SAT EXAM PREPARATION 2025-2026

Question:

There are 66 calories in 15 grams of grated Parmesan cheese, and 59% of those calories are from fat.
When measuring Parmesan cheese, 5 grams is equal to 1 tablespoon.

Which of the following is closest to the number of calories from fat per tablespoon of grated Parmesan cheese?

Options:

• A) 3

• B) 8

• C) 9

• D) 13

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Solution:

Step 1: Calculate total fat calories in 15 grams

$$\text{Fat calories} = 59\% \text{ of } 66 = 0.59 \times 66 = 38.94 \approx 39 \text{ calories}$$

Step 2: Find fat calories per gram

$$\frac{39 \text{ calories}}{15 \text{ grams}} = 2.6 \text{ calories per gram}$$

Step 3: Find fat calories per 1 tablespoon (which is 5 grams)

$$2.6 \times 5 = 13 \text{ calories from fat per tablespoon}$$

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✅ Correct Answer: D) 13

This is the closest value to the actual fat calories per tablespoon.

Question:

The base of a tree has 10 mushrooms growing from its roots.
The mushroom population doubles every 5 days.

What type of function best models the relationship between the mushroom population and time?

Options:

• A) Decreasing exponential

• B) Decreasing linear

• C) Increasing exponential

• D) Increasing linear

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Solution:

Let’s understand what’s happening:

• The starting population is 10 mushrooms.

• The population doubles every 5 days, which means it multiplies by 2 repeatedly over time.

This is a classic example of exponential growth, where the population is increasing over time, not decreasing.

Why not linear?

• Linear growth adds a fixed amount each time.

• Exponential growth multiplies (like doubling), so the rate of increase itself increases over time.

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✅ Correct Answer: C) Increasing exponential

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Bonus (Equation form):

The function could be modeled as:

$$M(t)=10\cdot {\mathrm{2^}}^{t\mathrm{/}5}$$

Where:

• $M(t)$ is the number of mushrooms after $t$ days,

• 10 is the initial count,

• The exponent $t\mathrm{/}5$ reflects doubling every 5 days.

  Question:
  

  Given the quadratic equation:

  $$x^2 + bx + c = 0$$

  where $b$ and $c$ are constants.

  If:

  $$\frac{-b+\sqrt{b^2-4c}}{2}=18\text{and}\frac{-b-\sqrt{b^2-4c}}{2}=10$$

  what is one possible value of $\mathrm{x\; ?}$

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  Solution:

  These two expressions are the quadratic formula results for the roots of the equation:

  $$x = \frac{-b \pm \sqrt{b^2 - 4c}}{2}$$So, the two solutions are:

  $$x_1 = 18, \quad x_2 = 10$$
  

  Thus, one possible value of x is:

  ✅ Answer: $\boxed{10}$ or $\boxed{18}$

  
  

Question:

Solve:

$$3y^2 - 4 = 5y + 1$$

Which of the following is a solution to the equation above?

Options:
A) $\frac{5 - \sqrt{85}}{6}$
B) $\frac{5 - \sqrt{85}}{3}$
C) 5
D) $5 + \sqrt{85}$

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Solution:

Step 1: Start with the given equation

$$3y^2 - 4 = 5y + 1$$

Step 2: Move all terms to one side to set the equation to 0

$$3y^2 - 5y - 5 = 0$$

Now we solve this quadratic using the quadratic formula:

$$y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(-5)}}{2(3)}$$ $$= \frac{5 \pm \sqrt{25 + 60}}{6} = \frac{5 \pm \sqrt{85}}{6}$$

So the two solutions are:

$$y = \frac{5 + \sqrt{85}}{6} \quad \text{or} \quad y = \frac{5 - \sqrt{85}}{6}$$

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✅ Correct Answer: A) $\frac{5 - \sqrt{85}}{6}$

Here is the question text and a full step-by-step solution based on the image:

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Question:

A rocket is launched vertically from ground level. The rocket reaches a maximum height of 46.36 meters above the ground after 2.4 seconds, and falls back to the ground after 4.8 seconds.

Which equation best represents the height $s$, in meters, of the rocket $u$ seconds after it is launched?

Options:

• A) $s = -8.05u^2 + 38.64u$

• B) $s = -4.8u^2 + 46.36u$

• C) $s = -u^2 + 46.36$

• D) $s = 8.05u^2 - 38.64u$

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Solution:

We are told:

• The rocket starts at ground level → initial height = 0

• Maximum height is 46.36 meters at u = 2.4 seconds

• It returns to ground at u = 4.8 seconds

• So the vertex of the parabola is at $u = 2.4$, and the parabola opens downward

General form of a quadratic equation:

$$s = au^2 + bu + c$$

Since it starts at ground level: $c = 0$

Let’s use the vertex form:

$$s = a(u - h)^2 + k$$

Where:

• $h = 2.4$ (time of max height)

• $k = 46.36$ (maximum height)

$$s = a(u - 2.4)^2 + 46.36$$

We also know that at $u = 0$, $s = 0$ (ground level). Plug into the equation:

$$0 = a(0 - 2.4)^2 + 46.36 \Rightarrow 0 = a(5.76) + 46.36 \Rightarrow a = -\frac{46.36}{5.76} \approx -8.05$$

Now plug $a$ into the standard form:

$$s = -8.05u^2 + (2 \cdot 8.05 \cdot 2.4)u = -8.05u^2 + 38.64u$$

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✅ Correct Answer: A) $s = -8.05u^2 + 38.64u$

• The rocket is launched from ground level, so the initial height is $0$.

• Using the general form of a quadratic equation for height,

  $$s(t) = -au^2 + bu + c$$

• Since the initial height is 0, $c = 0$

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At the maximum height:

$$s(2.4) = -a(2.4)^2 + 2.4b = 46.36$$

When it hits the ground:

$$s(4.8) = -a(4.8)^2 + 4.8b = 0$$

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From these conditions, solve the system of equations:

$$-23.04a + 4.8b = 0 \quad \text{(1)}$$

This gives:

$$b = 4.8a$$

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Substitute $b = 4.8a$ into:

$$-a(2.4)^2 + 2.4(4.8a) = 46.36 \Rightarrow -5.76a + 11.52a = 46.36 \Rightarrow 5.76a = 46.36 \Rightarrow a = 8.05$$

Then:

$$b = 4.8 \times 8.05 = 38.64$$

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Thus, the equation is:

$$s(u)=-8.05u^2+38.64u$$