# Nakafa Framework: LLM

URL: https://nakafa.com/en/exercises/high-school/snbt/quantitative-knowledge/try-out/set-8

Exercises: Try Out - Set 8: 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: "12/25/2025",
};

The common fraction form of the repeating decimal <InlineMath math="0.323323323..." /> is...


### Choices

- [ ] $$\frac{1}{3}$$
- [ ] $$\frac{32}{99}$$
- [ ] $$\frac{23}{99}$$
- [ ] $$\frac{232}{999}$$
- [x] $$\frac{323}{999}$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 1",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

Let <InlineMath math="x = 0.323323323..." />. Since there are three repeating digits (<InlineMath math="323" />), multiply both sides by <InlineMath math="1000" />.

<MathContainer>
  <BlockMath math="1000x = 323.323323..." />
</MathContainer>

Next, subtract the original equation <InlineMath math="x = 0.323323..." /> from this equation:

<MathContainer>
  <BlockMath math="\begin{aligned} 1000x &= 323.323323... \\ x &= 0.323323... \\ \hline 999x &= 323 \end{aligned}" />
</MathContainer>

Thus, we get:

<MathContainer>
  <BlockMath math="x = \frac{323}{999}" />
</MathContainer>

So, the common fraction form of the number is <InlineMath math="\frac{323}{999}" />.


---

## Exercise 2

### Question

export const metadata = {
  title: "Problem 2",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

<InlineMath math="42, 40, 39, 36, 34, 30, 27, 22, ..., ..." />


### Choices

- [ ] $$21, 15$$
- [ ] $$20, 14$$
- [ ] $$19, 13$$
- [x] $$18, 12$$
- [ ] $$18, 11$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 2",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

The sequence above consists of two alternating sub-sequences (skipping one number).

#### Pattern 1 (Odd positions)

<InlineMath math="42, 39, 34, 27, ..." />

The differences between consecutive terms are:

<BlockMath math="\begin{aligned} 39 - 42 &= -3 \\ 34 - 39 &= -5 \\ 27 - 34 &= -7 \end{aligned}" />

The pattern is subtracting consecutive odd numbers (<InlineMath math="-3, -5, -7" />). Thus, the next difference is <InlineMath math="-9" />. Next term: <InlineMath math="27 - 9 = 18" />.

#### Pattern 2 (Even positions)

<InlineMath math="40, 36, 30, 22, ..." />

The differences between consecutive terms are:

<BlockMath math="\begin{aligned} 36 - 40 &= -4 \\ 30 - 36 &= -6 \\ 22 - 30 &= -8 \end{aligned}" />

The pattern is subtracting consecutive even numbers (<InlineMath math="-4, -6, -8" />). Thus, the next difference is <InlineMath math="-10" />. Next term: <InlineMath math="22 - 10 = 12" />.

So, the next two numbers are <InlineMath math="18" /> and <InlineMath math="12" />.


---

## Exercise 3

### Question

export const metadata = {
  title: "Problem 3",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

If <InlineMath math="A" /> is <InlineMath math="22\frac{1}{4}\%" /> of <InlineMath math="25" /> and <InlineMath math="B" /> is <InlineMath math="\frac{1}{8}" /> of <InlineMath math="32" />, then...


### Choices

- [ ] $$A = B$$
- [ ] $$A = 2B$$
- [x] $$A > B$$
- [ ] $$A < B$$
- [ ] $$A = \frac{1}{2}B$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 3",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

Note that:

<MathContainer>
  <BlockMath math="22\frac{1}{4}\% = 22.25\%" />
</MathContainer>

Then the value of <InlineMath math="A" /> is:

<MathContainer>
  <BlockMath math="A = \frac{22.25}{100} \times 25 = \frac{22.25}{4} = 5.5625" />
</MathContainer>

And the value of <InlineMath math="B" /> is:

<MathContainer>
  <BlockMath math="B = \frac{1}{8} \times 32 = 4" />
</MathContainer>

Thus, <InlineMath math="A > B" />.


---

## Exercise 4

### Question

export const metadata = {
  title: "Problem 4",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

<InlineMath math="x_0" /> is the average of data <InlineMath math="x_1, x_2, ..., x_{10}" />. If the data changes following the pattern <InlineMath math="\frac{x_1}{2} + 3, \frac{x_2}{2} + 5, \frac{x_3}{2} + 7" />, and so on, then the average value becomes...


### Choices

- [ ] $$x_0 + 11$$
- [ ] $$x_0 + 12$$
- [ ] $$\frac{1}{2}x_0 + 11$$
- [x] $$\frac{1}{2}x_0 + 12$$
- [ ] $$\frac{1}{2}x_0 + 21$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 4",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

The new data follows the pattern:

<MathContainer>
  <BlockMath math="\frac{x_1}{2} + 3, \frac{x_2}{2} + 5, \frac{x_3}{2} + 7, ..." />
</MathContainer>

The new average can be obtained by:

<MathContainer>
  <BlockMath math="\bar{x} = \frac{\left(\frac{x_1}{2} + 3\right) + \left(\frac{x_2}{2} + 5\right) + \left(\frac{x_3}{2} + 7\right) + ... + \left(\frac{x_{10}}{2} + (2(10)+1)\right)}{10}" />
</MathContainer>

First, let's find the sum of the arithmetic series <InlineMath math="3, 5, 7, ..." />. This sequence has the first term <InlineMath math="a = 3" /> and difference <InlineMath math="b = 2" />. The 10th term is:

<MathContainer>
  <BlockMath math="U_{10} = 3 + 9(2) = 21" />
</MathContainer>

The sum of the first 10 terms is:

<MathContainer>
  <BlockMath math="S_{10} = \frac{10}{2}(3 + 21) = 5(24) = 120" />
</MathContainer>

The new average can be written as:

<MathContainer>
  <BlockMath math="\begin{aligned} \bar{x} &= \frac{\frac{1}{2}(x_1 + x_2 + ... + x_{10}) + 120}{10} \\ &= \frac{1}{2}\frac{(x_1 + x_2 + ... + x_{10})}{10} + \frac{120}{10} \\ &= \frac{1}{2}x_0 + 12 \end{aligned}" />
</MathContainer>

Thus, the new average is <InlineMath math="\frac{1}{2}x_0 + 12" />.


---

## Exercise 5

### Question

export const metadata = {
    title: "Problem 5",
    authors: [{ name: "Nabil Akbarazzima Fatih" }],
    date: "12/25/2025",
};

There are two boxes, both containing black and white balls. The number of balls in each box is <InlineMath math="20" />. One ball is taken from each box, and the probability that both balls are black is <InlineMath math="\frac{3}{10}" />. If the probability that both balls are white is <InlineMath math="\frac{m}{n}" />, then <InlineMath math="m + n = \dots" />


### Choices

- [ ] $$3$$
- [x] $$6$$
- [ ] $$7$$
- [ ] $$12$$
- [ ] $$14$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 5",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

Let:

- <InlineMath math="P(h_1) =" /> Probability of black in box 1
- <InlineMath math="P(h_2) =" /> Probability of black in box 2
- <InlineMath math="P(p_1) =" /> Probability of white in box 1
- <InlineMath math="P(p_2) =" /> Probability of white in box 2

Given the probability of getting black balls from both boxes is <InlineMath math="\frac{3}{10}" />. Since taking balls from box 1 and box 2 are independent events:

<MathContainer>
  <BlockMath math="\begin{aligned} P(h_1 \cap h_2) &= P(h_1) \times P(h_2) \\ \frac{3}{10} &= \frac{n(h_1)}{20} \times \frac{n(h_2)}{20} \\ \frac{3}{10} &= \frac{n(h_1) \times n(h_2)}{400} \\ n(h_1) \times n(h_2) &= 120 \end{aligned}" />
</MathContainer>

We need to find a pair of factors of <InlineMath math="120" /> where both are less than or equal to <InlineMath math="20" /> (maximum number of balls). The possible pair is <InlineMath math="12" /> and <InlineMath math="10" /> (or vice versa). Let <InlineMath math="n(h_1) = 12" /> and <InlineMath math="n(h_2) = 10" />.

Then the number of white balls is:

- Box 1: <InlineMath math="20 - 12 = 8" /> white balls
- Box 2: <InlineMath math="20 - 10 = 10" /> white balls

The probability that both balls are white is:

<MathContainer>
  <BlockMath math="\begin{aligned} P(p_1 \cap p_2) &= P(p_1) \times P(p_2) \\ &= \frac{8}{20} \times \frac{10}{20} \\ &= \frac{2}{5} \times \frac{1}{2} \\ &= \frac{1}{5} \end{aligned}" />
</MathContainer>

Given <InlineMath math="\frac{m}{n} = \frac{1}{5}" />. So <InlineMath math="m = 1" /> and <InlineMath math="n = 5" />. The value of <InlineMath math="m + n = 1 + 5 = 6" />.


---

## Exercise 6

### Question

export const metadata = {
  title: "Problem 6",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

Set <InlineMath math="A" /> has <InlineMath math="10" /> members. The number of subsets of <InlineMath math="A" /> that have more than <InlineMath math="3" /> members is...


### Choices

- [ ] $$512$$
- [ ] $$564$$
- [ ] $$624$$
- [ ] $$720$$
- [x] $$848$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 6",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

The order of members in a set does not matter, so we use the concept of combinations.

The total number of subsets of a set with <InlineMath math="n" /> members is <InlineMath math="2^n" />. Given <InlineMath math="n = 10" />, the total number of subsets is:

<BlockMath math="2^{10} = 1024" />

We are asked to find the number of subsets with **more than <InlineMath math="3" />** members. This means we are looking for the sum of subsets with 4, 5, 6, ..., 10 members.

An easier way is to use the complement principle: Total subsets minus subsets with **<InlineMath math="0" />, <InlineMath math="1" />, <InlineMath math="2" />, or <InlineMath math="3" />** members.

The combination formula is:

<BlockMath math="C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}" />

Let's calculate them one by one.

#### Subsets with 0 members

<BlockMath math="C(10, 0) = \frac{10!}{0!10!} = 1" />

#### Subsets with 1 member

<BlockMath math="C(10, 1) = \frac{10!}{1!9!} = 10" />

#### Subsets with 2 members

<BlockMath math="C(10, 2) = \frac{10!}{2!8!} = \frac{10 \times 9}{2 \times 1} = 45" />

#### Subsets with 3 members

<BlockMath math="C(10, 3) = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120" />

The number of subsets with <InlineMath math="\leq 3" /> members is:

<BlockMath math="1 + 10 + 45 + 120 = 176" />

Thus, the number of subsets with **more than <InlineMath math="3" />** members is:

<BlockMath math="\begin{aligned} \text{Total} - (\leq 3) &= 1024 - 176 \\ &= 848 \end{aligned}" />

So, the number of subsets of <InlineMath math="A" /> that have more than <InlineMath math="3" /> members is <InlineMath math="848" />.


---

## Exercise 7

### Question

export const metadata = {
    title: "Problem 7",
    authors: [{ name: "Nabil Akbarazzima Fatih" }],
    date: "12/25/2025",
};

The operation <InlineMath math="\odot" /> on the set of integers is defined by <InlineMath math="a \odot b = 3ab - 2(1 + b)" />. The value of <InlineMath math="(5 \odot 2) \odot 3 = \dots" />


### Choices

- [ ] $$96$$
- [ ] $$110$$
- [ ] $$128$$
- [ ] $$156$$
- [x] $$208$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 7",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

Based on the definition of the operation <InlineMath math="\odot" />, we will calculate the value of <InlineMath math="(5 \odot 2) \odot 3" /> step by step. First, we calculate the part inside the parentheses, which is <InlineMath math="5 \odot 2" /> with <InlineMath math="a = 5" /> and <InlineMath math="b = 2" />.

<BlockMath math="\begin{aligned} 5 \odot 2 &= 3(5)(2) - 2(1 + 2) \\ &= 30 - 2(3) \\ &= 30 - 6 \\ &= 24 \end{aligned}" />

Next, we operate the result with <InlineMath math="3" />. Now we calculate <InlineMath math="24 \odot 3" /> with <InlineMath math="a = 24" /> and <InlineMath math="b = 3" />.

<BlockMath math="\begin{aligned} 24 \odot 3 &= 3(24)(3) - 2(1 + 3) \\ &= 216 - 2(4) \\ &= 216 - 8 \\ &= 208 \end{aligned}" />

Thus, the value of <InlineMath math="(5 \odot 2) \odot 3" /> is <InlineMath math="208" />.


---

## Exercise 8

### Question

export const metadata = {
  title: "Problem 8",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

Three years ago, the ratio of Ali's age to Tika's age was <InlineMath math="4:5" />. If the ratio of their ages three years from now is <InlineMath math="5:6" />, then the difference in their current ages is...


### Choices

- [ ] $$2$$ years
- [ ] $$3$$ years
- [ ] $$4$$ years
- [ ] $$5$$ years
- [x] $$6$$ years

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 8",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

Let <InlineMath math="A" /> be Ali's current age and <InlineMath math="T" /> be Tika's current age. Given that the ratio of Ali's age to Tika's age three years ago was <InlineMath math="4:5" />.

<BlockMath math="\begin{aligned} \frac{A - 3}{T - 3} &= \frac{4}{5} \\ 5(A - 3) &= 4(T - 3) \\ 5A - 15 &= 4T - 12 \\ 5A - 4T &= 3 \quad \dots(1) \end{aligned}" />

Three years from now, the ratio of their ages will be <InlineMath math="5:6" />.

<BlockMath math="\begin{aligned} \frac{A + 3}{T + 3} &= \frac{5}{6} \\ 6(A + 3) &= 5(T + 3) \\ 6A + 18 &= 5T + 15 \\ 6A - 5T &= -3 \quad \dots(2) \end{aligned}" />

From equations <InlineMath math="(1)" /> and <InlineMath math="(2)" />, we perform elimination to find the value of <InlineMath math="A" />. Multiply equation <InlineMath math="(1)" /> by <InlineMath math="5" /> and equation <InlineMath math="(2)" /> by <InlineMath math="4" />.

<BlockMath math="\begin{aligned} 25A - 20T &= 15 \\ 24A - 20T &= -12 \\ \hline A &= 27 \end{aligned}" />

Ali's current age is <InlineMath math="27" /> years. Next, we substitute <InlineMath math="A = 27" /> into equation <InlineMath math="(1)" /> to find <InlineMath math="T" />.

<BlockMath math="\begin{aligned} 5(27) - 4T &= 3 \\ 135 - 4T &= 3 \\ 4T &= 132 \\ T &= 33 \end{aligned}" />

Thus, Tika's current age is <InlineMath math="33" /> years. The difference in their ages is:

<BlockMath math="33 - 27 = 6 \text{ years}" />


---

## Exercise 9

### Question

export const metadata = {
  title: "Problem 9",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

Line <InlineMath math="h" /> passes through point <InlineMath math="T(-1,3)" /> and has a gradient <InlineMath math="m" />. For <InlineMath math="h" /> to intersect the graph <InlineMath math="y = -x^2" /> at two distinct points, it must be that...


### Choices

- [ ] $$m > 6$$
- [ ] $$-2 < m < 6$$
- [ ] $$-6 < m < 2$$
- [ ] $$m \leq -2 \lor m \geq 6$$
- [x] $$m < -2 \lor m > 6$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 9",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

The equation of line <InlineMath math="h" /> is <InlineMath math="y = mx + c" />. Given <InlineMath math="y = 3" /> and <InlineMath math="x = -1" />, then:

<BlockMath math="\begin{aligned} 3 &= -m + c \\ c &= m + 3 \end{aligned}" />

Thus, the line equation can be written as <InlineMath math="y = mx + (m + 3)" />. Since line <InlineMath math="h" /> intersects the graph <InlineMath math="y = -x^2" /> at two distinct points, then:

<BlockMath math="\begin{aligned} mx + (m + 3) &= -x^2 \\ x^2 + mx + (m + 3) &= 0 \end{aligned}" />

For the line to intersect at two distinct points, the discriminant must satisfy <InlineMath math="D > 0" />.

<BlockMath math="\begin{aligned} m^2 - 4(m + 3) &> 0 \\ m^2 - 4m - 12 &> 0 \\ (m + 2)(m - 6) &> 0 \end{aligned}" />

The roots are <InlineMath math="m = -2" /> and <InlineMath math="m = 6" />. By testing intervals, we obtain the solution set:

<BlockMath math="m < -2 \lor m > 6" />


---

## Exercise 10

### Question

export const metadata = {
    title: "Problem 10",
    authors: [{ name: "Nabil Akbarazzima Fatih" }],
    date: "12/25/2025",
};

For <InlineMath math="(2m + 1)x^2 - 4(m + 1)x + m > -4" /> to hold for every real <InlineMath math="x" />, the values of <InlineMath math="m" /> must satisfy...


### Choices

- [ ] $$m < 0 \lor m > \frac{1}{2}$$
- [ ] $$-\frac{1}{2} < m < \frac{1}{2}$$
- [x] $$0 < m < \frac{1}{2}$$
- [ ] $$0 \leq m < \frac{1}{2}$$
- [ ] $$m < -\frac{1}{2} \lor m > 0$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 10",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

Given the inequality:

<BlockMath math="(2m + 1)x^2 - 4(m + 1)x + m > -4" />

Rearrange it into the standard quadratic form by moving <InlineMath math="-4" /> to the left side:

<BlockMath math="(2m + 1)x^2 - 4(m + 1)x + (m + 4) > 0" />

For the quadratic function to be positive for every real <InlineMath math="x" /> (positive definite), its graph must lie entirely above the <InlineMath math="x" />-axis. The conditions are:

1. The coefficient of <InlineMath math="x^2" /> must be positive (<InlineMath math="a > 0" />).
2. The discriminant must be negative (<InlineMath math="D < 0" />).

**Condition 1: <InlineMath math="a > 0" />**

<BlockMath math="\begin{aligned} 2m + 1 &> 0 \\ 2m &> -1 \\ m &> -\frac{1}{2} \quad \dots \text{(1)} \end{aligned}" />

**Condition 2: <InlineMath math="D < 0" />**

Recall that <InlineMath math="D = b^2 - 4ac" />. With <InlineMath math="a = 2m + 1" />, <InlineMath math="b = -4(m + 1)" />, and <InlineMath math="c = m + 4" />:

<BlockMath math="\begin{aligned} (-4(m + 1))^2 - 4(2m + 1)(m + 4) &< 0 \\ 16(m^2 + 2m + 1) - 4(2m^2 + 9m + 4) &< 0 \end{aligned}" />

Divide both sides by <InlineMath math="4" /> to simplify:

<BlockMath math="\begin{aligned} 4(m^2 + 2m + 1) - (2m^2 + 9m + 4) &< 0 \\ 4m^2 + 8m + 4 - 2m^2 - 9m - 4 &< 0 \\ 2m^2 - m &< 0 \\ m(2m - 1) &< 0 \end{aligned}" />

The roots are <InlineMath math="m = 0" /> and <InlineMath math="m = \frac{1}{2}" />. Since the inequality is "&lt;", the solution lies between the roots:

<BlockMath math="0 < m < \frac{1}{2} \quad \dots \text{(2)}" />

**Conclusion**

We find the intersection of conditions <InlineMath math="(1)" /> and <InlineMath math="(2)" />:
- Condition <InlineMath math="(1)" />: <InlineMath math="m > -\frac{1}{2}" />
- Condition <InlineMath math="(2)" />: <InlineMath math="0 < m < \frac{1}{2}" />

The intersection of both regions is:

<BlockMath math="0 < m < \frac{1}{2}" />


---

## Exercise 11

### Question

export const metadata = {
  title: "Problem 11",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

Ali will depart from city A to city B at <InlineMath math="09.15" /> driving a car with an average speed of <InlineMath math="60" /> km/h. At the same time, Tika heads to city A from city B in her vehicle at a speed of <InlineMath math="70" /> km/h. The distance between city A and city B is <InlineMath math="221" /> km. If they take the same route, they will cross paths at...


### Choices

- [ ] $$10.01$$
- [ ] $$10.20$$
- [ ] $$10.36$$
- [x] $$10.57$$
- [ ] $$11.02$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 11",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

Let:
- <InlineMath math="J_A =" /> Distance traveled by Ali until they meet
- <InlineMath math="J_T =" /> Distance traveled by Tika until they meet
- <InlineMath math="v_A =" /> Ali's speed <InlineMath math="= 60" /> km/h
- <InlineMath math="v_T =" /> Tika's speed <InlineMath math="= 70" /> km/h

Since Ali and Tika depart at the same time, the time they take to meet is also the same. Let <InlineMath math="t =" /> time.

Then:

<BlockMath math="\begin{aligned} J_A &= v_A \times t = 60t \\ J_T &= v_T \times t = 70t \end{aligned}" />

Since the total distance is <InlineMath math="221" /> km, then:

<BlockMath math="\begin{aligned} J_A + J_T &= 221 \\ 60t + 70t &= 221 \\ 130t &= 221 \\ t &= 1.7 \text{ hours} \end{aligned}" />

So the time required until they meet is:

<BlockMath math="1.7 \text{ hours} = 1 \text{ hour } 42 \text{ minutes}" />

Thus, they will meet at:

<BlockMath math="09.15 + 1 \text{ hour } 42 \text{ minutes} = 10.57" />


---

## Exercise 12

### Question

export const metadata = {
  title: "Problem 12",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

The average mathematics exam score of a class is <InlineMath math="7.5" />. The ratio of male to female students in the class is <InlineMath math="3 : 4" />. If the average score of female students is <InlineMath math="1.75" /> higher than the average score of male students, then the average score of female students is...


### Choices

- [ ] $$6.25$$
- [ ] $$6.50$$
- [ ] $$7.50$$
- [ ] $$7.75$$
- [x] $$8.25$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 12",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

Let:
- <InlineMath math="\bar{x}_K =" /> Class average score
- <InlineMath math="n_L =" /> Number of male students
- <InlineMath math="n_P =" /> Number of female students
- <InlineMath math="\bar{x}_L =" /> Average score of male students
- <InlineMath math="\bar{x}_P =" /> Average score of female students

Given:
- <InlineMath math="\bar{x}_K = 7.5" />
- <InlineMath math="\frac{n_L}{n_P} = \frac{3}{4} \implies n_L = 0.75 n_P" />
- <InlineMath math="\bar{x}_P = 1.75 + \bar{x}_L" />

The class average score can be obtained as follows:

<BlockMath math="\bar{x}_K = \frac{\bar{x}_P \cdot n_P + \bar{x}_L \cdot n_L}{n_P + n_L}" />

Substitute the known values:

<BlockMath math="\begin{aligned} 7.5 &= \frac{(1.75 + \bar{x}_L)n_P + \bar{x}_L \cdot 0.75n_P}{n_P + 0.75n_P} \\ 7.5 &= \frac{1.75n_P + \bar{x}_L n_P + 0.75\bar{x}_L n_P}{1.75n_P} \\ 7.5 &= \frac{1.75n_P + 1.75\bar{x}_L n_P}{1.75n_P} \\ 7.5 &= \frac{1.75n_P(1 + \bar{x}_L)}{1.75n_P} \\ 7.5 &= 1 + \bar{x}_L \\ \bar{x}_L &= 6.5 \end{aligned}" />

Therefore, the average score of female students is:

<BlockMath math="\bar{x}_P = 1.75 + 6.5 = 8.25" />


---

## Exercise 13

### Question

export const metadata = {
  title: "Problem 13",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

A fabric company requires <InlineMath math="20" /> workers to complete an order in <InlineMath math="12" /> days. If the number of orders remains unchanged but only <InlineMath math="15" /> people can work on the order, then the additional time required to complete the order is ... days


### Choices

- [x] $$4$$
- [ ] $$8$$
- [ ] $$10$$
- [ ] $$14$$
- [ ] $$16$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 13",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

The fewer workers there are, the longer it takes to complete the work. Thus, this is an inverse proportion.

Let <InlineMath math="x =" /> days required.

<BlockMath math="\begin{aligned} \frac{20}{15} &= \frac{x}{12} \\ x &= 16 \end{aligned}" />

So, the time required to complete the order if only <InlineMath math="15" /> people are working is <InlineMath math="16" /> days. Therefore, the additional time required is:

<BlockMath math="16 - 12 = 4 \text{ days}" />


---

## Exercise 14

### Question

export const metadata = {
  title: "Problem 14",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

A box contains <InlineMath math="5" /> red balls, <InlineMath math="4" /> white balls, and <InlineMath math="3" /> yellow balls. From the box, <InlineMath math="3" /> balls are drawn simultaneously at random. The probability of drawing <InlineMath math="2" /> red balls and <InlineMath math="1" /> white ball is...


### Choices

- [ ] $$\frac{1}{11}$$
- [x] $$\frac{2}{11}$$
- [ ] $$\frac{4}{11}$$
- [ ] $$\frac{3}{5}$$
- [ ] $$\frac{5}{7}$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 14",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

The number of ways to choose red balls is:

<BlockMath math="C(5,2) = \frac{5!}{2!3!} = 10" />

The number of ways to choose white balls is:

<BlockMath math="C(4,1) = \frac{4!}{3!1!} = 4" />

The total number of possible ways to choose the balls is:

<BlockMath math="C(12,3) = \frac{12!}{9!3!} = 220" />

Let <InlineMath math="A =" /> The event of drawing <InlineMath math="2" /> red balls and <InlineMath math="1" /> white ball.

The probability of drawing <InlineMath math="2" /> red balls and <InlineMath math="1" /> white ball is:

<BlockMath math="\begin{aligned} P(A) &= \frac{C(5,2)C(4,1)}{C(12,3)} \\ &= \frac{10 \cdot 4}{220} \\ &= \frac{40}{220} \\ &= \frac{2}{11} \end{aligned}" />


---

## Exercise 15

### Question

export const metadata = {
    title: "Problem 15",
    authors: [{ name: "Nabil Akbarazzima Fatih" }],
    date: "12/25/2025",
};

Given that

<BlockMath math="x = \frac{1}{2015} - \frac{2}{2015} + \frac{3}{2015} - \frac{4}{2015} + \dots - \frac{2014}{2015}" />

The value of <InlineMath math="x" /> that satisfies the equation is...


### Choices

- [x] $$-\frac{1007}{2015}$$
- [ ] $$-\frac{1008}{2015}$$
- [ ] $$\frac{1}{2015}$$
- [ ] $$\frac{1007}{2015}$$
- [ ] $$\frac{1008}{2015}$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 15",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

In addition or subtraction operations, we can use the associative property to simplify the calculation.

<BlockMath math="\begin{aligned} x &= \frac{1}{2015} - \frac{2}{2015} + \frac{3}{2015} - \frac{4}{2015} + \dots - \frac{2014}{2015} \\ &= \left(\frac{1}{2015} - \frac{2}{2015}\right) + \left(\frac{3}{2015} - \frac{4}{2015}\right) + \dots + \left(\frac{2013}{2015} - \frac{2014}{2015}\right) \\ &= \left(-\frac{1}{2015}\right) + \left(-\frac{1}{2015}\right) + \dots + \left(-\frac{1}{2015}\right) \end{aligned}" />

There are <InlineMath math="\frac{2014}{2} = 1007" /> groups, so:

<BlockMath math="x = -\frac{1}{2015} (1007) = -\frac{1007}{2015}" />


---

## Exercise 16

### Question

export const metadata = {
  title: "Question 16",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

Consider the following number sequence:

<BlockMath math="0.5, \ 3, \ 8, \ 18, \ 38, \ 78, \ \dots, \ 318" />

The correct number to complete the pattern is ...


### Choices

- [ ] $$120$$
- [x] $$158$$
- [ ] $$160$$
- [ ] $$168$$
- [ ] $$200$$

### Answer & Explanation

export const metadata = {
  title: "Solution to Question 16",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

The given number sequence is:

<BlockMath math="0.5, \ 3, \ 8, \ 18, \ 38, \ 78, \ \dots, \ 318" />

Let's analyze the difference between consecutive terms:

<MathContainer>
  <BlockMath math="3 - 0.5 = 2.5" />
  <BlockMath math="8 - 3 = 5" />
  <BlockMath math="18 - 8 = 10" />
  <BlockMath math="38 - 18 = 20" />
  <BlockMath math="78 - 38 = 40" />
</MathContainer>

Observing the pattern of differences (<InlineMath math="2.5, \ 5, \ 10, \ 20, \ 40" />), we can see that each difference is obtained by multiplying the previous difference by <InlineMath math="2" /> (a common ratio of <InlineMath math="2" />).

Therefore, the next difference after <InlineMath math="40" /> is:

<BlockMath math="40 \times 2 = 80" />

Thus, the number that fills the blank is:

<BlockMath math="78 + 80 = 158" />

To verify this pattern, let's calculate the subsequent term and check if it matches the last known term (<InlineMath math="318" />). The next difference after <InlineMath math="80" /> would be <InlineMath math="80 \times 2 = 160" />.

<BlockMath math="158 + 160 = 318" />

The calculation matches the last term in the problem.

Therefore, the correct number to complete the pattern is <InlineMath math="158" />.


---

## Exercise 17

### Question

export const metadata = {
  title: "Question 17",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

The decimal value closest to the largest value among the fractions <InlineMath math="\frac{34}{33}, \frac{6.1}{6}, \frac{56}{55}, \frac{12.5}{12.4}, \frac{26}{25}" /> is...


### Choices

- [ ] $$1.181$$
- [ ] $$1.030$$
- [ ] $$1.124$$
- [x] $$1.040$$
- [ ] $$1.110$$

### Answer & Explanation

export const metadata = {
  title: "Solution to Question 17",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

The fractions can be rewritten in the form <InlineMath math="1 + \frac{1}{x}" /> as follows:

<MathContainer>
  <BlockMath math="\frac{34}{33} = 1 \frac{1}{33} = 1 + \frac{1}{33}" />
  <BlockMath math="\frac{6.1}{6} = \frac{61}{60} = 1 \frac{1}{60} = 1 + \frac{1}{60}" />
  <BlockMath math="\frac{56}{55} = 1 \frac{1}{55} = 1 + \frac{1}{55}" />
  <BlockMath math="\frac{12.5}{12.4} = \frac{125}{124} = 1 \frac{1}{124} = 1 + \frac{1}{124}" />
  <BlockMath math="\frac{26}{25} = 1 \frac{1}{25} = 1 + \frac{1}{25}" />
</MathContainer>

So, all numbers have the form <InlineMath math="1 + \frac{1}{x}" />. To find the largest value, we need to find the fraction where <InlineMath math="\frac{1}{x}" /> is largest. The value of <InlineMath math="\frac{1}{x}" /> increases as the denominator <InlineMath math="x" /> decreases.

From the five fractions above, the values of <InlineMath math="x" /> are <InlineMath math="33, 60, 55, 124," /> and <InlineMath math="25" /> respectively. The smallest <InlineMath math="x" /> is <InlineMath math="25" />.

Therefore, the largest value is <InlineMath math="\frac{26}{25}" />.

Let's calculate its decimal value:

<BlockMath math="\frac{26}{25} = \frac{26 \times 4}{25 \times 4} = \frac{104}{100} = 1.04" />

The decimal value is <InlineMath math="1.040" />.


---

## Exercise 18

### Question

export const metadata = {
  title: "Question 18",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

A hemispherical water container with a radius of <InlineMath math="15 \text{ cm}" /> is full of water. All the water in the container is transferred into a cylindrical container with a radius equal to the radius of the hemisphere. The height of the water in the cylindrical container is...


### Choices

- [x] $$10 \text{ cm}$$
- [ ] $$12 \text{ cm}$$
- [ ] $$18 \text{ cm}$$
- [ ] $$20 \text{ cm}$$
- [ ] $$22 \text{ cm}$$

### Answer & Explanation

export const metadata = {
  title: "Solution to Question 18",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

Since the water in the hemispherical container is transferred to the cylindrical container, the volume of water remains the same.

<MathContainer>
  <BlockMath math="V_{\text{hemisphere}} = V_{\text{cylinder}}" />
  <BlockMath math="\frac{1}{2} \cdot \frac{4}{3} \cdot \pi \cdot r^3 = \pi \cdot r^2 \cdot t" />
</MathContainer>

Since the radius of the sphere is the same as the radius of the cylinder, we can simplify the equation by dividing both sides by <InlineMath math="\pi \cdot r^2" />:

<MathContainer>
  <BlockMath math="\frac{2}{3} \cdot r = t" />
</MathContainer>

Given that the radius <InlineMath math="r = 15 \text{ cm}" />. We substitute this value into the equation:

<MathContainer>
  <BlockMath math="\frac{2}{3} \cdot 15 = t" />
  <BlockMath math="t = 10 \text{ cm}" />
</MathContainer>

Thus, the height of the water in the cylinder is <InlineMath math="10 \text{ cm}" />.


---

## Exercise 19

### Question

export const metadata = {
  title: "Question 19",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

In a class, there are <InlineMath math="30" /> students participating in extracurricular activities. There are <InlineMath math="12" /> students who join the scouts extracurricular and <InlineMath math="15" /> students who join the futsal extracurricular, and <InlineMath math="7" /> students who join both. The number of students who do not join either is...


### Choices

- [ ] $$3$$
- [x] $$10$$
- [ ] $$13$$
- [ ] $$19$$
- [ ] $$22$$

### Answer & Explanation

export const metadata = {
  title: "Solution to Question 19",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

Let:
- <InlineMath math="n(S)" /> = Total number of students in the class
- <InlineMath math="n(P)" /> = Number of students in scouts
- <InlineMath math="n(F)" /> = Number of students in futsal

Given:

<MathContainer>
  <BlockMath math="n(S) = 30" />
  <BlockMath math="n(P) = 12" />
  <BlockMath math="n(F) = 15" />
  <BlockMath math="n(P \cap F) = 7" />
</MathContainer>

The number of students who do not participate in either (<InlineMath math="n(P \cup F)^c" />) can be found using the universal set formula:

<MathContainer>
  <BlockMath math="n(S) = n(P) + n(F) - n(P \cap F) + n(P \cup F)^c" />
</MathContainer>

Substitute the known values:

<MathContainer>
  <BlockMath math="30 = 12 + 15 - 7 + n(P \cup F)^c" />
  <BlockMath math="30 = 20 + n(P \cup F)^c" />
  <BlockMath math="n(P \cup F)^c = 30 - 20" />
  <BlockMath math="n(P \cup F)^c = 10" />
</MathContainer>

Thus, the number of students who do not join scouts and futsal extracurriculars is <InlineMath math="10" />.


---

## Exercise 20

### Question

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

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

<Graph
  title={<>Triangles <InlineMath math="Q" /> and <InlineMath math="R" /></>}
  description={
    <>
      Two triangles, <InlineMath math="Q" /> and <InlineMath math="R" />, with
      numbers on their sides and inside.
    </>
  }
/>

The value <InlineMath math="4" /> inside triangle <InlineMath math="Q" /> is the result of arithmetic operations on all numbers outside triangle <InlineMath math="Q" />. Using the same operation pattern, the correct value inside triangle <InlineMath math="R" /> is...


### Choices

- [ ] $$2$$
- [x] $$4$$
- [ ] $$5$$
- [ ] $$7$$
- [ ] $$1$$

### Answer & Explanation

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

In triangle <InlineMath math="Q" />, the value <InlineMath math="4" /> is obtained using the following arithmetic operation:

<MathContainer>
  <BlockMath math="22 - 20 + 2 = 4" />
</MathContainer>

The pattern used is (Left Number <InlineMath math="-" /> Right Number) <InlineMath math="+" /> Bottom Number <InlineMath math="=" /> Center Number.

Using the same operation pattern, the value inside triangle <InlineMath math="R" /> can be obtained as follows:

<MathContainer>
  <BlockMath math="13 - 12 + 3 = 4" />
</MathContainer>

Thus, the value inside triangle <InlineMath math="R" /> is <InlineMath math="4" />.


---
