Today’s article “OAT Test Prep: Quantitative Reasoning” was written by Salus 2014 Optometry Student Dale Paynter. All of the questions have been created by Dale and not extracted from other sources.
Welcome to the Quantitative Reasoning strategy article! The QR section of the OAT is the final section of the exam, and covers a variety of mathematical topics. These include fractions and percentages, probability, averages, rates, geometry, trigonometry, and algebra. Word problems are very common throughout this section. As always, we have plenty of free, exclusive practice problems for you!
The primary complaint of students on this section is that they don’t seem to have enough time. The best way to prepare for this section, then, is to practice doing math problems both quickly and accurately. And on test day, just as with any OAT section (except for Reading Comprehension) I advise you to skip any questions that give you too much trouble, and to only go back to them at the end once you have completed all of the easier problems. Get the points that you know you can get! Another way to help save time is to use shortcuts where appropriate, and throughout my QR practice problems, I will highlight useful shortcuts that I see. A secondary problem that many students face on test day is the fatigue factor. The QR section is at the end of a long, stressful examination, and a number of students have had trouble maintaining their stamina through this section. The best advice to help deal with this potential problem is to be well-rested for your OAT exam, to take the optional 15 minute break, and to do as many full-length practice OAT tests as possible during your preparation!
Before we go further, there is a major change coming for this section. Starting May 4, 2010, an on-screen calculator will be allowed on this section. This should help those of you who struggle with doing multiplication and division quickly, but remember that unless you set the problem up correctly, the calculator will be of no help to you. So still practice as much as you can!
Now for the practice problems: Note that * always denotes multiplication, and the topics are in bold.
I realize that most of these problems should appear extremely simple; the important thing is to be able to do them rapidly.
45x/64 = 9/3x
What is x?
I immediately notice that 9 is a factor of 45, and therefore that I can simplify the problem by dividing both sides by 9. This results in 5x/64 = 1/3x Next, I will cross-multiply, meaning I will multiply both sides of the equation by the denominator of each fraction. So I will multiply both sides by 64 and by 3x which will result in 15 x squared = 64 I can then immediately take the square root of both sides, which will give me the square root of 15 * x = 8 (remember, the 8 can be either positive or negative). This will then give the final answer of plus/minus 8 over the square root of 15 being equal to x.
1/5 * 5/8 + 3/10 * 5/7 – 2/3 = ?
This problem highlights three concepts: The order of operations (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction), multiplying fractions, and finally adding or subtracting fractions. It turns out to be a bit tedious, so in case you need it, I give a detailed walkthrough of this problem at the end of this article.
5% of x = 54
What is x?
Well, we know what 5% of x is, and we want to know what 100% of x is. In order to go from 5% to 100% we must multiply by a factor of 20, because 100/5 = 20. So all we need to do is multiply both sides of the equation by 20. This will give us 100% of x = 54*20 = 1080 = x
What percent of an hour is one second?
Well, the way to approach problems of this nature is to first find out what fraction of an hour a second is. Then, you convert this fraction into a fraction with a denominator of 100. This will directly give you the percentage. So, what fraction of an hour is a second? There are 60 seconds in one minute, and 60 minutes in an hour, so a second is 1/60 of a minute, which is 1/60 of an hour. 1/60 * 1/60 = 1/3600 meaning that a second is 1/3600 of an hour. All we need to do now is to convert 1/3600 into x/100 and x will be the percentage. So you can set up an equation 1/3600 = x/100 and simply multiply both sides by 100, resulting in 100/3600 or 1/36 being equal to x. So a second is equal to 1/36 of a percent of an hour.
If I have a bag of 20 marbles, with 4 red marbles and 5 blue marbles, then what are the odds that if I randomly take out 2 marbles (without replacement), that I will have exactly one red and one blue marble?
Well, to do this problem we need to find the probability of each possible way of achieving the desired result, and adding those probabilities together. There are two ways that I can end up with one red and one blue marble. I could first pick out a red marble, and then a blue one, or I could first pick out a blue marble, and then a red one. We will find the probability of each of these events, and then add them together to find the total probability of ending up with one red and one blue marble.
What is the probability of a red marble followed by a blue marble? Well, I first need to pick a red marble, which is 4/20 and then need a blue marble which is 5/19 (there are only 19 marbles left for the second event). Remember the critical and/or rule for probability theory. If you need BOTH events (i.e. first and second) to happen, you must multiply the probability of the two events to find the probability of both happening. If you need one OR the other (i.e. first or second) to happen, then you will add the probability of the two events to find the probability of either happening. These rules are for independent events. I will multiply 4/20 by 5/19 giving a 20/380 chance of first picking a red marble and then picking a blue marble. You can make the problem a bit easier by reducing 4/20 to 1/5 before multiplying, which means you would multiply 1/5 * 5/19 = 1/19.
What is the probability of a blue marble followed by a red marble? I first need to pick a blue marble which is 5/20 and then need to pick a red marble which is 4/19 so I have 5/20 * 4/19 = 20/380 chance of first picking a blue marble and then picking a red marble. This was the same as the probability of the first possibility, which may have been obvious to you. You could also have reduced 5/20 to 1/4 prior to multiplying, which then results in 1/4* 4/19 = 1/19.
I will now add the probabilities of blue first, red second and red first, blue second because I need either one OR the other to occur (again, following the and/or rule). 20/380 + 20/380 = 40/380 which reduces to 2/19 so our final answer is 2/19.
Thanks for reading, and more practice problems for Quantitative Reasoning covering the remaining topics will be up soon!
Walkthrough of Problem 2:
First, we’ll follow the order of operations by multiplying the correct fractions. 1/5 * 5/8 = 5/40 = 1/8. A slightly quicker way to perform this operation is by noticing that the 5 from the fraction 1/5 will cancel with the 5 from the fraction 5/8 which will result in 1/1 * 1/8 which is obviously equal to 1/8. The next multiplication is 3/10 * 5/7 = 15/70 = 3/14 Again, there is a small shortcut in that the 5 from 5/7 will partially cancel with the 10 in 3/10 resulting in 3/2 * 1/7 = 3/14. Now we can rewrite the original problem as 1/8 + 3/14 – 2/3. In order to do this problem, we must first find a common multiple between the denominators (bottom numbers of the fractions). It doesn’t technically have to be the least common multiple, although that is the most efficient number to use. So to add 1/8 to 3/14 we will be finding the least common multiple between 8 and 14. This will end up being 56. I found this number by testing the multiples of 14 against the multiplication table for 8 (which we should all have memorized). Is 14 a multiple of 8? No. How about 28? Again, no. 42? Not quite. But once I reach 56, I know that 8 * 7 = 56 therefore 56 is a multiple of 8. So we will convert 1/8 and 3/14 into fractions of 56 in order to add them together. 1/8 = 7/56 because I am multiplying the top and the bottom by 7 because 8 * 7 = 56 while 3/14 = 12/56 because I am multiplying the top and bottom by 4 because 4 * 14 = 56 So now we simply have 7/56 + 12/56 = 19/56. Now we have to subtract 2/3 from 19/56 so again we must find the least common multiple between the denominators. Using the same method as before, we will find that the least common multiple is 168. 19/56 will convert to 57/168 and 2/3 will convert to 112/168. So we finish the problem up by subtracting 112/168 from 57/168 which gives us an answer of negative 55/168
This problem ended up being somewhat more tedious than most problems on the OAT, simply because on most OAT problems there tends to be more possible cancellations that result in much easier numbers to work with.