The MCAT is a test that assesses your science knowledge as well as your ability to read critically. However, many of the scientific questions, particularly those involving physics and chemistry, may ask you to perform calculations that may require the use of a calculator.
Is it, however, permitted to use a calculator on exam day? What math on the MCAT will you be tested for?
This complete guide will walk you through a series of arithmetic abilities, from understanding concepts to using methods to pass the exam without using a calculator.
Does the MCAT have Math?
Is Math on the MCAT? Yes, absolutely.
The MCAT challenges you to complete math questions based on chemistry, physics, and statistical reasoning knowledge, among other areas. One of the MCAT’s problems is that you will not be permitted to use any type of calculator on Test Day.
Although there are very few heavy calculations on the current MCAT, there are numerous questions that need you to demonstrate your ability to handle some math.
No calculators imply that all calculations, regardless of the MCAT part, must be done mentally. The MCAT will not assess you on tough math, and the most of answer selections will be placed far enough apart numerically that you may be able to skip complex calculations entirely if you have excellent estimate abilities.
Math is essential on the MCAT. Consider yourself in a circumstance where you are given a calculation question that you know how to solve but cannot do it perfectly in the available time.
MCAT test students sometimes underestimate the mathematical skills necessary and end themselves in such situations. Math skills, like most other skills on this exam, may be developed with adequate familiarity and practice.
What math is on the MCAT?
As previously said, you are not allowed to use a calculator on the test day, there are questions on the test that will involve mathematical calculations, especially in the context of chemistry and physics. By strengthening your mathematics skills, you will greatly boost your ability to answer these exam problems.
While many of these math skills cover fundamental addition, subtraction, multiplication, and addition, this part gives a few extra topics to review for the MCAT, such as logarithm principles and fraction manipulation. We’ve included many practice problems at the conclusion of this article for you to try.
#1 Estimating and rounding numbers
Estimating and rounding numbers are important MCAT skills because they allow you to solve issues more effectively. This section offers crucial reminders to help you estimate and round.
Multiplication
Assume we’re attempting to multiply 3.64 x 7.35. How can we accurately approximate this result in a short period of time without a calculator?
When multiplying numbers with multiple nonzero digits, rounding might provide us with more acceptable results. We can substantially simplify our mathematical calculations by rounding these figures to one decimal place or a full integer.
It may be wise to look at the answer selections first to decide how much to round numbers. If the answer options are close together, we should round to one fewer digit to make our estimates more exact. We even can get away with rounding more if our answer options are more apart.
We can additionally correct for rounding for each digit to keep our results accurate. If we round a number up, we need to round the other number down to get closer to the exact value. For example, if we round 3.64 to 3.6, we should round 7.35 to 7.4 as well.
Assume we choose rounds 7.35 to 7.3 instead. How does this compare to the right answer and the answer given by 7.4?
3.64 x 7.35 = 26.754
3.6 x 7.4 = 26.64 => 26.6
3.6 x 7.3 = 26.28 => 26.3
When we apply 7.3, our calculation deviates quite much from the correct answer. So, we will come closer to the correct answer of 26.754 by accounting for rounding and applying the digit 7.4.
Division
Estimating and rounding division questions are comparable to estimating and rounding multiplication issues. As with multiplication, we initially examine the answer options to decide how much to round.
However, in the division, we apply other changes to accommodate for rounding. To maintain our numbers proportionate, we should move the dividend and divisor in opposite directions.
Let us suppose we wish to divide 20.97 by 2.78. If we choose to round 20.97 to 21, we should also round 2.78 to 3. This yields a result of 7, which is close to the true answer of 7.54.
Example problem
Let’s go over an example of a mathematics issue that may appear on the MCAT.
6.19 g of solid hydrochloric acid is added to 0.0500 L of water. What is the molarity of the end result?
Molarity is expressed as moles per liter. To determine the molarity of this solution; first, determine the number of moles of hydrochloric acid (HCl). First, we estimate HCl’s molar mass.
1.008g/mol+35.45g/mol=36.458g/mol
To calculate molarity, divide this number by the volume of the solution.
6.19g/mol(36.458g/mol HCl)
6g/mol(36g/mol HCl)=0.1667g/mol
Molarity is calculated by dividing this number by the volume of the solution.
0.1667g/mol HCl0.0500L=3.33M HCl
As a result, the estimated molarity of the solution is predicted to be 3.33M HCl.
#2 Logarithms
Logarithm rules
Another mathematical topic that is likely to occur on the MCAT is logarithms. Because these equations are the inverse of an exponential function, they follow the same rules. Here are some key principles and operations to grasp while working with logarithms:
- The log of 1 is always 0: logx1 = 0
- The base number’s log always equals 1: logxx = 1
- The log of 2 variables equals the total of the logs of the individual factors as long as the base numbers are the same: logn(x y) = logn(x) + logn(y)
- Similarly, the log of a fraction is the difference between the logs of the dividend and divisor: logn(x y) = logn(x) – logn(y)
Common logarithmic applications
One example of a common logarithm is with base “e.” These logarithms are referred to as natural logarithms and are represented by the symbol ln ().
Base 10 logarithms are also quite prevalent. You should be quite comfortable with and approximating logarithms in base 10.
It is vital to know that the prefix p stands for –log on the MCAT. pH, for example, equals -log[H+], while pOH equals -log[OH-].
Decibels are also a popular logarithmic use on the MCAT. Decibels (dB) are a unit of measurement for sound level. It can be determined using the following formula:
dB = 10log(II0)
where I = intensity of the sound, in Wm2
I0 = threshold of the lowest audible sound to a healthy human, equal to 10-2Wm2
The Henderson-Hasselbalch formula is a popular logarithmic application on the MCAT. This formula, which can be obtained from the equilibrium constant expression for the dissociation of a weak acid, enables us to do buffer calculations. Here is the Henderson-Hasselbalch equation:
pH = pKa+log(A–HA)
Example problem
Let’s look at an example of how logarithms may be used.
The pH of an HF solution (pKa = 3.2) is 4.8. What is the acid-to-conjugate-base ratio in the HF solution?
All of these values are related by the Henderson-Hasselbalch equation. We may use it to calculate the ratio.
pH = pKa+log(A–HA)
To begin, we isolate our ratio on one side of the equation.
pH = pKa+log(A–HA)
-1 (pH- pKa) =-1 log(A–HA)
pKa-pH =log(HAA–)
We may now enter our pKa and pH values.
3.2-4.8=log(HAA–)
-1.6=log(HAA–)
To calculate the ratio, we must use the exponential function, which is the inverse of the logarithm function. Because log() is the notation for base 10, we may add 10 to the numbers on both sides of the equation. Raise the number 10 to a logarithm with its own base, and the logarithm function on that side of the equation is essentially “canceled.”
10-1.6=HAA–
In this solution, the ratio of HF to its conjugate base F– is thus predicted to be 10-1.6. We may further estimate this value by identifying the range in which it should fall. While we don’t know the precise value of 10-1.6, we do know the precise values of 10-1 and 10-2:
10-2<10-1.6<10-1
1100<10-1.6<110
0.01<10-1.6<0.1
As a result, we’d seek for an answer choice that’s between 0.01 and 0.1.
#3 Methods for eliminating answer options
When dealing with calculation difficulties, it’s typically advisable to calculate an accurate solution first, then check to see if your result fits one of the answer choices. When you’re short on time, remember that you only need to choose one of four possible answers! Below are some pointers to assist you to increase your speed and accuracy while eliminating answer choices.
Scientific notation is a way of expressing numbers that include a significand and an exponent.
- Any real number with an absolute value between 1 and 10, but not including 10, is significant.
- Exponents are in base 10 and can be any whole number (i.e., negative, zero, or positive).
Significant figures
Using significant figure calculations, we can frequently remove answer options. When answer choices are presented in scientific notation, it is helpful to focus on the number of significant numbers in the significand.
Significant numbers indicate the accuracy of measurement depending on the measuring instrument. They must be taken into account in formulas. Significant numbers can be determined using the following rules:
- All numbers are significant between the initial nonzero digit on the left and the final nonzero digit on the right.
- When a number contains a decimal, the zeroes to the right of the final nonzero digit are important. When there is no decimal, the zeroes are meaningless. (For example, the number 4300.00 contains six significant numbers, yet the number 4300 has just two.)
- Leading zeros, or zeros that come before the first nonzero digit, have no meaning. (For example, 0.010 has two significant numbers.)
Calculating with substantial figures necessitates the application of special principles.
- Begin by noticing the place of the leftmost decimal point while doing addition and subtraction computations. After that, we proceed to solve the issue, keeping as many numbers as possible until the very conclusion of our calculation. Following this, we round the computed answer to the decimal point set in the first step.
- To begin multiplication and division calculations, determine which factor, divisor, or dividend has the fewest significant digits. We then solve the issue, keeping as many digits as possible till the finish. Finally, we round the calculated answer to the number of significant digits of the first-step factor, divisor, or dividend.
Exponents
Exponents are also a way to eliminate answer options. Exponential functions follow the same laws as logarithmic functions since they are the inverse of logarithmic functions:
A0=1
AxAy=Ax+y
AxAy=Ax-y
(Ax)y=Axy
(AB)x=AxBx
A-x=1Ax
When answer options are stated in scientific notation, the exponent’s base number is always 10. To exclude answer choices for questions regarding extremely little or very big numbers, we can examine the exponent of this base number 10.
For instance, if a question asks how many molecules are in a few moles of stuff, we should estimate a very big number. The answer will most likely contain an exponent raised to the 23rd power rather than the 5th power in this situation.
Example problem
Let’s put this information to use by solving a practice issue.
Calculate the molarity of 0.424 mol HI in 2.1 L of water.
First, determine which number contains the fewest significant digits. The amount of moles of HI has fewer significant numbers than the solution volume. Because the volume includes two significant figures, our final solution will have two as well.
We can now solve the issue without rounding up until the finish.
0424 mol Hl 2.1L=0.201M
This solution is then rewritten in scientific notation. We begin by adjusting the significand and multiplying by the appropriate base ten exponents. We multiply by 10-1 since we moved the decimal one point to the right.
0.2019M2.01910-1M
To get our final result, we round this estimated value into two significant figures.
0.2019M2.010-1M
We can be sure to exclude any response options with significant positive integers in the exponent when removing answer options.
#4 Fractions
You may be asked to answer fractional equations on the MCAT. Fractions are frequently used in the thin lens equation and the solution of optical systems.
Subtraction and addition
We apply our operations to the numerators and maintain the denominator when adding and subtracting fractions. Before continuing with our calculations, we must remember to identify a common denominator. This is critical for maintaining the balance of a fraction.
We seek for the least common multiple of the denominators we’re dealing with to find a common denominator. The lowest number in which both denominators are factors is known as the least common multiple.
For example, suppose we want to combine the following fractions:
23+19
We must determine the lowest integer in which both 3 and 9 are factors. Because both 3 and 9 are factors of 9, the least common multiple is 9.
2/3 is now rewritten as a fraction with a denominator of 9. To keep the fraction proportionate, we must do the identical procedure on the top and bottom. The denominator must be multiplied by three to provide the desired result of nine, therefore we multiply the top by three to obtain six. We may begin adding after translating 2/3 to 6/9.
69+19=6+19=79
Division and multiplication
We multiply the numerators and denominators to get our result using multiplication. We don’t need to identify a common denominator before multiplying in this example.
Nevertheless, before multiplying, we may sometimes reduce our calculations. We start by seeing if either element can be simplified on its own. Then we test if any fraction’s numerators can be simplified using denominators from other fractions, or vice versa. After that, we’ll start calculating.
If we’re multiplying the following fractions, for example:
271416
We may begin by reducing 14/16 to 7/8. The following is a rewrite of our multiplication problem:
2778
From this, we will simplify by dividing the denominator of one fraction by the numerator of another. You should note that the number 7 divides both the denominator and the numerator of the first and second fractions.
Furthermore, the number 2 divides both the numerator and denominator of the first and second fractions.
We now have the following multiplication issue to solve after simplifying these fractions:
1114
We combine all values on the top half of the fraction together and all values on the bottom half of the fraction together in fraction multiplication:
1114=1114=14
The correct answer is 1/4.
Multiplication and division are comparable operations. We multiply by the reciprocal of the divisor while dividing. For instance, consider the following fractions:
1347
Multiply the first fraction by the reciprocal of the second fraction to finish (74). Then, like with multiplication, we will proceed as follows:
1347=1374=1734=712
The correct answer is 7/12.
Thin lens equation application
On the MCAT, the thin lens equation is popular fraction math and physics application.
The thin lens equation connects three geometric optics lengths: focal length, object distance, and image distance.
- The focal length (f) of a lens or mirror is the distance between the focal point and the center.
- The object length (o) is the distance between the object and the lens or mirror’s center.
- The image distance (i) refers to the distance between the picture and the lens or mirror’s center.
The following is the thin lens equation:
1f=1o+1i
where f = focal length
o=object distance
and i=image distance
Each variable’s sign includes information about its position.
- Positive focal lengths are found in concave mirrors and converging lenses, whereas negative focal lengths are found in convex mirrors and diverging lenses.
- When the object is in front of the mirror or lens, positive object distances occur; when the item is behind the mirror or lens, negative object distances occur. The object distance is positive in practically all MCAT-related applications.
- The image is in front of the mirror or behind a lens if the image distance is positive. A real image is always inverted and is referred to as such.
- If the image distance is negative, the picture is in front of a lens or behind a mirror. This type of picture is a virtual image that is always upright.
Remember to apply the same procedures to both sides of an algebraic equation, such as the thin lens equation. For example, if we remove 4 from one side of the equation, we must likewise subtract 4 from the other. This maintains the equation’s equilibrium. Failure to do so may result in an inaccurate answer!
Example problem
Let’s try out the thin lens equation in practice.
Assume a 4 cm focal length converging lens is placed 6 cm distant from an item. Determine if this lens produces a real or virtual image.
Begin by jotting down an equation you’re already familiar with: the thin lens equation.
1f=1o+1i
This problem requires us to calculate the image distance, or i. By removing 1/o from both sides of the equation, we may isolate that variable on one side of the equation.
1i=1f–1o
We can now evaluate the formula using our previously calculated values for f=4 cm and o=6 cm. Then, we rearrange these fractions with a common denominator to remove these values. In this case, we use 4×6 = 24.
1i=14cm-1–16cm-1
1i=624cm-1–424cm-1
Continue assessing the expression:
1i=6-424cm-1
1i=224cm-1
This gets us a little closer to figuring out i. To get an exact number for I we must simplify these formulas without using fractions. Fortunately, we can cross-multiply here: multiply the numerator of the left-hand fraction by the denominator of the right-hand fraction and vice versa.
124=2i
24=2i
242=i
i=12 cm
We know the image is real since our image distance is positive i > 0.
#5 Questions and answers based on passages
Students were required to conduct a titration experiment using acetic acid as an analyte and sodium hydroxide as a titrant in a college laboratory.
A burette was filled with 100 mL of 0.200M NaOH for the experiment. An Erlenmeyer flask containing 50 mL of unknown strength acetic acid was put beneath the burette. The concentration of the given acetic acid (pKa = 4.7) had to be determined by the students.
While doing the assigned experiment, one of the students develops the titration curve below.
Question 1: At the equivalence point of the student’s titration, how many moles of NaOH were contained in the burette?
- A) 0.029 mol
- B) 0.03 mol
- C) 0.01 mol
- D) 0.014 mol
The correct answer is D.
The Erlenmeyer flask was filled with 28 mL NaOH at the equivalency point, keeping 72 mL in the burette 100mL – 28mL = 72 mL. The following equations are used to calculate the number of moles in 72 mL NaOH.
72 mL1 L1000 mL0.200 mol/L=0.0144 mol
We must round our answer to two significant figures because one of the original values (72 mL) contains two significant figures. 0.014 mol is the right answer.
Question 2: To make the titration solution, how many grams of NaOH were added to 100 mL of water?
- A) 0.5 g
- B) 0.8 g
- C) 1.2 g
- D) 0.3 g
The correct answer is B.
Our methods are identical to those used in the last problem. In this example, we must find out the number of grams in one mole of NaOH in addition to the total volume of NaOH in mL.
The atomic mass of Na is 22.99 grams per mol, which we should round up to 23 grams per mol.
The atomic mass of O is 15.99 grams per mol, which we’ll adjust up to 16 grams per mol.
The atomic mass of H is 1.01 grams per mol, which we should round to 1 gram per mol.
NaOH has a molar mass of 23 + 16 + 1 = 40 g/mol.
At this point, we can apply dimensional analysis and the solution’s known molarity to figure out how many grams of NaOH to add:
100 mL1 L1000 mL0.200 mol/L40g/mol=0.8 g
Question 3: At the titration’s equivalence point, what is the approximate ratio of [HA] to [A-]?
- A) 103
- B) 10-4.3
- C) 103
- D) 110-3.3
The correct answer is B.
To get the ratio, we will apply the Henderson-Hasselbalch equation. The pH is roughly 9 at the equivalency point. Acetic acid has a pKa value of 4.7. The ratio of [HA] to [A-], or [HA]/[A-], is required in this question.
pH = pKa+log(A–HA)
pH-pKa=log(A–HA)
-1 (pH- pKa) =-1 log(A–HA)
pKa-pH =log(HAA–)
4.7-9=log(HAA–)
-4.3=log(HAA–)
10-4.3=HAA–
Question 4: Is the concentration of acetic acid after the student has added 14 mL NaOH larger than the concentration of acetic acid after the student has added 28 mL NaOH by how many orders of magnitude?
- A) 6
- B) 7
- C) 106
- D) 107
The correct answer is B.
Around 14 mL NaOH, the pH is 2, and around 28 mL NaOH, the pH is 9. The difference between these values is 7, which signifies the order of magnitude difference between the two concentrations (choice D is correct). Instead of the order of magnitude, Answer C reveals the actual ratio of these differences.
Question 5: Assume that instead of acetic acid, the same concentration of formic acid (pKa = 3.7) was used in this experiment. What effect would this have on the pH of the equivalency point?
- A) The equivalency point would move up
- B) The equivalence point would fall
- C) There is no change
- D) Either A or C
The correct answer is B.
To solve this question, keep in mind that pKa is equal to –log(Ka); so, the greater the Ka value, the lower the pKa value.
Formic acid has a greater Ka value than acetic acid because its pKa is lower. Because formic acid is a stronger acid, its conjugate base will be weaker and will not react as easily with water to make formic acid and hydroxide. The pH of the solution will fall as a result of the decreased generation of hydroxide (B).
#6 Standalone questions and answers
Question 1: Assume that a solution in a beaker has the following elements:
[HA] = 0.001, [A-] = 0.1 and pKa = 3.2
What is the pH of this solution, approximately?
- A) 10-1.2
- B) 10-5.2
- C) 1.2
- D) 5.2
The correct answer is D.
To find the pH, we can utilize the Henderson-Hasselbalch equation. Three of the Henderson-Hasselbalch equation’s values are explicitly provided in the question stem.
pH = pKa+log(A–HA)
pH = 3.2+log(0.10.001)
pH = 3.2+log(100)
pH = 3.2+2=5.2
Question 2: How many significant figures are there in the year 2090.10?
- A) 3
- B) 4
- C) 5
- D) 6
The correct answer is D.
Count all three numbers to the left of the decimal point as significant using the criteria for significant figures. Since they are between two nonzero integers, the two zeroes in the hundreds and ones locations are likewise important. The hundredths place zero is next to a nonzero number, making it significant. There are six significant figures in all.
Question 3: In scientific notation, what is the number 909.90?
- A) 9.099 x 104
- B) 9.0990 x 104
- C) 9.099 x 102
- D) 9.0990 x 102
The correct answer is D.
The significand’s absolute value in scientific notation is less than ten. As a result, we rewrite 909.90 by shifting the decimal point two positions to the left. We increase the significand by a factor of 100 to compensate for this modification. D is the only option that does this while keeping the number of significant figures from the original notation.
Question 4: Sound A has a decibel level of 200 and Sound B has a decibel level of 40. How many orders of magnitude greater is Sound A than Sound B?
- A) 5
- B) 16
- C) 50
- D) 160
The correct answer is B.
We can solve this question by utilizing the decibels equation. We can determine the intensity of each sound and then the ratio of both intensities to compare two distinct sound levels.
The intensity of sound A will be denoted as IA, while the intensity of sound B will be denoted as IB.
dB = 10 log(II0)
200 = 10 log(IAI0)
20 = log(IAI0)
1020=IAI0
I01020=IA
Then, 40 =10 log(IBI0)
4 =log(IBI0)
104=IBI0
I0104=IB
Thus, I01020I0104= IAIB
1016=IAIB
Sound A is 16 orders of magnitude higher than Sound B, according to the IA/IB ratio of 1016.
Question 5: A student is requested to make a hypochlorous acid sample weighing 43.6 g. (HClO). How much HClO should the student prepare?
- A) 0.237 mol HOCl
- B) 0.672 mol HOCl
- C) 0.831 mol HOCl
- D) 1.20 mol HOCl
The correct answer is C.
We use the molar mass of HOCl to calculate the number of moles in this given mass. First, we determine HOCl’s molar mass.
- H has a molar mass of 1.01 g/mol. This can be rounded to 1 g/mol.
- O has a molar mass of 15.99 g/mol. This can be rounded to 16 g/mol.
As a result, the molar mass of HClO can be estimated as:
1 g/mol H+16 g/mol O+35.45g/mol Cl=52.45g/mol HOCl
To figure out how many moles are in a sample with a known mass, divide the mass by the molar mass.
43.6g HOCl(52.45g/mol HOCl)
We can now rule out option D since 43.6 is less than the number being divided by (52.47), implying that the answer to this division issue must be less than 1.
We can round down our numbers to the closest 10 because the response choices are so widely apart. This should result in a ratio that is easy to compute.
40g HOCl(50g/mol HOCl)=0.8 mol HOCl
Figure choice C gives 0.831 mol, which is quite close to this approximated answer.
Conclusion
Does this ultimate guide seem to be a little long? But we think it’s worth your time to read it carefully. Math on the MACT is fundamental but challenging because you are even not allowed to bring a calculator to the test room.
Don’t worry! To ace the MCAT math, you should practice doing math by hand and remember to pace yourself.
Besides, our detailed math skills concepts and example problems above will be powerful materials for your study and pass the math test with flying colors. Best of luck!
Don’t forget to take our free MCAT practice test at Medtutor to get familiarized with the format as well as the questions of the actual exam to strengthen your knowledge and skills, as a result, enhancing your chance to pass the MCAT exam with a high score on your first attempt. Good luck to you!
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