The Digital SAT has introduced a new era of standardized testing, bringing with it a slew of adaptive challenges, especially in the math section. For those gearing up for this exam, understanding the hardest Digital SAT math questions can be a game-changer. Let's delve into the 10 toughest questions on Digital SAT compiled from the 2023 and 2024 examinations that are making waves among high scorers and first-timers alike.

**10 of the Hardest Digital SAT Math Questions**

Before diving into the solutions, it's crucial to give each problem your best effort. Be efficient with your use of the calculator wherever possible, and once you’ve worked through a question, compare your method to the provided solution to see if you've used the most efficient approach. This self-review is key to refining your strategy and enhancing your problem-solving skills for the Digital SAT.

Our team at Thriving Scholars has handpicked these questions after a careful review of all the DIGITAL SAT past papers and discussions with our students. This selection ensures that you are working on the most pertinent and challenging problems, helping you to prepare effectively for the demands of the Digital SAT.

#### Problem 1

(Dec'23 Digital SAT)

**Solution:**

1. Start with the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

where a = 1, b = -12, and c = 7.

2. Substitute the coefficients into the formula:

x = (-( -12 ) ± √( (-12)² - 4**1** 7 )) / (2 * 1)

x = (12 ± √(144 - 28)) / 2

x = (12 ± √(116)) / 2

x = 6 ± √(29)

3. Identify the constants j and k from the solution format j - √(k):

Here, the solution fits the form j - √(k), which gives us j = 6 and k = 29.

4. Calculate jk:

jk = 6 * 29 = 174

**Conclusion:**

The value of jk is 174.

---

#### Problem 2

(Dec'23 Digital SAT)

**Solution:**

1. Set up 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 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.

**Conclusion:**

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

#### Problem 3

(Dec'23 Digital SAT)

**Solution:**

Identity that sin(θ) = cos(90° - θ).

1. Equating angles using the identity:

a° = 90° - b°

a = 90 - b

2. Setting up the relationship between a and b:

We know a is 70% greater than b, so:

a = b + 0.7b = 1.7b

3. Solving for a and b:

Setting the two equations for a equal:

1.7b = 90 - b

2.7b = 90

b = 90 / 2.7

b = 33.33 (approximately)

Now substitute back to find a:

a = 1.7 * 33.33

a = 56.67 (approximately)

**Conclusion:**

The value of a, based on the given conditions, is approximately 56.67 degrees.

#### Problem 4

(Dec'23 Digital SAT)

**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:

1. Calculate the cost for the first 28 people:

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

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

2. Subtract the cost for the first 28 people from the total paid:

Total paid = $447

Remaining amount = $447 - $252 = $195

3. Determine the cost per additional person beyond 28 people:

Each additional person costs $15.

4. Calculate how many additional people attended:

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

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

5. Calculate the total number of people in attendance:

Total people = First 28 people + Additional people

Total people = 28 + 13 = 41

**Conclusion:**

There were 41 people in total in attendance at the arcade.

#### Problem 5

(May'24 Digital SAT)

**Solution:**

Start by defining the variables and expressions:

1. Concentration formula:

C = n / V

where C is the concentration, n is the number of moles, and V is the volume in cubic centimeters.

2. Setting up the problem:

Both containers have the same number of moles, 144 moles. The concentration in the larger container is 3 moles per cubic centimeter.

3. Calculating the volume of the larger container:

Clarge = n / Vlarge

3 = 144 / Vlarge

Vlarge = 144 / 3

Vlarge = 48 cubic centimeters

4. Relation of sizes:

The edge length of the larger container is 4 times that of the smaller container. If s is the edge length of the smaller container, then:

Vlarge = (4s)³

48 = 64s³

s³ = 48 / 64

s³ = 3 / 4

Vsmall = s³

Vsmall = 3 / 4 cubic centimeters

5. Finding the difference in volumes:

Difference = Vlarge - Vsmall

Difference = 48 - 3 / 4

Difference = 48 - 0.75

Difference = 47.25 cubic centimeters

**Conclusion:**

The positive difference between the larger container's volume and the smaller container's volume is 47.25 cubic centimeters.

#### Problem 6

(May'24 Digital SAT)

**Solution:**

To find the area of the region that lies within the circle but outside the inscribed equilateral triangle, you can follow these calculations:

1. Calculate the area of the circle:

Formula for the area of a circle is:

A_circle = πr²

Given the radius (r = 8 inches, the area of the circle is:

A_circle = π(8²) = 64π square inches

2. Calculate the side length of the equilateral triangle:

In an equilateral triangle inscribed in a circle, the side length \(s\) is related to the radius \(r\) by:

s = √3 × r

Thus, s = √3 × 8 = 8√3 inches

3. Calculate the area of the equilateral triangle:

The formula for the area of an equilateral triangle with side length \(s\) is:

A_triangle = (√3/4) s²

Substituting \(s = 8√3\):

A_triangle = (√3/4) (8√3)² = (√3/4) × 192 = 48√3 square inches

4. Calculate the area of the region outside the triangle within the circle:

A_region = A_circle - A_triangle

A_region = 64π - 48√3 square inches

Approximating π ≈ 3.1416 and √3 ≈ 1.732,

A_region = 64 × π - 48 × 1.732 ≈ 200.96 - 83.136 ≈ 117.9 square inches

**Conclusion:**

Rounding to the nearest tenth, the area of the region that lies within the circle but outside the equilateral triangle is approximately 117.9 square inches. This method allows for a straightforward subtraction of the triangle's area from the circle's to find the remaining area.

#### Problem 7

(March'24 Digital SAT)

**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 center and radius:

1. 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.

2. 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.

3. Form 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.

4. Identify the circle's center and radius:

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

5. Find the side length and area of the inscribed square:

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

side = 2 √(15).

6. Calculate the area of the square:

Area = s² = (2 √(15))² = 60.

**Conclusion:**

The area of the square inscribed in the circle is 60 square units. This simple breakdown uses basic algebraic steps to isolate the circle's properties and deduce the square's dimensions based on the circle's diameter.

#### Problem 8

(March'24 Digital SAT)

**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 = 0.92^3

Calculate 0.92 raised to the power of 3:

0.92^3 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

**Conclusion:**

The value of p that satisfies the equation, when rounded, is closest to 22.

#### Problem 9

(Oct'23 Digital SAT)

**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:

1. 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

2. 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

3. 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) * s*² 8

16 = (1/3) * s*²* * 8

16 = (8/3) * s²

s²= (16 * 3) / 8

s²= 6

s = √((6)

s approximately equals 2.45 inches

4. 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

5. Calculate the perimeter of the base of Pyramid B:

Perimeter = 4 * Side length

Perimeter = 4 * 17.15

Perimeter approximately equals 68.6 inches

**Conclusion:**

The perimeter of the square base of Square pyramid B is approximately 68.6 inches, rounded to the nearest tenth. This calculation provides the required length around the base of the pyramid, which is inscribed in a square.

#### Problem 10

(Nov'23 Digital SAT)

**Solution:**

1. Rearrange the linear equation to express y:

40x - 4y = a

y = 10x - a/4

2. 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

3. Combine terms to form a new quadratic equation:

Move all terms to one side:

3x^2 - 15x + a/4 - 12 = 0

4. 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

a = 123

**Conclusion:**

The value of a that makes the line tangent to the parabola is 123. This means the system of equations has exactly one distinct real solution when a is 123.

**What Makes These Questions Hard?** If you've just completed these problems and are evaluating your scores, remember, it's completely okay if you didn't perform as well as you hoped. These are genuinely hard problems designed to challenge and stretch your mathematical skills to their limits. The difficulty of these questions largely stems from their adaptive nature and the deeper understanding of concepts they require. This format aims to stretch a student's abilities by adapting in real-time to their performance, presenting problems that are tailored to push their limits.

**Preparation Tips** Preparing for these questions demands a robust study plan that goes beyond traditional methods. Utilizing advanced prep tools like the Desmos calculator can provide the needed edge. Practice under timed conditions and familiarize yourself with the calculator's advanced features to enhance efficiency.

**The Role of Coaching in Mastering Digital SAT Math** While self-study can be effective, personalized coaching can exponentially increase the efficiency of your SAT prep. **Rajat Sir**, known for helping students achieve a perfect 800 in digital SAT Math, provides expert guidance on leveraging tools like the Desmos calculator and mastering shortcut hacks. His tailored approach in both individual and group classes ensures that every student finds strategies that work best for them. Consider taking a **free demo class** with Rajat Sir to experience firsthand how structured coaching can help demystify the most challenging aspects of the digital SAT and optimize your study time.

The Digital SAT, with its adaptive nature, aims to push you into new areas of learning and problem-solving. Don’t be discouraged by setbacks; instead, use them as stepping stones to develop and strengthen your understanding. Remember, mastering these tough questions is a journey, and every attempt brings you one step closer to success. Keep practicing, keep learning, and stay persistent!

## Comentarios