Set 1

A farmer builds a chicken coop with a height of $3 \text{ m}$ and a rectangular base area of $256 \text{ m}^2$ . The walls of the coop are made of whole plywood panels that are not cut. The size of one plywood panel is $80 \text{ cm} \times 300 \text{ cm}$ .

The width of the land available to build the coop is $7.5 \text{ m}$ , and the minimum width so that the chickens do not experience stress is $5 \text{ m}$ . The price of one plywood panel is $\text{Rp}40,000.00$ . The cost to purchase panels with these conditions is ....

$\text{Rp}4,400,000.00$

$\text{Rp}4,560,000.00$

$\text{Rp}4,640,000.00$

$\text{Rp}4,720,000.00$

$\text{Rp}4,800,000.00$

Explanation

We will determine the coop dimensions and the number of panels needed.

Coop Dimension Analysis

Given:

Coop height $= 3 \text{ m}$ .
Base area $= 256 \text{ m}^2$ .
Panel size $= 80 \text{ cm} \times 300 \text{ cm} = 0.8 \text{ m} \times 3 \text{ m}$ .

Since the panel height matches the coop height ( $3 \text{ m}$ ), panels are installed vertically. To avoid cutting, the length and width of the coop must be multiples of the panel width, which is $0.8 \text{ m}$ .

Width constraint ( $w$ ):

5 \leq w \leq 7.5

We look for multiples of $0.8$ within that range:

$0.8 \times 7 = 5.6$
$0.8 \times 8 = 6.4$
$0.8 \times 9 = 7.2$

Check corresponding length ( $l$ ) so that area is $256$ and $l$ is also a multiple of $0.8$ :

If $w = 5.6$ , then $l = \frac{256}{5.6} \approx 45.71$ (Not a multiple of $0.8$ ).
If $w = 6.4$ , then $l = \frac{256}{6.4} = 40$ . Check multiple: $\frac{40}{0.8} = 50$ (Valid).
If $w = 7.2$ , then $l = \frac{256}{7.2} \approx 35.56$ (Not a multiple of $0.8$ ).

So, the possible coop dimensions are width $6.4 \text{ m}$ and length $40 \text{ m}$ .

Calculating Number of Panels

Number of panels for width side:

\frac{6.4}{0.8} = 8 \text{ panels}

Number of panels for length side:

\frac{40}{0.8} = 50 \text{ panels}

Total panels needed to surround the coop (2 width sides + 2 length sides):

\text{Total} = 2 \times (8 + 50) = 2 \times 58 = 116 \text{ panels}

Calculating Cost

The price of one panel is $\text{Rp}40,000.00$ .

\text{Cost} = 116 \times 40,000

\text{Cost} = 4,640,000

Thus, the required cost is $\text{Rp}4,640,000.00$ .

Rani has an empty rectangular plot of land behind her house measuring $1 \text{ m} \times 0.35 \text{ m}$ that can be planted with various types of fruit seedlings. Rani bought a package of fruit seedlings consisting of 2 strawberry seedlings, 1 orange seedling, and 4 mango seedlings for $\text{Rp}45,000.00$ . Planting one strawberry seedling requires a land area of $10 \text{ cm} \times 5 \text{ cm}$ , and planting one orange seedling requires twice the land area needed for one strawberry seedling.

It is known that planting one mango seedling requires a land area of $5 \text{ cm} \times 2.5 \text{ cm}$ . If Rani wants to buy two packages of fruit seedlings and fertilizer for the land to be planted with seedlings at a price of $\text{Rp}100,000.00/\text{m}^2$ , then the total amount of money Rani has to spend is ....

$\text{Rp}95,000.00$

$\text{Rp}96,000.00$

$\text{Rp}97,000.00$

$\text{Rp}98,000.00$

$\text{Rp}99,000.00$

Explanation

We will calculate the total cost required by Rani, which consists of the purchase price of the seedling packages and the cost of fertilizer for the land used.

Determining the Number of Seedlings

Rani buys 2 packages of fruit seedlings. Each package contains 2 strawberry seedlings, 1 orange seedling, and 4 mango seedlings. Thus, the total seedlings obtained from 2 packages are:

Strawberry: $2 \times 2 = 4$ seedlings.
Orange: $2 \times 1 = 2$ seedlings.
Mango: $2 \times 4 = 8$ seedlings.

Calculating the Required Land Area

We calculate the land area for each type of seedling:

Strawberry Seedlings

The land size for one strawberry seedling is $10 \text{ cm} \times 5 \text{ cm}$ .

L_{\text{strawberry}} = 10 \times 5 = 50 \text{ cm}^2

Total area for 4 strawberry seedlings:

L_{\text{total strawberry}} = 4 \times 50 = 200 \text{ cm}^2

Orange Seedlings

The land size for an orange seedling is twice the size of the land for a strawberry seedling. In this context, the length and width dimensions are each doubled ( $20 \text{ cm} \times 10 \text{ cm}$ ).

L_{\text{orange}} = 20 \times 10 = 200 \text{ cm}^2

Total area for 2 orange seedlings:

L_{\text{total orange}} = 2 \times 200 = 400 \text{ cm}^2

Mango Seedlings

The land size for one mango seedling is $5 \text{ cm} \times 2.5 \text{ cm}$ .

L_{\text{mango}} = 5 \times 2.5 = 12.5 \text{ cm}^2

Total area for 8 mango seedlings:

L_{\text{total mango}} = 8 \times 12.5 = 100 \text{ cm}^2

Total Land Area

L_{\text{total}} = 200 + 400 + 100

L_{\text{total}} = 700 \text{ cm}^2

We convert the area unit to square meters ( $1 \text{ m}^2 = 10,000 \text{ cm}^2$ ):

L_{\text{total}} = \frac{700}{10,000} = 0.07 \text{ m}^2

Calculating Total Cost

Seed Cost

The price of 1 package is $\text{Rp}45,000.00$ . For 2 packages:

\text{Seed Cost} = 2 \times 45,000 = \text{Rp}90,000.00

Fertilizer Cost

The price of fertilizer is $\text{Rp}100,000.00/\text{m}^2$ . For land with an area of $0.07 \text{ m}^2$ :

\text{Fertilizer Cost} = 0.07 \times 100,000 = \text{Rp}7,000.00

Total Expenditure

\text{Total} = 90,000 + 7,000

\text{Total} = 97,000

So, the amount of money Rani has to spend is $\text{Rp}97,000.00$ .

Mbox laptop store offers a $20\%$ discount for purchasing series $A$ laptops. Purchases at the store can be paid in monthly installments (with equal amounts, starting from the second month) with a simple interest of $6\%$ per year. The down payment required is at least $10\%$ of the initial price, and every subsequent minimum multiple will get an interest reduction of $1\%$ per year, up to a maximum of $2\%$ per year.

Juju buys a series $B$ laptop for $\text{Rp}20,000,000.00$ by installments. If Juju pays a down payment of $10\%$ , the interest Juju has to pay per month is ....

$\text{Rp}80,000.00$

$\text{Rp}90,000.00$

$\text{Rp}100,000.00$

$\text{Rp}110,000.00$

$\text{Rp}120,000.00$

Explanation

We will calculate the monthly interest Juju has to pay based on the laptop price, down payment, and applicable interest rate. Note that the $20\%$ discount only applies to series $A$ laptops, while Juju buys a series $B$ laptop, so the normal price applies.

Calculating Down Payment

Juju buys a laptop for $\text{Rp}20,000,000.00$ and pays a down payment of $10\%$ .

\text{Down Payment} = 10\% \times 20,000,000 = 2,000,000

So, the down payment paid is $\text{Rp}2,000,000.00$ .

Calculating Remaining Loan Amount

The remaining loan amount (principal) is the initial price minus the down payment.

\text{Remaining Loan} = 20,000,000 - 2,000,000 = 18,000,000

The remaining amount to be paid in installments is $\text{Rp}18,000,000.00$ .

Determining Interest Rate

The normal simple interest is known to be $6\%$ per year. There is a provision for interest reduction if the down payment is greater than $10\%$ (subsequent multiples). Since Juju only pays the minimum down payment of $10\%$ , she does not get an interest reduction.

The applicable interest rate remains $6\%$ per year.

Calculating Monthly Interest

Annual interest is calculated from the remaining loan:

\text{Annual Interest} = 6\% \times 18,000,000

Since the question asks for monthly interest, we divide the annual interest by 12:

\text{Monthly Interest} = \frac{6\%}{12} \times 18,000,000

\text{Monthly Interest} = 0.5\% \times 18,000,000

\text{Monthly Interest} = 90,000

So, the interest Juju has to pay per month is $\text{Rp}90,000.00$ .

In December 2024, Jeje bought a series $A$ laptop for $\text{Rp}12,000,000.00$ and paid in 15 installments with the minimum down payment to get the lowest interest rate. The amount of down payment Jeje has to pay is ....

$\text{Rp}1,200,000.00$

$\text{Rp}2,400,000.00$

$\text{Rp}2,880,000.00$

$\text{Rp}3,600,000.00$

$\text{Rp}4,000,000.00$

In celebration of its 22nd anniversary, the store offers an installment plan without a down payment with the lowest installment interest rate. Jojo buys a series $A$ laptop for $\text{Rp}15,000,000.00$ by installments. So that Jojo's monthly installment is not more than $\text{Rp}600,000.00$ , Jojo must pay in installments at least ... times.

Explanation

We will determine the minimum number of installments so that Jojo's monthly payment does not exceed the specified limit.

Calculating Price After Discount

Jojo buys a series $A$ laptop for $\text{Rp}15,000,000.00$ . Since there is a $20\%$ discount, the price to be paid is:

\text{Discounted Price} = (100\% - 20\%) \times 15,000,000

\text{Discounted Price} = 80\% \times 15,000,000 = 12,000,000

The principal loan amount is $\text{Rp}12,000,000.00$ .

Determining Interest and Principal Installment

In this anniversary promo, Jojo gets the offer with the lowest installment interest. Based on store rules, the minimum interest obtainable is $4\%$ per year (maximum reduction of $2\%$ from the base interest of $6\%$ ).

We calculate the monthly interest component:

\text{Monthly Interest} = \frac{4\%}{12} \times 12,000,000

\text{Monthly Interest} = \frac{1}{300} \times 12,000,000 = 40,000

The monthly installment amount consists of the principal installment and interest:

\text{Installment} = \frac{\text{Loan Principal}}{n} + \text{Monthly Interest}

Where $n$ is the number of installments.

Calculating Minimum Installment Duration

It is known that the monthly installment must not exceed $\text{Rp}600,000.00$ .

\frac{12,000,000}{n} + 40,000 \le 600,000

\frac{12,000,000}{n} \le 560,000

n \ge \frac{12,000,000}{560,000}

n \ge 21.42

Since the number of installments must be an integer, Jojo must pay in installments at least 22 times.

At the annual special event, the store offers an installment program without a down payment, with the first, second, third, fourth, and subsequent monthly installments being $\text{Rp}2,000,000.00$ , $\text{Rp}1,800,000.00$ , $\text{Rp}1,600,000.00$ , $\text{Rp}1,400,000.00$ , and so on until zero rupiah for purchasing a series $B$ laptop priced at $\text{Rp}10,000,000.00$ . With this purchase method, the price of the series $B$ laptop is ... percent more expensive compared to the initial price.

Explanation

We will calculate the total cost to be paid through this special installment scheme and compare it with the initial price of the laptop.

Installment Pattern

The installments paid form a decreasing arithmetic sequence:

First installment ( $a$ ) = $2,000,000$
Second installment = $1,800,000$
Difference ( $b$ ) = $1,800,000 - 2,000,000 = -200,000$

The installments continue until the value approaches zero (the last valid payment cannot be zero rupiah as it adds no cost, so we calculate until the last positive payment).

We determine the number of installments ( $n$ ) assuming the last installment ( $U_n$ ) is $200,000$ (one step before zero):

U_n = a + (n-1)b

200,000 = 2,000,000 + (n-1)(-200,000)

We divide both sides by $200,000$ to simplify:

1 = 10 + (n-1)(-1)

1 = 10 - n + 1

n = 10

So, there are 10 installments.

Calculating Total Payment

The total payment is the sum of the arithmetic series ( $S_{10}$ ):

S_n = \frac{n}{2}(a + U_n)

S_{10} = \frac{10}{2}(2,000,000 + 200,000)

S_{10} = 5(2,200,000)

S_{10} = 11,000,000

The total paid is $\text{Rp}11,000,000.00$ .

Calculating Price Increase Percentage

The initial price of the series $B$ laptop is $\text{Rp}10,000,000.00$ . Price difference = $11,000,000 - 10,000,000 = 1,000,000$ .

Percentage price increase:

\text{Percentage} = \frac{\text{Difference}}{\text{Initial Price}} \times 100\%

\text{Percentage} = \frac{1,000,000}{10,000,000} \times 100\%

\text{Percentage} = 10\%

So, the price of the series $B$ laptop is $10\%$ more expensive compared to the initial price.

A farmer rents 3 types of tractors to plow his rice field. If only tractor 1 is used, it takes $3$ days to plow the field. If only tractor 2 is used, it takes $4 \frac{1}{5}$ days. Meanwhile, if only tractor 3 is used, it takes $3.5$ days. The rental price for tractor 1 is $\text{Rp}54,000.00$ per day, tractor 2 is $\text{Rp}27,000.00$ per day, and the rental price for tractor 3 is $\text{Rp}45,000.00$ per day. Tractor rental does not have to be for a full day with proportional costs.

If on the first day the farmer uses only tractors 1 and 3, the portion of the field that can be plowed by both tractors is ... part.

A cuboid-shaped cardboard box has a base perimeter of $6x \text{ cm}$ and a height of $\left(5 - \frac{1}{3}x\right) \text{ cm}$ where $x$ is a real number.

If the width of the box's base is $(4x - 5) \text{ cm}$ , the length of the base is ... cm.

A cuboid-shaped cardboard box has a base perimeter of $6x \text{ cm}$ and a height of $\left(5 - \frac{1}{3}x\right) \text{ cm}$ where $x$ is a real number.

If the base of the box is rectangular with a width of $(4x - 5) \text{ cm}$ and $A$ is the set containing all real numbers $x$ that satisfy the dimensions of the cuboid, the correct set $A$ is ....

$\{x \mid x < 5, x \in \mathbb{R}\}$

$\{x \mid x > \frac{5}{4}, x \in \mathbb{R}\}$

$\{x \mid \frac{5}{4} < x < 5, x \in \mathbb{R}\}$

$\{x \mid x < 15, x \in \mathbb{R}\}$

$\{x \mid 5 < x < 15, x \in \mathbb{R}\}$

A cuboid-shaped cardboard box has a base perimeter of $6x \text{ cm}$ and a height of $\left(5 - \frac{1}{3}x\right) \text{ cm}$ where $x$ is a real number.

If the base of the box is rectangular with a width of $(4x - 5) \text{ cm}$ and $A$ is the set containing all real numbers $x$ that satisfy the dimensions of the cuboid, the maximum surface area of the box is ... $\text{cm}^2$ .

Explanation

We will find the value of $x$ that maximizes the surface area of the box, and then calculate that surface area.

Determining Surface Area Formula

The surface area of a cuboid ( $L$ ) can be calculated with the formula:

L = 2 \times \text{Base Area} + \text{Base Perimeter} \times \text{Height}

From previous questions, we know the box dimensions in terms of $x$ :

Length ( $p$ ) = $5 - x$
Width ( $l$ ) = $4x - 5$
Base Perimeter ( $K$ ) = $6x$
Height ( $t$ ) = $5 - \frac{1}{3}x$

Substitute these dimensions into the surface area formula:

L(x) = 2(p \times l) + K \times t

L(x) = 2((5 - x)(4x - 5)) + 6x\left(5 - \frac{1}{3}x\right)

Simplifying the Surface Area Function

We expand the equation above:

Base Area Part:

2((5 - x)(4x - 5)) = 2(20x - 25 - 4x^2 + 5x)

= 2(-4x^2 + 25x - 25)

= -8x^2 + 50x - 50

Perimeter times Height Part:

6x\left(5 - \frac{1}{3}x\right) = 30x - 2x^2

Total Surface Area:

L(x) = (-8x^2 + 50x - 50) + (30x - 2x^2)

L(x) = -10x^2 + 80x - 50

Finding the Maximum Value

The function $L(x)$ is a quadratic function with a negative $x^2$ coefficient, so its parabolic graph opens downwards and has a maximum value. The maximum value is reached when its first derivative equals zero ( $L'(x) = 0$ ).

L'(x) = -20x + 80

-20x + 80 = 0

20x = 80

x = 4

We check if $x = 4$ satisfies set $A$ (positive dimension condition) found previously, which is $\frac{5}{4} < x < 5$ . Since $1.25 < 4 < 5$ , the value $x = 4$ is valid.

Calculating Maximum Surface Area

Substitute $x = 4$ into the function $L(x)$ :

L(4) = -10(4)^2 + 80(4) - 50

L(4) = -10(16) + 320 - 50

L(4) = -160 + 320 - 50

L(4) = 110

So, the maximum surface area of the box is $110 \text{ cm}^2$ .

A building construction project is completed in 50 working days with the help of 80 workers. Kardi, as one of the workers, makes a plaster mixture of cement and sand every day with a ratio of $1:5$ . To assist Kardi's work, a truck is used to transport sand. The truck has a bed size with a length of $4 \text{ meters}$ , a width of $2 \text{ meters}$ , and a height of $1.5 \text{ meters}$ . The sand delivered by the truck is purchased at a price of $\text{Rp}300,000.00$ per cubic meter.

The price of sand transported in one full truck bed is ....

$\text{Rp}3,000,000.00$

$\text{Rp}3,200,000.00$

$\text{Rp}3,400,000.00$

$\text{Rp}3,600,000.00$

$\text{Rp}3,800,000.00$

A building construction project is completed in 50 working days with the help of 80 workers. Kardi, as one of the workers, makes a plaster mixture of cement and sand every day with a ratio of $1:5$ .

To make the plaster material, the amount of cement required by Kardi to be mixed with $0.24 \text{ cubic meters}$ of sand is ... kg. (Use sand density $1,400 \text{ kg/m}^3$ ).

It is known that in a day Kardi can plaster 10 walls, while his colleague, Anton, can plaster 6 walls of the same size.

If Kardi and Anton work together to plaster 56 walls, the time required for both of them to complete the task is ... days.

A building construction project is completed in 50 working days with the help of 80 workers.

If the project is carried out by 64 workers assuming each worker has the same working ability, the project can be completed in ... days.

A building construction project is completed in 50 working days with the help of 80 workers.

If the project is to be completed in 40 days assuming each worker has the same ability, the number of additional workers needed so that the project can be completed as planned is ... people.

Consider the following apple mass table.

Name	Mass (g)	Mass (g)	Mass (g)	Mass (g)	Mass (g)	Mass (g)
Ferdian	$157$	$163$	$182$	$171$	$167$	$x$
Belle	$180$	$175$	$162$	$154$	$161$	$170$

Ferdian and Belle are each collecting $6$ apples. The mass of the apples is shown in the table above. The average mass of apples collected by Ferdian is $3 \text{ grams}$ greater than the average mass of apples collected by Belle.

The value of $x$ is ....

Consider the following apple mass table.

Name	Mass (g)	Mass (g)	Mass (g)	Mass (g)	Mass (g)	Mass (g)
Ferdian	$157$	$163$	$182$	$171$	$167$	$x$
Belle	$180$	$175$	$162$	$154$	$161$	$170$

Given that the value of $x$ is $180$ . Based on the information above, determine whether each of the following statements is TRUE or FALSE. Order your answer from the first statement to the third.

$25\%$ of the smallest apples have a mass of $< 162 \text{ grams}$ .
$25\%$ of the largest apples have a mass of $> 180 \text{ grams}$ .
$50\%$ of the smallest apples have a mass of $< 170 \text{ grams}$ .

Explanation

We will combine all apple mass data from Ferdian and Belle, sort it, and then evaluate each statement.

Combining and Sorting Data

Ferdian's apple mass data (with $x = 180$ ):

157, 163, 182, 171, 167, 180

Belle's apple mass data:

180, 175, 162, 154, 161, 170

Combined data (total $12$ apples):

157, 163, 182, 171, 167, 180, 180, 175, 162, 154, 161, 170

Sorted data from smallest to largest:

154, 157, 161, 162, 163, 167, 170, 171, 175, 180, 180, 182

Total data ( $n$ ) = $12$ .

Evaluating First Statement

Statement: $25\%$ of the smallest apples have a mass of $< 162 \text{ grams}$ .

We calculate $25\%$ of the total data:

25\% \times 12 = 3 \text{ items}

The $3$ smallest data points are $154$ , $157$ , $161$ . All these values are less than $162$ grams.

Conclusion: First statement is TRUE.

Evaluating Second Statement

Statement: $25\%$ of the largest apples have a mass of $> 180 \text{ grams}$ .

We calculate $25\%$ of the total data:

25\% \times 12 = 3 \text{ items}

The $3$ largest data points are $182$ , $180$ , $180$ . Not all of these are greater than $180$ (since two values are exactly $180$ ).

Conclusion: Second statement is FALSE.

Evaluating Third Statement

Statement: $50\%$ of the smallest apples have a mass of $< 170 \text{ grams}$ .

We calculate $50\%$ of the total data:

50\% \times 12 = 6 \text{ items}

The $6$ smallest data points are $154$ , $157$ , $161$ , $162$ , $163$ , $167$ . The largest value in this group is $167$ , which is less than $170$ .

Conclusion: Third statement is TRUE.

So, the sequence of truth is TRUE - FALSE - TRUE.

Consider the following apple mass table.

Name	Mass (g)	Mass (g)	Mass (g)	Mass (g)	Mass (g)	Mass (g)
Ferdian	$157$	$163$	$182$	$171$	$167$	$x$
Belle	$180$	$175$	$162$	$154$	$161$	$170$

Given that the value of $x$ is $180$ . If they then exchange apples so that Ferdian carries the heavy apples and Belle carries the light ones, the difference in the mass of apples carried by Ferdian and Belle is ....

$90 \text{ grams}$

$92 \text{ grams}$

$94 \text{ grams}$

$96 \text{ grams}$

$98 \text{ grams}$

A coffee shop is known to be able to sell 200 cups of iced coffee in a normal day at a price of $\text{Rp}12,000.00$ per cup. After conducting various surveys, the coffee shop owner concludes that for every decrease in the price of iced coffee by $\text{Rp}1,000.00$ , the number of iced coffees sold will increase by 20 cups per day. In order not to lose money, the minimum daily income of the shop is $\text{Rp}2,100,000.00$ to cover the shop's operational costs.

The number of iced coffees sold per day when the price of iced coffee becomes $\text{Rp}9,000.00$ is ....

If the shop owner wants to increase the number of iced coffee sales, but with the same income as a normal day, the number of iced coffees sold is ... cups.

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