Set 2

Explanation

Let the first number be $A$ and the second number be $B$.

From the problem, we get the following system of equations:

1. Twice the first number minus three times the second number equals $14$:

   $$2A - 3B = 14$$

2. The first number multiplied by $k$ plus twice the second number equals $4$:

   $$Ak + 2B = 4$$

3. The value of $k$ is equal to the sum of the first and second numbers:

   $$k = A + B$$

   With the condition $k > 0$.

Solution Steps

First, substitute equation $(3)$ into equation $(2)$:

Factor the terms containing $B$:

Note that $4 - A^2$ is a difference of squares, so it can be factored as $(2 - A)(2 + A)$:

Since $2 + A$ is the same as $A + 2$, we can divide both sides by $A + 2$ (assuming $A \neq -2$):

Next, substitute the value of $B$ into equation $(1)$:

So, the value of the first number is $4$.

Verification

We can verify the result:

If $A = 4$, then $B = 2 - 4 = -2$.

The value of $k = A + B = 4 + (-2) = 2$.

The condition $k > 0$ is satisfied since $2 > 0$.

Check equation $(1)$: $2(4) - 3(-2) = 8 + 6 = 14$ (Correct).

Check equation $(2)$: $4(2) + 2(-2) = 8 - 4 = 4$ (Correct).

Explanation

Let the first number be $A$ and the second number be $B$.

From the problem, we get the following information:

1. Twice the first number minus three times the second number equals $14$:

   $$2A - 3B = 14$$

2. The first number multiplied by $k$ plus twice the second number equals $4$:

   $$Ak + 2B = 4$$

3. The value of the first number is $2$ more than $k$ (meaning $A = k + 2$ or $k = A - 2$).

Solution Steps

Substitute $k = A - 2$ into equation $(2)$:

Rearrange the equation to solve for $B$:

Next, substitute this value of $B$ into equation $(1)$:

Multiply both sides by $2$ to eliminate the denominator:

Factor the quadratic equation:

This gives two possible values for $A$:

Determining the Largest Value of the Second Number

We find the value of $B$ for each value of $A$ using equation $2A - 3B = 14 \Rightarrow 3B = 2A - 14 \Rightarrow B = \frac{2A - 14}{3}$.

Case 1: For $A = 4$

Case 2: For $A = -\frac{10}{3}$

Compare the two obtained values of $B$: $-2$ and $-\frac{62}{9}$.

The value $-2$ is larger than $-\frac{62}{9}$.

Thus, the largest possible value of the second number is $-2$.

Explanation

First, we need to determine the function $g(x)$.

Given $f(x) = 2x + 5$ and $(f \circ g)(x) = 2x^2 - 6x + 9$.

Based on the definition of function composition $(f \circ g)(x) = f(g(x))$, we have:

Now, let's check the truth of each statement.

Statement 1

The value of $g(1)$:

Statement $(1)$ is True.

Statement 2

To determine the number of intersection points of the quadratic function $g(x) = x^2 - 3x + 2$ with the x-axis, we check the discriminant ($D = b^2 - 4ac$).

Since $D > 0$, the function $g$ intersects the x-axis at two distinct points, not one.

Statement $(2)$ is False.

Statement 3

Finding the inverse of $f(x) = 2x + 5$:

Let $f(x) = y$, then:

So the inverse is:

Statement $(3)$ is False (it should be minus $\frac{5}{2}$, not plus).

Statement 4

Summing $f(x)$ and $g(x)$:

Statement $(4)$ is True.

Conclusion

The correct statements are statements $(1)$ and $(4)$. Thus, there are 2 true statements.

Explanation

To determine the passing percentage, we must check two passing conditions for each student:

1. The Science score must be higher than the median Science score of all students.
2. The average exam score (Math + Science) divided by $2$ must be greater than $85$.

Determining the Median Science Score

The Science scores are: $78, 82, 87, 85, 79, 88, 81, 80$.

Sort the data from smallest to largest: $78, 79, 80, 81, 82, 85, 87, 88$.

Since there are $8$ data points, the median is the average of the 4th and 5th data points:

Students who satisfy the first condition (Science Score $> 81.5$) are:

• Boni ($82$)
• Cindy ($87$)
• Dita ($85$)
• Fani ($88$)

The other students (Ayu, Erfan, Gilang, Herman) have Science scores below the median, so they automatically do not pass.

Checking the Average Score Condition

Now we check the second condition for students who passed the first condition:

1. Boni:

   $$\text{Average} = \frac{92 + 82}{2} = \frac{174}{2} = 87$$

   Since $87 > 85$, Boni Passes.

2. Cindy:

   $$\text{Average} = \frac{90 + 87}{2} = \frac{177}{2} = 88.5$$

   Since $88.5 > 85$, Cindy Passes.

3. Dita:

   $$\text{Average} = \frac{86 + 85}{2} = \frac{171}{2} = 85.5$$

   Since $85.5 > 85$, Dita Passes.

4. Fani:

   $$\text{Average} = \frac{88 + 88}{2} = \frac{176}{2} = 88$$

   Since $88 > 85$, Fani Passes.

So, there are $4$ students who pass out of a total of $8$ students.

Calculating Quantity $P$

Comparing $P$ and $Q$

Given $Q = 65\%$.

Thus:

Explanation

The first step is to find the value of $a$ by summing up all students and equating it to the known total students, which is $150$.

Calculating the Value of $a$

The total number of students is the sum of students from all subjects in all classes:

Solve the equation:

Calculating Total Social Studies Students

Now we calculate the number of students who like Social Studies (IPS):

Calculating Quantity $P$

Quantity $P$ is the percentage of total students with Social Studies as their favorite subject out of the total students:

Comparing $P$ and $Q$

Given $Q = 50\%$.

Thus:

Explanation

To form a distinct four-digit odd number from the digits $1, 2, 3, 4, 5, 6$, we can use the filling slots method.

There are $4$ digit positions: Thousands, Hundreds, Tens, and Units.

Step 1: Filling the Units Position

For the number to be odd, the units digit must be an odd number. From the available digits ($1, 2, 3, 4, 5, 6$), the odd numbers are $1, 3, 5$.

So, there are $3$ choices for the units position.

Step 2: Filling the Thousands Position

After one digit is used for the units position, there are $6 - 1 = 5$ digits remaining for the thousands position (since the digits must be distinct).

So, there are $5$ choices for the thousands position.

Step 3: Filling the Hundreds Position

After two digits are used (for units and thousands), there are $6 - 2 = 4$ digits remaining.

So, there are $4$ choices for the hundreds position.

Step 4: Filling the Tens Position

After three digits are used, there are $6 - 3 = 3$ digits remaining.

So, there are $3$ choices for the tens position.

Total Calculation

The number of numbers that can be formed is the product of the number of choices for each position:

So, there are $180$ numbers that can be formed.

Explanation

The first step is to solve the given linear inequality in one variable.

Inequality:

Group the terms containing the variable $a$ on the left side and the constants on the right side:

Divide both sides by $2$:

It is given that $a$ is an integer greater than 7 ($a > 7$).

So we are looking for an integer that satisfies both of the following conditions:

1. $a \leq 8$
2. $a > 7$

This means $7 < a \leq 8$.

The only integer that satisfies this interval is:

Thus, the value of $a$ is $8$.

Explanation

Let's check the truth of each statement.

Statement 1

We will find the inverse function of $f(x) = \frac{2x + 1}{3x - 5}$.

Let $f(x) = y$, then:

So, the inverse function is:

Statement 1 is True.

Statement 2

The range of the inverse function $f^{-1}(x)$ is equal to the domain of the original function $f(x)$.

The function $f(x) = \frac{2x + 1}{3x - 5}$ is defined if the denominator is not equal to zero:

This means the domain of $f(x)$ is all real numbers except $\frac{5}{3}$.

Therefore, the range of $f^{-1}(x)$ is all real numbers except $\frac{5}{3}$.

So, the value $y = \frac{5}{3}$ is indeed not a member of the range of $f^{-1}(x)$.

Statement 2 is True.

Statement 3

We check if $x = \frac{1}{2}$ satisfies $f(x) = f^{-1}(x)$.

Calculate $f\left(\frac{1}{2}\right)$:

Calculate $f^{-1}\left(\frac{1}{2}\right)$:

Since $-\frac{4}{7} \neq -7$, then $x = \frac{1}{2}$ does not satisfy the equation.

Statement 3 is False.

Conclusion

The correct sequence of truth values is True, True, False.

Explanation

Given the matrix equation:

Where $A = \begin{pmatrix} 6 & 3 \\ b & 14 \end{pmatrix}$ and $B = \begin{pmatrix} -1 & 2 \\ 4 & 5 \end{pmatrix}$.

If both sides are multiplied by $M^{-1}$ from the right, then:

Substitute the known matrices:

Perform matrix multiplication on the right side:

From the equality of two matrices, we obtain:

Element at row 1 column 1:

Element at row 2 column 1:

Now we check the truth of each statement.

Statement 1

States that $a = 2$. Since the calculated value is $a = 0$, this statement is False.

Statement 2

States that $a + b = 15$. Calculate $a + b$:

This statement is True.

Statement 3

States that the determinant of the inverse matrix $M$ is $3$. The inverse matrix $M$ is $M^{-1} = \begin{pmatrix} a & 1 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 3 & 2 \end{pmatrix}$.

The determinant of $M^{-1}$ is:

Since the determinant result is $-3$, not $3$, this statement is False.

Conclusion

The correct sequence of truth values is False, True, False.

Explanation

The sample space ($S$) is the selection of 3 numbers from the set of whole numbers less than $10$, which is $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$. The number of elements in this set is $n = 10$.

The number of ways to choose 3 numbers from 10 is:

Let's check each statement.

Statement 1

Set of even numbers: $\{0, 2, 4, 6, 8\}$ (5 numbers). Number of ways to choose 3 even numbers from 5:

Probability:

Is $\frac{1}{12} > \frac{1}{20}$?

Since $\frac{5}{60} > \frac{3}{60}$, statement 1 is True.

Statement 2

The number $3$ must be selected. So we just need to choose 2 more numbers from the remaining 9.

Probability:

Statement 2 is True.

Statement 3

Factors of 9 in the set are $\{1, 3, 9\}$ (3 numbers). Numbers that are not factors of 9 are $\{0, 2, 4, 5, 6, 7, 8\}$ (7 numbers).

The probability that at least one factor of 9 is selected is $1$ minus the probability that no factor of 9 is selected.

Number of ways to choose 3 numbers that are not factors of 9:

Probability of no factor of 9:

Probability of at least one:

Is $\frac{17}{24} < \frac{1}{2}$? $\frac{1}{2} = \frac{12}{24}$. Since $\frac{17}{24} > \frac{12}{24}$, statement 3 is False.

Statement 4

Odd prime numbers in the set: $\{3, 5, 7\}$ (3 numbers). Numbers that are not odd primes: $\{0, 1, 2, 4, 6, 8, 9\}$ (7 numbers).

The event "the selected numbers are not odd prime numbers" means all three numbers are chosen from the set of non-odd-prime numbers.

Number of ways to choose 3 numbers from 7 non-odd-prime numbers:

Probability:

Is $\frac{7}{24} < \frac{1}{3}$? $\frac{1}{3} = \frac{8}{24}$. Since $\frac{7}{24} < \frac{8}{24}$, statement 4 is True.

Conclusion

The correct statements are 1, 2, and 4. There are 3 true statements.

Explanation

We are asked to find the interval of $x$ values where the function value $f(x)$ is greater than or equal to $5$ ($f(x) \geq 5$).

Let's check each part of the function one by one:

1. Interval $0 \leq x < 2$

   In this interval, $f(x) = 3$. Is $3 \geq 5$? No. So, this interval does not satisfy the condition.

2. Interval $2 \leq x < 5$

   In this interval, $f(x) = 5$. Is $5 \geq 5$? Yes. So, the interval $[2, 5)$ satisfies the condition.

3. Interval $5 \leq x < 8$

   In this interval, $f(x) = 7$. Is $7 \geq 5$? Yes. So, the interval $[5, 8)$ satisfies the condition.

4. Interval $8 \leq x < 10$

   In this interval, $f(x) = 9$. Is $9 \geq 5$? Yes. So, the interval $[8, 10)$ satisfies the condition.

Combining Intervals

We combine all intervals that satisfy the condition:

Note that the right end of the first interval ($5$) matches the left end of the second interval, and the right end of the second interval ($8$) matches the left end of the third interval. Since the connection is continuous (closed at the start, open at the end, then closed again at the next start), we can merge them into a single continuous interval.

Thus, the interval of $x$ values that satisfies the condition is $x \in [2, 10)$.