# Nakafa Framework: LLM

URL: https://nakafa.com/en/exercises/high-school/snbt/quantitative-knowledge/try-out/set-9

Exercises: Try Out - Set 9: Real exam simulation to sharpen your skills and build confidence.

---

## Exercise 1

### Question

export const metadata = {
  title: "Problem 1",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

import { Graph } from "../graph";

<Graph title="Geometry Visualization" description="Large square, inscribed circle, and small square." />

If the area of the large square is <InlineMath math="256 \text{ cm}^2" />, then the circumference of the circle is...


### Choices

- [ ] $$7\pi\text{ cm}$$
- [ ] $$10\pi\text{ cm}$$
- [ ] $$12\pi\text{ cm}$$
- [ ] $$14\pi\text{ cm}$$
- [x] $$16\pi\text{ cm}$$

### Answer & Explanation

export const metadata = {
  title: "Solution to Problem 1",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

Given the area of the large square is <InlineMath math="256 \text{ cm}^2" />.

#### Finding the Side Length of the Large Square

The area of a square (<InlineMath math="A" />) is given by <InlineMath math="A = s^2" />.

<MathContainer>
  <BlockMath math="s^2 = 256" />
  <BlockMath math="s = \sqrt{256} = 16 \text{ cm}" />
</MathContainer>

#### Finding the Diameter of the Circle

Notice that the circle is inscribed in the large square, tangent to its sides. Therefore, the diameter of the circle (<InlineMath math="d" />) is equal to the side length of the large square.

<MathContainer>
  <BlockMath math="d = s = 16 \text{ cm}" />
</MathContainer>

#### Calculating the Circumference of the Circle

The circumference of a circle (<InlineMath math="C" />) is given by <InlineMath math="C = \pi d" />.

<MathContainer>
  <BlockMath math="C = \pi \cdot 16 = 16\pi \text{ cm}" />
</MathContainer>

Therefore, the circumference of the circle is <InlineMath math="16\pi \text{ cm}" />.


---

## Exercise 2

### Question

export const metadata = {
  title: "Problem 2",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

import { Graph } from "../graph";

<Graph
    title="Geometry Visualization"
    description={
        <>
            Square <InlineMath math="ABCD" /> and a circle passing through points{" "}
            <InlineMath math="A" /> and <InlineMath math="D" />, tangent to side{" "}
            <InlineMath math="BC" />.
        </>
    }
/>

Square <InlineMath math="ABCD" /> has a side length of <InlineMath math="24 \text{ cm}" />. A circle passes through points <InlineMath math="A" /> and <InlineMath math="D" />, and is tangent to side <InlineMath math="BC" />. The area of the circle is...


### Choices

- [ ] $$144\pi\text{ cm}^2$$
- [x] $$225\pi\text{ cm}^2$$
- [ ] $$256\pi\text{ cm}^2$$
- [ ] $$336\pi\text{ cm}^2$$
- [ ] $$425\pi\text{ cm}^2$$

### Answer & Explanation

export const metadata = {
  title: "Solution to Problem 2",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

import { Graph } from "../graph";

<Graph
    mode="answer"
    title="Solution Visualization"
    description={
        <>
            Circle with center <InlineMath math="F" />, radius{" "}
            <InlineMath math="r" />, and auxiliary lines to calculate the radius.
        </>
    }
/>

Let <InlineMath math="F" /> be the center of the circle, so <InlineMath math="AF = FD = FG = r" /> (the radius of the circle).

Given the side length of the square <InlineMath math="AD = 24 \text{ cm}" />. Since <InlineMath math="E" /> is the midpoint of <InlineMath math="AD" /> (based on the symmetry of the circle passing through <InlineMath math="A" /> and <InlineMath math="D" />), then:

<MathContainer>
  <BlockMath math="AE = ED = \frac{1}{2} AD = 12 \text{ cm}" />
</MathContainer>

#### Finding the Radius of the Circle

We will find the radius of the circle (<InlineMath math="r" />) using the Pythagorean theorem on the right-angled triangle <InlineMath math="\triangle AEF" />.
Note that the length <InlineMath math="EF" /> is the distance from the center of the circle to side <InlineMath math="AD" />.
Since the circle is tangent to side <InlineMath math="BC" /> at point <InlineMath math="G" />, and the total distance from side <InlineMath math="AD" /> to side <InlineMath math="BC" /> is the side length of the square (<InlineMath math="24 \text{ cm}" />), then <InlineMath math="EF = AB - r = 24 - r" />.

Using the Pythagorean theorem on <InlineMath math="\triangle AEF" />:

<MathContainer>
  <BlockMath math="AF^2 = AE^2 + EF^2" />
  <BlockMath math="r^2 = 12^2 + (24 - r)^2" />
  <BlockMath math="r^2 = 144 + (576 - 48r + r^2)" />
  <BlockMath math="r^2 = r^2 - 48r + 720" />
  <BlockMath math="48r = 720" />
  <BlockMath math="r = \frac{720}{48} = 15 \text{ cm}" />
</MathContainer>

#### Calculating the Area of the Circle

After finding the radius <InlineMath math="r = 15 \text{ cm}" />, we can calculate the area:

<MathContainer>
  <BlockMath math="A_o = \pi r^2" />
  <BlockMath math="A_o = \pi (15)^2" />
  <BlockMath math="A_o = 225\pi \text{ cm}^2" />
</MathContainer>

Therefore, the area of the circle is <InlineMath math="225\pi \text{ cm}^2" />.


---

## Exercise 3

### Question

export const metadata = {
  title: "Problem 3",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

import { Graph } from "../graph";

<Graph
  title="Lines and Angles"
  description={
    <>
      Two parallel lines <InlineMath math="PQ" /> and <InlineMath math="RS" />{" "}
      cut by a zig-zag line forming angles <InlineMath math="25^\circ" />,{" "}
      <InlineMath math="135^\circ" />, and <InlineMath math="\alpha" />.
    </>
  }
/>

If <InlineMath math="PQ" /> is parallel to <InlineMath math="RS" />, then the value of <InlineMath math="\alpha" /> is...


### Choices

- [ ] $$45^\circ$$
- [ ] $$65^\circ$$
- [x] $$70^\circ$$
- [ ] $$75^\circ$$
- [ ] $$80^\circ$$

### Answer & Explanation

export const metadata = {
  title: "Solution to Problem 3",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

import { Graph } from "../graph";

<Graph
  mode="answer"
  title="Solution Visualization"
  description={
    <>
      Auxiliary line divides angle <InlineMath math="\alpha" /> into{" "}
      <InlineMath math="\beta" /> and <InlineMath math="\gamma" />.
    </>
  }
/>

We draw an auxiliary line parallel to lines <InlineMath math="PQ" /> and <InlineMath math="RS" /> passing through the vertex of angle <InlineMath math="\alpha" />.
This line divides angle <InlineMath math="\alpha" /> into two parts, <InlineMath math="\beta" /> and <InlineMath math="\gamma" />.

#### Finding the Value of <InlineMath math="\beta" />

Notice that angle <InlineMath math="\beta" /> and the <InlineMath math="25^\circ" /> angle are alternate interior angles.
Since the lines are parallel, their measures are equal:

<MathContainer>
  <BlockMath math="\beta = 25^\circ" />
</MathContainer>

#### Finding the Value of <InlineMath math="\gamma" />

Angle <InlineMath math="\gamma" /> and the <InlineMath math="135^\circ" /> angle are consecutive interior angles.
The sum of these two angles is <InlineMath math="180^\circ" />:

<MathContainer>
  <BlockMath math="\gamma + 135^\circ = 180^\circ" />
  <BlockMath math="\gamma = 180^\circ - 135^\circ = 45^\circ" />
</MathContainer>

#### Calculating the Value of <InlineMath math="\alpha" />

The value of <InlineMath math="\alpha" /> is the sum of <InlineMath math="\beta" /> and <InlineMath math="\gamma" />:

<MathContainer>
  <BlockMath math="\alpha = \beta + \gamma" />
  <BlockMath math="\alpha = 25^\circ + 45^\circ" />
  <BlockMath math="\alpha = 70^\circ" />
</MathContainer>

Therefore, the value of <InlineMath math="\alpha" /> is <InlineMath math="70^\circ" />.


---

## Exercise 4

### Question

export const metadata = {
    title: "Problem 4",
    authors: [{ name: "Nabil Akbarazzima Fatih" }],
    date: "12/26/2025",
};

The average test score of class <InlineMath math="A" /> is <InlineMath math="\bar{X}_A" /> and the average test score of class <InlineMath math="B" /> is <InlineMath math="\bar{X}_B" />. After the scores of both classes are combined, the average score becomes <InlineMath math="\bar{X}" />. If <InlineMath math="\bar{X}_A : \bar{X}_B = 9 : 10" /> and <InlineMath math="\bar{X} : \bar{X}_A = 84 : 81" />, then the ratio of the number of students in class <InlineMath math="A" /> to class <InlineMath math="B" /> is...


### Choices

- [ ] $$1 : 2$$
- [x] $$2 : 1$$
- [ ] $$3 : 4$$
- [ ] $$4 : 5$$
- [ ] $$4 : 3$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 4",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

Let <InlineMath math="n_A" /> be the number of students in class A and <InlineMath math="n_B" /> be the number of students in class B.

Given the ratio of the average scores of class A and class B is:

<MathContainer>
  <BlockMath math="\frac{\bar{X}_A}{\bar{X}_B} = \frac{9}{10} \iff \bar{X}_B = \frac{10}{9} \bar{X}_A" />
</MathContainer>

And the ratio of the combined average to the average of class <InlineMath math="A" /> is:

<MathContainer>
  <BlockMath math="\frac{\bar{X}}{\bar{X}_A} = \frac{84}{81} \iff \bar{X} = \frac{84}{81} \bar{X}_A" />
</MathContainer>

The ratio of the number of students in class A to class B can be found using the combined average formula:

<MathContainer>
  <BlockMath math="\bar{X} = \frac{\bar{X}_A \cdot n_A + \bar{X}_B \cdot n_B}{n_A + n_B}" />
</MathContainer>

Substitute the values of <InlineMath math="\bar{X}" /> and <InlineMath math="\bar{X}_B" /> in terms of <InlineMath math="\bar{X}_A" />:

<MathContainer>
  <BlockMath math="\frac{84}{81} \bar{X}_A = \frac{\bar{X}_A \cdot n_A + \left(\frac{10}{9} \bar{X}_A\right) \cdot n_B}{n_A + n_B}" />
</MathContainer>

Divide both sides by <InlineMath math="\bar{X}_A" /> (since <InlineMath math="\bar{X}_A \neq 0" />):

<MathContainer>
  <BlockMath math="\frac{84}{81} = \frac{n_A + \frac{10}{9} n_B}{n_A + n_B}" />
</MathContainer>

Cross-multiply:

<MathContainer>
  <BlockMath math="84(n_A + n_B) = 81\left(n_A + \frac{10}{9} n_B\right)" />
  <BlockMath math="84n_A + 84n_B = 81n_A + 90n_B" />
</MathContainer>

Group like terms:

<MathContainer>
  <BlockMath math="84n_A - 81n_A = 90n_B - 84n_B" />
  <BlockMath math="3n_A = 6n_B" />
</MathContainer>

Simplify the equation to find the ratio <InlineMath math="n_A : n_B" />:

<MathContainer>
  <BlockMath math="\frac{n_A}{n_B} = \frac{6}{3} = \frac{2}{1}" />
</MathContainer>

Therefore, the ratio of the number of students in class A to class B is **2 : 1**.


---

## Exercise 5

### Question

export const metadata = {
  title: "Problem 5",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

Inside a box there are <InlineMath math="1" /> red ball, <InlineMath math="6" /> green balls, and <InlineMath math="2" /> white balls. If <InlineMath math="7" /> balls are to be drawn without replacement, then the probability that the number of green balls drawn is twice the number of white balls drawn is...


### Choices

- [ ] $$\frac{1}{12}$$
- [ ] $$\frac{1}{4}$$
- [x] $$\frac{5}{12}$$
- [ ] $$\frac{1}{2}$$
- [ ] $$\frac{2}{3}$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 5",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

Let <InlineMath math="A" /> be the event that the number of green balls drawn is twice the number of white balls drawn.

**Given:**

There are <InlineMath math="9" /> balls in a box (<InlineMath math="1" /> red + <InlineMath math="6" /> green + <InlineMath math="2" /> white = <InlineMath math="9" />).
<InlineMath math="7" /> balls will be drawn at random. The total number of ways to draw the balls is:

<MathContainer>
  <BlockMath math="n(S) = C^9_7 = \frac{9!}{7!(9-7)!} = \frac{9 \times 8}{2 \times 1} = 36" />
</MathContainer>

We want to find the probability that the number of green balls is twice the number of white balls. We divide this into two cases.

**Case 1**

If <InlineMath math="1" /> white ball is drawn, then there must be <InlineMath math="2" /> green balls (since green = 2 times white).
To reach a total of <InlineMath math="7" /> balls drawn, the remaining balls to be drawn must be <InlineMath math="7 - 1 - 2 = 4" /> balls. These must be red balls.
However, there is only <InlineMath math="1" /> red ball available. Therefore, **Case 1 is impossible**.

**Case <InlineMath math="2" />**

If <InlineMath math="2" /> white balls are drawn, then there must be <InlineMath math="4" /> green balls.
To reach a total of <InlineMath math="7" /> balls drawn, the remaining balls to be drawn must be <InlineMath math="7 - 2 - 4 = 1" /> ball. This must be a red ball.
This case is possible because there are <InlineMath math="2" /> white balls, <InlineMath math="6" /> green balls, and <InlineMath math="1" /> red ball available.

The number of ways to draw for this case is:

<MathContainer>
  <BlockMath math="n(A) = C^2_2 \cdot C^6_4 \cdot C^1_1" />
  <BlockMath math="= 1 \cdot 15 \cdot 1" />
  <BlockMath math="= 15" />
</MathContainer>

**Conclusion**

Thus, the probability of event <InlineMath math="A" /> is:

<MathContainer>
  <BlockMath math="P(A) = \frac{n(A)}{n(S)} = \frac{15}{36} = \frac{5}{12}" />
</MathContainer>

Therefore, the probability that the number of green balls drawn is twice the number of white balls drawn is <InlineMath math="\frac{5}{12}" />.


---

## Exercise 6

### Question

export const metadata = {
  title: "Problem 6",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

If <InlineMath math="1 < x < y < 2" /> and <InlineMath math="z = x - y" />, then <InlineMath math="z" /> is between the values...


### Choices

- [ ] $$0$$ and $$2$$
- [ ] $$1$$ and $$2$$
- [x] $$-1$$ and $$0$$
- [ ] $$-2$$ and $$2$$
- [ ] $$-2$$ and $$1$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 6",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

Given <InlineMath math="1 < x < y < 2" />. We want to find the range of values for <InlineMath math="z = x - y" />.

Since <InlineMath math="x < y" />, the difference <InlineMath math="x - y" /> must be negative (<InlineMath math="z < 0" />).

Let's analyze the limits of <InlineMath math="z" />:

1.  **Maximum value of <InlineMath math="z" />**

    The value of <InlineMath math="z" /> will be maximum (closest to <InlineMath math="0" />) when the difference between <InlineMath math="x" /> and <InlineMath math="y" /> is as small as possible.

    Since <InlineMath math="x" /> and <InlineMath math="y" /> are in the interval <InlineMath math="(1, 2)" /> and <InlineMath math="x < y" />, their difference can be arbitrarily close to <InlineMath math="0" /> but never reaches <InlineMath math="0" /> (because <InlineMath math="x \neq y" />).

    Thus, the upper bound of <InlineMath math="z" /> is <InlineMath math="0" />.

2.  **Minimum value of <InlineMath math="z" />**

    The value of <InlineMath math="z" /> will be minimum (most negative) when <InlineMath math="x" /> is as small as possible and <InlineMath math="y" /> is as large as possible.

    -   <InlineMath math="x" /> approaches <InlineMath math="1" /> (lower bound).
    -   <InlineMath math="y" /> approaches <InlineMath math="2" /> (upper bound).

    Therefore:

    <MathContainer>
      <BlockMath math="z \approx 1 - 2 = -1" />
    </MathContainer>

    Thus, the lower bound of <InlineMath math="z" /> is <InlineMath math="-1" />.

**Conclusion**

The value of <InlineMath math="z" /> is between <InlineMath math="-1" /> and <InlineMath math="0" />.

In mathematical notation: <InlineMath math="-1 < z < 0" />.

Therefore, <InlineMath math="z" /> is between the values <InlineMath math="-1" /> and <InlineMath math="0" />.


---

## Exercise 7

### Question

export const metadata = {
  title: "Problem 7",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

Seven years ago, the sum of the ages of the older sibling and the younger sibling was <InlineMath math="26" /> years. Four years from now, twice the age of the older sibling will be equal to twice the age of the younger sibling plus eight years.

The current age of the younger sibling is...


### Choices

- [ ] $$14$$ years
- [ ] $$17$$ years
- [x] $$18$$ years
- [ ] $$20$$ years
- [ ] $$22$$ years

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 7",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

Let:
-   <InlineMath math="a =" /> current age of the younger sibling
-   <InlineMath math="k =" /> current age of the older sibling

#### First Condition Analysis

Seven years ago, the sum of the ages of the older sibling and the younger sibling was <InlineMath math="26" /> years.

The mathematical model is:

<MathContainer>
  <BlockMath math="(k - 7) + (a - 7) = 26" />
  <BlockMath math="k + a - 14 = 26" />
  <BlockMath math="k + a = 40 \quad \dots (1)" />
</MathContainer>

#### Second Condition Analysis

Four years from now, twice the age of the older sibling is equal to twice the age of the younger sibling plus eight years.

The mathematical model is:

<MathContainer>
  <BlockMath math="2(k + 4) = 2(a + 4) + 8" />
  <BlockMath math="2k + 8 = 2a + 8 + 8" />
  <BlockMath math="2k - 2a = 8" />
  <BlockMath math="k - a = 4 \quad \dots (2)" />
</MathContainer>

#### Finding the Younger Sibling's Age

We eliminate equation <InlineMath math="(1)" /> and <InlineMath math="(2)" /> to find the value of <InlineMath math="a" />.

<MathContainer>
  <BlockMath math="\begin{aligned} k + a &= 40 \\ k - a &= 4 \\ \hline 2a &= 36 \\ a &= 18 \end{aligned}" />
</MathContainer>

> We subtract the second equation from the first: <InlineMath math="(k+a) - (k-a) = 2a" /> and <InlineMath math="40 - 4 = 36" />.

Thus, the current age of the younger sibling is <InlineMath math="18" /> years.

Therefore, the younger sibling is <InlineMath math="18" /> years old.


---

## Exercise 8

### Question

export const metadata = {
  title: "Problem 8",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

<MathContainer>
  <BlockMath math="\frac{(8 + \sqrt{5}) + (2\sqrt{5} + 1)}{\sqrt{5} + 3} = \dots" />
</MathContainer>


### Choices

- [ ] $$2$$
- [x] $$3$$
- [ ] $$5$$
- [ ] $$10$$
- [ ] $$14$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 8",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

#### Simplifying Algebraic Expression

We simplify the numerator by grouping like terms (integers with integers, radicals with radicals).

<MathContainer>
  <BlockMath math="\begin{aligned} \frac{(8 + \sqrt{5}) + (2\sqrt{5} + 1)}{\sqrt{5} + 3} &= \frac{(8 + 1) + (\sqrt{5} + 2\sqrt{5})}{\sqrt{5} + 3} \\ &= \frac{9 + 3\sqrt{5}}{\sqrt{5} + 3} \end{aligned}" />
</MathContainer>

Next, we factor the numerator by taking out the number <InlineMath math="3" />:

<MathContainer>
  <BlockMath math="\begin{aligned} \frac{9 + 3\sqrt{5}}{\sqrt{5} + 3} &= \frac{3(3 + \sqrt{5})}{\sqrt{5} + 3} \end{aligned}" />
</MathContainer>

Notice that <InlineMath math="(3 + \sqrt{5})" /> is equal to <InlineMath math="(\sqrt{5} + 3)" />. Thus, we can divide both the numerator and the denominator by this value:

<MathContainer>
  <BlockMath math="\frac{3(3 + \sqrt{5})}{\sqrt{5} + 3} = 3" />
</MathContainer>

Therefore, the final result is <InlineMath math="3" />.


---

## Exercise 9

### Question

export const metadata = {
  title: "Problem 9",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

For the function <InlineMath math="f(x) = \frac{\sqrt{5 - x}}{\sqrt{x^2 + x - 6}}" /> to be defined in its domain, the domain of function <InlineMath math="f" /> is...


### Choices

- [ ] $$Df = \{x | x \leq 5\}$$
- [ ] $$Df = \{x | 2 < x \leq 5\}$$
- [ ] $$Df = \{x | x < -3 \text{ or } 2 < x < 5\}$$
- [x] $$Df = \{x | x < -3 \text{ or } 2 < x \leq 5\}$$
- [ ] $$Df = \{x | x < -3 \text{ or } 2 \leq x \leq 5\}$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 9",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

To find the domain of the function, we need to ensure that all components of the function are defined (values inside square roots must be non-negative, and the denominator cannot be zero).

The function <InlineMath math="f(x) = \frac{\sqrt{5 - x}}{\sqrt{x^2 + x - 6}}" /> has two main conditions:

#### Condition 1: Numerator

The value inside the square root in the numerator must be non-negative:

<MathContainer>
  <BlockMath math="5 - x \geq 0" />
  <BlockMath math="-x \geq -5" />
  <BlockMath math="x \leq 5" />
</MathContainer>

#### Condition 2: Denominator

The value inside the square root in the denominator must be positive (since the square root in the denominator cannot be zero):

<MathContainer>
  <BlockMath math="x^2 + x - 6 > 0" />
  <BlockMath math="(x + 3)(x - 2) > 0" />
</MathContainer>

The zeros of this inequality are <InlineMath math="x = -3" /> and <InlineMath math="x = 2" />.

Using test points, we find the solution region for the denominator: <InlineMath math="x < -3" /> or <InlineMath math="x > 2" />.

#### Domain Intersection

The domain of function <InlineMath math="f" /> is the intersection of the two conditions above:
1.  <InlineMath math="x \leq 5" />
2.  <InlineMath math="x < -3" /> or <InlineMath math="x > 2" />

The intersection is:
-   For <InlineMath math="x < -3" />, it satisfies <InlineMath math="x \leq 5" />.
-   For <InlineMath math="x > 2" />, it is bounded by <InlineMath math="x \leq 5" />, becoming <InlineMath math="2 < x \leq 5" />.

Therefore, the domain of function <InlineMath math="f" /> is:

<MathContainer>
  <BlockMath math="Df = \{x | x < -3 \text{ or } 2 < x \leq 5\}" />
</MathContainer>


---

## Exercise 10

### Question

export const metadata = {
  title: "Problem 10",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

If the ratio of the volumes of two cubes is <InlineMath math="1 : 27" />, then the ratio of the surface areas of the two cubes is...


### Choices

- [ ] $$1 : 3$$
- [ ] $$1 : 4$$
- [ ] $$1 : 8$$
- [x] $$1 : 9$$
- [ ] $$2 : 3$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 10",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

#### Finding the Side Length Ratio

The volume of a cube (<InlineMath math="V" />) is directly proportional to the cube of its side length (<InlineMath math="s" />). Given that the volume ratio is <InlineMath math="1 : 27" />.

<MathContainer>
  <BlockMath math="\frac{V_1}{V_2} = \left(\frac{s_1}{s_2}\right)^3" />
  <BlockMath math="\frac{1}{27} = \left(\frac{s_1}{s_2}\right)^3" />
  <BlockMath math="\frac{s_1}{s_2} = \sqrt[3]{\frac{1}{27}}" />
  <BlockMath math="\frac{s_1}{s_2} = \frac{1}{3}" />
</MathContainer>

So, the ratio of the side lengths of the two cubes is <InlineMath math="1 : 3" />.

#### Finding the Surface Area Ratio

The surface area of a cube (<InlineMath math="L" />) is directly proportional to the square of its side length.

<MathContainer>
  <BlockMath math="\frac{L_1}{L_2} = \frac{6s_1^2}{6s_2^2}" />
  <BlockMath math="\frac{L_1}{L_2} = \left(\frac{s_1}{s_2}\right)^2" />
  <BlockMath math="\frac{L_1}{L_2} = \left(\frac{1}{3}\right)^2" />
  <BlockMath math="\frac{L_1}{L_2} = \frac{1}{9}" />
</MathContainer>

Thus, the ratio of the surface areas of the two cubes is <InlineMath math="1 : 9" />.


---

## Exercise 11

### Question

export const metadata = {
  title: "Problem 11",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

The probabilities of Ali, Bima, and Dika passing the National Exam are <InlineMath math="0.75" />, <InlineMath math="0.8" />, and <InlineMath math="0.85" /> respectively.

The probability that only one person passes among the three is...


### Choices

- [ ] $$0.085$$
- [x] $$0.095$$
- [ ] $$0.85$$
- [ ] $$0.95$$
- [ ] $$0.075$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 11",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

Let:
-   <InlineMath math="P(A)" /> = Probability Ali passes
-   <InlineMath math="P(B)" /> = Probability Bima passes
-   <InlineMath math="P(D)" /> = Probability Dika passes

Given:

<MathContainer>
  <BlockMath math="P(A) = 0.75 \rightarrow P(A^c) = 1 - 0.75 = 0.25" />
  <BlockMath math="P(B) = 0.8 \rightarrow P(B^c) = 1 - 0.8 = 0.2" />
  <BlockMath math="P(D) = 0.85 \rightarrow P(D^c) = 1 - 0.85 = 0.15" />
</MathContainer>

If only one person passes, there are three mutually exclusive possibilities:

#### Ali passes, Bima and Dika fail

<MathContainer>
  <BlockMath math="P(A) \cdot P(B^c) \cdot P(D^c) = 0.75 \cdot 0.2 \cdot 0.15" />
  <BlockMath math="= 0.0225" />
</MathContainer>

#### 2. Bima passes, Ali and Dika fail

<MathContainer>
  <BlockMath math="P(B) \cdot P(A^c) \cdot P(D^c) = 0.8 \cdot 0.25 \cdot 0.15" />
  <BlockMath math="= 0.03" />
</MathContainer>

#### Dika passes, Ali and Bima fail

<MathContainer>
  <BlockMath math="P(D) \cdot P(A^c) \cdot P(B^c) = 0.85 \cdot 0.25 \cdot 0.2" />
  <BlockMath math="= 0.0425" />
</MathContainer>

#### Total Probability

Thus, the probability that only one person passes is the sum of these three possibilities:

<MathContainer>
  <BlockMath math="0.0225 + 0.03 + 0.0425 = 0.095" />
</MathContainer>


---

## Exercise 12

### Question

export const metadata = {
  title: "Problem 12",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

A team consists of <InlineMath math="12" /> people who can finish a job in <InlineMath math="15" /> days. If the job needs to be completed faster in <InlineMath math="9" /> days, the percentage increase in workers needed to finish the job is...


### Choices

- [ ] $$25.0\%$$
- [ ] $$37.5\%$$
- [ ] $$50.0\%$$
- [ ] $$62.5\%$$
- [x] $$66.7\%$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 12",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

This problem can be solved using inverse proportion.

Given:

<MathContainer>
  <BlockMath math="12 \text{ people} \rightarrow 15 \text{ days}" />
  <BlockMath math="x \text{ people} \rightarrow 9 \text{ days}" />
</MathContainer>

#### Calculating the Number of Workers Needed

Thus, we can write:

<MathContainer>
  <BlockMath math="\frac{12}{x} = \frac{9}{15}" />
  <BlockMath math="x = \frac{12 \cdot 15}{9}" />
  <BlockMath math="x = 20" />
</MathContainer>

So, the number of workers needed to finish in <InlineMath math="9" /> days is <InlineMath math="20" /> workers.

#### Calculating the Percentage Increase

Thus, the percentage increase in workers is:

<MathContainer>
  <BlockMath math="\frac{20 - 12}{12} \times 100\% = \frac{8}{12} \times 100\% = 66.7\%" />
</MathContainer>


---

## Exercise 13

### Question

export const metadata = {
  title: "Problem 13",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

A set of data has an average of <InlineMath math="24" /> and a range of <InlineMath math="12" />. If each data value is subtracted by <InlineMath math="a" /> and then the result is divided by <InlineMath math="b" />, it produces new data with an average of <InlineMath math="10" /> and a range of <InlineMath math="6" />. The values of <InlineMath math="a" /> and <InlineMath math="b" /> are...


### Choices

- [ ] $$14 \text{ and } 2$$
- [ ] $$12 \text{ and } 2$$
- [ ] $$8 \text{ and } 2$$
- [x] $$4 \text{ and } 2$$
- [ ] $$2 \text{ and } 2$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 13",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

#### Properties of Average and Range

-   **Average**: Its value changes if each data point is added, subtracted, multiplied, or divided.
-   **Range**: Its value **only** changes if each data point is multiplied or divided (it does not change if added or subtracted).

#### Finding the Value of <InlineMath math="b" />

The initial range is <InlineMath math="12" />. Then, each data value is subtracted by <InlineMath math="a" /> and divided by <InlineMath math="b" />. Since the range only changes when multiplied or divided, the new range becomes:

<MathContainer>
  <BlockMath math="j_{new} = \frac{j_{initial}}{b}" />
  <BlockMath math="6 = \frac{12}{b}" />
  <BlockMath math="b = \frac{12}{6}" />
  <BlockMath math="b = 2" />
</MathContainer>

#### Finding the Value of <InlineMath math="a" />

The initial average is <InlineMath math="24" />. Then, each data value is subtracted by <InlineMath math="a" /> and divided by <InlineMath math="b" />, so the new average becomes:

<MathContainer>
  <BlockMath math="\bar{x}_{new} = \frac{\bar{x}_{initial} - a}{b}" />
  <BlockMath math="10 = \frac{24 - a}{b}" />
</MathContainer>

Substitute the value <InlineMath math="b = 2" />:

<MathContainer>
  <BlockMath math="10 = \frac{24 - a}{2}" />
  <BlockMath math="20 = 24 - a" />
  <BlockMath math="a = 24 - 20" />
  <BlockMath math="a = 4" />
</MathContainer>

Thus, the values are <InlineMath math="a = 4" /> and <InlineMath math="b = 2" />.


---

## Exercise 14

### Question

export const metadata = {
  title: "Problem 14",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

The correct order of the numbers below from smallest to largest is...

<MathContainer>
  <BlockMath math="1.45; \frac{17}{14}; 123\%; \sqrt{12}; \frac{5}{3}" />
</MathContainer>


### Choices

- [x] $$\frac{17}{14}; 123\%; 1.45; \frac{5}{3}; \sqrt{12}$$
- [ ] $$\frac{17}{14}; 123\%; 1.45; \sqrt{12}; \frac{5}{3}$$
- [ ] $$\frac{5}{3}; \frac{17}{14}; 123\%; 1.45; \sqrt{12}$$
- [ ] $$\frac{5}{3}; 123\%; 1.45; \sqrt{12}; \frac{17}{14}$$
- [ ] $$123\%; \frac{5}{3}; 1.45; \sqrt{12}; \frac{17}{14}$$

### Answer & Explanation

export const metadata = {
  title: "Solution 14",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

#### Converting to Fractions

To determine the order from smallest to largest, we convert all numbers into fractions with a common denominator.

<MathContainer>
  <BlockMath math="1.45 = \frac{145}{100}" />
  <BlockMath math="\frac{17}{14}" />
  <BlockMath math="123\% = \frac{123}{100}" />
  <BlockMath math="\frac{5}{3}" />
  <BlockMath math="\sqrt{12} \approx 3.46" />
</MathContainer>

#### Finding a Common Denominator

We use <InlineMath math="2100" /> as the common denominator (LCM of <InlineMath math="100" />, <InlineMath math="14" />, and <InlineMath math="3" />).

<MathContainer>
  <BlockMath math="1.45 = \frac{145 \times 21}{2100} = \frac{3045}{2100}" />
  <BlockMath math="\frac{17}{14} = \frac{17 \times 150}{2100} = \frac{2550}{2100}" />
  <BlockMath math="123\% = \frac{123 \times 21}{2100} = \frac{2583}{2100}" />
  <BlockMath math="\frac{5}{3} = \frac{5 \times 700}{2100} = \frac{3500}{2100}" />
</MathContainer>

For <InlineMath math="\sqrt{12}" />, since its value is approximately <InlineMath math="3.46" />, if converted to a fraction over 2100, it would be very large (much larger than the others which are all below 2).

<MathContainer>
  <BlockMath math="\sqrt{12} \approx \frac{7274}{2100}" />
</MathContainer>

#### Conclusion

Based on the numerators with the common denominator:
<MathContainer>
  <BlockMath math="2550 < 2583 < 3045 < 3500 < 7274" />
</MathContainer>

So the order is:
<MathContainer>
  <BlockMath math="\frac{2550}{2100}; \frac{2583}{2100}; \frac{3045}{2100}; \frac{3500}{2100}; \sqrt{12}" />
</MathContainer>

Returning to the original forms:
<MathContainer>
  <BlockMath math="\frac{17}{14}; 123\%; 1.45; \frac{5}{3}; \sqrt{12}" />
</MathContainer>


---

## Exercise 15

### Question

export const metadata = {
  title: "Problem 15",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

If <InlineMath math="x" /> is <InlineMath math="\frac{3}{4}" /> of <InlineMath math="34" /> and <InlineMath math="y" /> is <InlineMath math="46\%" /> of <InlineMath math="86" />, then...


### Choices

- [ ] $$x > y$$
- [x] $$x < y$$
- [ ] $$x = y$$
- [ ] $$2x = y$$
- [ ] $$x = 4y$$

### Answer & Explanation

export const metadata = {
  title: "Solution 15",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

#### Calculating the Value of <InlineMath math="x" />

Given <InlineMath math="x = \frac{3}{4}" /> of <InlineMath math="34" />, which means:

<MathContainer>
  <BlockMath math="x = \frac{3}{4} \times 34" />
  <BlockMath math="x = 25.5" />
</MathContainer>

#### Calculating the Value of <InlineMath math="y" />

Given <InlineMath math="y = 46\%" /> of <InlineMath math="86" />, then:

<MathContainer>
  <BlockMath math="y = 46\% \times 86" />
  <BlockMath math="y = \frac{46}{100} \times 86" />
  <BlockMath math="y = 39.56" />
</MathContainer>

#### Conclusion

Based on the calculations above, we get <InlineMath math="x = 25.5" /> and <InlineMath math="y = 39.56" />.

Consequently, <InlineMath math="x < y" />.


---

## Exercise 16

### Question

export const metadata = {
    title: "Question 16",
    authors: [{ name: "Nabil Akbarazzima Fatih" }],
    date: "12/26/2025",
};

The fractional form of the repeating decimal <InlineMath math="3.15931593..." /> is?


### Choices

- [ ] $$\frac{1170}{37}$$
- [x] $$\frac{3510}{1111}$$
- [ ] $$\frac{1177}{111}$$
- [ ] $$\frac{1170}{999}$$
- [ ] $$\frac{1177}{999}$$

### Answer & Explanation

export const metadata = {
  title: "Solution 16",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

Let <InlineMath math="x = 3.15931593..." />.

Since there are <InlineMath math="4" /> repeating digits (which are <InlineMath math="1593" />), we multiply <InlineMath math="x" /> by <InlineMath math="10000" />:

<MathContainer>
  <BlockMath math="10000x = 31593.15931593..." />
</MathContainer>

Next, we subtract the original equation <InlineMath math="x" /> from this equation:

<MathContainer>
  <BlockMath math="\begin{aligned} 10000x &= 31593.1593... \\ x &= \phantom{3159}3.1593... \\ \hline 9999x &= 31590 \end{aligned}" />
</MathContainer>

Thus, we obtain:

<MathContainer>
  <BlockMath math="x = \frac{31590}{9999}" />
</MathContainer>

We simplify the fraction by dividing both the numerator and the denominator by <InlineMath math="9" />:

<MathContainer>
  <BlockMath math="x = \frac{31590 \div 9}{9999 \div 9} = \frac{3510}{1111}" />
</MathContainer>


---

## Exercise 17

### Question

export const metadata = {
    title: "Question 17",
    authors: [{ name: "Nabil Akbarazzima Fatih" }],
    date: "12/26/2025",
};

A rectangle with length <InlineMath math="p" /> and width <InlineMath math="l" /> has an area of <InlineMath math="32" />. If the perimeter of the rectangle is <InlineMath math="6l" />, what is the value of <InlineMath math="p" />?


### Choices

- [ ] $$4$$
- [ ] $$5$$
- [ ] $$6$$
- [ ] $$7$$
- [x] $$8$$

### Answer & Explanation

export const metadata = {
  title: "Solution 17",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

#### Finding the Relation between <InlineMath math="p" /> and <InlineMath math="l" />

Given that the perimeter of the rectangle is <InlineMath math="6l" />. The formula for the perimeter of a rectangle is <InlineMath math="2(p + l)" /> or <InlineMath math="2p + 2l" />. Thus:

<MathContainer>
  <BlockMath math="\begin{aligned} 6l &= 2p + 2l \\ 6l - 2l &= 2p \\ 4l &= 2p \\ 2l &= p \end{aligned}" />
</MathContainer>

#### Finding the Value of <InlineMath math="l" />

Next, it is known that the area of the rectangle is <InlineMath math="32" />. The formula for the area is <InlineMath math="p \times l" />. We substitute <InlineMath math="p = 2l" /> into the area formula:

<MathContainer>
  <BlockMath math="\begin{aligned} \text{Area} &= 32 \\ p \times l &= 32 \\ (2l) \times l &= 32 \\ 2l^2 &= 32 \\ l^2 &= 16 \\ l &= 4 \end{aligned}" />
</MathContainer>

#### Calculating the Value of <InlineMath math="p" />

After obtaining the value <InlineMath math="l = 4" />, we substitute it back into the equation <InlineMath math="p = 2l" />:

<MathContainer>
  <BlockMath math="p = 2(4) = 8" />
</MathContainer>

So, the value of <InlineMath math="p" /> is <InlineMath math="8" />.


---

## Exercise 18

### Question

export const metadata = {
  title: "Question 18",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

A rectangular prism has a volume of <InlineMath math="12x^2" />, length <InlineMath math="3x" />, and width <InlineMath math="2x" />. What is the height of the rectangular prism?


### Choices

- [ ] $$1$$
- [x] $$2$$
- [ ] $$3$$
- [ ] $$4$$
- [ ] $$5$$

### Answer & Explanation

export const metadata = {
  title: "Solution 18",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

Given the volume of the rectangular prism is <InlineMath math="V = 12x^2" />, the length is <InlineMath math="p = 3x" />, and the width is <InlineMath math="l = 2x" />.

The formula for the volume of a rectangular prism is:

<MathContainer>
  <BlockMath math="V = p \times l \times t" />
</MathContainer>

We substitute the known values into the formula:

<MathContainer>
  <BlockMath math="\begin{aligned} 12x^2 &= (3x) \times (2x) \times t \\ 12x^2 &= 6x^2 \times t \end{aligned}" />
</MathContainer>

Next, we solve for <InlineMath math="t" />:

<MathContainer>
  <BlockMath math="t = \frac{12x^2}{6x^2} = 2" />
</MathContainer>

So, the height of the rectangular prism is <InlineMath math="2" />.


---

## Exercise 19

### Question

export const metadata = {
  title: "Question 19",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

The average biology score in a class of <InlineMath math="50" /> students is <InlineMath math="70" />. If the average score of male students is <InlineMath math="76" /> and the number of female students is <InlineMath math="4" /> more than the number of male students, then the average score of female students in the class is...


### Choices

- [ ] $$64$$
- [x] $$64.88$$
- [ ] $$65.09$$
- [ ] $$65.20$$
- [ ] $$65.34$$

### Answer & Explanation

export const metadata = {
  title: "Solution 19",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

#### Finding the Number of Male and Female Students

It is known that the total number of students is <InlineMath math="50" />.
Let <InlineMath math="n_L" /> be the number of male students and <InlineMath math="n_P" /> be the number of female students.
It is also known that the number of female students is <InlineMath math="4" /> more than the number of male students, so:

<MathContainer>
  <BlockMath math="n_P = n_L + 4" />
</MathContainer>

Since the total number of students is <InlineMath math="50" />, then:

<MathContainer>
  <BlockMath math="\begin{aligned} n_L + n_P &= 50 \\ n_L + (n_L + 4) &= 50 \\ 2n_L + 4 &= 50 \\ 2n_L &= 46 \\ n_L &= 23 \end{aligned}" />
</MathContainer>

So the number of female students is:

<MathContainer>
  <BlockMath math="n_P = 23 + 4 = 27" />
</MathContainer>

#### Calculating Total Scores

Given that the average score of male students (<InlineMath math="\bar{x}_L" />) is <InlineMath math="76" />. Then the total score of male students is:

<MathContainer>
  <BlockMath math="\sum x_L = n_L \times \bar{x}_L = 23 \times 76 = 1748" />
</MathContainer>

Given that the average score of the entire class (<InlineMath math="\bar{x}_{total}" />) is <InlineMath math="70" />. Then the total score of the entire class is:

<MathContainer>
  <BlockMath math="\sum x_{total} = 50 \times 70 = 3500" />
</MathContainer>

#### Calculating the Average Score of Female Students

The total score of female students can be found by subtracting the total score of male students from the total score of the class:

<MathContainer>
  <BlockMath math="\sum x_P = 3500 - 1748 = 1752" />
</MathContainer>

Thus, the average score of female students (<InlineMath math="\bar{x}_P" />) is:

<MathContainer>
  <BlockMath math="\bar{x}_P = \frac{\sum x_P}{n_P} = \frac{1752}{27} \approx 64.88" />
</MathContainer>

So, the average score of female students is <InlineMath math="64.88" />.


---

## Exercise 20

### Question

export const metadata = {
    title: "Problem 20",
    authors: [{ name: "Nabil Akbarazzima Fatih" }],
    date: "12/26/2025",
};

In a box, there are <InlineMath math="8" /> yellow balls and <InlineMath math="3" /> blue balls. If two balls are taken simultaneously at random, determine the expected frequency of obtaining two blue balls from <InlineMath math="110" /> trials!


### Choices

- [ ] $$3$$
- [ ] $$5$$
- [x] $$6$$
- [ ] $$9$$
- [ ] $$11$$

### Answer & Explanation

export const metadata = {
  title: "Solution 20",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

First, we calculate the probability of picking two blue balls. The total number of balls in the box is the sum of yellow and blue balls.

<MathContainer>
  <BlockMath math="n(S) = 8 + 3 = 11" />
</MathContainer>

The number of ways to choose <InlineMath math="2" /> balls from a total of <InlineMath math="11" /> balls can be calculated using combinations.

<MathContainer>
  <BlockMath math="C^{11}_2 = \frac{11!}{2!(11-2)!} = \frac{11 \times 10 \times 9!}{2 \times 1 \times 9!} = \frac{110}{2} = 55" />
</MathContainer>

Next, we calculate the number of ways to choose <InlineMath math="2" /> blue balls from the <InlineMath math="3" /> available blue balls.

<MathContainer>
  <BlockMath math="n(A) = C^3_2 = \frac{3!}{2!(3-2)!} = \frac{3 \times 2!}{2! \times 1!} = 3" />
</MathContainer>

The probability of obtaining two blue balls (<InlineMath math="P(2B)" />) is the ratio of the number of ways to pick two blue balls to the total number of ways to pick any two balls.

<MathContainer>
  <BlockMath math="P(2B) = \frac{n(A)}{n(S)} = \frac{3}{55}" />
</MathContainer>

The expected frequency (<InlineMath math="F_h" />) is the product of the probability of the event and the number of trials. The number of trials is <InlineMath math="110" />.

<MathContainer>
  <BlockMath math="F_h = P(2B) \times N" />
  <BlockMath math="F_h = \frac{3}{55} \times 110" />
  <BlockMath math="F_h = 3 \times 2 = 6" />
</MathContainer>

Thus, the expected frequency of obtaining two blue balls is <InlineMath math="6" />.


---