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The SAT Maths section could make or break your undergraduate college acceptance. It can seem challenging, but it gives very high returns with the right preparation. Adaptive testing plays a huge role in the types of questions asked within the set syllabus, and thus, we must learn how it works to make the best of it.

Let’s go through the SAT math exam pattern & math syllabus in detail:

SAT Math Exam Pattern

The Maths section is digitalised and divided into 2 modules. The first module contains a mix of difficulty levels- easy, medium and hard. Depending on how you perform in the first module, the difficulty level of the second module changes. 

The questions are mostly MCQ’s however, some questions require you to type in the answer instead (grid-ins). You have a total of 70 minutes (35 minutes per module) to answer a total of 44 questions (22 per module). 

SAT Math Syllabus

Maths questions are divided into 4 categories in each module, and are arranged from easiest to hardest, gradually increasing in complexity. They challenge you to answer application-based questions with a focus on critical thinking and analysis. 

This progressive pattern of testing ensures that questions are asked from every domain, with an appropriate difficulty level. 

The number of questions from each section is given below, along with the syllabus and the type of questions asked:

30% of Maths questions are contextual. (Science, social studies or real-world-scenario-based) These test your ability to apply mathematical skills and concepts practically.  You are allowed to use a calculator during the test. 

The maths section is divided into the four domains of Algebra, Data Analysis, Geometry, and Trigonometry. Each section further has a different number of questions, thus making it easier for you to gauge what is important. 

The specific topics on which questions are asked according to their domains are:

• Algebraic concepts are easy to grasp, however, some of the most difficult practical questions come from this section. These are of the linear functions and linear inequalities in 1 or 2 variable varieties.
• Advanced math takes Algebra a step further but introduces abstract topics requiring a deep in-depth knowledge of the systems of equations in 2 variables and nonlinear functions specifically.
• Problem Solving has a major arithmetic component, such as Ratios, rates, proportional relationships, and units, making it a highly scoring domain. The Data Analysis part, however, can be challenging as it entails the interpretation and application of concepts. It primarily includes valuating statistical claims: observational studies and experiments.
• Geometry is one of the easiest sections, with easy concepts. It has area and volume formula-based questions that require memorisation and practice. Trigonometry can be lengthy, which at times can cause panic when there is less time. 

Digital SAT maths isn’t very difficult if you give it enough time and practise consistently. However, some questions can be really hard, and you need to scratch your head to get them right- something that is challenging when there’s a time crunch. To help you gain an understanding of what you’re up against, here are 15 of the hardest SAT questions with their solutions, that cover the domains of Algebra, Data Analysis, Geometry, and Trigonometry. 

Domain Algebra

Domain algebra consists of problems of the types- Linear equations in 1 and 2 variables, Linear functions, Systems of 2 linear equations in 2 variables and Linear inequalities in 1 or 2 variables. 

Questions from this domain are framed to confuse the student, and thus, it is important to learn how to navigate word-play during the test.

Let’s go through some examples:

Q1: (Dec’23 Digital SAT) 

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The function f is defined by the given equation. For what value of k does f(k)=6k+3?

Solution:

1. Formulate the Equation:

• Substitute f(k) with 6k + 3 in the equation of the function:

|8 - 5k| = 6k + 3

2. Consider the absolute value:

The equation |8 - 5k| = 6k + 3 can be split into two cases because of the absolute value:

Case 1: 8 - 5k = 6k + 3

• Combine like terms:

8 - 3 = 11k

5 = 11k

k = 5/11

• Check if 5/11 satisfies the original absolute condition:

8 - 5 * (5/11) = 8 - 25/11 = 63/11

6 * (5/11) + 3 = 30/11 + 33/11 = 63/11

So, k = 5/11 works for this case.

Case 2: 8 - 5k = -(6k + 3)

• Expand and simplify:

8 - 5k = -6k - 3

8 + 3 = -6k + 5k

11 = -k

k = -11

• Check if -11 satisfies the original absolute condition:

8 - 5 * (-11) = 8 + 55 = 63

6 * (-11) + 3 = -66 + 3 = -63

63≠-63, so k = does not work.

Thus, the values of k that satisfy f(k) = 6k + 3 are k = 5/11.

Q2: (Dec’ 23 Digital SAT)

For groups of 28 or more people, an arcade charges $9 per person for the first 28 people and $15 for each additional person. How many total people were in attendance if the group paid $447 to go to the arcade?

Solution:

Let's first calculate the total cost for the first 28 people and then determine how many additional people could attend based on the remaining amount:

• Calculate the cost for the first 28 people:

Each of the first 28 people pays $9, so:

Cost for 28 people = 28 * $9 = $252

• Subtract the cost for the first 28 people from the total paid:

Total paid = $447

Remaining amount = $447 - $252 = $195

• Determine the cost per additional person beyond 28 people:

Each additional person costs $15.

• Calculate how many additional people attended:

Number of additional people = Remaining amount / Cost per additional person

Number of additional people = $195 / $15 = 13

• Calculate the total number of people in attendance:

Total people = First 28 people + Additional people

Total people = 28 + 13 = 41

Thus, there were 41 people in total in attendance at the arcade.

Q3: If 3x−y=12, what is the value of 8x/2y?

A) 212
B) 44
C) 82
D) The value cannot be determined from the information given.

Solution: 

• One approach is to express 8x/2y

So that the numerator and denominator are expressed with the same base. Since 2 and 8 are both powers of 2, substituting 23 for 8 in the numerator of 8x/2y gives

(23)x/2y

• which can be rewritten as 23x/2y

Since the numerator and denominator/tor of have a common base, this expression can be rewritten as 2(3x−y). In the question, it states that 3x−y=12, so one can substitute 12 for the exponent, 3x−y, which means that

8x/2y=212

The final answer is A.

Q4: The incomplete table above summarises the number of left-handed students and right-handed students by gender for the eighth-grade students at Keisel Middle School. There are 5 times as many right-handed female students as there are left-handed female students, and there are 9 times as many right-handed male students as there are left-handed male students. If there are a total of 18 left-handed students and 122 right-handed students in the school, which of the following is closest to the probability that a right-handed student selected at random is female? (Note: Assume that none of the eighth-grade students are both right-handed and left-handed.)

A) 0.410
B) 0.357
C) 0.333
D) 0.250 

Solution: 

•  In order to solve this problem, you should create two equations using two variables (x and y) and the information you're given.

 Let x be the number of left-handed female students and let y be the number of left-handed male students. 

• Using the information given in the problem, the number of right-handed female students will be 5x, and the number of right-handed male students will be 9y. Since the total number of left-handed students is 18 and the total number of right-handed students is 122, the system of equations below must be true:

x+y=18

5x+9y=122

• When you solve this system of equations, you get x=10 and y=8. 

Thus, 5*10, or 50, of the 122 right-handed students are female. Therefore, the probability that a right-handed student selected at random is female is 50122, which to the nearest thousandth is 0.410.

The final answer is A.

Domain Advanced Math

Domain Advanced Math needs you to have a high accuracy, speed, and understanding of logs, and graphs. It revolves around some of the most difficult concepts, and questions that require serious thinking are formed from it.

Let’s have a look at a few questions that will help you ascertain what to expect:

Q5: (8-i) / (3 - 2i) 

If the expression above is rewritten in the form a+bi, where a and b are real numbers, what is the value of a? (Note: i=−1)

Solution: 

To rewrite (8-i) / (3 - 2i) in the standard form a + bi, multiply both the numerator and denominator by the conjugate of the denominator, which is (3 + 2i).

• Multiply the numerator and denominator:

[(8 - i) * (3 + 2i)] / [(3 - 2i) * (3 + 2i)]

• Expand the numerator

(8)(3) + (8)(2i) - (i)(3) - (i)(2i)
= 24 + 16i - 3i - (-1)(2)
= 24 + 16i - 3i + 2
= 26 + 13i

• Expand the denominator:

(3)(3) - (2i)(2i)
= 9 - (-4)
= 9 + 4
= 13

• Simplify the fraction:

(26 + 13i) / 13 = 26/13 + 13i/13 = 2 + i

Thus, when (8-i) / (3 - 2i) is rewritten in the standard form a + bi, the value of a is 2.

Q6: For a polynomial p(x), the value of p(3) is −2. Which of the following must be true about p(x)?

A) x−5 is a factor of p(x).
B) x−2 is a factor of p(x).
C) x+2 is a factor of p(x).
D) The remainder when p(x) is divided by x−3 is −2.

Solution: 

If the polynomial p(x) is divided by a polynomial of the form x+k (which accounts for all of the possible answer choices in this question), the result can be written as

p(x)/x+k=q(x)+r/x+k

where q(x) is a polynomial and r is the remainder. Since x+k is a degree-1 polynomial (meaning it only includes x1 and no higher exponents), the remainder is a real number.

Therefore, p(x) can be rewritten as p(x)=(x + k)q(x) r, where r is a real number.

The question states that p(3)=−2, so it must be true that

−2=p(3)=(3+k)q(3)+r

• Now we can plug in all the possible answers. If the answer is A, B, or C, r will be 0, while if the answer is D, r will be −2.

A. −2=p(3)=(3+(−5))q(3)+0
−2=(3−5)q(3)
−2=(−2)q(3)

This could be true, but only if q(3)=1

B. −2=p(3)=(3+(−2))q(3)+0
−2=(3−2)q(3)
−2=(−1)q(3)

This could be true, but only if q(3)=2

C. −2=p(3)=(3+2)q(3)+0
−2=(5)q(3)

This could be true, but only if q(3)=−25

D. −2=p(3)=(3+(−3))q(3)+(−2)
−2=(3−3)q(3)+(−2)
−2=(0)q(3)+(−2)

This will always be true no matter what q(3) is.

Of the answer choices, the only one that must be true about p(x) is D, that the remainder when p(x) is divided by x−3 is -2.

Thus, the final answer is D.

Q7: The function f(x)=x3−x2−x−11/4 is graphed in the xy-plane above. If k is a constant such that the equation f(x)=k has three real solutions, which of the following could be the value of k?

Solution:

The equation f(x)=k gives the solutions to the system of equations

y=f(x)=x3−x2−x−11/4

and

y=k

• A real solution of a system of two equations corresponds to a point of intersection of the graphs of the two equations in the xy-plane.

The graph of y=k is a horizontal line that contains the point (0,k) and intersects the graph of the cubic equation three times (since it has three real solutions). 

• Given the graph, the only horizontal line that would intersect the cubic equation three times is the line with the equation y=−3, or f(x)=−3. Therefore, k is −3.

The final answer is D.

Q8: (March'24 Digital SAT)

f(x) = 5(0.92)³Ë£
The function f is defined by the given equation. The equation can be rewritten as f(x) = 5(1 - p/100)Ë£, where p is a constant. Which of the following is closest to the value of p? 

A) 8
B) 12
C) 22
D) 24

Solution:

The function f(x) is defined as follows:

f(x) = 5(1 - p/100)x

And this function is equivalent to:

f(x) = 5(0.92)3x

To find the value of p, set the expressions inside the parentheses equal to each other:

1 - p/100 = 0.923

Calculate 0.92 raised to the power of 3:

0.923 is approximately 0.778688

Solve for p:

1 - 0.778688 = p/100

0.221312 = p/100

Multiply both sides by 100 to solve for p:

p = 0.221312 * 100

p = 22.1312

Since p needs to be an integer, round p to the nearest integer within the given choices:

p ≈ 22
Thus, the answer is p=22.

Q9: (Nov 23 Digital SAT)

y = 3x² - 5x - 12 

40x - 4y = a

In the given system of equations, a is a positive constant. The system has exactly one distinct real solution. What is the value of a?

Solution:

• Rearrange the linear equation to express y:

  40x - 4y = a

  y = 10x - a/4

• Substitute y in the quadratic equation:

  Replace y in y = 3x^2 - 5x - 12 with the expression from the linear equation:

  10x - a/4 = 3x^2 - 5x - 12

• Combine terms to form a new quadratic equation:

  Move all terms to one side:

  3x^2 - 15x + a/4 - 12 = 0

• Find the condition for tangency (Discriminant = 0):

  The discriminant of a quadratic equation ax^2 + bx + c = 0 is given by:

  Discriminant = b^2 - 4ac

  Apply this to our equation:

  a = 3, b = -15, c = a/4 - 12

  Calculate the discriminant:

  (-15)^2 - 4 3 (a/4 - 12) = 0

  Solve for a:

  225 - 3(a - 48) = 0

  225 - 3a + 144 = 0

  369 = 3a

Thus, the answer is that a = 123.

Domain Problem Solving and Data Analysis

This domain tests your reasoning skills and ability to statistically analyse mathematical data. More Arithmetically oriented, the domain problem solving and data analysis section tests percentages and ratios, and evidence-based decision making. The following examples illustrate the kind of questions you might expect.

Q10: A gear ratio r:s is the ratio of the number of teeth of two connected gears. The ratio of the number of revolutions per minute (rpm) of two gear wheels is s:r. In the diagram below, if Gear A is rotated by the motor at a rate of 100 rpm, what is the number of revolutions per minute for Gear C?

A) 50
B) 110
C) 200
D) 1,000

Solution:

Because gears A and C do not connect directly, but instead through gear B, we should first try to figure out the rotational relationship between A and B (at 100 rpm) before applying that to B and C. 

Because B is larger than A (and has more gears), A is going to rotate fully multiple times before B rotates once. 

• How many times? Here, it’s helpful to consider a ratio. 

A has 20 gears. 

B has 60 gears. 

So A is going to have to rotate three times before B rotates once. (20 goes into 60 three times.) 

Therefore, the ratio of rotation between A and B is 3 : 1. 

• Apply the same method to figure out the ratio between B and C. 

B has 60 gears. 

C has 10 gears. 

Here, B only has to rotate a sixth of its distance for C to rotate once, so the ratio of rotation between B and C is 1 : 6.

• Now we take the number of RPMs the problem gives us, start with the gear on the left and multiply through with our ratios. 

So if Gear A rotates 100 times per minute, Gear B will rotate a third of that distance… 

So we divide 100 by 3. 

• Because we know Gear C rotates six times as fast as Gear B, we then take our answer and multiply it by 6. 

So we get (100)(â…“)(6). 

Which gives us 200 rpm.

Thus, the final answer is C.

Q11: If x is the average (arithmetic mean) of m and 9, y is the average of 2m and 15, and z is the average of 3m and 18, what is the average of x, y, and z in terms of m?

A) m+6
B) m+7
C) 2m+14
D) 3m+21

Solution:

Since the average (arithmetic mean) of two numbers is equal to the sum of the two numbers divided by 2, the equations x=(m+9)/2, y=(2m+15)/2, z=(3m+18)/2 are true. 

• The average of x, y, and z is given by (x+y+z)/3. Substituting the expressions in m for each variable (x, y, z) gives

[(m+9)/2 + (2m+15)/2 + (3m+18)/2] /3

• This fraction can be simplified to m+7.

The final answer is B.

Domain Geometry and Trigonometry

This section in SAT tests your knowledge of core geometric and trigonometric principles.  The concepts here often connect to real-world applications and require you to be precise in your calculations. The following questions will help you get an idea of what to expect.

Q12: (March’24 Digital SAT)

3x2+18x+3y2−6y−15=0

The equation above gives the graph of a circle in the xy plane. If the circle is inscribed in a square, what is the area of the square?

Solution:

To solve the problem of finding the area of the square inscribed in the circle represented by the given equation, we first need to rewrite and simplify the circle's equation to find its centre and radius:

• Simplify the equation:

Start with the equation 3x² + 18x + 3y² - 6y - 15 = 0.

Divide every term by 3 to simplify:

x² + 6x + y² - 2y - 5 = 0.

• Complete the square for both x and y:

For x:

x² + 6x -> Complete the square by adding and subtracting (6/2)² = 9.

(x + 3)² - 9.

For y:

y² - 2y -> Complete the square by adding and subtracting (-2/2)² = 1.

(y - 1)² - 1.

•  From the standard circle equation:

Substitute the completed squares back into the equation:

(x + 3)² - 9 + (y - 1)² - 1 - 5 = 0.

Combine and rearrange:

(x + 3)² + (y - 1)² = 15.

• Identify the circle's centre and radius:

The circle is centred at (-3, 1) with a radius of √15).

• Find the side length and area of the inscribed square:

The side of the square equals the diameter of the circle:

Side = 2 √(15).

• Calculate the area of the square:

Area = s² = (2 √(15))² = 60.

Thus, the area of the square inscribed in the circle is 60 square units. 

Q13: Points A and B lie on a circle with radius 1, and arc AB has a length of π/3. What fraction of the circumference of the circle is the length of arc AB?

Solution:

To figure out the answer to this question, you'll first need to know the formula for finding the circumference of a circle.

• The circumference, C, of a circle is C=2πr, where r is the radius of the circle. For the given circle with a radius of 1, the circumference is C=2(π)(1), or C=2π.

To find what fraction of the circumference the length of AB⌢ is, divide the length of the arc by the circumference, which gives π/3 ÷ 2π. This division can be represented by=1/6.

• The fraction 1/6 can also be rewritten as 0.166 or 0.167.

The final answer is 1/6, 0.166, or 0.167.

Q14: A grain silo is built from two right circular cones and a right circular cylinder with internal measurements represented by the figure above. Of the following, which is closest to the volume of the grain silo, in cubic feet?

A) 261.8
B) 785.4
C) 916.3
D) 1047.2

Solution: 

The volume of the grain silo can be found by adding the volumes of all the solids of which it is composed (a cylinder and two cones). 

The silo is made up of a cylinder (with height 10 feet and base radius 5 feet) and two cones (each with height 5 ft and base radius 5 ft)

• The formulas given at the beginning of the SAT Math section are
  Volume of a Cone (V=â…“(πr2h)) and Volume of a Cylinder (V=πr2h).

These can be used to determine the total volume of the silo. Since the two cones have identical dimensions, the total volume, in cubic feet, of the silo is given by

• Vsilo=π(52)(10)+(2)(1/3)π(52)(5)=(4/3)(250)π

which is approximately equal to 1,047.2 cubic feet.

The final answer is D.

Q15: (Digital SAT October 23)

In geometry, a square pyramid is a pyramid that has a square base and four lateral faces. Square pyramid A is similar to Square pyramid B. The table gives the volumes, in cubic inches, of the two square pyramids. If the height of Square Pyramid A is 8 inches, what is the perimeter of the square base of Square Pyramid B, in inches? (Round your answer to the nearest tenth of an inch.)

Solution:

To solve this problem, we need to find the perimeter of the square base of Square Pyramid B, given the volumes of both pyramids and the height of Square Pyramid A. Here's how we can approach this using simple text math notation:

• Establish the volume ratio:

  Volume of Square pyramid A = 16 cubic inches

  Volume of Square pyramid B = 5488 cubic inches

  Volume ratio = Volume of B / Volume of A = 5488 / 16 = 343

•  Calculate the ratio of side lengths (Scale factor):

  Volume ratio = k³ (where k is the scale factor between the pyramids)

  343 = k³

  k = cube root of 343

  k = 7

•  Determine the side length of the base of Pyramid A:

  Volume of a pyramid = (1/3) Base Area Height

  Let s be the side length of the square base of Pyramid A.

  Volume of A = (1/3) s² 8

  16 = (1/3) s² 8

  16 = (8/3) * s²

  s²= (16 * 3) / 8

  s²= 6

  s = √((6)

  s approximately equals 2.45 inches

•  Calculate the side length of the base of Pyramid B:

  Side length of B = 7 * Side length of A

  Side length of B = 7 * 2.45

  Side length of B approximately equals 17.15 inches

•  Calculate the perimeter of the base of Pyramid B:

  Perimeter = 4 * Side length

  Perimeter = 4 * 17.15

  Thus, the perimeter approximately equals 68.6 inches.

SAT questions are usually not very difficult if you have a good understanding of the syllabus and question pattern. The problem mainly arises when you can’t untangle what’s important in a question, because it has been framed to confuse you. Such questions require interpretation, which is difficult to do in a time crunch. Tricky questions blend several topics and require critical thinking, which also consumes time.

Students often combat these problems by finding efficient techniques that reduce the number of steps and involve strategic thinking. This knowledge of what ‘works’ for you is entirely subjective and can only be understood by appearing for and analysing mocks. Over time, you will be able to identify which types of questions trip you up and identify patterns of questions in which you do well. 

The main purpose is to minimise time per question while maintaining a high accuracy rate. Making formula sheets can help in quick recollection and save time. Knowing when to use the calculator and when to calculate in your head is also very necessary and helpful in minimising the number of steps. Sometimes, even just plugging in the values can get you the answer more quickly. It is necessary to note, though, that all these shortcuts can only help you if you practise applying them regularly.

It is better for you to aim high, fail, learn, and then succeed as compared to aiming low and getting complacent. With the former, there is more capacity to learn quicker, adapt faster,r and diversify yourself. Multiple fortes maximise scores and help in getting differentiated. 

Visibility

Adaptive testing is a game changer. Universities will pay attention to the types of questions you attempt and their difficulty levels. Solving harder questions will get you visibility. 

Speed

The Maths SAT is essentially a speed test. Practising hard questions will help you save precious time, thus helping you stay in the right frame of mind. 

Learn test tricks

If you can solve hard ones, the easy ones will be a piece of cake. Hard questions will help you address multiple concepts at once, accustomising your brain to thinking from different angles. The more you solve, the better the tricks you’ll discover too.

The digital Math SAT is a computer-based, adaptive exam. Most students who have appeared for tests in the pen and paper mode throughout their schooling struggle with adapting to the digital paper pattern. 

Following the tips below can help make this transition easier. 

Familiarization

Practising with the Bluebook App helps you understand the test layout and adaptive mechanism. Work within your ability level while gradually challenging yourself to improve.

Master the Calculator

It is crucial to make the best use of this tool. Know when to use it and when not to. The provided Desmos graphing calculator or your own approved device. 

Practice time management

Strengths first, weaknesses later. Attempt higher weightage domains- Algebra and Advanced Math on priority. Use the highlighting features available to maximise efficiency with good accuracy in the test. 

Review and Analyse Mocks

Understand performance pie charts and graphs with deep analysis. Go over alternative methods of solving (Mental maths over the calculator saves time). Target weak areas and work on them.

Scoring well on the SAT can make a world of difference in your study abroad journey. So far, СÓÅÊÓƵ has helped 30,000+ students attain excellent scores, live up to their potential, and move to the US, UK, and Australia for UG studies. The digital SAT math is tricky, but with consistent practice, you can ace it! Speak with our best advisors to know how you can make the most of your time and free resources. For the speediest help, real-time insights on mocks, and advice in any other domain, check out СÓÅÊÓƵ Premium.