Mathematics Study Guide for the HESI Exam

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General Information

You thought you were going into a science-related profession, so why are you being asked to do math? Love it or hate it, nurses use mathematics for a number of calculations in nearly every area of their job. From Urology to Gastroenterology, nurses use math to conduct dimensional analysis, read and analyze patient charts, and carry out day-to-day functions.

For this test, you will have access to a very simple, on-screen calculator for the math questions. You may not bring and/or use your own calculator, so don’t rely on being able to do much beyond the four basic operations (add, subtract, multiply, and divide) on a calculator. This test should not require you to do anything beyond those, anyway, though.

Here are some basic math concepts you should understand before taking the mathematics section of the HESI exam.

Computation with Whole Numbers

Computation with whole numbers includes basic operations such as addition, subtraction, multiplication, and division. It relies upon the concept of place value in our number system. You might remember the charts from school showing how each digit in a number, for example 7,654,321, has a place value, as shown in the following chart:

Millions Hundred Thousands Ten Thousands Thousands Hundreds Tens Ones
7 6 5 4 3 2 1

When reading this number aloud, you would say, “Seven million, six hundred fifty-four thousand, three hundred and twenty-one.” This system also allows us to borrow and carry digits when computing addition and subtraction because our number system relies upon powers of ten in what is called a base ten system.

For example, the number 45 can be understood to be 4 tens and 5 ones, or 45 ones. Similarly, the number 7,200 can be called 72 hundreds or 7 thousands and 2 hundreds.

Note: In this guide, we use the terms “borrow” and “carry.” In modern math courses, both of these are covered under the term “regroup.” The words mean the same.


Addition is the most basic computation we can do with whole numbers. It is the process of combining two numbers into one. Traditionally, the easiest way to do this is by writing the two numbers in column format, or vertically, so that each place value is aligned, and adding first the ones values, then tens values, then hundreds, and so on until the problem is complete. In other words, when setting up your problem, be sure to align it so that the rightmost numbers―the ones values―are on top of each other.


Susana and Tim are coworkers at a hospital. They each purchased boxes of bandages for the hospital dispensary without talking to each other. Susana bought 5,800 boxes while Tim purchased 3,500 boxes. How many boxes did they wind up with, in total?

In this problem, the two numbers we are going to add together are 5,800 and 3,500. We write them in column format, making sure the ones place values are aligned, and then add down.

\(\begin{align} 5800& \\ \underline{+\quad3500}& \\ 9300& \\ \end{align}\)

Notice how when we add the hundreds column, 8 hundreds plus 5 hundreds is 13 hundreds or 1 thousand and 3 hundreds. We put the 3 down as our answer for the hundreds column and carry the 1 over into the thousands column. Our final answer is 9,300 boxes of bandages.


Cameron has $8,726,904 in his bank account. At the end of the next week, his paycheck is $791 after taxes. How much money is in his bank account after he deposits the check?

Again, we want to make sure the ones values sit on top of each other when we set up this problem.

\[\begin{align} 8,726,904& \\ \underline{+\quad\quad 791}& \\ 8,727,695 \end{align}\]

Cameron has $8,727,695 in his account after his paycheck is deposited.


Subtraction is the idea of finding the difference between two numbers. Much like addition, we traditionally solve subtraction problems by aligning the ones position into a column and taking away some number from the top. If the bottom digit is larger than the top digit, we will have to borrow from the next column. Let’s look at an example.


Jason has 174 pounds of apples for sale at his farmer’s market stall. He sells 48 pounds by the end of the day. How many pounds of apples does he have left for sale the next day?

\(\require{cancel} \quad\quad\;6\;14\)
\(\begin{align} \quad 1 \cancel{7} \cancel{4} \\ \underline{- 4\,8} & \\ 1\,2\,6& \\ \end{align}\)

Jason has 126 pounds of apples left to sell. Since we can’t subtract 8 from 4, we have to borrow from the tens column. The 7 tens is the same as 6 tens and 10 ones (really!). The 10 ones are moved over to the ones column to join the 4 ones for a total of 14 ones. We leave the 6 tens in the tens column. From here we can easily subtract down.

Note that for subtraction problems, the order of the numbers matters. We start with 174 pounds, and then take away 48 pounds. Pay attention to what number you start with in a word problem and what is subtracted from it.


A bag of saline solution originally containing 975 mL has delivered 190 mL to the patient. How much saline is left in the bag?

The original amount of saline we started with is 975 mL, so that number will be on top. We subtract 190 mL, so that will be placed on the bottom as we set up the subtraction problem.

\(\require{cancel} \;8\;17\)
\(\begin{align} \cancel{9}\cancel{7} 5 \\ \underline{- 1\,9\,0}& \\ 7\,8\,5& \\ \end{align}\)

Since we cannot take 9 tens from 7 tens, we have to borrow from the 9 (sometimes called “regrouping”) in the hundreds column. The 9 hundreds becomes 8 hundreds and 10 tens. The 10 tens combine with our 7 tens to become 17 tens. Now we can subtract down each column. The amount of saline left in the bag is 785 mL.


Multiplication is a shortcut for repeated addition. We can look at this visually with the following table:

2 + 2 + 2
X   X   X
X   X   X

This table shows a familiar multiplication problem, 3\(\; \times\;\)2. If we repeatedly add by twos, we find that 2 + 2 + 2 = 6 or, in other words, 2, added 3 times, equals 6. This is also the basis for skip counting. An example of skip counting by twos is in the common cheer heard at sports rallies: “Two, four, six, eight, who do we appreciate?”

This is nice for smaller, familiar multiplication problems. What happens when you have to multiply large numbers, like 360 times 21? Then, a column format similar to what we used for addition and subtraction is helpful. It’s easiest to place the larger number on top for your calculations.

The first step is to multiply the ones digit from the bottom number by each digit of the top number, working from right (the ones column) to left.

\[\begin{align} 360& \\ \underline{\times\quad 21}& \\ 126& \\ \end{align}\]

In this particular multiplication problem, we start by multiplying the 1 in 21 by 360, digit by digit, from right to left.

Our next step is to multiply the tens digit from the bottom number by the top number. Before we do that, we’re going to insert a placeholder, 0, to show that the 2 in 21 indicates we’re multiplying by 2 tens.

\[\begin{align} 360& \\ \underline{\times\quad 21}& \\ 360& \\ \underline{+7200}& \\ 7560& \\ \end{align}\]

The final step is to add the products, or answers to a multiplication problem, together. It may help to think of this as though the first step is multiplying 360 by 1 and the second step is multiplying 360 by 20. Since multiplication is repeated addition, the final step to find the answer in 360 times 21 is to add the two previous steps together.

Note also that we had to carry a one over to the hundreds column after multiplying 2 by 6 to get 12. This carried number is then added to our answer when we multiply the 3 from 360 by the 2 in 21.

Before we look at a word problem example, ask yourself, “How does the place value system help us do long multiplication?”


Jada works at a popular mobile phone and technology store. She receives 245 boxes each filled with 27 brand-new phones to be released for sale the next week. How many phones do they have in stock total?

One way to visualize this problem is to picture one box that you open up to see 27 phones packed inside. As you open up each new box, you have to add another 27 phones to your inventory. But with 245 boxes, that can take quite a while. So we’re going to work it out in column format.

We put the larger number, 245, on top. Step one is to multiply the ones column from the bottom number, 7, by 245.

\(\begin{align} 245& \\ \underline{\times\;27}& \\ 1715& \\ \end{align}\)

Again, we have to carry over a number after multiplying 5 by 7 and 4 by 7.

The next step is to multiply the tens column of the bottom number, 2, by 245. Don’t forget your placeholder zero, because the 2 in 27 is really 2 tens.

\(\quad\quad 1\)
\(\begin{align} 245& \\ \underline{\times\quad 27}& \\ 1715& \\ \underline{+4900}& \\ 6615& \\ \end{align}\)

Our final step is to add 1715 + 4900. Our answer is that Jada has 6,615 shiny new phones to sell in her store.

Another example:

Suppose Jada is able to sell all 6,615 of the phones. Each phone sold for $124. How much money did she take in selling the phones?

Every time Jada sells a phone, she receives $124. But it’s hard to keep track when she sells 6,615 phones! We’re going to help her out by multiplying 6,615 by $124.

This problem is only different from our previous examples in that the bottom number, 124, has three digits instead of just two. That means that we will have to add another step where we multiply the hundreds column, 1, by 6,615. Since it is 1 hundred, we’ll have to add in two placeholder zeros.

The first step is to multiply 4 ones by 6615, as shown in the following equation:

\(\begin{align} 6615& \\ \underline{\times\quad 124}& \\ 26460& \\ \end{align}\)

The next step is to multiply the tens column of the bottom number, 2, by 6,615. Don’t forget a placeholder zero.

\(\begin{align} 6615& \\ \underline{\times\quad 124}& \\ 26460& \\ 132300& \\ \end{align}\)

The third step is to multiply the hundreds column of the bottom number, 1, by 6,615. We use two placeholder zeros (bolded below) to show that we’re multiplying by 1 hundred.

\[\begin{align} 6615& \\ \underline{\times\quad 124}& \\ 26460 & \\ 13230\bf{0}& \\ \underline{+\,6615\bf{0}\bf{0}}& \\ 820260& \\ \end{align}\]

The final step is to add up the three multiplications we did. The total amount Jada collected from selling all the new phones is $820,260.

Check with a Calculator:

Get a calculator. Double-check the answer to the given example by following these steps:

  1. Multiply 6,615 by 4.

  2. Multiply 6,615 by 20.

  3. Multiply 6,615 by 100.

  4. Add together your answers from steps one to three. Is it the same as the answer we got above?


If multiplication is repeated addition, then division is repeated subtraction. We start off with some number and what we want to know is how many times we would have to subtract another number from it until we arrive at zero. The number we start with is the dividend, while the number we repeatedly subtract is the divisor. In division problems, the answer is called the quotient. In other words, dividend ÷ divisor = quotient.

Again, this is easy to visualize and do with smaller numbers but when it comes to larger numbers, it’s not so practical. We have a method for this―long division. This looks different from addition, subtraction, and multiplication problems. Let’s take a look at an example.


Alexis has 4,630 milliliters of a chemical. Each time she does an experiment, she uses 125 milliliters. How many experiments can she perform?

\[\require{enclose} \quad\quad\quad\;\; 37R5 \\[-3pt] \begin{array}{r} 125 \enclose{longdiv}{4630} \\[-3pt] \underline{-375}\phantom{0} \\[-3pt] 880 \\[-3pt] \underline{-875} \\[-3pt] \quad \; \phantom{8}\phantom{7}5 \\[-3pt] \end{array}\]

The work above shows how a long division problem works.

  • The divisor, 125, sits outside the long division symbol, while the dividend resides underneath it.

  • The quotient, as it is worked out, is placed on top of the symbol, with each of its digits corresponding to the place value of the dividend (ones values over ones values, tens values over tens values, etc.)

  • The “R” indicates a remainder, or that the dividend is not divided by the divisor into equal parts, but that some amount remains left over. In this example, the remainder is 5.

In the first step for this problem, we look at how many times the divisor, 125, goes into 463 from the dividend. Why 463? That is the smallest amount that can still be divisible by 125. The number 46 is too small, while 4,630 is too large. It goes in 3 times, so we write that up top, over the last digit we included (3).

Then we multiply 1253 = 375, and write that below the dividend, aligning the digits to 463. We subtract down to find 88 remains.

The ones digit from the dividend is carried down so that we are now working with 880. How many times does 125 go into 880? It goes in evenly 7 times, so we write that above in our quotient.

Again, 1257 = 875. Subtracting 880 – 875 = 5, which is our remainder, since it is too small for 125 to be divided into it.

Thus, Alexis can do 37 experiments with 125 milliliters of chemical each, but she will have 5 milliliters left over.


Manuel is a dispensary technician in a hospital. He is sorting 872 smocks from a new shipment into 16 groups evenly for each department’s use. How many smocks will each department receive and how many will be left over?

First, identify which number is your divisor and dividend. We start with 872 smocks, so that is the dividend and goes under the long division symbol. The divisor, or how many groups we’re splitting them into evenly, is 16, and this number goes to the left of the division symbol.

\[\require{enclose} \begin{array}{r} \; \; 5 \; \; \\[-3pt] 16 \enclose{longdiv}{872} \\[-3pt] \underline{80}\phantom{2} \\[-3pt] 7 \phantom{0} \end{array}\]

The first step is to divide 16 into 87 as many times as it can go evenly. Since this is 5, we write 5 as the greatest digit in our quotient. Since 165 = 80, we place that below the dividend and subtract. We are left with 7.

\[\require{enclose} \quad\quad\;\; 54R8 \\[-3pt] \begin{array}{r} 16 \enclose{longdiv}{872} \\[-3pt] \underline{80}\phantom{2} \\[-3pt] 72 \\[-3pt] \underline{-\,64} \\[-3pt] \phantom{6}8 \\[-3pt] \end{array}\]

The next step is to bring down the 2. Then 72 divided by 16 is 4, which is placed in the ones place of the quotient. Since 164 = 64, we subtract that from 72 and are left with 8. Since 8 is too small to be divided by 16, that is the remainder and we have finished.

Manuel will have sixteen groups of 54 smocks evenly, but 8 smocks will be left over.


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