Set 2

Explanation

We are given three different land areas: $120\text{ m}^2$, $180\text{ m}^2$, and $240\text{ m}^2$. The goal is to divide these lands into plots of equal size, maximizing the area of each plot.

This problem can be solved by finding the Greatest Common Divisor (GCD) of the three numbers. The GCD will give us the largest value that evenly divides all three land areas.

First, we perform prime factorization for each area:

To determine the GCD, we take the common prime factors with the lowest exponent:

• Factor 2: the lowest exponent is $2^2$.
• Factor 3: the lowest exponent is $3^1$ (or 3).
• Factor 5: the lowest exponent is $5^1$ (or 5).

Thus, the GCD is:

Therefore, the largest possible area for each plot of land is $60\text{ m}^2$.

Explanation

Our goal is to find the area of each plot in three locations (A, B, and C) with the condition that the number of plots in each location must be the same and maximized.

Since the number of plots must be the same and maximized, we need to find the Greatest Common Divisor (GCD) of the land areas: $120$, $180$, and $240$. This GCD value will represent the number of plots to be created.

Let's perform prime factorization for the three numbers:

The GCD is obtained by multiplying the common prime factors with the lowest exponent:

• Factor 2: the lowest exponent is $2^2$.
• Factor 3: the lowest exponent is $3^1$ (or 3).
• Factor 5: the lowest exponent is $5^1$ (or 5).

So, the number of plots to be created in each location is 60 plots.

Next, we calculate the area of each plot for each location by dividing the total land area by the number of plots ($60$):

Location A:

Location B:

Location C:

Thus, the area of each plot from each location respectively is $2\text{ m}^2$, $3\text{ m}^2$, and $4\text{ m}^2$.

Explanation

We are given information regarding the original price of the motorcycle, the discount percentage, and the tax applied to that discount.

Given:

• Original price = $\text{Rp}75,000,000.00$
• Discount = $10\%$
• Tax on discount = $2.5\%$

First, we calculate the discount amount:

Next, we calculate the price of the motorcycle after the discount (before tax):

The problem states that there is a $2.5\%$ tax applied on the discount. This means we calculate $2.5\%$ of the discount amount ($\text{Rp}7,500,000.00$) and add it to the payable amount.

Finally, we calculate the total price the Father has to pay by adding the price after discount and the tax:

Thus, the final motorcycle price the Father has to pay is $\text{Rp}67,687,500.00$.

Explanation

First, we calculate the total capital (buying price) for $100$ durians.

Next, we calculate the total revenue from sales. The sales are divided into two parts: the first $70$ fruits and the remaining $30$ fruits.

Revenue from the first 70 fruits:

Revenue from the remaining 30 fruits:

Thus, the total revenue is:

Since the total revenue is greater than the total capital, the dealer makes a profit. The profit amount is the difference between revenue and capital:

Finally, we calculate the profit percentage relative to the initial capital:

So, the dealer makes a profit of $13.75\%$.

Explanation

The basic principle used here is Archimedes' principle. When balls are inserted into a cylinder full of water, water spills out. The volume of the spilled water is equal to the total volume of the balls inserted. Thus, the volume of water remaining in the cylinder is the difference between the initial volume of the cylinder and the total volume of the balls.

Volume of the Cylinder (Initial Water Volume):

Volume of One Ball:

Determining the Number of Balls (n):

It is given that the volume of water remaining is $64\pi \text{ cm}^3$. We use the equation:

Dividing both sides by $\pi$ to simplify the calculation:

Since the number of balls $n$ must be an integer and cannot exceed the cylinder's capacity, we take the nearest integer rounded down, which is $17$. If we were to insert $18$ balls, the total volume of the balls would exceed the volume of the cylinder ($18 \times 288\pi = 5184\pi > 5120\pi$), which is impossible.

Thus, the number of balls inserted is $17$ balls.

Explanation

To find the distance from point $Q$ to line $BD$, we will form triangle $BDQ$ and calculate its height from the vertex $Q$. The first step is to calculate the lengths of all three sides of this triangle.

Calculating Side $BD$ (Base Diagonal)

Line $BD$ is the diagonal of the cube's base with edge length $s = 4$.

Calculating Side $DQ$

Point $Q$ is at the midpoint of $FG$. We can find the length of $DQ$ by considering the right-angled triangle $DHQ$. First, we calculate the length of $HQ$ on the top face of the cube:

Next, use the Pythagorean theorem on triangle $DHQ$:

Calculating Side $BQ$

The length of $BQ$ can be calculated from the right-angled triangle $BFQ$, with $BF = 4$ and $FQ = 2$:

Calculating Distance from $Q$ to Line $BD$

Now we have triangle $BDQ$ with sides $BD = 4\sqrt{2}$, $DQ = 6$, and $BQ = 2\sqrt{5}$. Let $O$ be the perpendicular projection of $Q$ onto line $BD$.

Let the length $DO = x$, then the length $OB = 4\sqrt{2} - x$. The height of the triangle $QO$ (let's call it $y$) is the distance we are looking for. Using the two formed right-angled triangles, we obtain the equation:

Substituting the side values:

After finding $x = 3\sqrt{2}$, we substitute it back to find $y$:

Thus, the distance from point $Q$ to line $BD$ is $3\sqrt{2}$ units.

Explanation

This problem can be solved using the concept of an Arithmetic Sequence, where Mr. Brikson's salary increases by a constant amount each year. We need to determine the starting term and the common difference based on the age and salary information.

Analyzing the Information

We know that this year, when Mr. Brikson is $40$ years old, his salary is $\text{Rp}15,750,000.00$. The annual salary increase (common difference) is $\text{Rp}750,000.00$.

We need to find his salary when he is $54$ years old.

Calculating the Time Difference

The first step is to calculate the time elapsed between the target age of $54$ years and the current age of $40$ years.

This implies that Mr. Brikson's salary will undergo $14$ increments.

Calculating Salary at Age $54$

The salary at age $54$ is determined by adding the accumulated increase over $14$ years to the base salary.

Let's calculate the total increase:

So the total salary becomes:

Thus, Mr. Brikson's monthly salary when he is $54$ years old will be $\text{Rp}26,250,000.00$.

Explanation

Identifying Production Patterns

This problem applies the concept of an Arithmetic Sequence, where pan production increases by a constant amount each month. We have information about production in the $3\text{rd}$ month and the $6\text{th}$ month.

Given:

Determining Production Increase and Initial Production

In an arithmetic sequence, the $n\text{-th}$ term is given by $U_n = a + (n-1)b$. We can form two linear equations to find the initial value $a$ and the common difference $b$.

The difference between these two equations gives the value of $b$:

So, production increases by $250$ units each month. Next, substitute the value of $b$ into equation $(1)$ to find $a$:

Production in the first month is $1,000$ units.

Determining the Target Month

We need to find the $n\text{-th}$ month where production exceeds $10,000$ units.

Simplify the inequality:

Since $n$ must be an integer greater than $37$, the first month that satisfies the condition is the $38\text{th}$ month.

Note that in the $37\text{th}$ month, production exactly reaches $10,000$ units, so it has not yet exceeded the target.

Explanation

Collision Condition

A collision between the rocket and the aircraft carrier occurs when both are at the same position at the same time. Therefore, we must equate the rocket's position function with the aircraft carrier's position function.

Determining the Collision Time

We rearrange the quadratic equation into standard form to solve for $t$.

Next, we factorize this quadratic equation. We need to find two numbers that multiply to $-4000$ and add up to $60$. These numbers are $100$ and $-40$.

From this factorization, we obtain two possible values for $t$:

Since time ($t$) cannot be negative, the valid value is $40$ seconds. Thus, the collision occurs exactly at $40$ seconds.

Explanation

Analyzing Probability for the First Box

In the first box, there are a total of $4 + 3 = 7$ balls. We want to draw a red ball, of which there are $4$. The probability of drawing a red ball from the first box is:

Analyzing Probability for the Second Box

In the second box, there are a total of $7 + 3 = 10$ balls. We want to draw a white ball, of which there are $3$. The probability of drawing a white ball from the second box is:

Calculating the Combined Probability

Since drawing balls from the two boxes are independent events (the event in the first box does not affect the second box), the probability of both events occurring together is calculated by multiplying their individual probabilities.

Thus, the probability of drawing a red ball from the first box and a white ball from the second box is $\frac{12}{70}$.

Explanation

Identifying the Information

It is known that the total number of eggs in the box is $40$. Of these, $3$ are rotten. We are asked to find the probability of selecting a non-rotten egg.

Calculating the Number of Non-Rotten Eggs

The first step is to determine the number of favorable outcomes, which is selecting an egg in good condition. Since there are $3$ rotten eggs out of a total of $40$, the number of non-rotten eggs is the difference.

Calculating the Probability

The probability of an event is calculated by dividing the number of favorable outcomes by the total number of possible outcomes (sample space).

Thus, the probability of selecting one non-rotten egg is $\frac{37}{40}$.

Explanation

Determining the Time of Impact

The impact occurs when both objects are at the same position. Therefore, we need to equate the position equations of the rocket and the aircraft carrier to find the time of the event.

Next, we factor the quadratic equation to find the value of $t$.

From the factorization above, we obtain two solutions: $t = -100$ or $t = 40$. Since time cannot be negative, we take the positive value.

Calculating the Position of Impact

After finding the time of impact, we can calculate the position by substituting $t = 40$ into either position equation. We use the aircraft carrier's equation as it is simpler.

Finally, we convert the unit from meters to kilometers.

Thus, the position when the rocket exactly hits the aircraft carrier is $2.4$ km.

Explanation

Analyzing the Track

The track is circular with the checkpoint sequence $A \to B \to C \to D \to A$. Since the distance between checkpoints is equal, we can count the number of segments the runner has traversed. Starting from $A$, the runner has passed checkpoint $C$ $3$ times.

1. Reaching $C$ for the 1st time: $A \to B \to C$ ($2$ segments).
2. Reaching $C$ for the 2nd time: $C \to D \to A \to B \to C$ ($4$ segments).
3. Reaching $C$ for the 3rd time: $C \to D \to A \to B \to C$ ($4$ segments).

The total segments traversed is the sum of these stages.

Calculating Time per Segment

It is known that the time required to cover $10$ segments is $12$ minutes. We can determine the time taken for one segment.

Calculating Total Time to Reach C for the 4th Time

To reach checkpoint $C$ for the fourth time, the runner must complete another full lap from their last position (at the 3rd $C$). The additional path is $C \to D \to A \to B \to C$, which consists of $4$ segments.

The additional time required is:

Thus, the total time from the start to reaching checkpoint $C$ for the fourth time is the sum of the initial time and the additional time.

Therefore, the time required is $16.8$ minutes.

Explanation

Determining Time per Segment

Based on the previous track analysis, it is known that to pass checkpoint $C$ $3$ times, the runner covers a total of $10$ normal track segments in $12$ minutes. Thus, the travel time for one normal segment is:

Calculating Time for the New Path

We are asked to find the travel time from $A$ to $D$ given that the distance from $C$ to $D$ is widened to $2$ times the original length.

The normal route from $A$ to $D$ follows the path:

The breakdown of segments is:

1. $A \to B$: $1$ normal segment.
2. $B \to C$: $1$ normal segment.
3. $C \to D$: Since it is widened $2$ times, this is equivalent to $2$ normal segments.

The total effective segments to be covered is:

Thus, the total time required is:

Therefore, the time required for the runner to go from checkpoint $A$ to $D$ is $4.8$ minutes.

Explanation

Problem Analysis

We can model this problem as a system of linear equations with two variables. Let the price of meat per kg be $x$ and the price of fish per kg be $y$ (in thousands of rupiah).

Formulating the Equations

From the data of Shop A, the sale of $80$ kg of meat and $20$ kg of fish results in $2960$ thousand rupiah:

Simplify this equation by dividing both sides by $20$:

From the data of Shop B, the sale of $70$ kg of meat and $40$ kg of fish results in $3040$ thousand rupiah:

Simplify this equation by dividing both sides by $10$:

Solving the System of Equations

We will eliminate variable $y$ to find the value of $x$. Multiply equation $(1)$ by $4$ so that the coefficient of $y$ matches equation $(2)$:

Subtract this resulting equation by equation $(2)$:

Determining the Price of Fish

Substitute the value $x = 32$ into equation $(1)$ to find $y$:

The value $y = 20$ means the price of fish is $20$ thousand rupiah or Rp20,000.00.

Explanation

Modeling the Problem into Equations

Let $x$ be the daily wage and $y$ be the overtime wage. Based on the given information, we can form a system of linear equations with two variables:

Equation $(1)$ comes from Mr. Budi working $6$ days with $4$ overtime days, while equation $(2)$ comes from Mr. Agus working $5$ days with $2$ overtime days.

Solving the System of Equations

First, simplify equation $(1)$ by dividing both sides by $2$:

Now we have two simpler equations:

We can eliminate variable $y$ by subtracting equation $(3)$ from equation $(2)$: