SAT Math Problem Types and Solutions

1. Solving for Coefficients with Regression

Summary: This problem type involves finding the values of coefficients in an equation by rewriting it algebraically to match a given expression. A more efficient approach using Desmos regression is presented.

Highlights:

Traditional Method: Algebraically rewrite one side of the equation to match the other, which can be complex and error-prone.
Desmos Regression: Use Desmos to fit coefficients by specifying a function and letting Desmos find values that satisfy the equation at multiple points.

Steps:

Set the left-hand side of the equation equal to f(x) and the right-hand side equal to g(x) in Desmos.
Type f(k) ~ g(k) to run a regression, where k is a constant.
Specify multiple values for k (e.g., 1, 2, 3, 4) to ensure the functions match at several points. The number of points needed equals the number of coefficients you're solving for.
Avoid using values of k that make the denominator of any fraction equal to zero, as this results in undefined expressions.
Read the values of the coefficients (a, b, c, etc.) directly from Desmos.
Calculate the required expression (e.g., a + b + c).

Example:

Solve for a, b, and c in the equation (expression provided in the original transcript).

f(x) = (expression)
g(x) = (expression)
Enter f(k) ~ g(k) with k = 1, 2, pi, 4 (avoiding k = 3).
Desmos will provide the values for a, b, and c.

2. Percentage Problems

Summary: This problem type requires careful interpretation of percentage language, particularly "more than."

Highlights:

"X% more than Y": Means Y + (X/100) * Y, or (1 + X/100) * Y.
Converting percentages to decimals: Divide the percentage by 100 (move the decimal place two spots to the left).

Steps:

Translate the word problem into algebraic equations.
When dealing with "x% more than", remember to add 1 to the percentage before multiplying. For example, "540% more than B" translates to A = 6.4B.
Substitute equations to eliminate variables and find the desired relationship.

Example:

A is 540% more than B, and B is 0.02% of C. If A is k% of C, find k.

A = 6.4B
B = 0.0002C
A = (k/100) * C
Substitute B in the first equation: A = 6.4 * (0.0002C) = 0.00128C
Equate the coefficients: k/100 = 0.00128
Solve for k: k = 0.128

3. Integer Constant Problems

Summary: These problems involve integer constants and require determining which expressions must also be integers.

Highlights:

Integer Definition: Positive or negative whole numbers (including zero). Fractions and decimals are not integers.
Integer Operations: The sum and product of integers are always integers. The division of integers may or may not be an integer.
Coefficient Comparison: Expand expressions and compare coefficients to establish relationships between the integer constants.

Steps:

Expand any binomial expressions using FOIL.
Compare coefficients to form equations.
Evaluate each answer choice:
- Check if the expression can be rewritten in terms of known integers.
- Consider the properties of integers under addition, subtraction, multiplication, and division.
Be wary of constant divided by integer because that is generally how this topic is tested on the SAT.

Example:

Given that h, k, q, and z are all integer constants and (hx + k)(qx + z) = ax^2 + bx + 35, determine which of the following must be an integer:

35/k
kq + hz
a/q

Expand: hq x^2 + (kq + hz)x + kz = ax^2 + bx + 35
Compare coefficients:
- hq = a
- kq + hz = b
- kz = 35
Evaluate:
1. 35/k = z (Since kz = 35 and z is an integer, 35/k must be an integer).
2. kq + hz = b (Since k, q, h, and z are integers, their products and sum must also be an integer).
3. a/q = h (Since hq = a and h is an integer, a/q must be an integer).
Therefore, all three expressions must be integers.

4. Quadratic Functions and Vertex Form

Summary: This type of problem presents a quadratic function and asks you to derive properties based on given information, requiring knowledge of quadratic formulas and relationships.

Highlights:

Vertex Location: If f(a) = f(b), the x-coordinate of the vertex (h) is the midpoint between a and b: h = (a + b) / 2.
Quadratic Formulas: Memorize relationships such as:
- h = -b / 2a
- Sum of solutions = -b / a (also equals 2h)
Vertex Form: f(x) = a(x - h)^2 + k, where (h, k) is the vertex.

Steps:

Use f(a) = f(b) to find the x-coordinate of the vertex (h).
Apply quadratic formulas to determine the relationships between coefficients and roots.
Rewrite the equation in vertex form to find the values of a, h, and k.
Use vertex form to easily find coefficients and answer questions about the graph.

Example:

Given f(x) = -3x^2 + bx + c and f(-4) = f(8), determine which of the following must be true:

r + s = 4, where r and s are x-intercepts
b < 0
c < k, where k is the y-coordinate of the vertex.

h = (-4 + 8) / 2 = 2
Sum of solutions (r + s) = 2h = 4
h = -b / (2a) => 2 = -b / (-6) => b = 12 (so b > 0)
Vertex form: f(x) = -3(x - 2)^2 + k = -3(x^2 - 4x + 4) + k = -3x^2 + 12x - 12 + k
c = -12 + k => c < k
Therefore, 1 and 3 are true.

5. Consecutive Even Integer Problems

Summary: This type of problem requires setting up inequalities based on the properties of consecutive even integers.

Highlights:

Consecutive Even Integers: Can be represented as x, x + 2, x + 4, etc.
Translating words to inequalities: Pay close attention to phrases like "at least" (≥), "more than" (>), "less than" (<), and "at most" (≤).

Steps:

Define the consecutive even integers as x, x + 2, x + 4, etc.
Translate the word problem into an algebraic inequality.
Solve the inequality for x.
Use the solution to find the minimum or maximum value of the required expression (remembering the even integer constraint).

Example:

Five times the first of three consecutive even integers is at least 9 more than the sum of the second and third integers. Find the minimum value for the second integer.

Integers: x, x + 2, x + 4
Inequality: 5x ≥ 9 + (x + 2) + (x + 4)
Solve: 5x ≥ 2x + 15 => 3x ≥ 15 => x ≥ 5
Since x must be an even integer, the smallest possible value for x is 6.
Therefore, the minimum value for the second integer (x + 2) is 6 + 2 = 8.

6. Right Triangle Trigonometry and Angle Relationships

Summary: These problems involve right triangles, trigonometric ratios, and angle relationships (especially complementary angles).

Highlights:

SOH CAH TOA: sin(θ) = Opposite / Hypotenuse, cos(θ) = Adjacent / Hypotenuse, tan(θ) = Opposite / Adjacent
Complementary Angles: If A + B = 90°, then sin(A) = cos(B) and cos(A) = sin(B).
Altitude in Right Triangles: This sets up similar triangles.

Steps:

Draw a diagram of the triangle.
Identify the relevant angles and sides.
Apply trigonometric ratios (SOH CAH TOA) to relate angles and sides.
Use the complementary angle relationship to relate sines and cosines.

Example:

In right triangle ABC, angle B is 90°, AB = 12, and D is a point on AC. Determine which of the following must be true:

BC = 12 * tan(CAB)
cos(ABD) - sin(DBC) = 0
ABD = BDC

tan(CAB) = BC/AB = BC/12 => BC = 12 * tan(CAB)
ABD + DBC = 90° => cos(ABD) = sin(DBC) => cos(ABD) - sin(DBC) = 0

Therefore, statements 1 and 2 must be true.

7. Altitudes in Right Triangles and 30-60-90 Triangles

Summary: This type of problem involves altitudes drawn in right triangles, creating similar triangles and often involving 30-60-90 triangles.

Highlights:

Altitude: A line segment from a vertex perpendicular to the opposite side.
30-60-90 Triangle Ratios: If the side opposite 30° is x, then the side opposite 60° is x√3 and the hypotenuse is 2x.

Steps:

Draw a diagram with the altitude.
Identify the similar triangles formed by the altitude.
Determine the angles in the triangles (often 30°, 60°, and 90°).
Apply the 30-60-90 triangle ratios to find the lengths of the sides.

Example:

In right triangle ABC, angle B is 90°, BC = 10, and BD is an altitude. If angle ABD is 60°, find the length of AD.

Triangle ABD and BDC are 30-60-90 triangles.
In triangle BDC, BC (hypotenuse) = 10, so BD (side opposite 30°) = 5 and DC (side opposite 60°) = 5√3.
In triangle ABD, BD (side opposite 60°) = 5√3, so AD (side opposite 30°) = 5√3 / √3 = 5, then Multiply AD by rad 3 once again, 5 * rad 3 * rad 3 = 15.

8. Circle Equations and Tangent Lines

Summary: This type of problem combines circle equations with the properties of tangent lines. While the circle equation is useful, Desmos can often provide a shortcut.

Highlights:

Circle Equation: (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius.
Tangent Line: A line that touches the circle at only one point. The radius drawn to the point of tangency is perpendicular to the tangent line.
Perpendicular Lines: Perpendicular lines have slopes that are negative reciprocals of each other.

Steps:

Enter the circle equation into Desmos to visualize the circle.
Use Desmos functions to find the circle's center and radius by manually plotting two points on the circle and finding the midpoint of the diameter.
Determine the slope of the radius that intersects the tangent line (negative reciprocal of the tangent line's slope).
Use the slope and the center of the circle to find the coordinates of the point of tangency.

Example:

A circle has the equation (equation given in transcript). A line tangent to the circle has a slope of 3/4. Find the y-coordinate of the point where the line is tangent to the circle.

Enter the equation into Desmos and plot the circle.
Find the center and radius using Desmos functions for midpoint and distance.
Slope of the radius = -4/3.
The radius has a slope of -4/3 and goes to point (a,b).
The y-coordinate is -5.

9. Pyramid Volume and Surface Area

Summary: This complex problem combines the volume and surface area formulas for a square pyramid with geometric relationships.

Highlights:

Pyramid Volume: (1/3) * length * width * height = (1/3) * base^2 * height.
Slant Height: The distance from the apex of the pyramid to the midpoint of a base edge.
Pythagorean Theorem: Relates the height, half the base, and the slant height: (base/2)^2 + height^2 = slant height^2.
Surface Area: Area of Base + 4(1/2 * base * slant height).

Steps:

Visualize drawing of pyramid (This is very important as described in the video).
Express the volume in terms of the base and height (V = (1/3) * base^2 * height).
Use the Pythagorean theorem to relate the height, half the base, and the slant height.
Solve the system of equations (volume equation and Pythagorean theorem equation) using Desmos.
Use the integer constraint to determine the correct base and height values.
Calculate the surface area using the base and slant height.

Example:

A square right pyramid has a volume of 400 cubic inches and a slant height of 13 inches. If the length, width, and height are integers, find the surface area.

V = (1/3) * base^2 * height = 400.
(base/2)^2 + height^2 = 13^2.
Enter the two equations into Desmos:
- (1/3) * x^2 * y = 400.
- (x/2)^2 + y^2 = 169.
Desmos provides solutions. Use the integer solution base=10 and height=12.
Area of Base + 4(1/2 * base * slant height).
Surface area = 10^2 + 4(1/21013) = 100 + 260 = 360 square inches.

10. Frequency Chart Analysis: Mean, Median, Range, and Standard Deviation

Summary: This type of problem involves analyzing frequency charts and how data manipulations affect statistical measures.

Highlights:

Frequency Chart Interpretation: Understand that frequencies represent how many times a particular value appears in the data set.
Mean: The average value. In a symmetric distribution, the mean is equal to the median.
Median: The middle value when the data is ordered.
Range: The difference between the maximum and minimum values.
Standard Deviation: A measure of the spread of the data. The larger the standard deviation, the more spread out the data is.

Steps:

Understand the concept of each of the terms.
If a data set is multiplied by X, then the mean, median, range, and standard deviation all will change by a factor of X.

Example:

Data set A is summarized by a frequency chart. Data set B is created by multiplying all values in data set A by 2. Determine which of the following must be true:

The difference between the mean and median of data set B is 0.
The standard deviation of data set B is equal to the standard deviation of data set A.
The range of data set B is double the range of data set A.

If all data points are multiplied by a factor, the mean, median, range, and standard deviation are multiplied by the same factor.