3\), we have:
\[a^3 + b^3 = 2^3 + 3^3 = (2 + 3)((2)^2 - (2)(3) + (3)^2)\] \[= 5(4 - 6 + 9) = 5(7) = 35\]Thus, with the given values, \(a^3 + b^3 = 35\).
Finding the Difference of Cubes
This formula shows that the difference between two cubes can be factored into the product of a difference and a trinomial:
\[a^3 - b^3 = (a - b)(a^2 + ab + b^2)\]Using the formula for \(a = 5\) and \(b = 2\), we have:
\[a^3 - b^3 = 5^3 - 2^3 = (5 - 2)((5)^2 + (5)(2) + (2)^2)\] \[= 3(25 + 10 + 4) = 3(39) = 117\]Thus, with the given values, \(a^3 - b^3 = 117\).
Squaring a Trinomial
A trinomial is an expression with three terms. To expand \((a + b + c)^2\), square each term and then add the products of each pair of terms, doubled:
\[(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac\]Using this formula for \(a = x\), \(b = 1\), and \(c = 2\), we have:
\[(a + b + c)^2 = (x + 1 + 2)^2 = x^2 + 1^2 + 2^2 + 2(x)(1) + 2(x)(2) + 2(1)(2)\] \[= x^2 + 1 + 4 + 2x + 4x + 4 = x^2 + 6x + 9\]Thus, with the given values, \((a + b + c)^2 = x^2 + 6x + 9\).
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