Set 4

Explanation

We will compare the total volume of oil $A$ received in April (30 days) and May (31 days). Note that the delivery schedule for oil $A$ is every $3$ days.

Here is the detailed comparison:

Notice that in May, there are $11$ shipments because May $31$ is also a scheduled delivery date (3 days after the 28th).

Thus, the difference in volume for oil $A$ is:

So, the difference in the volume of oil brand $A$ received is $65$ liters.

Explanation

First, let's identify the problem constraints:

1. Total budget: Rp$400.000,00$.
2. Total items: $4$ bouquets.
3. Mandatory purchase: At least $1$ Snack Bouquet (Rp$70.000,00$).

The first step is to subtract the price of the mandatory item from the total budget to find the remaining funds.

Now we have a remaining budget of Rp$330.000,00$ to buy $3$ more bouquets (since the desired total is $4$ items, and $1$ is already accounted for).

Let's check possible combinations to spend Rp$330.000,00$ on $3$ items. Looking at the price list, a Money Bouquet costs Rp$130.000,00$.

If we buy $2$ Money Bouquets:

The remaining money after buying 2 Money Bouquets is:

This remaining Rp$70.000,00$ is exactly enough to buy $1$ more Snack Bouquet.

So, the breakdown of our total purchase is:

The total items purchased are:

This combination fits both the budget and the item count requirement.

Explanation

We focus on the product "Minyak mu" which has an initial price of Rp$80.000,00$ per $5$ liters. We need to find the price difference after discount at two supermarkets: Indoapril and Betamidi.

Here is the comparison of the final price calculations:

From the table above, it is clear that the price of "Minyak mu" at Indoapril is Rp$60.000,00$ and at Betamidi is Rp$56.000,00$.

The price difference between the two can be easily calculated:

So, the price difference of "Minyak mu" at these two supermarkets is Rp$4.000,00$.

Explanation

Problem Visualization

Dimension Analysis

Given the prism dimensions:

• Length $p = 2l$
• Height $t = l$
• Width $l$
• Space diagonal $DF = 9$ cm

Calculation

The formula for the space diagonal is $d^2 = p^2 + l^2 + t^2$.

Substitute the known values:

Solve for $l$:

Rationalize the denominator:

Thus, the width of the prism is $\frac{3\sqrt{6}}{2}$ cm.

Explanation

Problem Visualization

Determining the Width

Use the space diagonal $DF$ to find $l$ with $d^2 = p^2 + l^2 + t^2$.

Calculating Area of Plane $ADFG$

Plane $ADFG$ is a rectangle with sides $AD = l$ and the front-face diagonal $AF$.

Compute $AF$ from the right triangle on the front face:

So the area is:

Thus, the area of plane $ADFG$ is $\frac{27\sqrt{5}}{2}$ cm².

Explanation

We can model this problem using the concept of a Geometric Series. The THR Bima receives each year forms a geometric sequence because each year the amount doubles from the previous year.

Given Bima's monthly salary is $Rp3.000.000,00$. The first step is to determine the first term ($a$), which is the THR amount in the first year:

Since the THR always doubles from the previous year, the common ratio ($r$) of this sequence is $2$. We need to find the total THR over $5$ years, which corresponds to calculating the sum of the first $5$ terms ($S_5$) of this geometric series.

The formula for the sum of the first $n$ terms of a geometric series is $S_n = \frac{a(r^n - 1)}{r - 1}$. Substituting the values $a = 300,000$, $r = 2$, and $n = 5$ into the formula:

Thus, the total THR Bima receives after working for $5$ years is $Rp9.300.000,00$.

Explanation

Yanuar's running training pattern forms a geometric sequence because the distance doubles every week. This means the common ratio ($r$) of the sequence is $2$.

The key information we have is the distance in the $3\text{rd}$ week ($U_3$), which is $4\text{ km}$. Using the formula for the $n\text{-th}$ term of a geometric sequence, $U_n = ar^{n-1}$, we can backtrack to find the initial distance in the first week ($a$):

Now that we know Yanuar started by running $1\text{ km}$ in the first week, we can calculate the total distance over $5$ weeks ($S_5$) using the geometric series sum formula $S_n = \frac{a(r^n - 1)}{r - 1}$:

Thus, the total distance covered by Yanuar during the training is $31\text{ km}$.

Explanation

This problem can be solved using the concept of a Linear Function or an Arithmetic Sequence, because the land price increase is constant (fixed) every year.

The initial land price ($a$) is given as $10,000,000$ and the annual price increase ($b$) is $1,200,000$. We want to find the land price after $5$ years.

If we model this as a linear function where $y$ is the land price and $x$ is the number of years, the equation is:

Substitute $x = 5$ to calculate the land price after $5$ years:

Thus, the land price in $5$ years will be $Rp16.000.000,00$ per square meter.

Explanation

Given the total length of the wire is $20\text{ m}$. This length represents the perimeter of the rectangular chicken coop. Let the length be $p$ and the width be $l$, we can write the perimeter equation:

Next, we want to maximize the area of the coop ($L$). The area of a rectangle is $L = p \times l$. Substitute the equation $l = 10 - p$ into the area formula:

This equation is a quadratic function whose curve is a downward-opening parabola (since the coefficient of $p^2$ is negative), meaning it has a maximum value at its vertex. The coordinate of the vertex ($p_{max}$) can be found using the formula $-\frac{b}{2a}$:

After finding the length $p = 5\text{ m}$, we can determine the width $l$:

Thus, for the area to be maximum, the length and width of the coop must be $5\text{ m}$ and $5\text{ m}$ respectively.

Explanation

Since each married couple must ride in the same car, we can treat each couple as a single unit. There are a total of $5$ couple units to be distributed into $2$ cars.

The capacity of each car is $6$ passengers, which is equivalent to $3$ married couples. Therefore, the number of couples in a single car cannot exceed $3$. The possible distributions of couples for Car A and Car B that satisfy the capacity constraint are:

1. Car A carries $3$ couples, Car B carries $2$ couples.
2. Car A carries $2$ couples, Car B carries $3$ couples.

For the first case, we choose $3$ couples out of the $5$ available to go into Car A. The remaining couples automatically go to Car B.

For the second case, we choose $2$ couples out of the $5$ available to go into Car A.

The total number of ways to arrange the passengers is the sum of both possibilities.

Explanation

The probability of selecting a student who wears a watch is determined by the ratio of students wearing watches to the total number of students. Let's calculate these components first.

Male students wearing watches:

Female students wearing watches:

Thus, the total number of students wearing watches ($n(A)$) and the total number of students ($n(S)$) are:

The probability of selecting a student who wears a watch is:

Therefore, the probability is $\frac{1}{2}$.

Explanation

The first step is to determine the total daily allowance budget for all children. We need to calculate the total monthly allowance based on the salary percentage, then divide it by the number of effective school days.

Total daily allowance:

Next, we use the allowance proportion ratio. It is known that child $A$ receives Rp$4,000$ with a proportion of $4$, while the total allowance is Rp$20,000$ with a total proportion of $4 + 3 + x$.

Determining the value of $x$:

Therefore, the value of $x$ is $13$.

Explanation

The concept used is Uniform Linear Motion. Since both cars are moving towards each other to meet, the total distance covered by both cars is equal to their initial separation distance.

The relationship between the distances covered can be written as:

We substitute the speed values of each car to find the time $t$:

Thus, they will pass each other after $5$ seconds.

Explanation

To solve this efficiently, we use the concept of work rate, where rate is inversely proportional to time. Let the work rates of each person be $v_A$, $v_B$, and $v_C$.

Based on the problem, we can set up three combined rate equations:

Instead of finding each value individually, we can sum all three equations. Notice that each variable appears twice.

From the equation above, we get their total combined work rate:

Since the total work rate is $\frac{1}{10}$ of the job per day, the time required to complete 1 full job is the reciprocal.

Thus, they need $10$ days.

Explanation

We start by defining variables for speed and time. Let $v_1$ be the speed of the first car, so the speed of the second car is $v_2 = v_1 + 15$. We know the difference in their travel times is 1 hour.

Since time is distance divided by speed ($t = \frac{s}{v}$), we can set up an equation based on this time difference:

Next, we solve this equation to find $v_1$. Multiply both sides by $v_1(v_1 + 15)$ to eliminate the denominators:

We factor the quadratic equation:

Since speed cannot be negative, we take $v_1 = 75$ km/h. Thus, the speed of the second car is:

Finally, we calculate the average speed of the two cars:

Explanation

To solve this problem easily, we use the principle of Sum of Values Balance. The combined total score must equal the sum of the total scores of each group.

Let $n$ be the initial number of students. We can set up an equation based on the total values as follows:

Next, we solve this simple linear equation. Notice how we group the terms containing $n$.

Thus, the number of students before being combined is $40$.

Explanation

To solve this problem, we need to understand the statistical properties regarding data transformation. Remember the key principles:

1. Mean changes according to all arithmetic operations (multiplication, division, addition, subtraction).
2. Range changes only with multiplication or division, but remains unchanged with addition or subtraction.

First, let's find the value of $p$ using the range information. Since the initial range of $6$ becomes $9$ after being multiplied by $p$, we have:

Next, we find the value of $q$ using the mean information. The mean is affected by both the multiplication by $p$ and the subtraction by $q$.

Finally, calculate the requested value using $p$ and $q$.

Thus, the value of $2p + q$ is $7$.

Explanation

Let $A$ be the hilltop, $B$ the base, $C$ the initial boat position, and $D$ the pier position. The height of the hill is $AB = h$. Use the right triangle $ABC$ with the elevation angle at $C$.

Using the elevation angle at $D$, apply cosine to find $AD$.