The SAT Math phase gives various issues to test your problem-fixing and reasoning abilities. To assist you ace this part of the exam, we have compiled a comprehensive guide that breaks down the maths topics included at the exam.
We’ll additionally offer designated motives and techniques to tackle the 20 hardest SAT Schedule questions.
What Is the SAT Math Section?
The SAT Math phase evaluates students’ mathematical talents through more than one-choice and grid-in questions. It contains subsections: one where a calculator is permitted and some other where it isn’t always.
With fifty eight questions to be completed in 80 minutes, this section of the SAT challenges students to apply mathematical reasoning and trouble-solving strategies underneath time constraints.
What Topics Are Covered in the Math Section of SAT?
The Math segment covers several subjects, along with algebra, geometry, trigonometry, and possibility. Students can particularly expect questions about linear equations, quadratic equations, features, graphs, geometry concepts (angles, triangles, circles, and polygons), trigonometric functions, and identities.
The SAT Math segment might also include questions related to information and statistics interpretation, which calls for students to research and interpret graphs, tables, and charts.
20 Hardest SAT Math Questions
Here’s our listing of the maximum difficult SAT Educational Institutes in Dubai math questions with answers that will help you put together for this exam.
Question 1
Simplify the expression: 3×2-5x+2x-2
Solution
To simplify this expression, we can use lengthy departments.
3x-2+2 divided by way of x-2
=3x+1
Therefore, the simplified expression is 3x+1.
Question 2
If f(x) = x2-4/x-2, what is the cost of f(2)?
Solution
To find the fee of f(2), we substitute x=2 into the feature f(x)
f(2)= 22-four/2-2
4-four/zero
Since the department by means of zero is undefined, f(2) is undefined.
Question 3
In the xy-aircraft, the factor (-2, 3) is reflected throughout the x-axis to provide factor P. Point P is then translated three units to the left to produce point Q. What are the coordinates of factor Q?
Solution
When a factor is pondered throughout the x-axis, the y-coordinate adjustments sign. So, the y-coordinate of factor P is -three.
To translate a point 3 units to the left, subtract 3 from the x-coordinate.
Thus, the coordinates of factor Q are (-five, -3).
Question 4
If sin x= three/5 , and x is in quadrant II, what is the price of cos x?
Solution
In quadrant II, each sine and cosine are fantastic.
Since x=3/five, and we understand that sin2x+cos2x=1, we are able to use pythagorean identity to locate cosx.
cos2x=1-sin2x
=1-(three/five)2
1-9/25
sixteen/25
Since cosine is high quality in quadrant II, cosx=√16/25=four/5
Question 5
A rectangle has an area of 24 square units. If its period is 3 gadgets longer than its width, what’s the perimeter of the rectangle?
Solution
Let the width of the rectangle be w devices. Then, its period is w+ three units
The region of a rectangle is given through the method A= period X width
So, we’ve the equation 24=(w+three X w)
Expanding and rearranging, we get the quadratic equation w2+3w-24=0
Solving for w, we discover w=4 (due to the fact that width can’t be bad)
Therefore, the length of the rectangle is four+three=7 gadgets
The perimeter of the rectangle is 2 X (4+7)=22 devices
Question 6
Triangle ABC is equilateral, and triangle DEF is equilateral. The ratio of the location of triangle ABC to the area of triangle DEF is sixteen:nine. What is the ratio of the fringe of triangle ABC to the fringe of triangle DEF?
Solution
In an equilateral triangle, all sides are equal, and all angles are 60∘.
The ratio of the areas of comparable figures is identical to the rectangular of the ratio in their corresponding side lengths.
So, the ratio of the aspect lengths of triangle ABC to triangle DEF is √16/nine= 4/3
Question 7
A circle with a radius of 5 is inscribed in a rectangular shape. What is the vicinity of the shaded area?
Solution
To find the place of the shaded location, we need to subtract the place of the circle from the region of the rectangular.
The location of the rectangular is same to the facet length squared, so 102=100 square devices
The place of the circle is π2= π x fifty two=25π rectangular gadgets
Therefore, the location of the shaded place is 100-25π rectangular gadgets
Question 8
The ratio of the period of a rectangle to its width is four:3. If the period is elevated by 2 units and the width is decreased by 1 unit, the ratio will become 2:1. What is the perimeter of the original rectangle?
Solution
Let the length of the unique rectangle be 4x and the width be 3x
According to the given ratios, we’ve got the equation 4x+2/3x-1=2/1
Solving for x, we discover x=1
Therefore , the unique length is four X 1= four units, and the original width is 3 X 1=3 units
The perimeter of the original rectangle is 2 X (four+three)=14 devices
Question 9
In a right triangle, the period of the altitude interested in the hypotenuse is identical to half of the period of the hypotenuse. What is the degree of the extreme attitude shaped by way of the altitude and the hypotenuse?
Solution
Let the duration of the hypotenuse be 2x gadgets, and the period of the altitude be x gadgets. By the Pythagorean theorem, the lengths of the opposite two facets are x units each.
Therefore, the triangle is an isosceles right triangle, and the intense angle shaped via the altitude and the hypotenuse is forty five∘
Question 10
A rectangle has a fringe of 30 devices. If its period is twice its width, what are the size of the rectangle?
Solution
Let’s denote the width of the rectangle as w devices. Since the period is two times the width, the duration l can be expressed as 2w devices
The perimeter of a rectangle is given by using P = 2l +2w. Substituting the given values, we get:
30=2(2w)+2w
30=4w+2w
30=6w
Dividing both aspects by 6
w=30/6
w=5
Now that we’ve got the cost of w, allow’s find the duration l, that’s two times the duration
l=2w=2(5)=10
So, the duration of the rectangle is 10 devices
Question 11
A series is defined recursively as follows: a1=three and a πn+1=1/2an+1 for n>1. What is the cost of a10?
Solution
To locate the price of a10 inside the series, let’s begin by using locating the values of a2, a3, and so on until a10 the use of this system
a2=1/2a1+1
a3=1/2a2+1
a4=1/2a3+1
…
a10=1/2a9+1
Since we have a1=three, we will start with this fee
a2= half of(three)+1=3/2+1=five/2
a3= half of(five/2)+1=5/4+1=9/four
Continuing this procedure, we sooner or later locate
a10= half of(9/2)+1=nine/four+1=134
So, the value of a10 inside the collection is thirteen/4
Question 12
If f(x)=2x-3/x+1, what’s the cost of f-1(1)?
Solution
To find f-1(1), we first need to locate the fee of x such that f(x) = 1
Given that f(x)=2x-three/x+1, we set f(x) = 1 and remedy for x
1=2x-three/x+1
To remedy this equation, we cross-multiply
2x-three=x+1
Subtract x from each sides
x-3=1
Add three to both aspects
x=4
Now that we’ve determined the cost of x, we can substitute it into the inverse feature f-1(x) to discover the corresponding price
f-1(1)= 4+1=5
Therefore f-1(1)=5
Question 13
In a right triangle, the length of the altitude interested in the hypotenuse is eight units, and one leg of the triangle is 15 units. What is the period of the opposite leg?
Solution
To resolve this problem, we’ll use the geometric belongings of right triangles concerning altitudes drawn to the hypotenuse.
Given:
Length of the altitude drawn to the hypotenuse (peak) = 8 gadgets
Length of 1 leg of the triangle (base) = 15 devices
Let’s denote the period of the opposite leg (unknown side) as x
According to the geometric property, the fabricated from the lengths of the segments of the hypotenuse cut up by using an altitude is same. Therefore, we will write the equation
eight X x= 15 X (x+8)
Now, permit’s resolve for x
8x= 15x+one hundred twenty
8x-15x=one hundred twenty
-7x=120
x=-a hundred and twenty/7
However, the length of a aspect can not be terrible, so we discard the negative solution.
Therefore, the length of the opposite leg of the right triangle is x=120/7units
Question 14
Given a triangle ABC wherein AB=8, BC =15, and AC=17, discover the measure of angle A
Solution
To find the degree of angle A in triangle ABC, we can use the Law of Cosines
The Law of Cosines states that for any triangle with facets a, b, and c and angle C contrary side c, the subsequent equations holds
c2=a2+b2-2abcosC
In this situation, a=BC=15, b=AC=17, and c=AB=8
Let’s plug these values into the Law of Cosines to discover the cosine of perspective A
82=152+172-2(15)(17) cos A
sixty four=225+289-510 cos A
64=514-510 cos A
510 cos A=514-sixty four
510 cos A=450
cos A = 450510
cos A= 4551
Now, we are able to use the inverse cosine feature to locate the measure of perspective A
A= cos-1(45/51)
A cos-1 (0.882)
A29.Forty seven∘
Therefore, the measure of angle A in triangle ABC is approximately 29.Forty seven∘
Question 15
If cos x=four/5 and x is in quadrant IV, what is the cost of sin x?
Solution
Since x is in quadrant IV, cosine is tremendous, however sine is negative. Given that cos x=45, we want to find the cost of sin x
sin2x=1-cos2x
sin2x=1-452
sin2x=1-2025
sin2x= -2024
Since sine is bad in quadrant IV, we take the bad square root
- sin x= -√-2024
- sin x = -√-1 X 2024
- sin x = -√2024i
Therefore, the cost of sin x is -√2024i in quadrant IV
Question 16
The sum of the first n phrases of an arithmetic collection is 2n2+n. What is the nth time period of the collection?
Solution
To discover the nth term of an arithmetic series, we want to use the formula for the sum of the primary nth term of an mathematics sequence.
The system for the sum of the primary n terms of an arithmetic sequence is given via
Sn=n/2(a1+an)
Where Sn is the sum of the first n terms, a1 is the primary phrases, an is the nth time period, and n is the number of phrases
In this problem, the sum of the primary n phrases is given as 2n2+n. So, we have
2n2+n=n/2(a1+an)
We can rewrite this equation as:
4n2+2n= n(a1+an)
4n+2=a1+an
Since the series is arithmetic, the distinction between consecutive phrases is constant. So, an-a1=(n-1)d, where d is the commonplace distinction
Now, we will locate the price of d by way of subtracting the primary time period from the second time period
d=a2-a1= (a1+ (n-1)d)-a1
d=a1+ (n-1)d-a1
d=nd
1-n
So, d=1
Now that we’ve the common distinction, we are able to locate the nth term by means of substituting d=1 into the equation an=a1+ (n-1)d
an= a1+ (n-1)d
an=a1+ (n-1)(1)
an= a1+ n-1
Now, we want to locate a1 in phrases of n. We recognise that a1 is the primary time period, so it’s miles the time period whilst n=1. Substituting n=1 into the sum method
2(1)2+1=2+1+3
Thus, a1=three
Finally, substituting a1=3 into the components for the nth time period:
an=3+n-1
an= 2+n
Therefore, the nth term of the series is 2+n
Question 17
A rectangular and an equilateral triangle have identical perimeters. If the aspect length of the rectangular is 12, what is the facet length of the equilateral triangle?
Solution
Let’s denote the facet length of the equilateral triangle as s
The perimeter of the rectangular is equal to the sum of the lengths of its 4 sides, so it’s four X 12=48
The perimeter of an equilateral triangle is equal to the sum of the lengths of its 3 facets, so it is 3s
Since the edges of the square and the equilateral triangle are identical, we have the equation:
forty eight=3s
Now, let’s remedy for s
s=48/three
s=sixteen
Therefore, the side length of the equilateral triangle is sixteen
Question 18
In a circle with a radius of 6, what is the duration of an arc that subtends a critical perspective of 120o?
Solution
To find the duration of an arc that subtends a primary perspective of 120o in a circle with a radius of 6, we use the formula:
Arc length = n/360 X 2πr
Where,
n is the degree of the important angle in degrees.
r is the radius of the circle.
In this case, n = 120o and r=6
Let’s plug those values into the components:
- Arc duration =a hundred and twenty/360 X 2π X 6
- Arc period = 1/3 X 2 X 22/7 X 6
- Arc period = 1/three X 44/7 X 6
- Arc duration =1/three X 26/forty seven
- Arc period=88/7
So, the duration of the arc that subtends a relevant perspective of one hundred twenty∘ in a circle with radius 6 (the usage of π = 22/7) is 88/7units
Question 19
If f(x)= x2-4x+five, what’s the value of f(2)?
Solution
To discover the price of f(2) for the feature f(x) = x2-4x+five, we replacement x=2 into the characteristic
f(2)=(2)2-four(2)+five
f(2)=4-8+5
f(2)=1
Therefore, the value of f(2) is 1
Question 20
A car travels from Town A to Town B at a mean pace of 60 mph and returns to Town A at a mean speed of 40 mph. What is the common velocity for the round ride?
Solution
To locate the common velocity for the round trip, we are able to use the components for average pace
Average velocity= Total distance/Total time
Let’s denote:
d as the one-manner distance between Town A and Town B.
t1 as the time taken to journey from Town A to Town B.
t2 because the time taken to go back from Town B to Town A
Since the space traveled is the identical for both legs of the experience, d is the same for each.
For the primary leg of the ride:
t1= d/Speed= d/60
For the second one leg of the ride:
t2= d/Speed= d/forty
The total time for the spherical experience is t1+t2
Total Time= d/60+d/40=second+3d/one hundred twenty=5d/120=d/24
Now, allow’s discover the total distance traveled:
Total distance = d+d=2nd
Now, we will use the components for average velocity to locate the average velocity for the spherical experience:
Average pace= Total Distance/Total Time= second/d24=2 X 2= 48 mph
Therefore, the average speed for the round trip is forty eight mph
No Calculator Allowed SAT Math Questions
Take a look at these Math questions that don’t require a calculator to resolve.
Question 1
In a bag, there are 8 red balls, 5 blue balls, and 3 green balls. If a ball is randomly selected from the bag, what is the chance that it’s far both purple or blue?
Solution
To discover the possibility of selecting both a crimson or blue ball, we first want to decide the overall number of balls in the bag, after which locate the wide variety of red and blue balls.
Total range of balls = eight (purple) + 5 (blue) + 3 (green) = 16
Number of red and blue balls = eight (crimson) + five (blue) = thirteen
The possibility of selecting either a purple or blue ball is given via the ratio of the wide variety of pink and blue balls to the entire number of balls:
Probability = Number of pink and blue balls / Total number of balls
Probability = 13/sixteen
Therefore, the opportunity of selecting either a pink or blue ball is thirteen/16
Question 2
A rectangle has a period that is three instances its width. If the fringe of the rectangle is 48 inches, what is the duration of the rectangle?
Solution
Let’s denote:
L because the duration of the rectangle.
W because the width of the rectangle.
We’re for the reason that the length of the rectangle is three times its width, so we will write the equation:
L=3W
The perimeter of a rectangle is given by using the components:
P=2(L+W)
We’re also given that the perimeter of the rectangle is forty eight inches, so we will write the equation:
forty eight=2(L+W)
Now, we’ll substitute L=3W into the fringe equation:
48 =2(3W+W)
48=2(4W)
forty eight=8W
Now, we’re going to remedy for W
W=48/6
W=6
Now that we’ve discovered the width of the rectangle, we can locate the duration the usage of the equation L=3W
L=3(6)
L=18
Therefore, the length of the rectangle is eighteen inches
SAT Math Questions Where Calculators Are Allowed
Unlike those wherein calculators aren’t required, those questions require important questioning, knowing formulas, and calculations. Here are a few SAT math examples in this category:
Question 1
The sum of the measures of the indoors angles of a polygon is 1980 ranges. How many facets does the polygon have?
Solution
To discover the variety of facets of the polygon, we can use the system for the sum of the measures of the interior angles of a polygon:
Sum of interior angles = (n-2) X 180o
Where n is the variety of facets of the polygon.
We’re for the reason that the sum of the measures of the interior angles of the polygon is 1980o, so we are able to write the equation:
1980= (n-2) X 180
Now, let’s resolve for n
1980 = 180n – 360
1980 +360 = 180n
2340= 180n
n= 2340/a hundred and eighty
n=13
Therefore, the polygon has 13 sides
Question 2
A circle has a radius of five inches. What is the area of the world fashioned by using a critical attitude of 60∘?
Solution
To find the region of the world shaped through a central attitude of 60∘ in a circle with a radius of five inches, we will use the formula for the location of a area:
Area of area = Central attitude/360∘ X πr2
Substituting π=22/7, Central perspective = 60∘, and r = five, we’ve
- Area of zone 60/360 X (22/7) X (five)2
- Area of quarter= 1/6 X (22/7)(five)2
- Area of zone=1/6 X 22/7 X 25
- Area of area=22 X 25/7 X 6
- Area of quarter=550/forty two
- Area of quarter = 13.10
- Therefore, the vicinity of the sector shaped by a important perspective of 60∘ in a circle with a radius of 5 inches is 13.10 square inches
Tips on How to Tackle Difficult Math Questions
Solving maths questions may be hard, however by understanding the query, figuring out key concepts, and knowing how to paint backward, you can approach with self belief. Here are strategies that will help you address tough math questions
- Understand the Question: Read the query carefully, figuring out key information and what is being asked. Break down complicated issues into smaller, more conceivable parts.
- Identify Key Concepts: Recognizing the mathematical standards or formulas had to clear up the problem. Review relevant formulation and concepts earlier than attempting to clear up the question.
- Work Backwards: Sometimes, beginning from the solution choices and operating backward can help you eliminate wrong options or slim down possibilities.
- Draw Diagrams or Visual Aids: Visualizing the hassle by means of drawing diagrams or sketches can offer clarity and assist you better understand the hassle.
- Practice, Practice, Practice: Practice solving numerous difficult questions to enhance your problem-fixing abilties and construct confidence. Review your errors and examine them. Try our mock SAT checks that will help you improve your skills immensely!
- Stay Calm and Focused: Difficult math questions can be intimidating, however retaining a calm and focused mind lets you approach the trouble with readability.
If you comply with those techniques, you may be ready to tackle the maximum difficult SAT Math questions with ease. Remember to practice these strategies regularly, and don’t be afraid to seek assistance when need.
