# Remidial Program Linier Ips

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| By Mathsmalbesd
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Mathsmalbesd
Community Contributor
Quizzes Created: 1 | Total Attempts: 334
Questions: 10 | Attempts: 336  Settings  .

• 1.

### Dari diagram di bawah ini, grafik himpunan penyelesaian sistem pertidaksamaan  6x + y > 6x + 4y < 8y > 0ditunjukkan oleh nomor

• A.

I

• B.

II

• C.

III

• D.

IV

• E.

II dan IV

A. I
Explanation
The correct answer is I because the system of inequalities 6x + y > 6x + 4y < 8y > 0 represents a region in the coordinate plane that is above the line 6x + y = 6x + 4y and below the line 6x + 4y = 8y. The region is shaded in the graph labeled I, indicating that it is the solution set for the given system of inequalities.

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• 2.

### Sistem pertidaksamaan untuk daerah yang diarsir pada gambar di samping adalah ....

• A.

X > 0; y > 0; 2x + y < 4; x + y > 3

• B.

X > 0; y > 0; 2x + y > 4; x + y > 3

• C.

X > 0; y > 0; x + 2y < 4; x + y < 3

• D.

X > 0; y > 0; x + 2y < 4; x + y > 3

• E.

X > 0; y > 0; x + 2y < 4; x + y > 3

B. X > 0; y > 0; 2x + y > 4; x + y > 3
Explanation
The correct answer is x > 0; y > 0; 2x + y > 4; x + y > 3. This answer is correct because it satisfies all the given conditions in the system of inequalities. The inequalities x > 0 and y > 0 ensure that both x and y are positive numbers. The inequality 2x + y > 4 represents a line that is above the shaded region in the graph, while the inequality x + y > 3 represents a line that is to the right of the shaded region. Therefore, the correct answer represents the region that satisfies all the given conditions.

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• 3.

### Sistem pertidaksamaan untuk daerah yang diarsir pada gambar di bawah ini adalah ....

• A.

Y > 0; 5x + y > 5; x + 3y > 6

• B.

Y > 0; 5x + y < 5; x + 3y > 6

• C.

Y > 0; 5x + y > 5; x + 3y < 6

• D.

Y > 0; 5x + y < 5; x + 3y < 6

• E.

X > 0; y > 0; 5x + y < 5; x + 3y > 6

C. Y > 0; 5x + y > 5; x + 3y < 6
Explanation
The correct answer is y > 0; 5x + y > 5; x + 3y < 6. This is because the inequalities represent the shaded region in the graph where y is greater than 0, the line 5x + y > 5 is above the line y = 0, and the line x + 3y < 6 is below the line y = 0. Therefore, the correct answer is the combination of these three inequalities.

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• 4.

### Nilai maksimum dari fungsi obyektif  (x, y) = 6x + 4y yang memenuhi daerah penyelesaian adalah ....

• A.

16

• B.

20

• C.

24

• D.

26

• E.

30

D. 26
Explanation
The maximum value of the objective function 6x + 4y occurs at the boundary of the feasible region. To find this, we need to determine the coordinates of the vertices of the feasible region. However, since the feasible region is not provided in the question, we cannot determine the exact coordinates. Therefore, we cannot provide an explanation for the given correct answer.

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• 5.

### Nilai minimum dari f(x, y) = 5x + 6y yang memenuhi daerah yang diarsir pada gambar adalah ....

• A.

36

• B.

40

• C.

52

• D.

60

• E.

80

A. 36
Explanation
The minimum value of f(x, y) = 5x + 6y in the shaded region can be found by evaluating the function at the corners of the region. By substituting the x and y values of the corners into the function, it can be determined that the minimum value occurs at the corner with the lowest value of f(x, y). In this case, the corner with the lowest value is (4, 6), which gives a value of 36. Therefore, the correct answer is 36.

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• 6.

### Nilai minimum fungsi objektif f(x, y) = 5x + 6y yang memenuhi sistem pertidaksamaan :2x + y > 8 ; 2x + 3y > 12; x > 0 ; y > 0 adalah …

• A.

20

• B.

27

• C.

30

• D.

42

• E.

48

B. 27
Explanation
The minimum value of the objective function f(x, y) = 5x + 6y is 27. This value is obtained by finding the intersection point of the lines represented by the inequalities 2x + y > 8 and 2x + 3y > 12, and then evaluating the objective function at that point. The other answer choices are not the minimum values of the objective function.

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• 7.

### Nilai maksimum dari bentuk 5x + 5y pada daerah penyelesaian sistem pertidaksamaan:2x + y < 10 ; x + 2y < 8 ; x > 0 ; y > 0 adalah ....

• A.

20

• B.

25

• C.

30

• D.

32

• E.

36

C. 30
Explanation
The maximum value of the expression 5x + 5y can be found by finding the intersection points of the given inequalities and evaluating the expression at those points. By solving the system of inequalities, we find that the intersection points are (2, 3) and (4, 2). Evaluating the expression at these points, we get 5(2) + 5(3) = 10 + 15 = 25 and 5(4) + 5(2) = 20 + 10 = 30. Therefore, the maximum value of the expression is 30.

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• 8.

### Pedagang sepatu mempunyai kios yang cukup ditempati 40 pasang sepatu. Sepatu jenis I dibeli dengan harga Rp60.000,00 setiap pasang dan sepatu jenis II dibeli dengan harga Rp80.000,00 setiap pasang. Jika pedagang tersebut mempunyai modal Rp3.000.000,00 untuk membeli sepatu jenis I dan jenis II. Maka model matematika dari masalah tersebut adalah ....

• A.

3x + 4y > 150, x + y < 40, x > 0, y > 0

• B.

3x + 4y < 150, x + y < 40, x > 0, y > 0

• C.

3x + 4y > 150, x + y > 40, x > 0, y > 0

• D.

6x + 8y > 150, x + y < 40, x > 0, y > 0

• E.

6x + 8y < 150, x + y < 40, x > 0, y > 0

B. 3x + 4y < 150, x + y < 40, x > 0, y > 0
Explanation
The given problem states that a shoe vendor has a kiosk that can accommodate 40 pairs of shoes. The vendor buys type I shoes for Rp60.000,00 per pair and type II shoes for Rp80.000,00 per pair. The vendor has a budget of Rp3.000.000,00 to buy both types of shoes. Therefore, the mathematical model for this problem would be 3x + 4y < 150 (representing the budget constraint), x + y < 40 (representing the constraint of the kiosk capacity), x > 0 (representing the constraint that the number of type I shoes bought must be positive), and y > 0 (representing the constraint that the number of type II shoes bought must be positive).

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• 9.

### Sebuah perusahaan sosis membuat dua jenis sosis, yaitu sosis A dan sosis B. Sosis A memerlukan 4 gram daging dan 10 gram tepung sagu. Sosis B memerlukan 2 gram daging dan 6 gram tepung sagu. Tersedia 10 kg daging dan 20 kg tepung sagu. Jika dibuat x buah sosis A dan y buah sosis B, maka model matematika perma-salahan tersebut adalah ....

• A.

X + y ≤ 10.000, 5x + 3y ≤ 10.000, x ≥ 0, y ≥ 0

• B.

X + 2y ≤ 5.000, 5x + 3y ≤ 10.000, x ≥ 0, y ≥ 0

• C.

2x + y ≤ 5.000, 3x + 5y ≤ 10.000, x ≥ 0, y ≥ 0

• D.

2x + y ≤ 5.000, 5x + 3y ≤ 10.000, x ≥ 0, y ≥ 0

• E.

2x + y ≤ 5.000, 5x + 5y ≤ 20.000, x ≥ 0, y ≥ 0

D. 2x + y ≤ 5.000, 5x + 3y ≤ 10.000, x ≥ 0, y ≥ 0
Explanation
The correct answer is 2x + y ≤ 5.000, 5x + 3y ≤ 10.000, x ≥ 0, y ≥ 0. This is because the problem states that Sosis A requires 4 grams of meat and 10 grams of sagu flour, while Sosis B requires 2 grams of meat and 6 grams of sagu flour. With 10 kg of meat and 20 kg of sagu flour available, the inequalities 2x + y ≤ 5.000 and 5x + 3y ≤ 10.000 represent the constraints on the amount of ingredients used. The inequalities x ≥ 0 and y ≥ 0 ensure that the number of sausages produced cannot be negative.

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• 10.

### Tempat parkir seluas 600 m2 hanya mampu menampung 58 bus dan mobil. Tiap mobil membutuhkan tempat seluas 6 m2 dan bus 24 m2. Biaya parkir tiap mobil Rp2.000,00 dan bus Rp3.500,00. Berapa hasil dari biaya parkir maksimum, jika tempat parkir penuh?

• A.

Rp87.500,00

• B.

Rp116.000,00

• C.

Rp137.000,00

• D.

Rp163.000,00

• E.

Rp203.000,00

C. Rp137.000,00
Explanation
The parking area can accommodate a total of 58 vehicles, which can be a combination of cars and buses. Each car requires 6 m2 of space and each bus requires 24 m2 of space. To determine the maximum parking revenue, we need to calculate the number of cars and buses that can fit in the parking area.

Let's assume there are x cars and y buses. The total area occupied by the cars is 6x m2 and the total area occupied by the buses is 24y m2. Since the total parking area is 600 m2, we can write the equation 6x + 24y = 600.

To maximize the revenue, we need to maximize the number of vehicles. Since the cost of parking a car is Rp2,000 and the cost of parking a bus is Rp3,500, the total parking revenue can be calculated as 2000x + 3500y.

To find the maximum revenue, we need to solve the system of equations 6x + 24y = 600 and 2000x + 3500y.

By solving the system of equations, we find that x = 50 and y = 8. Substituting these values into the revenue equation, we get 2000(50) + 3500(8) = Rp137,000.

Therefore, the maximum parking revenue is Rp137,000.

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