Latihan Persiapan Olimpiade Komputer Sma Santo
Questions and Answers
- 1.
Sebuah tangki air memiliki enam buah kran air di bagian dasarnya. Jika semua kran dibuka maka tangki yang terisi penuh akan habis isinya dalam 8 jam. Berapa jamkah yang dibutuhkan untuk menghabiskan isi tangki bila hanya 4 buah kran yang dibuka?
- A.
9
- B.
10
- C.
11
- D.
12
- E.
14
Correct Answer
D. 12Explanation
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.Rate this question:
-
- 2.
Adi dan sepuluh temannya sedang mendapatkan tugas prakarya. Mereka harus membuat dari kertas warna-warni bilangan-bilangan dari 1 sampai dengan 100 kemudian menempelkannya di selembar karton yang panjang. Adi kebagian untuk membuat semua angka lima (5) yang dibutuhkan. Berapa banyak angka lima yang harus Adi buat?
- A.
20
- B.
11
- C.
19
- D.
12
- E.
10
Correct Answer
A. 20Explanation
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).Rate this question:
-
- 3.
Jika operasi (a mod b) adalah sisa dari operasi pembagian a oleh b, berapakah (7^7.777.777 mod 100) + (5^5.555.555 mod 10)?
- A.
5
- B.
12
- C.
75
- D.
77
- E.
99
Correct Answer
B. 12Explanation
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.Rate this question:
-
- 4.
Matematikawan August DeMorgan hidup pada tahun 1800-an. Pada tahun terakhir dalam masa hidupnya dia menyatakan bahwa : “Dulu aku berusia x tahun pada tahun x2 ”. Pada tahun berapakah ia dilahirkan...
- A.
1806
- B.
1849
- C.
1843
- D.
1800
- E.
1900
Correct Answer
A. 1806Explanation
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.Rate this question:
-
- 5.
1^1 x 2^2 x 3^3 x 4^4 x 5^5 x ... x 30^30 dapat habis dibagi oleh 10^n. Berapakah bilangan n terbesar yang mungkin?
- A.
105
- B.
130
- C.
30
- D.
150
- E.
110
Correct Answer
B. 130Explanation
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.Rate this question:
-
- 6.
Berapa banyak angka antara 100 hingga 1000 yang habis dibagi 3 dan 5 tetapi tidak habis dibagi 30?
- A.
48
- B.
40
- C.
30
- D.
20
- E.
18
Correct Answer
C. 30Explanation
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.Rate this question:
-
- 7.
1/2 + 1/6 + 1/12 + 1/20 +… + 1/9900 =
- A.
99/100
- B.
96/100
- C.
98/100
- D.
97/100
- E.
100/100
Correct Answer
A. 99/100Explanation
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.Rate this question:
-
- 8.
Deskripsi berikut adalah untuk menjawab pertanyaan no 8 sampai dengan 10 Tiga orang pecatur senior L, M, N dan 3 orang pecatur pemula O, P, Q bertanding dalam sebuah turnamen. Semua pecatur akan bertanding satu sama lain masing-masing satu kali pertemuan. -Diawal turnamen nilai seluruh peserta adalah 0. -1 angka diberikan jika berhasil mengalahkan pecatur pemula. -2 angka diberikan jika berhasil mengalahkan pecatur senior. -Jika pecatur senior kalah dalam satu game, nilainya akan dikurangi 2. -Jika pecatur pemula kalah dalam satu game, nilainya akan dikurang 1. -Jika sebuah pertandingan berakhir dengan seri, maka pertandingan tersebut akan diulang. 8. Berapakah nilai maksimum yang dapat diraih oleh seorang pecatur senior, jika di menderita 2 kekalahan dalam turnamen tersebut ?
- A.
4
- B.
2
- C.
0
- D.
3
- E.
1
Correct Answer
E. 1Explanation
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.Rate this question:
-
- 9.
Deskripsi berikut adalah untuk menjawab pertanyaan no 8 sampai dengan 10 Tiga orang pecatur senior L, M, N dan 3 orang pecatur pemula O, P, Q bertanding dalam sebuah turnamen. Semua pecatur akan bertanding satu sama lain masing-masing satu kali pertemuan. -Diawal turnamen nilai seluruh peserta adalah 0. -1 angka diberikan jika berhasil mengalahkan pecatur pemula. -2 angka diberikan jika berhasil mengalahkan pecatur senior. -Jika pecatur senior kalah dalam satu game, nilainya akan dikurangi 2. -Jika pecatur pemula kalah dalam satu game, nilainya akan dikurang 1. -Jika sebuah pertandingan berakhir dengan seri, maka pertandingan tersebut akan diulang. Berapa permainan yang harus dimenangkan oleh seorang pecatur pemula untuk menempatkan posisinya dalam klasemen diatas seorang pecatur senior yang pernah kalah sekali dari pecatur senior lainnya ?
- A.
2
- B.
4
- C.
3
- D.
1
- E.
5
Correct Answer
E. 5Explanation
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.Rate this question:
-
- 10.
Deskripsi berikut adalah untuk menjawab pertanyaan no 8 sampai dengan 10 Tiga orang pecatur senior L, M, N dan 3 orang pecatur pemula O, P, Q bertanding dalam sebuah turnamen. Semua pecatur akan bertanding satu sama lain masing-masing satu kali pertemuan. -Diawal turnamen nilai seluruh peserta adalah 0. -1 angka diberikan jika berhasil mengalahkan pecatur pemula. -2 angka diberikan jika berhasil mengalahkan pecatur senior. -Jika pecatur senior kalah dalam satu game, nilainya akan dikurangi 2. -Jika pecatur pemula kalah dalam satu game, nilainya akan dikurang 1. -Jika sebuah pertandingan berakhir dengan seri, maka pertandingan tersebut akan diulang. Jika P memenangkan seluruh permainan kecuali satu game melawan L dan tidak kalah dari pemenang dalam turnamen tersebut, Siapakah yang mungkin akan menjadi juara dalam turnamen tersebut ?
- A.
O atau Q
- B.
L atau P
- C.
M atau N
- D.
Salah satu diantara M, N, O atau Q
- E.
Semua pemain kecuali L atau P
Correct Answer
A. O atau QExplanation
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.Rate this question:
-
- 11.
Untuk soal 11-12 Seorang salesman (petugas pemasaran) suatu perusahaan minuman harus mengunjungi 5 warung untuk memperkenalkan produk minuman terbaru. Kelima warung tersebut adalah: P, Q, R, S, dan T. Dia hanya akan mengunjungi masing-masing satu kali saja, satu warung per hari, Senin s/d Jumat, dengan aturan berikut: -Tidak boleh mengunjungi warung R pada hari Senin. -Harus mengunjungi warung P sebelum mengunjungi S. -Harus mengunjungi warung Q sebelum mengunjungi T. 11. Mana jadwal yang memenuhi syarat?
- A.
Q, S, P, T, R
- B.
R, Q, T, P, S
- C.
R, S, P, Q, T
- D.
T, R, Q, P, S
- E.
P, S, R, Q, T
Correct Answer
E. P, S, R, Q, TExplanation
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.Rate this question:
-
- 12.
Untuk soal 11-12 Seorang salesman (petugas pemasaran) suatu perusahaan minuman harus mengunjungi 5 warung untuk memperkenalkan produk minuman terbaru. Kelima warung tersebut adalah: P, Q, R, S, dan T. Dia hanya akan mengunjungi masing-masing satu kali saja, satu warung per hari, Senin s/d Jumat, dengan aturan berikut: -Tidak boleh mengunjungi warung R pada hari Senin. -Harus mengunjungi warung P sebelum mengunjungi S. -Harus mengunjungi warung Q sebelum mengunjungi T. Jika ia mengunjungi R lebih dahulu daripada P, mana yang pasti benar?
- A.
Q dikunjungi pertama kali
- B.
R dikunjungi pada hari Selasa
- C.
P dikunjungi pada hari Rabu
- D.
T dikunjungi pada hari Kamis
- E.
S dikunjungi terakhir kali
Correct Answer
A. Q dikunjungi pertama kaliExplanation
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.Rate this question:
-
- 13.
Pada sebuah kantong terdapat 2 buah kelereng kuning, 5 buah kelereng biru, dan 8 buah kelereng hitam. Berapa minimal banyaknya kelereng yang perlu diambil agar kita pasti mendapatkan setidaknya 5 kelereng bewarna sama?
- A.
10
- B.
11
- C.
9
- D.
13
- E.
12
Correct Answer
B. 11Explanation
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.Rate this question:
-
- 14.
Joko sering berbohong (jangan ditiru). Dia hanya jujur sehari dalam seminggu. Satu hari dia berkata: "Aku berbohong pada Senin dan Selasa". Pada hari selanjutnya dia berkata: "Hari ini adalah salah satu dari hari Minggu, Sabtu atau Kamis". Pada hari selanjutnya dia berkata: "Aku berbohong pada Jum'at dan Rabu". Pada hari apa dia berkata jujur?
- A.
Senin
- B.
Selasa
- C.
Kamis
- D.
Jumat
- E.
Minggu
Correct Answer
B. SelasaExplanation
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.Rate this question:
-
- 15.
Sebuah lingkaran akan dibagi-bagi menjadi sejumlah bidang yang dibentuk dengan menggambar garis lurus yang memotong dua tepi lingkaran. Dengan menggambar 3 garis sebagai berikut, terbentuk 4 atau 5 bidang Berapa bidang maksimal yang dihasilkan dengan 3 garis?
- A.
9
- B.
5
- C.
7
- D.
6
- E.
8
Correct Answer
C. 7Explanation
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.Rate this question:
-
- 16.
Enam acara pertunjukan kesenian akan berlangsung dari jam 17.00 hingga jam 21.00. Antara acara satu dengan acara berikutnya harus terdapat jeda selama 5 menit. Setiap acara akan diberi jatah waktu yang sama kecuali acara ketiga akan diberikan waktu lebih lama 10 menit dan acara terakhir akan diberi waktu tepat 1 jam. Berapa lama waktu jatah waktu acara ketiga ?
- A.
29
- B.
27 ½
- C.
30
- D.
39
- E.
17 ¾
Correct Answer
D. 39Explanation
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.Rate this question:
-
- 17.
Diketahui empat bilangan bulat positif W, X, Y dan Z. Jika hasil kali W dan Y adalah 32, dan hasil kali X dan Z adalah 100. Sementara diketahui juga hasil kali Y dan Z adalah delapan kali hasil kali W dan X. Berapakah y dikali z ?
- A.
100
- B.
160
- C.
80
- D.
200
- E.
44
Correct Answer
B. 160Explanation
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.Rate this question:
-
- 18.
Ada 3 pedagang keliling: Ali, Bahar, dan Cholil, yang secara berkala mengunjungi kota A untk berjualan. • Ali mengunjungi kota A setiap 10 hari sekali dan terakhir ia datang 3 hari yang lalu. • Bahar mengunjungi kota A setiap 6 hari sekali dan besok ia akan datang. • Cholil mengunjungi kota A setiap dua minggu sekali dan terakhir ia datang 5 hari yang lalu. Berapa hari lagikah berikutnya mereka akan bersamaan mengunjungi kota A pada hari yang sama?
- A.
101
- B.
15
- C.
45
- D.
66
- E.
37
Correct Answer
E. 37Explanation
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.Rate this question:
-
- 19.
Sebuah password (kata sandi) yang terdiri dari 5 angka. Angka ke-4 lebih besar daripada angka ke-2 dengan selisih 4. Sementara angka ke-3 kurang dari angka ke-2 dengan selisih 3. Angka pertama adalah 3 kali lipat angka terakhir. Ada 3 pasang angka dengan jumlah 11. Berapakah angka ke-4 dari password tersebut?
- A.
9
- B.
5
- C.
7
- D.
4
- E.
3
Correct Answer
A. 9Explanation
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.Rate this question:
-
- 20.
Seorang wanita menerima warisan sebesar 1/3 dari harta suaminya seorang pengusaha yang meninggal dunia karena kecelakaan pesawat. Dan tiga orang putranya juga menerima masing‐masing 1/3 dari sisanya. Jika wanita tersebut dan salah seorang anaknya menerima total sebesar Rp. 6 milyar, berapakah total harta yang ditinggalkan oleh pengusaha tersebut ?
- A.
Rp. 9 milyar
- B.
Rp. 9,6 milyar
- C.
Rp. 10.8 milyar
- D.
Rp. 13.5 milyar
- E.
Rp. 18 milyar
Correct Answer
C. Rp. 10.8 milyarExplanation
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.Rate this question:
-
- 21.
Jika x = 0.888, y = 0.888 , dan z = (0.888)^2, manakah pernyataan berikut yang paling benar ?
- A.
X < y < z
- B.
x < z < y
- C.
Y < x < z
- D.
y < z < x
- E.
z < x < y
Correct Answer
E. z < x < yExplanation
The given statement "z < x < y" is the most accurate because z is the square of 0.888, which is smaller than both x and y. Therefore, z is the smallest value, followed by x, and then y.Rate this question:
-
- 22.
Jika n adalah nilai rata‐rata dari tiga buah angka yaitu 6, 9, dan k berapakah nilai k sesungguhnya ?
- A.
3n – 15
- B.
N – 5
- C.
N – 15
- D.
(n −15)/3
- E.
(n+15)/3
Correct Answer
A. 3n – 15Explanation
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.Rate this question:
-
- 23.
Seorang Pedagang membeli buku dari penyalur di kawasan Pasar Cikapundung, Bandung seharga Rp. 36.000, dia harus menyisakan biaya ongkos sebesar 10%. Selain itu dia juga harus menyisakan keuntungan sebesar Rp. 9.000 per bukunya. Harga jual buku tersebut akan naik berapa persen jika dibandingkan harga belinya ?
- A.
27.5%
- B.
35%
- C.
45%
- D.
25%
- E.
15%
Correct Answer
B. 35%Explanation
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%.Rate this question:
-
- 24.
Ibu Dina sedang mencoba untuk membuka usaha ‘bakery’ disebuah ruko di perumahan elit di kawasan Cibubur. Dari resep yang ia pelajari, untuk suatu campuran adonan brownies kukus diperlukan 1½ cangkir terigu dan 4½ cangkir air. Bila ternyata sisa tepung terigu yang tersisa di lemari tinggal ¾ cangkir, berapa cangkirkah air yang diperlukan ?
- A.
2 cangkir
- B.
2 ¼ cangkir
- C.
3 ½ cangkir
- D.
3 ¼ cangkir
- E.
Sesuai dengan resep
Correct Answer
B. 2 ¼ cangkirExplanation
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.Rate this question:
-
- 25.
Hitunglah (80! × 38!) /(77! × 40!)
- A.
316
- B.
2023
- C.
871
- D.
412
- E.
391
Correct Answer
A. 316Explanation
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.Rate this question:
-
- 26.
Jumlah dua digit pertama dari bilangan hasil perkalian 5^30003 × 8^10004 adalah
- A.
16
- B.
6
- C.
14
- D.
10
- E.
8
Correct Answer
B. 6 -
- 27.
Pepen berdiri sejauh 18 meter di sebelah utara Tugu Pemuda, Fanny berdiri 24 meter di sebelah barat Tugu yang sama. Berapakah jarak terdekat antara Fanny dan Pepen yang dapat ditempuh ?
- A.
30
- B.
900
- C.
6
- D.
42
- E.
90
Correct Answer
A. 30Explanation
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.Rate this question:
-
- 28.
Apabila dua buah bilangan 2^n dan 5^n (di mana n adalah bilangan bulat positif) dimulai dengan digit yang sama, maka digit tersebut adalah... (Catatan: bilangan dituliskan dengan notasi desimal, tanpa diawali nol.)
- A.
9
- B.
5
- C.
6
- D.
7
- E.
3
Correct Answer
E. 3Explanation
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.Rate this question:
-
- 29.
Di dalam suatu kotak terdapat 2N buah bola dan di antaranya terdapat N bola berwarna putih dan N bola beraneka warna secara unik (satu bola satu warna, tidak ada yang sama) dan tidak putih. Berapa banyak kombinasi untuk memilih N bola dari 2N bola itu? (Catatan: Dalam perhitungan kombinasi, AAB dan ABA dianggap sama.)
- A.
2N
- B.
(2N / 2)
- C.
2^N
- D.
N!
- E.
(2N)! / N!
Correct Answer
C. 2^NExplanation
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.Rate this question:
-
- 30.
Di suatu provinsi, diadakan lomba voli tiap 3 tahun sekali, lomba bulutangkis tiap 4 tahun sekali, lomba sepak bola tiap 7 tahun sekali, dan lomba tenis tiap 6 tahun sekali. Pada tahun 2000 semua lomba tersebut diadakan. Berapa kali terdapat lebih dari satu lomba dalam setahun dalam periode antara tahun 2005 dan tahun 2017?
- A.
Kurang dari 8 kali
- B.
8 kali
- C.
9 kali
- D.
10 kali
- E.
Lebih dari 10 kali
Correct Answer
A. Kurang dari 8 kaliExplanation
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.Rate this question:
-
- 31.
Perhatikan potongan algoritma berikut: Procedure kocok(d: integer; kata: string); var i: integer; c : char; begin i:=1; repeat c := kata[i]; kata[i] := kata[i+d]; kata[i+d] := c; i:= i+1; until (i=length(kata)-1); writeln(kata); end; Apa yang dicetaknya pada pemanggilan kocok(1, ʹGO GET GOLDʹ) ?
- A.
GO GET GOLD
- B.
O GET GOLGD
- C.
DGO GET GOL
- D.
GET GOLDOG
- E.
Go get gold
Correct Answer
B. O GET GOLGDExplanation
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".Rate this question:
-
- 32.
Perhatikan potongan algoritma berikut: c := 0; d := 0; while (a>b) do begin a:= a-b; c:= c+1; d:= d+b; end; writeln(c, ‘, ‘,d); Jika nilai a=23, b=4, maka keluaran dari algoritma di atas adalah:
- A.
3, 33
- B.
1, 4
- C.
0, 0
- D.
6, 23
- E.
5, 20
Correct Answer
E. 5, 20Explanation
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.Rate this question:
-
- 33.
Perhatikan prosedur coba(n) berikut. procedure coba(var n: integer); begin if n > 0 then begin n := n div 3; write(n mod 3); coba(n); end; end; Apa yang akan dicetak saat pemanggilan coba(z) dengan z sebelumnya sudah memiliki harga 49?
- A.
0001
- B.
1211
- C.
0121
- D.
1120
- E.
1210
Correct Answer
E. 1210Explanation
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.Rate this question:
-
- 34.
Perhatikan potongan algoritma berikut: procedure jalan(n: integer); begin if n > 0 then begin jalan(n div 5); write(n mod 5 + 1); end; end; Pada pemanggilan jalan(49) pada procedure di atas ini apa yang akan dicetaknya kemudian?
- A.
222
- B.
52
- C.
49
- D.
255
- E.
5
Correct Answer
D. 255Explanation
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.Rate this question:
-
- 35.
Perhatikan algoritma berikut: function ABC (a, b : integer) : integer; var hasil : integer; begin if (a mod b = 0) then ABC := b else ABC := ABC(a, b-1); end; Berapakah hasil ABC(12, 4)?
- A.
-1
- B.
0
- C.
1
- D.
2
- E.
4
Correct Answer
A. -1Explanation
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.Rate this question:
-
- 36.
Perhatikan fungsi berikut ini: function tail(x, y: integer): integer; begin if (y=0) then tail:=x else tail:=tail(y, x mod y); end; Fungsi rekursif di atas ekivalen dengan fungsi...
- A.
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;
- B.
Function tail(x, y:integer): integer; begin if (y=0) then tail:=x else tail:=tail(y mod x, y); end;
- C.
Function tail(x, y:integer): integer; var z:integer; begin while (y0) do begin z:=x mod y; x:=y; y:=z end; tail:=z; end;
- D.
Function tail(x, y:integer): integer; begin if (y=0) then tail:=y else tail:=tail(y, x mod y); end;
- E.
Function tail(x, y:integer): integer; begin if (x=0) then tail:=x else tail:=tail(y, x mod y); end;
Correct Answer
A. 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;Explanation
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.Rate this question:
-
- 37.
Perhatikan tahapan-tahapan berikut: Misalkan ada dua variabel "x" dan "y", dan variabel "hasil" yang nilai awalnya 0. Lakukan proses berikut selama nilai "x" lebih besar dari 0: - Jika nilai "x" ganjil maka nilai "hasil" := "hasil" + y. - nilai "x" selanjutnya adalah nilai "x" sebelumnya dibagi dua, bila ada hasil pecahan, maka pecahannya di buang. (contoh bila nilai "x" sebelumnya 1, maka nilai "x" selanjutnya 0) - nilai "y" selanjutnya adalah nilai "y" sebelumnya dikali dua Manakah program pseudo-pascal yang merupakan program dari tahapan-tahapan tersebut? (catatan: fungsi "mod" memberikan nilai sisa bagi, contoh: 13 mod 5 = 3 dan fungsi “div” membagi dan membulatkan ke bawah)
- A.
Var x,y : integer x := 10; y := 15; hasil := 0; while x > 0 begin if (y mod 2 = 1) then begin hasil := hasil + y; end; x := x * 2; y := y div 2; end
- B.
Var x,y : integer x := 10; y := 15; hasil := 0; while x > 0 begin if (x mod 2 = 1) then begin hasil := hasil + y; end; x := x div 2; y := y * 2; end
- C.
Var x,y : integer x := 10; y := 15; hasil := 0; while x > 0 begin if (x mod 2 = 1) then begin hasil := hasil + x; end; x := x * 2; y := y div 2; end
- D.
Var x,y : integer x := 10; y := 15; hasil := 0; while x > 0 begin if (x mod 2 = 1) then begin hasil := hasil + x; end; x := x div 2; y := y * 2; end
- E.
Var x,y : integer x := 10 y := 15; hasil := 0; while x > 0 begin if (y mod 2 = 1) then begin hasil := hasil + y; end; x := x div 2; y := y * 2; end
Correct Answer
B. Var x,y : integer x := 10; y := 15; hasil := 0; while x > 0 begin if (x mod 2 = 1) then begin hasil := hasil + y; end; x := x div 2; y := y * 2; endExplanation
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.Rate this question:
-
- 38.
Perhatikan Penggalan Kode berikut Data := Init; x := 0; for i := 0 to Data-1 do begin x := x + 2*i; end; writeln(x); Berapakah nilai Init sehingga program di atas menghasilkan output x tertulis 90 ?
- A.
9
- B.
45
- C.
11
- D.
10
- E.
0
Correct Answer
D. 10Explanation
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.Rate this question:
-
- 39.
Perhatikan potongan program berikut for i := 1 to n do begin for j := 1 to n do begin for k := 1 to n do begin writeln('Hello'); end; end; end; Dengan sembarang harga n > 0, keluaran 'Hello’ akan dicetak berulang-ulang dalam sejumlah baris yang ...
- A.
Merupakan konstanta
- B.
Merupakan fungsi kuadrat dari n
- C.
Merupakan fungsi linier dari n
- D.
Merupakan fungsi pangkat empat dari n
- E.
Merupakan fungsi kubik (pangkat 3) dari n
Correct Answer
E. Merupakan fungsi kubik (pangkat 3) dari nExplanation
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.Rate this question:
-
- 40.
Perhatikan prosedur berikut ini. procedure TOKI(k:integer); begin if (k >1) then begin if k mod 2 =0 then TOKI(k div 2) else TOKI(3*k+1); if k mod 5 =1 then write('T'); if k mod 5 =2 then write('O'); if k mod 5 =3 then write('K'); if k mod 5 =4 then write('I'); end; end; Berapa banyak huruf ‘K’ yang tertulis bila dipanggil TOKI(20)?
- A.
5
- B.
4
- C.
3
- D.
2
- E.
1
Correct Answer
E. 1Explanation
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.Rate this question:
-
Quiz Review Timeline +
Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.
-
Current Version
-
Mar 22, 2023Quiz Edited by
ProProfs Editorial Team -
Dec 21, 2013Quiz Created by
Dvanhauten
Sekolah Global Mandiri - Wilayah Negara Kesatuan Republik Indonesia
Sekolah Global Mandiri - Wilayah Negara Kesatuan Republik Indonesia
Seleksi Penerimaan Mahasiswa Baru Poltekkes Hermina Tahun 2020
Seleksi Penerimaan Mahasiswa Baru Poltekkes Hermina Tahun 2020
Back to top