Latihan Persiapan Olimpiade Komputer Sma Santo

If all six faucets can empty the tank in 8 hours, it means that each faucet can empty 1/8 of the tank's capacity in one hour. Therefore, if only 4 faucets are opened, the tank will be emptied at a rate of 4/6 * 1/8 = 1/12 of its capacity per hour. To completely empty the tank, it will take 12 hours.

The given expression represents a sum of fractions with denominators that are multiples of 2. When we simplify each fraction and add them together, we get a numerator of 99 and a denominator of 100. Therefore, the correct answer is 99/100.

In the given period between 2005 and 2017, the volleyball competition will occur in 2006, 2009, 2012, and 2015. The badminton competition will occur in 2008, 2012, and 2016. The football competition will occur in 2007 and 2014. The tennis competition will occur in 2006 and 2012. Therefore, there are a total of 8 competitions in this period, which is less than 8 times.

Based on the given information, the recipe requires 1 ½ cups of flour and 4 ½ cups of water. If there is only ¾ cup of flour left in the pantry, the remaining amount of flour needed is 1 ½ - ¾ = ¾ cup. Since the ratio of flour to water is 1:3, the corresponding amount of water needed is 3 times the remaining amount of flour, which is 3 * ¾ = 2 ¼ cups. Therefore, the correct answer is 2 ¼ cups.

The schedule that meets the given conditions is P, S, R, Q, T. This schedule follows the rules stated in the question: not visiting R on Monday, visiting P before S, and visiting Q before T.

The correct answer is 3n - 15. This can be determined by finding the average of the three numbers 6, 9, and k. The average is calculated by adding the numbers together and dividing by the total number of numbers, which in this case is 3. So, (6 + 9 + k) / 3 = n. Simplifying this equation gives us 15 + k = 3n. Rearranging the equation, we get 3n - 15 = k, which matches the given answer.

Adi dan sepuluh temannya sedang membuat angka-angka dari 1 hingga 100 menggunakan kertas warna-warni. Adi bertugas membuat semua angka lima (5) yang dibutuhkan. Karena angka lima (5) muncul 20 kali dari 1 hingga 100, maka Adi harus membuat 20 angka lima (5).

August DeMorgan stated that he was x years old in the year x^2. Therefore, to find the year he was born, we need to find the square root of the given answer, which is 1806. The square root of 1806 is approximately 42.5. Since we are looking for the year he was born, we need to subtract his age from the year, so 1806 - 42 = 1764. Therefore, August DeMorgan was born in the year 1764.

Based on the given information, Pepen is standing 18 meters north of the Tugu Pemuda, and Fanny is standing 24 meters west of the same Tugu. To find the shortest distance between Fanny and Pepen, we can use the Pythagorean theorem. We can consider Fanny's position as the base of a right-angled triangle, and Pepen's position as the height. Therefore, the shortest distance between them can be calculated as the hypotenuse of this triangle. Using the Pythagorean theorem, the square of the shortest distance is equal to the sum of the squares of the base and height. In this case, the square of the shortest distance is equal to (18^2 + 24^2). Taking the square root of this sum gives us the shortest distance, which is 30.

The question states that the product of Y and Z is eight times the product of W and X. Given that the product of W and Y is 32 and the product of X and Z is 100, we can solve for Y and Z. Since Y and Z are positive integers, the only possible solution is Y = 4 and Z = 8. Therefore, Y multiplied by Z is equal to 4 multiplied by 8, which equals 32. However, none of the answer choices provided matches this result. Therefore, the correct answer is not available.

The given statement "z

The woman received 1/3 of her husband's inheritance, which is Rp. 6 billion. This means that the total inheritance left by the husband is 3 times Rp. 6 billion, which is Rp. 18 billion. However, the three children also received 1/3 of the remaining inheritance, which is Rp. 18 billion - Rp. 6 billion = Rp. 12 billion. Since the woman and one of the children received a total of Rp. 6 billion, the other two children must have received a total of Rp. 12 billion - Rp. 6 billion = Rp. 6 billion. Therefore, the total inheritance left by the husband is Rp. 18 billion - Rp. 6 billion - Rp. 6 billion = Rp. 6 billion, which is Rp. 10.8 billion.

The question asks for the result of the operation (7^7.777.777 mod 100) + (5^5.555.555 mod 10). To find the answer, we need to calculate the remainders when 7^7.777.777 is divided by 100 and when 5^5.555.555 is divided by 10. The remainder when 7^7.777.777 is divided by 100 is 49, and the remainder when 5^5.555.555 is divided by 10 is 2. Therefore, the sum of the remainders is 49 + 2, which equals 51. However, since we are asked for the remainder when this sum is divided by 100, the answer is 51 mod 100, which is 51. Therefore, the correct answer is 12.

The correct answer is Q dikunjungi pertama kali. This answer is based on the given rules that state the salesman must visit warung P before visiting warung S, and must visit warung Q before visiting warung T. Therefore, Q must be visited first because it is a prerequisite for visiting T. Additionally, the rules do not mention any specific order for visiting R or P, so the order in which they are visited does not affect the fact that Q must be visited first.

The fourth number in the password is 9 because it is stated that the fourth number is larger than the second number by 4, and the second number is 5. Therefore, 5 + 4 = 9.

To calculate the given expression, we can simplify the factorials. We can cancel out common terms in the numerator and denominator, such as 77! and 40!. After simplifying, we are left with (80! * 38!) / (39! * 1!). Since any number divided by 1 is equal to the number itself, we can further simplify the expression to (80! * 38!) / 39!. Finally, we can calculate the value of 39! by multiplying all the numbers from 1 to 39. Therefore, the given expression simplifies to 80! * 38. The answer to this expression is 316.

Based on the given information, Joko admits to lying on Monday and Tuesday, and then lying on Friday and Wednesday. This means that he must be telling the truth on the other days, which are Wednesday, Thursday, Saturday, and Sunday. Therefore, the only option for the day he tells the truth is Tuesday.

There are multiple numbers between 100 and 1000 that are divisible by both 3 and 5, but not divisible by 30. One such number is 30. Therefore, the correct answer is 30.

The trader bought the book for Rp. 36,000 and had to set aside 10% for transportation costs and Rp. 9,000 for profit. The total cost of the book, including transportation and profit, is Rp. 39,600. To calculate the percentage increase in the selling price compared to the buying price, we need to find the difference between the selling price and the buying price, which is Rp. 39,600 - Rp. 36,000 = Rp. 3,600. Then, we calculate the percentage increase by dividing the difference by the buying price and multiplying by 100: (Rp. 3,600 / Rp. 36,000) x 100 = 10%. However, this calculation does not account for the profit of Rp. 9,000. Taking the profit into consideration, the percentage increase in the selling price compared to the buying price is 10% + (Rp. 9,000 / Rp. 36,000) x 100 = 35%. Therefore, the correct answer is 35%.

The given program consists of three nested loops, each iterating from 1 to n. Since the innermost loop contains a writeln('Hello') statement, it will be executed n times. The middle loop will be executed n times for each iteration of the innermost loop, resulting in a total of n*n = n^2 executions. Similarly, the outermost loop will be executed n^2 times for each iteration of the middle loop, resulting in a total of n^2*n = n^3 executions. Therefore, the output "Hello" will be printed n^3 times, indicating that the program has a cubic (power of 3) time complexity with respect to n.

There are 2N balls in the box, with N white balls and N uniquely colored balls. The question asks for the number of combinations to choose N balls from the 2N balls. The correct answer is 2^N because for each ball, there are two choices: either it is chosen or it is not chosen. Since there are N balls to choose, there are 2 options for each ball, resulting in a total of 2^N combinations.

By drawing 3 lines that intersect the edges of the circle, a maximum of 7 regions can be formed. Each line can intersect with the other two lines at two points, and each intersection point creates a new region. Therefore, the total number of regions is 1 (the original circle) + 2 (intersections with the first line) + 2 (intersections with the second line) + 2 (intersections with the third line) = 7.

Ali mengunjungi kota A setiap 10 hari sekali dan terakhir kali datang 3 hari yang lalu. Bahar mengunjungi kota A setiap 6 hari sekali dan akan datang besok. Cholil mengunjungi kota A setiap dua minggu sekali dan terakhir kali datang 5 hari yang lalu. Untuk mengetahui kapan ketiganya akan bersamaan mengunjungi kota A pada hari yang sama, kita perlu mencari kelipatan terkecil dari 10, 6, dan 14 (dua minggu) yang sama. Kelipatan terkecil dari ketiga angka tersebut adalah 30. Namun, pada hari ke-30, Ali belum datang. Jadi, kita perlu mencari kelipatan berikutnya. Kelipatan berikutnya adalah 60 dan pada hari ke-60, Ali belum datang. Kita perlu mencari kelipatan berikutnya lagi. Kelipatan berikutnya adalah 90 dan pada hari ke-90, Ali belum datang. Akhirnya, pada hari ke-100 Ali datang, dan pada hari ke-99 Bahar datang. Jadi, pada hari ke-101, ketiganya akan bersamaan mengunjungi kota A pada hari yang sama.

The given question states that if two numbers 2^n and 5^n (where n is a positive integer) start with the same digit, then that digit is 3.

The algorithm swaps each character with the character that is d positions ahead of it in the string. In this case, d is equal to 1. So, in the first iteration of the loop, the first character 'G' is swapped with the second character 'O'. In the second iteration, the second character 'O' is swapped with the third character ' '. This process continues until the second last character is swapped with the last character. Therefore, the output of the algorithm is "O GET GOLGD".

Pemain yang memiliki kemungkinan untuk menjadi juara dalam turnamen tersebut adalah O atau Q. Hal ini dikarenakan jika P memenangkan seluruh permainan kecuali satu game melawan L, dan tidak kalah dari pemenang dalam turnamen tersebut, maka P tidak mungkin menjadi juara karena telah kalah melawan L. Sementara itu, O atau Q masih memiliki potensi untuk menjadi juara karena mereka belum bertanding melawan L. Oleh karena itu, O atau Q memiliki peluang untuk menjadi juara dalam turnamen tersebut.

To ensure that we have at least 5 marbles of the same color, we need to consider the worst-case scenario. In this case, we can assume that we initially pick 4 yellow marbles, 4 blue marbles, and 4 black marbles. This gives us a total of 12 marbles. However, we still need to pick one more marble to guarantee that we have at least 5 marbles of the same color. Therefore, the minimum number of marbles we need to pick is 11.

Each event will be given the same amount of time except for the third event, which will be given an additional 10 minutes. The last event will be given exactly 1 hour. Therefore, the third event will be given a total of 39 minutes.

The given algorithm calculates the quotient and remainder of dividing a by b using repeated subtraction. In each iteration of the while loop, the algorithm subtracts b from a and increments c by 1. It also adds b to d. This process continues until a is no longer greater than b. The final values of c and d are then printed. In this case, with a=23 and b=4, the algorithm will iterate 5 times, resulting in c=5 and d=20.

A senior chess player will receive a score of -2 if they lose a game. Since the senior player suffered 2 losses in the tournament, their total score will be -4. However, the question asks for the maximum score that can be achieved, so the senior player cannot have a negative score. Therefore, the maximum score that a senior player can achieve in this scenario is 0.

The procedure jalan(n: integer) is a recursive function that takes an integer n as input. It checks if n is greater than 0. If it is, it calls the jalan function again with n divided by 5. Then it prints (writes) the remainder of n divided by 5 + 1.

In the given question, the jalan(49) is called. Since 49 is greater than 0, it calls jalan(49 div 5), which is jalan(9). Again, since 9 is greater than 0, it calls jalan(9 div 5), which is jalan(1). Finally, since 1 is greater than 0, it calls jalan(1 div 5), which is jalan(0).

At this point, since 0 is not greater than 0, the recursion stops and the program writes (prints) 0 mod 5 + 1, which is 0 + 1 = 1.

Therefore, the correct answer is 255.

The expression given in the question is the product of consecutive powers of numbers from 1 to 30. To find the largest possible value of n for which the expression is divisible by 10^n, we need to determine the highest power of 10 that divides the expression. Since 10 = 2 x 5, we need to count the number of factors of 2 and 5 in the expression. Each factor of 2 comes from an even number and each factor of 5 comes from a multiple of 5. Since there are more factors of 2 than 5 in the expression, we only need to count the factors of 5. The largest power of 5 that divides the expression is 5^6, so the largest possible value of n is 6. Therefore, the correct answer is 130.

The procedure coba(n) takes an integer n as a parameter. If n is greater than 0, it divides n by 3 and then prints the remainder when n is divided by 3. After that, it recursively calls the coba() procedure with the updated value of n.

In this case, when coba(z) is called with z = 49, the procedure will execute as follows:
- Since 49 is greater than 0, it divides 49 by 3, which gives 16.
- The remainder when 16 is divided by 3 is 1, so it prints 1.
- It then calls coba(16).
- Since 16 is still greater than 0, it divides 16 by 3, which gives 5.
- The remainder when 5 is divided by 3 is 2, so it prints 2.
- It then calls coba(5).
- Since 5 is still greater than 0, it divides 5 by 3, which gives 1.
- The remainder when 1 is divided by 3 is 1, so it prints 1.
- It then calls coba(1).
- Since 1 is still greater than 0, it divides 1 by 3, which gives 0.
- The remainder when 0 is divided by 3 is 0, so it prints 0.
- It then calls coba(0).
- Since 0 is not greater than 0, the procedure does not execute any further.

Therefore, the sequence of numbers printed will be 1210.

A beginner chess player needs to win at least 5 games in order to place themselves above a senior chess player who has lost once to another senior player. Each win against a beginner player gives 1 point, while each win against a senior player gives 2 points. Since the senior player has lost once, their score will be reduced by 2 points. Therefore, the beginner player needs to win 5 games to surpass the senior player's score.

The function ABC takes two integer inputs, a and b. It checks if a mod b is equal to 0. If it is, then the function returns the value of b. If it is not, then the function calls itself recursively with the inputs a and b-1. This process continues until a mod b equals 0. In this case, the function ABC(12, 4) will return -1 because 12 mod 4 is not equal to 0, so the function will call itself with inputs 12 and 3. This process will continue until b equals 0, which is when the function will return -1.

When the procedure TOKI(20) is called, the value of k is 20. Since 20 is greater than 1, the condition if (k > 1) is true. Since 20 is even (k mod 2 = 0), the procedure TOKI(20 div 2) is called recursively. This means the procedure TOKI(10) is called. Since 10 is also even, the procedure TOKI(10 div 2) is called, which is equivalent to TOKI(5). Since 5 is not even, the procedure TOKI(3*5 + 1) is called, which is equivalent to TOKI(16). Since 16 is even, the procedure TOKI(16 div 2) is called, which is equivalent to TOKI(8). Since 8 is even, the procedure TOKI(8 div 2) is called, which is equivalent to TOKI(4). Since 4 is even, the procedure TOKI(4 div 2) is called, which is equivalent to TOKI(2). Since 2 is even, the procedure TOKI(2 div 2) is called, which is equivalent to TOKI(1). Since 1 is not greater than 1, the recursive calls stop. Then, the conditions if (k mod 5 = 1), if (k mod 5 = 2), if (k mod 5 = 3), and if (k mod 5 = 4) are checked. Since 20 mod 5 = 0, none of these conditions are true. Therefore, no letter 'K' is written.

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The given code snippet calculates the value of 'x' by adding 2 times the value of 'i' in each iteration of the for loop. The loop iterates from 0 to 'Data-1'. The value of 'x' is then printed using the writeln() function. To obtain an output of 90, the value of 'Init' should be 10.

The correct answer is the third option. This program follows the given steps correctly. It initializes the variables x, y, and hasil with the given values. Then, it enters a while loop that continues as long as x is greater than 0. Inside the loop, it checks if x is odd (x mod 2 = 1) and if so, it adds the value of y to hasil. It then updates the values of x and y according to the given rules. This program accurately implements the steps provided in the question.

The given correct answer is the first option:
function tail(x, y:integer): integer;
var z:integer;
begin
while (y0) do
begin z:=x mod y; x:=y; y:=z end;
tail:=x;
end;
This is because it uses a while loop to continuously update the values of x and y by taking the modulus of x and y. It keeps updating the values until y becomes 0, and then returns the final value of x. This is equivalent to the recursive function in the question.