# Nakafa Framework: LLM

URL: https://nakafa.com/en/exercises/high-school/snbt/quantitative-knowledge/try-out/set-9/4

Exercises: Try Out - Set 9: Real exam simulation to sharpen your skills and build confidence. - Problem 4

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## 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**.


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