Lesson 5: Mathematics
Complete IMAT GuideIntroduction to IMAT Mathematics
The mathematics section of the IMAT tests your logical reasoning, numerical agility, and deep understanding of fundamental concepts rather than rote memorization of obscure formulas. This comprehensive module covers the three most heavily tested pillars: Euclidean Geometry, Probability (including Combinatorics), and Statistics.
Mastery of these topics is absolutely crucial, as they frequently intertwine. For example, calculating the probability of a specific genetic outcome often requires combinatorial mathematics and binomial expansion. Let's build this foundation step-by-step.
📐 Part 1: Euclidean Geometry
Euclidean Geometry is the study of flat space and the relationships between points, lines, and shapes. We will focus strictly on the high-yield formulas and concepts that appear on the exam.
2. Triangles: Notable Points and Segments
A triangle is the most fundamental polygon. Inside any triangle, drawing specific line segments reveals geometric "centers" known as notable points. You must memorize which segment creates which point.
The perpendicular segment dropped from a vertex to the opposite side (or its extension). The intersection of all three altitudes is the Orthocenter. In an obtuse triangle, the orthocenter lies outside the triangle.
The segment joining a vertex to the exact midpoint of the opposite side. The intersection is the Centroid (Center of Mass). Crucial Property: The centroid divides each median exactly in a 2:1 ratio (vertex to side).
A ray that perfectly cuts an interior angle in half. The intersection is the Incenter. It is equidistant from all three sides and represents the exact center of the inscribed circle (incircle).
A line perpendicular to a side passing through its midpoint. The intersection is the Circumcenter. It is equidistant from all three vertices and represents the center of the circumscribed circle.
Visualizing the Geometric Segments
Challenge an IMAT Question!
Official Paper: 2021 - Q55
worked solution & explanation
Step 1 Triangles RST and RPQ are similar. Scale factor $k = PR/RS = 1.4$.
Step 2 Find leg $RQ = 5 \times 1.4 = 7$. Trap side $TQ = 7 - 5 = 2$.
Step 3 Hypotenuse $ST = \sqrt{2.5^2+5^2} = 2.5\sqrt{5}$. Hypotenuse $PQ = 1.4 \times 2.5\sqrt{5} = 3.5\sqrt{5}$.
Step 4 Perimeter = $PQ + TQ + ST + PS = 3.5\sqrt{5} + 2 + 2.5\sqrt{5} + 1 = 3 + 6\sqrt{5} = 3(1+2\sqrt{5})$.
3. Polygons & Circles
A regular polygon is a geometric figure with $n$ equal sides and $n$ equal angles. Circles represent the limit as $n \to \infty$.
The sum of the exterior angles is ALWAYS $360^\circ$, regardless of the number of sides. Imagine walking around the perimeter; you turn a full circle.
The sum of the interior angles is found by dividing the polygon into $(n-2)$ triangles.
From each of $n$ vertices, you can draw a diagonal to $(n-3)$ other vertices. We divide by 2 to prevent double-counting.
The perimeter is simply $2p = n \cdot l$.
Challenge an IMAT Question!
Official Paper: 2022 - Q59
worked solution & explanation
Step 1 Since larger radius is 1.5x, its Area scales by $1.5^2 = 2.25$. Large Area = $36 \times 2.25 = 81$.
Step 2 Sector Area formula: $\frac{\theta}{360} \times \text{Total Area} = \text{Sector Area} \implies \frac{\theta}{360} \times 81 = \frac{27}{8}$.
Step 3 Solve for $\theta$: $\theta = \frac{27}{8} \times \frac{360}{81} = \frac{1}{8\times 3} \times 360 = 360 / 24 = 15$.
Challenge an IMAT Question!
Official Paper: 2018 - Q54
worked solution & explanation
Step 1 Decompose the polygon. Two adjacent $90^\circ$ angles separated by a side of 6cm forms the base of a $6\times6$ square. Area of square part = $36$.
Step 2 Removing $90^\circ$ from the $150^\circ$ angles leaves a $60^\circ$ angle for the top triangle. By symmetry, the top is an equilateral triangle with side 6. Area = $\frac{\sqrt{3}}{4} 36 = 9\sqrt{3}$.
Step 3 Combine: $36 + 9\sqrt{3} = 9(4+\sqrt{3})$.
Challenge an IMAT Question!
Official Paper: 2013 - Q60
worked solution & explanation
Step 1 Triangle base is $2r$ and height is $r$. Area of triangle = $0.5 \cdot 2r \cdot r = r^2$.
Step 2 Area of quarter circle = $\pi r^2 / 4$.
Step 3 Subtract: $r^2 - \pi r^2 / 4 = r^2(1 - \pi/4)$.
Challenge an IMAT Question!
Official Paper: 2011 - Q80
worked solution & explanation
Step 1 Let square side = $x$. Semicircle is cut from one side, so its diameter is $x$ ($r = x/2$).
Step 2 Remaining Area = $x^2 - 0.5\pi(x/2)^2 = x^2 - \frac{\pi x^2}{8} = x^2(\frac{8-\pi}{8}) = 100$.
Step 3 Solve for $x$: $x^2 = \frac{800}{8-\pi} \implies x = \sqrt{\frac{400 \times 2}{8-\pi}} = 20\sqrt{\frac{2}{8-\pi}}$.
🎲 Part 2: Probability & Combinatorics
Probability theory is the mathematical framework for quantifying uncertainty and calculating likelihoods. In the IMAT, calculating complex probabilities often requires combining basic probability axioms with advanced counting techniques (Combinatorics). Because medical diagnostics and genetics heavily rely on probability, this section is universally high-yield.
4. Basic Concepts & Set Theory
Probability is defined mathematically on a scale between 0 (an absolute impossibility) and 1 (absolute certainty). The foundational theoretical formula is:
📝 Basic Example
If you roll a standard, fair 6-sided die, what is the probability of rolling a prime number?
1. Determine the total sample space (S): $S = \{1, 2, 3, 4, 5, 6\}$. Total outcomes = 6.
2. Determine the favorable outcomes (E): The prime numbers on a die are 2, 3, and 5. (Note: 1 is NOT a prime number). $E = \{2, 3, 5\}$. Favorable outcomes = 3.
3. Calculate: $P(E) = \frac{3}{6} = \frac{1}{2} = 50\%$
Set Theory Logic (AND / OR)
IMAT questions rarely ask for the probability of a single event. You must know how to combine probabilities using the rules of sets.
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The Union (OR Rule / Addition Rule)
The probability that event A OR event B occurs. If events are mutually exclusive (they cannot happen simultaneously, like drawing a card that is both a Heart and a Spade), you simply add them: $P(A \cup B) = P(A) + P(B)$.
If they are NOT mutually exclusive (they can overlap, like drawing a card that is a Heart AND a King), you must subtract the overlap so you don't double count:General Addition Rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ -
The Intersection (AND Rule / Product Rule)
The probability that event A AND event B occur simultaneously or sequentially. For independent events (where A does not affect B), you simply multiply them.
Product Rule for Independent Events: $P(A \cap B) = P(A) \times P(B)$
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The Complement Rule (NOT)
The probability that event A does NOT occur. Sometimes, calculating the probability of something NOT happening is vastly easier than calculating every way it COULD happen.
$P(\text{NOT A}) = 1 - P(A)$
📝 Complement Rule Example
You flip a fair coin 5 times. What is the probability of getting at least one Heads?
Trying to calculate exactly 1 Head, exactly 2 Heads, exactly 3 Heads... and adding them all up is extremely tedious and prone to error. Use the complement rule!
1. The only outcome that does NOT satisfy "at least one Head" is getting absolutely ZERO Heads (meaning getting 5 Tails in a row).
2. Probability of getting a Tail on one flip is $1/2$. Because flips are independent, $P(5 \text{ Tails}) = (1/2)^5 = 1/32$.
3. Calculate Complement: $P(\text{At least 1 Head}) = 1 - P(\text{0 Heads}) = 1 - 1/32 = \frac{31}{32}$
Challenge an IMAT Question!
Official Paper: 2012 - Q77
worked solution & explanation
Concept Conditional Probability $p = P(\text{Math} \mid \text{Male}) = P(\text{Math} \cap \text{Male}) / P(\text{Male})$.
Min Overlap (p = 0)
Max Overlap (p = 5/6)
Step 1 Assume total 15 students (LCM). Males = 6. Females = 9. Math = 5.
Step 2 Max overlap: All 5 math students are male. Then $p = 5/6$.
Step 3 Min overlap: All 5 math students are female. This is possible since 9 females exist. Then $p = 0/6 = 0$.
Challenge an IMAT Question!
Official Paper: 2023 - Q51
worked solution & explanation
Step 1 Identify valid targets (squares of primes $\le 36$): $2^2=4$, $3^2=9$, $5^2=25$.
Step 2 Find combinations. For 4: $(1,4), (4,1), (2,2) \implies 3$ ways. For 9: $(3,3) \implies 1$ way. For 25: $(5,5) \implies 1$ way.
| × | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 1 | 1 | 2 | 3 | 4 | 5 | 6 |
| 2 | 2 | 4 | 6 | 8 | 10 | 12 |
| 3 | 3 | 6 | 9 | 12 | 15 | 18 |
| 4 | 4 | 8 | 12 | 16 | 20 | 24 |
| 5 | 5 | 10 | 15 | 20 | 25 | 30 |
| 6 | 6 | 12 | 18 | 24 | 30 | 36 |
Step 3 Calculate probability. Total successful = 5. Total possible outcomes = 36. Prob = 5/36.
Challenge an IMAT Question!
Official Paper: 2015 - Q57
worked solution & explanation
Step 1 Extract initial sum: $a+b+c = 3 \times 8 = 24$.
Step 2 Create new sum: $(a+b+c) + 1 + 2 + 6 + 3 = 24 + 12 = 36$.
Step 3 New mean (now 4 items) = $36 / 4 = 9$.
Challenge an IMAT Question!
Official Paper: 2025 - Q50
worked solution & explanation
Concept The Cartesian product $A \times B$ is a fundamental set theory definition.
Step 1 By definition, $A \times B = \{(a, b) \mid a \in A \text{ and } b \in B\}$.
Notice The pairs must be ordered (meaning $(a,b) \ne (b,a)$). Options mentioning "unordered" are incorrect.
5. Independent vs. Dependent Events (With vs Without Replacement)
When calculating the probability of drawing items sequentially, you must pay extreme attention to the wording of the question regarding replacement.
📝 Dependent Events Example (Without Replacement)
A standard deck contains 52 cards (26 red, 26 black). You draw two cards randomly without replacement. What is the probability that both cards are red?
1. The probability the first card is red is simple: $26/52 = 1/2$.
2. Because you did NOT replace the first red card, the sample space has permanently changed. There are now only 25 red cards remaining out of 51 total cards.
3. The probability the second card is red (given the first was red) is $25/51$.
4. Multiply (Product Rule): $\frac{26}{52} \times \frac{25}{51} = \frac{1}{2} \times \frac{25}{51} = \frac{25}{102}$
Challenge an IMAT Question!
Official Paper: 2011 - Q76
worked solution & explanation
Step 1 Scenario 1: Move Star ($P=4/6=2/3$). Box B now has 3S, 1H. P(Star from B) = 3/4. Path probability = $(2/3) \times (3/4) = 1/2$.
Step 2 Scenario 2: Move Heart ($P=2/6=1/3$). Box B now has 2S, 2H. P(Star from B) = 2/4 = 1/2. Path probability = $(1/3) \times (1/2) = 1/6$.
Step 3 Combine branches: $1/2 + 1/6 = 3/6 + 1/6 = 4/6 = 2/3$.
Challenge an IMAT Question!
Official Paper: 2022 - Q57
worked solution & explanation
Step 1 Determine total sum: $5 \times 20 = 100$. Sequence: $a, b, 24, d, e$ (since median is 24).
Step 2 To maximize $e$, minimize all others. $a=1, b=1$. To maintain sort order, $d$ must be at least 24, so set $d=24$.
Step 3 $1 + 1 + 24 + 24 + e = 100 \implies 50 + e = 100 \implies e = 50$.
6. Combinatorics: Permutations & Combinations
When the sample space is huge, we cannot count outcomes manually. We use combinatorics. The fundamental question to ask yourself immediately upon reading a question is: "Does the order matter?"
Used when arranging objects in a strict sequence (e.g., assigning 1st, 2nd, and 3rd place in a race; creating a password; lining up books on a shelf). Selecting object A then B is treated mathematically as a completely different outcome than selecting B then A.
Used when selecting a group or committee from a larger pool where roles are identical (e.g., picking 3 pizza toppings from a menu of 10; selecting 4 students to form a study group). Selecting A then B is exactly the same as selecting B then A. We divide by $k!$ to erase the duplicate internal orderings.
📝 Combinations Example
A hospital ward has 8 available nurses. The head doctor needs to randomly select exactly 3 nurses to assist in an emergency surgery. How many different groups of 3 nurses can be formed?
Does order matter? No. Being chosen 1st, 2nd, or 3rd doesn't change the fact that you are in the surgery group. Therefore, we use Combinations ($C$). Total pool $n=8$, chosen $k=3$.
$$C(8, 3) = \frac{8!}{3!(8-3)!} = \frac{8!}{3! \times 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56 \text{ groups}$$
Anagrams (Permutations with Repetition)
A very common, highly specific IMAT question asks how many distinct words (anagrams) can be formed by rearranging the letters of a given word. If all letters are totally unique (e.g., "CAT"), the answer is simply the factorial of the length: $n!$ ($3! = 3 \times 2 \times 1 = 6$). However, if some letters repeat, swapping two identical letters creates an indistinguishable word. You must divide the total permutations by the factorial of the frequency of each repeating letter to remove these phantom duplicates.
1. Count total letters ($n$): There are 11 letters total. So the numerator is $11!$.
2. Identify repeating letters: The letter 'I' appears 2 times (so divide by $2!$). The letter 'A' appears 2 times (divide by $2!$).
Challenge an IMAT Question!
Official Paper: 2018 - Q57
worked solution & explanation
Step 1 Total marbles = $3 + 4 + 5 = 12$.
Step 2 Number of non-red marbles = $4 (\text{blue}) + 5 (\text{green}) = 9$.
Step 3 Probability = $9/12 = 3/4$.
Challenge an IMAT Question!
Official Paper: 2017 - Q59
worked solution & explanation
Step 1 Sums $\ge 10$ are 10, 11, 12.
Step 2 Pairs for 10: (4,6), (5,5), (6,4) [3 ways]. Pairs for 11: (5,6), (6,5) [2 ways]. Pairs for 12: (6,6) [1 way].
Step 3 Total successful outcomes = $3 + 2 + 1 = 6$. Total possible outcomes = 36. Probability = $6/36 = 1/6$.
8. Binomial Probability (Repeated Trials)
If you conduct an experiment multiple times (e.g., flipping a coin 5 times, or having 4 children where each has a 25% chance of inheriting a disease), you must use the Binomial Probability formula. This calculates the probability of getting exactly $k$ successes in exactly $n$ independent trials.
• $\binom{n}{k}$ calculates all the different possible orders the successes could happen.
• $p^k$ is the probability of the successes happening.
• $(1-p)^{n-k}$ is the probability of the failures happening.
📝 Binomial Genetics Example
Two parents are heterozygous carriers (Aa) for Cystic Fibrosis, a recessive disorder. They plan to have 3 children. What is the exact probability that exactly 2 of their 3 children will be born with the disease?
1. Identify variables: Total trials $n=3$. Desired successes (sick children) $k=2$. From a Punnett square of Aa x Aa, the probability of a child having the disease (aa) is $p = 1/4$. The probability of being healthy is $(1-p) = 3/4$.
2. Apply formula: $P(X=2) = \binom{3}{2} \cdot (1/4)^2 \cdot (3/4)^1$
3. Calculate: $\binom{3}{2} = 3$ (The sick children could be the 1st&2nd, 1st&3rd, or 2nd&3rd).
4. Final Math: $3 \times (\frac{1}{16}) \times (\frac{3}{4}) = \frac{9}{64}$
9. Conditional Probability & Bayes' Theorem
Conditional probability, denoted as $P(A|B)$, is the probability of event A occurring, given the strict condition that event B has already occurred. This effectively shrinks the mathematical sample space down to only the universe where B is true.
"The probability of the overlap (A and B happening together) divided by the probability of the new restricted universe (B)."
Tree Diagrams for Sequential Events
Bayes' Theorem & The False Positive Paradox
Bayes' Theorem allows us to mathematically reverse conditional probabilities. This is the absolute mathematical foundation of medical diagnostics. It answers the terrifying patient question: "Doctor, if my screening test result came back positive, what is the actual probability that I truly have the disease?"
IMAT Hack: The Hypothetical Population Tree
To solve Bayes' problems quickly without memorizing the complex expanded formula, use a Hypothetical Population of 10,000 people.
- Assume a population of exactly 10,000 people.
- Use the Prevalence $P(D)$ of the disease to split the 10,000 into Sick and Healthy branches. (e.g., 1% prevalence = 100 sick, 9,900 healthy).
- Use the Sensitivity of the test on the Sick branch to find the True Positives. (e.g., 99% sensitive = 99 true positives).
- Use the False Positive Rate on the Healthy branch to find the False Positives. (e.g., 5% false positive rate applied to 9,900 = 495 false positives).
- Answer = (True Positives) / (Total Positives). Here, $99 / (99 + 495) = 99 / 594 \approx 16\%$. Even with a highly accurate 99% test, if the disease is extremely rare, a positive result often means you are still overwhelmingly likely to be healthy.
Challenge an IMAT Question!
Official Paper: 2024 - Q50
worked solution & explanation
Concept Replacement ensures draws are independent events.
Step 1 $P(1\text{st Green}) = 7 / 10$.
Step 2 $P(2\text{nd Green}) = 7 / 10$ (state is reset).
Step 3 Multiply independent events: $(7/10) \times (7/10) = 49/100$.
Challenge an IMAT Question!
Official Paper: 2019 - Q54
worked solution & explanation
Step 1 Total mass = $3 \times 2.1 = 6.3$ kg.
Step 2 Heaviest baby = Lightest + Range = $1.8 + 0.7 = 2.5$ kg.
Step 3 Median = Total - Lightest - Heaviest = $6.3 - 1.8 - 2.5 = 2.0$ kg.
Challenge an IMAT Question!
Official Paper: 2020 - Q57
worked solution & explanation
Step 1 Sum of first 4 numbers = $4 \times 5 = 20$.
Step 2 Sum of 5 numbers = $5 \times 6 = 30$.
Step 3 The fifth number = $30 - 20 = 10$.
📊 Part 3: Statistics & Data Analysis
Before we can analyze large datasets, we must understand the mathematical logic governing ordered lists of numbers (Sequences) and their sums (Series).
10. Sequences & Sigma Notation
Growth by constant Addition (Linear). The mathematical difference ($d$) between any two consecutive terms is constant.
Sum of first $n$ terms (Arithmetic Series):
Growth by constant Multiplication (Exponential). The ratio ($r$) between consecutive terms is constant.
Sum of an infinite geometric series (only if $|r| < 1$):
11. Measures of Central Tendency
"Where is the exact center of the data?" Statistics offers three distinct mathematical answers. Knowing which one to use depends entirely on the shape of the data distribution.
The arithmetic average. It incorporates every precise mathematical value in the entire dataset. Because it uses magnitude, it is highly, dangerously sensitive to extreme outliers.
The exact physical middle value when data is ordered from smallest to largest. It completely ignores the magnitude of extreme values, making it highly robust to outliers (e.g., used for calculating average national housing prices).
The single most frequently occurring value in the dataset. It is the only measure of central tendency that can be used for non-numerical, categorical data (e.g., identifying the most common blood type).
Visualizing Right-Skewed Data
12. Measures of Dispersion (Spread)
Central tendency alone is dangerously incomplete. You must mathematically know how spread out or reliable the data is around that center.
Variance ($\sigma^2$)
Logic: It calculates the average of the squared mathematical distances of each individual data point from the mean. We square the differences so negative distances don't simply cancel out positive ones, and importantly, squaring heavily penalizes massive outliers.
Standard Deviation ($\sigma$)
Logic: Variance mathematically gives results in "units squared" (e.g., dollars squared), which is uninterpretable in the real world. Taking the square root elegantly returns the measurement back to the original, understandable scale.
📝 Mastery Practice: IMAT Mathematics Quiz
Test your comprehensive understanding of advanced geometry, deep probability logic, combinatorics, and statistical interpretation. These questions are meticulously styled after the rigorous, integrated, multi-step logic required for the mathematical reasoning section of the IMAT.