Are you desperately looking for SAT math questions? How many questions did you find for the topic you were looking for? Are they any good? Well, if you are scavenging for good practice questions you’ve just hit a gold mine! In this blog, we will show you how to solve SAT math problems through some quality math SAT questions along with their detailed explanations. Let’s go!

IN THIS BLOG:

## How to solve SAT Math problems?

We go by the adage ‘show not tell’! So in this section, we picked a sample math problem (a difficult one) and decided to solve it as simply and slowly as possible.

Question: A rectangular prism has a volume of 96 cubic units and a surface area of 98 square units. What is the length of the longest diagonal of the prism?

Solution: Let the dimensions of the prism be x, y, and z. Then we have: xyz = 96 — equation (1) 2xy + 2xz + 2yz = 98 — equation (2)

We can use the first equation to express one of the variables in terms of the other two. For example, z = 96/xy. Substituting this into equation (2), we get:

2xy + 2xz + 2yz = 2xy + 2x(96/xy) + 2y(96/xy) = 98

Simplifying, we get:

2x^2 + 192/xy + 2y^2 = 98

Multiplying both sides by xy, we get:

2x^3y + 192 + 2xy^3 = 98xy

Dividing both sides by 2, we get:

x^3y + xy^3 + 96 = 49xy

We can use the AM-GM inequality to obtain an upper bound for xy:

xy <= (x^2 + y^2)/2

So, we have:

x^3y + xy^3 + 96 <= x^3(x^2 + y^2)/2 + y^3(x^2 + y^2)/2 + 96

= (x^5 + y^5)/2 + 48(x^2 + y^2)

We want to find the maximum value of the longest diagonal of the rectangular prism, which is given by:

sqrt(x^2 + y^2 + z^2)

Substituting z = 96/xy, we get:

sqrt(x^2 + y^2 + (96/xy)^2)

We can use the AM-GM inequality again to obtain an upper bound for this expression:

sqrt(x^2 + y^2 + (96/xy)^2) <= (x^2 + y^2 + (96/xy)^2 + 2(x^2 + y^2 + (96/xy)^2)/xy)/2

= ((x^4 + y^4 + 9216/x^2y^2) + 2(x^3 + y^3 + 96))/2xy

Now, we can use the inequality we obtained earlier to get:

sqrt(x^2 + y^2 + (96/xy)^2) <= ((x^5 + y^5)/2 + 48(x^2 + y^2) + 2(x^3 + y^3 + 96))/2xy

= (x^5 + y^5 + 96xy + 96(x^2 + y^2))/2xy

Substituting xy = 96/z, we get:

sqrt(x^2 + y^2 + (96/xy)^2) <= (x^5 + y^5 + 96^3/z^3 + 96(x^2 + y^2))/192

We want to maximize this expression, subject to the constraint that xyz = 96. We can use Lagrange multipliers to do this:

f(x,y,z) = (x^5 + y^5 + 96^3/z^3 + 96(x^2 + y^2))/192

g(x,y,z) = xyz – 96

Setting the partial derivatives of f and g equal to each other, we get:

5x^4 + 192y^2/x – lambdayz = 0 — equation (1)

5y^4 + 192x^2/y – lambdaxz = 0 — equation (2)

288/ z^4 – lambdaxy = 0 — equation (3)

Multiplying equation (1) by y, equation (2) by x, and adding the resulting equations.

SUGGESTED READ: Things to know before you take the SAT Exam 2023

## Practice Questions For SAT Math

1. If x^3 + 3x^2y + 3xy^2 + y^3 = 64 and x + y = 4, what is the value of xy?
2. If f(x) = x^2 – 3x + 2, what is the value of f(-2) + f(1)?
3. If a + b = 5 and a^2 + b^2 = 29, what is the value of ab?
4. What is the sum of the solutions of equation 3x^2 – 2x – 1 = 0?
5. If f(x) = 2x + 3 and g(x) = 5x – 2, what is f(g(4))?
6. What is the 3rd term of the sequence?
7. What is the value of x that satisfies the equation (x – 2)/(x – 4) = (x – 5)/(x – 1)?

## Solutions To Practice Math SAT Questions

Math SAT Question 1: If x^3 + 3x^2y + 3xy^2 + y^3 = 64 and x + y = 4, what is the value of xy?

Solution: We can rewrite the expression x^3 + 3x^2y + 3xy^2 + y^3 as (x + y)^3 = 4^3 = 64. Substituting x + y = 4, we get x^3 + 3x^2y + 3xy^2 + y^3 = x^3 + y^3 + 3xy(x + y) = 64. Substituting x + y = 4 again, we can simplify the expression to x^3 + y^3 + 12xy = 64. We can then use the fact that (x + y)^2 = x^2 + 2xy + y^2 to rewrite the expression as (x^2 + y^2 + 2xy) + 10xy = 16, or (x + y)^2 – 2xy + 10xy = 16. Simplifying, we get 8xy = 16, or xy = 2.

Have questions about SAT Math preparation?

Math SAT Question 2: If f(x) = x^2 – 3x + 2, what is the value of f(-2) + f(1)?

Solution: We can find the value of f(-2) by substituting x = -2 into the expression for f(x): f(-2) = (-2)^2 – 3(-2) + 2 = 10. We can find the value of f(1) by substituting x = 1: f(1) = 1^2 – 3(1) + 2 = 0. Therefore, f(-2) + f(1) = 10 + 0 = 10.

Math SAT Question 3: If a + b = 5 and a^2 + b^2 = 29, what is the value of ab?

Solution: We can square the equation a + b = 5 to get a^2 + 2ab + b^2 = 25. Subtracting the equation a^2 + b^2 = 29 from this equation, we get 2ab = -4, or ab = -2.

Math SAT Question 4: What is the sum of the solutions of equation 3x^2 – 2x – 1 = 0?

Solution: We can use the formula for the sum of the roots of a quadratic equation: -b/a. In this case, a = 3 and b = -2. So, the sum of the solutions is -(-2)/3 = 2/3.

Math SAT Question 5: If f(x) = 2x + 3 and g(x) = 5x – 2, what is f(g(4))?

Solution: First, we need to evaluate g(4) which is 5(4) – 2 = 18. Then, we plug 18 into f(x) to get f(g(4)) = f(18) = 2(18) + 3 = 39.

The sum of the first n terms of an arithmetic sequence is 5n + 3, and the sum of the first 2n terms is 15n – 7.

Math SAT Question 6: What is the 3rd term of the sequence?

Solution: We can use the formula for the sum of an arithmetic sequence: Sn = n/2 (a1 + an). Let a1 be the first term and d be a common difference. Then we have:

n/2 (a1 + a1 + (n-1)d) = 5n + 3 — equation (1)

2n/2 (a1 + a1 + (2n-1)d) = 15n – 7 — equation (2)

Simplifying equation (1), we get:

a1 + (n-1)d = (10/n) + (6/n)

Simplifying equation (2), we get:

a1 + (2n-1)d = (15/2) – (7/2n)

Subtracting equation (1) from equation (2), we get:

n*d = (5/2) – (13/2n)

Substituting this value of d into equation (1), we get:

a1 = (20/n) + (3/n) – (n/2) = (23/n) – (n/2)

So, the 3rd term of the sequence is a1 + 2d = (23/n) – (n/2) + 2((5/2) – (13/2n))/n.

Math SAT Question 7: What is the value of x that satisfies the equation (x – 2)/(x – 4) = (x – 5)/(x – 1)?

Solution: Cross-multiplying, we get:

(x – 2)(x – 1) = (x – 5)(x – 4)

Expanding both sides, we get:

x^2 – 3x + 2 = x^2 – 9x + 20

Simplifying, we get:

6x = 18

So, x = 3.

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