philschmid/DMath huggingface dataset:
Category: Arithmetic calculation

Question (EN): It requires 30 pictures to play 1 second of the animated film called 'The Snow Prince'. How many pictures are necessary to make a 25-second screening?
Question (KO): 만화 영화 ‘눈의 왕자’를 1초 동안 상영하려면 그림이 30장 필요합니다. ‘눈의 왕자’를 25초 상영하려면 그림이 몇 장 필요할까요?

Answer: 750

Solution (Python):
var_a = 30
var_b = 25
var_c = var_a * var_b
print(int(var_c))
Category: Geometry

Question (EN): You want to create a rectangular field that has the same area as a triangular field with a base of 7.2 meters (m) and a height of 7 meters (m). If the width of a rectangular field is 4 meters (m), how many meters (m) should the length be?
Question (KO): 밑변이 7.2미터(m), 높이가 7미터(m)인 삼각형 모양의 밭과 넓이가 같은 직사각형 모양의 밭을 만들려고 합니다. 직사각형 모양 밭의 가로를 4미터(m)로 한다면 세로는 몇 미터(m)로 해야 합니까?

Answer: 6.3

Solution (Python):
var_a = 7.2
var_b = 7
var_c = var_a * var_b
var_d = 2
var_e = var_c / var_d
var_f = 4
var_g = var_e / var_f
print('{:.2f}'.format(round(var_g+1e-10,2)))
Category: Arithmetic calculation

Question (EN): At first, Jungkook had 8 apples. How many apples does Jungkook have in total when he is given 7 apples?
Question (KO): 정국이가 처음에 가지고 있던 사과는 8개입니다. 사과 7개를 선물받았을 때, 정국이가 가지고 있는 사과의 수는 총 몇 개입니까?

Answer: 15

Solution (Python):
var_a = 8
var_b = 7
var_c = var_a + var_b
print(int(var_c))
Category: Arithmetic calculation

Question (EN): Jungkook tries to divide 10 balls into boxes containing 5 balls each. How many boxes does Jungkook need?
Question (KO): 정국이는 10개의 공을 5개씩 나누어 담으려고 합니다. 정국이가 필요한 상자의 개수는 몇 개 입니까?

Answer: 2

Solution (Python):
var_a = 10
var_b = 5
var_c = var_a / var_b
print(int(var_c))
Category: Geometry

Question (EN): Haechan used all the ribbon tapes he had to make a rectangle of 7 centimeters (cm) long and 5 centimeters (cm) wide. If he used this ribbon tape again to make a square, how many centimeters (cm) would each side be?
Question (KO): 해찬이는 가지고 있는 리본 테이프 전부를 이용하여 가로가 7센티미터(㎝), 세로가 5센티미터(㎝)인 직사각형을 만들었습니다. 이 리본 테이프를 다시 사용하여 정사각형을 만든다면 한 변의 길이는 몇 센티미터(㎝)로 해야 합니까?

Answer: 6

Solution (Python):
var_a = 7
var_b = 5
var_c = var_a + var_b
var_d = 2
var_e = var_c * var_d
var_f = 4
var_g = var_e / var_f
print(int(var_g))
Category: Correspondence

Question (EN): If you divide a number by 1.2, subtract 22.5, multiply that number by 0.6, then add 10.5, you can get 30. Find out this number. 
Question (KO): 어떤 수를 1.2로 나눈 후 22.5를 빼고, 그 값에 0.6을 곱한 후 10.5를 더하면 30이 됩니다. 어떤 수를 구하시오.

Answer: 66

Solution (Python):
var_a = 30
var_b = 10.5
var_c = var_a - var_b
var_d = 0.6
var_e = var_c / var_d
var_f = 22.5
var_g = var_e + var_f
var_h = 1.2
var_i = var_g * var_h
print(int(var_i))
Category: Correspondence

Question (EN): Subtracting 1.9 from a number, multiplying by 1.5, adding 32 and then dividing by 2.5 results in 20. Calculate the number including the decimal point.
Question (KO): 어떤 수에서 1.9를 뺀 후 1.5를 곱하고, 32를 더한 값을 2.5로 나누면 20이 됩니다. 어떤 수를 소수점을 포함하여 구하시오.

Answer: 13.9

Solution (Python):
var_a = 20
var_b = 2.5
var_c = var_a * var_b
var_d = 32
var_e = var_c - var_d
var_f = 1.5
var_g = var_e / var_f
var_h = 1.9
var_i = var_g + var_h
print('{:.2f}'.format(round(var_i+1e-10,2)))
Category: Arithmetic calculation

Question (EN): At Seongmin's school, there are upper and lower math classes. The math average of the top 20 students in the upper class is 42 points, and the average of the bottom 20 students in the lower class is 38 points. What is the average math score of Seongmin's class?
Question (KO): 성민이네 학교는 수학 상반과 하반이 있다. 상반 20명의 수학 평균은 42점이고 하반 20명의 평균은 38점이다. 성민이네 반 전체 수학 평균은 얼마인지 구하시오.

Answer: 40

Solution (Python):
var_a = 20
var_b = 42
var_c = var_a * var_b
var_d = 20
var_e = 38
var_f = var_d * var_e
var_g = var_c + var_f
var_h = 20
var_i = 20
var_j = var_h + var_i
var_k = var_g / var_j
print(int(var_k))
Category: Possibility

Question (EN): Yeonji wants to make a three-digit number by arranging the arrows when they hit the targets which have 2, 4, 6, and 8 written on them. In this case, find the sum of all possible numbers.
Question (KO): 연지는 2, 4, 6, 8이 적인 과녁에 화살을 쏴 맞춘 순서대로 배열하여 세 자리 수를 만들고자 한다. 이 때, 만들 수 있는 모든 수의 합을 구하여라.

Answer: 35520

Solution (Python):
var_a = 2
var_b = 4
var_c = 6
var_d = 8
list_a= []
if "/" in str(var_d):
    var_d = eval(str(var_d))
list_a.append(var_d)
if "/" in str(var_c):
    var_c = eval(str(var_c))
list_a.append(var_c)
if "/" in str(var_b):
    var_b = eval(str(var_b))
list_a.append(var_b)
if "/" in str(var_a):
    var_a = eval(str(var_a))
list_a.append(var_a)
list_a.reverse()
var_e = 3
list_b = [str(i) for i in list_a]
list_b = list(itertools.product(list_b, repeat=var_e))
list_b = [''.join(num_list) for num_list in list_b]
list_b = [str_num for str_num in list_b if str_num[0] != '0']
list_b = [float(i) for i in list_b]
list_b = [float(i) for i in list_b]
var_f = sum(list_b)
print(int(var_f))
Category: Comparison

Question (EN): Find how many three-digit numbers 7A6 are less than 756. Each of the numbers from 0 to 9 can be in the place of A.
Question (KO): 어떤 세 자리 수는 백의 자리 숫자가 7이고 일의 자리 숫자가 6이다. 756보다 작은 수는 모두 몇 개인지 구하시오. 십의 자리에는 0부터 9까지의 숫자가 들어갈 수 있습니다.

Answer: 5

Solution (Python):
var_a = 7
var_b = 'A'
var_c = 6
list_a= []
if "/" in str(var_c):
    var_c = eval(str(var_c))
list_a.append(var_c)
if "/" in str(var_b):
    var_b = eval(str(var_b))
list_a.append(var_b)
if "/" in str(var_a):
    var_a = eval(str(var_a))
list_a.append(var_a)
list_a.reverse()
var_d=""
for i in list_a:
    i = str(i)
    var_d = var_d + i
ans_dict = dict()
var_d = str(var_d)
list_b = []
variable_candi = set(['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'])
for v in set(var_d):
    if v in variable_candi:
        ans_dict[v] = 0
candi = list(itertools.product('0123456789', repeat=len(ans_dict)))
for c in candi:
    temp = var_d
    for i, (k, _) in enumerate(ans_dict.items()):
        temp = temp.replace(k, str(c[i]))
    if len(var_d) == len(str(int(temp))):
        new_elem = int(temp)
        list_b.append(new_elem)
var_e = 756
list_c = []
for i in list_b:
    if i < var_e:
        list_c.append(i)
var_f = len(list_c)
print(int(var_f))
Category: Possibility

Question (EN): There is a list of numbers: 3, 5, 6, and 7. You are going to use 2 numbers from it to create a two-digit number. What is the possible smallest number that is a multiple of 7?
Question (KO): 3, 5, 6, 7 총 4장의 수 카드 중 2장을 사용하여 두 자리 수를 만들려고 합니다. 만들 수 있는 7의 배수 중 가장 작은 수는 무엇인가요?

Answer: 35

Solution (Python):
var_a = 3
var_b = 5
var_c = 6
var_d = 7
list_a= []
if "/" in str(var_d):
    var_d = eval(str(var_d))
list_a.append(var_d)
if "/" in str(var_c):
    var_c = eval(str(var_c))
list_a.append(var_c)
if "/" in str(var_b):
    var_b = eval(str(var_b))
list_a.append(var_b)
if "/" in str(var_a):
    var_a = eval(str(var_a))
list_a.append(var_a)
list_a.reverse()
var_e = 2
list_b = [str(i) for i in list_a]
list_b = list(itertools.permutations(list_b, var_e))
list_b = [''.join(num_list) for num_list in list_b]
list_b = [str_num for str_num in list_b if str_num[0] != '0']
list_b = [float(i) for i in list_b]
var_f = 7
list_c = []
var_f = int(var_f)
for i in list_b:
    i = int(i)
    if i % var_f == 0:
        list_c.append(i)
var_g = 1
list_d=list_c.copy()
list_d.sort()
var_h = list_d[var_g-1]
print(int(var_h))
Category: Correspondence

Question (EN): The three-digit number 7A7 rounded up to the tens place is 730. Find A.
Question (KO): 세 자리 수 7A7을 십의 자리까지 올림하면 730이 됩니다. A를 구하시오.

Answer: 2

Solution (Python):
var_a = '7A7'
ans_dict = dict()
var_a = str(var_a)
list_a = []
variable_candi = set(['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'])
for v in set(var_a):
    if v in variable_candi:
        ans_dict[v] = 0
candi = list(itertools.product('0123456789', repeat=len(ans_dict)))
for c in candi:
    temp = var_a
    for i, (k, _) in enumerate(ans_dict.items()):
        temp = temp.replace(k, str(c[i]))
    if len(var_a) == len(str(int(temp))):
        new_elem = int(temp)
        list_a.append(new_elem)
var_b = 730
list_b = []
for i in list_a:
    if i <= var_b:
        list_b.append(i)
var_c = 730
var_d = 10
var_e = var_c - var_d
list_c = []
for i in list_b:
    if i > var_e:
        list_c.append(i)
var_f = '7A7'
var_g = 'A'
var_f = str(var_f)
var_g = str(var_g)
unk_idx = var_f.index(var_g)
var_h = 0
for elem in list_c:
    elem = str(elem)
    var_h = int(elem[unk_idx])
print(int(var_h))
Category: Geometry

Question (EN): Find the value of dividing the area of a rectangle in half when the rectangle has a width of 25 centimeters (cm) and a height of 16 centimeters (cm).
Question (KO): 가로가 25센티미터(㎝)이고 높이가 16센티미터(㎝)인 직사각형의 넓이를 반으로 나눈 값을 구하시오.

Answer: 200

Solution (Python):
var_a = 25
var_b = 16
var_c = var_a * var_b
var_d = 2
var_e = var_c / var_d
print(int(var_e))
Category: Comparison

Question (EN): One tomato was planted in the 8th row from the front, 14th row from the back, 7th row from the left, and 13th row from the right. If you plant the same number of tomatoes in each row, how many tomatoes will there be in your garden?
Question (KO): 텃밭에 토마토 하나를 앞에서 8번째 줄, 뒤에서 14번째 줄, 왼쪽에서 7번째 열, 오른쪽에서 13번째 열에 심었습니다. 각 줄마다 같은 개수의 토마토를 심는 다고 할 때, 텃밭에 토마토 수는 모두 몇 개 입니까?

Answer: 399

Solution (Python):
var_a = 8
var_b = 14
var_c = var_a + var_b
var_d = 1
var_e = var_c - var_d
var_f = 7
var_g = 13
var_h = var_f + var_g
var_i = 1
var_j = var_h - var_i
var_k = var_e * var_j
print(int(var_k))
Category: Correspondence

Question (EN): The shop owner had arranged 21.52 liters (L) of milk, but only 12.64 liters (L) were sold. Find the amount of liters (L) of the remaining milk.
Question (KO): 마트주인은 우유를 21.52리터(L)를 진열해놓았지만 실제로는 12.64리터(L)가 팔렸다. 마트에서 팔리지 못하고 남은 우유는 몇 리터(L)인지 구하시오.

Answer: 8.88

Solution (Python):
var_a = 21.52
var_b = 12.64
var_c = var_a - var_b
print('{:.2f}'.format(round(var_c+1e-10,2)))
Category: Comparison

Question (EN): There are 5 fish A, B, C, D, and E. A is faster than B and slower than C. E is slower than D. B is faster than D. Which fish is the slowest?
Question (KO): 5마리의 물고기 (가), (나), (다), (라), (마)가 있습니다. (가)는 (나)보다 빠르고 (다)보다 느립니다. (마)는 (라)보다 느립니다. (나)는 (라)보다 빠릅니다. 가장 느린 물고기는 어느 것입니까?

Answer: E

Solution (Python):
var_a = 'A'
var_b = 'B'
var_c = 'C'
var_d = 'D'
var_e = 'E'
list_a= []
if "/" in str(var_e):
    var_e = eval(str(var_e))
list_a.append(var_e)
if "/" in str(var_d):
    var_d = eval(str(var_d))
list_a.append(var_d)
if "/" in str(var_c):
    var_c = eval(str(var_c))
list_a.append(var_c)
if "/" in str(var_b):
    var_b = eval(str(var_b))
list_a.append(var_b)
if "/" in str(var_a):
    var_a = eval(str(var_a))
list_a.append(var_a)
list_a.reverse()
var_f = 'A'
var_g = 'B'
var_h = '>'
var_i = 'A'
var_j = 'C'
var_k = '<'
var_l = 'E'
var_m = 'D'
var_n = '<'
var_o = 'B'
var_p = 'D'
var_q = '>'
list_b= []
if "/" in str(var_q):
    var_q = eval(str(var_q))
list_b.append(var_q)
if "/" in str(var_p):
    var_p = eval(str(var_p))
list_b.append(var_p)
if "/" in str(var_o):
    var_o = eval(str(var_o))
list_b.append(var_o)
if "/" in str(var_n):
    var_n = eval(str(var_n))
list_b.append(var_n)
if "/" in str(var_m):
    var_m = eval(str(var_m))
list_b.append(var_m)
if "/" in str(var_l):
    var_l = eval(str(var_l))
list_b.append(var_l)
if "/" in str(var_k):
    var_k = eval(str(var_k))
list_b.append(var_k)
if "/" in str(var_j):
    var_j = eval(str(var_j))
list_b.append(var_j)
if "/" in str(var_i):
    var_i = eval(str(var_i))
list_b.append(var_i)
if "/" in str(var_h):
    var_h = eval(str(var_h))
list_b.append(var_h)
if "/" in str(var_g):
    var_g = eval(str(var_g))
list_b.append(var_g)
if "/" in str(var_f):
    var_f = eval(str(var_f))
list_b.append(var_f)
list_b.reverse()
global item_name_index_dict
items_name_list = list_a.copy()
conditions = []
condition_list = list_b.copy()
temp_stack = []
for index_, cond_ in enumerate(map(str, condition_list)):
    if cond_ in ("<", ">", "="):
        operand_right = temp_stack.pop()
        operand_left = temp_stack.pop()
        if cond_ == "=":
            cond_ = "=="
        conditions.append(f"{operand_left} {cond_} {operand_right}")
    else:
        if not cond_.isdigit():
            cond_ = "{" + cond_ + "}"
        temp_stack.append(cond_)
item_name_index_dict = {}
for perm in itertools.permutations(range(1, len(items_name_list) + 1)):
    item_name_index_dict = dict(zip(items_name_list, perm))
    formatted_conditions = \
        [condition.format_map(item_name_index_dict) for condition in conditions]
    if all(map(eval, formatted_conditions)):
        break
list_c = list(item_name_index_dict.keys())
list_c.sort(key=item_name_index_dict.get, reverse=True)
var_r = len(list_c)
var_s = list_c[var_r-1]
print(var_s)
Category: Arithmetic calculation

Question (EN): There was 1 liter (L) of milk, of which Yeseul drank 0.1 liters (L), and Gayoung drank 0.2 liters (L) more than Yeseul. After a while, Hancho came and drank the milk, and as a result, there were 0.3 liters (L) left. How many liters (L) of milk did Hancho drink?
Question (KO): 우유 1리터(L)가 있었는데 이 중에서 예슬이가 0.1리터(L)를 마셨고, 가영이는 예슬이보다 0.2리터(L)를 더 많이 마셨습니다. 잠시 후에 한초가 와서 우유를 마셨더니 0.3리터(L)가 남았습니다. 한초가 마신 우유는 몇 리터(L)입니까?

Answer: 0.3

Solution (Python):
var_a = 1
var_b = 0.1
var_c = var_a - var_b
var_d = 0.1
var_e = 0.2
var_f = var_d + var_e
var_g = var_c - var_f
var_h = 0.3
var_i = var_g - var_h
print('{:.2f}'.format(round(var_i+1e-10,2)))
Category: Correspondence

Question (EN): If one of the 10 natural numbers with an average of 42 is set to 50, the average is 44. Find what this number was before fixing it.
Question (KO): 평균이 42인 10개의 자연수 중 한 수를 50으로 고치면 평균이 44가 된다고 한다. 고치기 전에 이 수는 얼마였는지 구하여라.

Answer: 30

Solution (Python):
var_a = 50
var_b = 44
var_c = 42
var_d = var_b - var_c
var_e = 10
var_f = var_d * var_e
var_g = var_a - var_f
print(int(var_g))
Category: Comparison

Question (EN): 22 people stand in a line in descending order according to their heights. When Youngchul is the 13th person when counted from the back, where will Youngchul be, when counted from the back, when all the people line up in ascending order? Answer using an ordinal number.
Question (KO): 22명의 사람들이 키가 큰 사람부터 순서대로 한 줄로 서 있습니다. 영철이는 뒤에서 13번째에 서 있을 때, 키가 작은 사람부터 줄을 서게 되면 영철이는 뒤에서 몇 번째에 서게 되는지 구하시오.

Answer: 10

Solution (Python):
var_a = 22
var_b = 13
var_c = var_a - var_b
var_d = 1
var_e = var_c + var_d
print(int(var_e))
Category: Correspondence

Question (EN): A+B+B=814.8. When you move the decimal point of B one place to the right, it becomes equivalent to A. Calculate the value of A-B including the decimal point.
Question (KO): A+B+B=814.8입니다. B의 소수점을 오른쪽으로 한 칸 옮기면 A와 같습니다. 소수점을 포함하여 A-B의 값을 구하시오.

Answer: 611.1

Solution (Python):
var_a = 814.8
var_b = 10
var_c = 2
var_d = var_b + var_c
var_e = var_a / var_d
var_f = 10
var_g = var_e * var_f
var_h = 814.8
var_i = 10
var_j = 2
var_k = var_i + var_j
var_l = var_h / var_k
var_m = var_g - var_l
print('{:.2f}'.format(round(var_m+1e-10,2)))
Category: Arithmetic calculation

Question (EN): How many two-digit natural numbers have both odd digits or both even digits?
Question (KO): 두 자리 자연수 중에서 두 자리 모두 홀수이거나 두 자리 모두 짝수인 수는 몇 개입니까?

Answer: 45

Solution (Python):
var_a = 11
var_b = 99
var_c = 1
list_a = [i for i in range(var_a, var_b + 1, var_c)]
list_b=[]
for i in list_a:
    var_d = 0
    i = int(i)
    while i//10 > 0:
        var_d = var_d + i%10
        i = i//10
    var_d = var_d + i%10
    list_b.append(var_d)
var_e = 2
list_c = []
var_e = int(var_e)
for i in list_b:
    i = int(i)
    if i % var_e == 0:
        list_c.append(i)
var_f = len(list_c)
print(int(var_f))
Category: Arithmetic calculation

Question (EN): Nine oil paintings and seven watercolor paintings are pasted on the walls of the classroom. How many pictures are on the walls of the classroom?
Question (KO): 교실 벽면에 유화가 9점, 수채화가 7점 붙어져 있습니다. 교실 벽면에 붙어져 있는 그림은 모두 몇 점입니까?

Answer: 16

Solution (Python):
var_a = 9
var_b = 7
var_c = var_a + var_b
print(int(var_c))
Category: Correspondence

Question (EN): Young-gi gave Suji 300 won more than 2/5 of the money he had, and saved 200 won less than 1/2 of the remaining money. If the remaining money is 800 won, how much money did Young-gi have at the beginning?
Question (KO): 영기는 가지고 있던 돈의 2/5보다 300원 많은 돈을 수지에게 주었고, 남은 돈의 1/2보다 200원 적은 돈을 저금했다. 남은돈이 800원이라면, 영기가 처음 가진 돈은 얼마인가?

Answer: 2500

Solution (Python):
var_a = 800
var_b = 200
var_c = var_a - var_b
var_d = 1
var_e = 0.5
var_f = var_d - var_e
var_g = var_c / var_f
var_h = 300
var_i = var_g + var_h
var_j = 1
var_k = 0.4
var_l = var_j - var_k
var_m = var_i / var_l
print(int(var_m))
Category: Possibility

Question (EN): We are trying to line up two fruits from among Apples, Peaches, Pears, Melons, and Strawberries. How many possible cases are there?
Question (KO): 사과, 복숭아, 배, 참외, 딸기 중에서 2가지의 과일을 뽑아 줄을 세우려고 합니다. 가능한 경우의 수는 모두 몇 가지인가요?

Answer: 20

Solution (Python):
var_a = 'Apples'
var_b = 'Peaches'
var_c = 'Pears'
var_d = 'Melons'
var_e = 'Strawberries'
list_a= []
if "/" in str(var_e):
    var_e = eval(str(var_e))
list_a.append(var_e)
if "/" in str(var_d):
    var_d = eval(str(var_d))
list_a.append(var_d)
if "/" in str(var_c):
    var_c = eval(str(var_c))
list_a.append(var_c)
if "/" in str(var_b):
    var_b = eval(str(var_b))
list_a.append(var_b)
if "/" in str(var_a):
    var_a = eval(str(var_a))
list_a.append(var_a)
list_a.reverse()
var_f = len(list_a)
var_g = 2
var_h = 1
var_f = int(var_f)
var_g = int(var_g)
for i, elem in enumerate(range(var_g)):
    var_h = var_h * (var_f-i)
print(int(var_h))
Category: Correspondence

Question (EN): There is a number that equals twice the sum of the quotient of 40 divided by 8 and the quotient of 34 divided by 12. Find this number. 
Question (KO): 어떤 수는 40을 8로 나눈 몫과 34을 12로 나눈 몫을 더한 값의 2배인 수입니다. 어떤 수를 구하세요.

Answer: 14

Solution (Python):
var_a = 40
var_b = 8
var_c = var_a // var_b
var_d = 34
var_e = 12
var_f = var_d // var_e
var_g = var_c + var_f
var_h = 2
var_i = var_g * var_h
print(int(var_i))
Category: Arithmetic calculation

Question (EN): Find how many multiples of 3 ends in 3 that are less than 1000.
Question (KO): 1000 이하의 3의 배수 중 끝이 3으로 끝나는 수는 모두 몇 개인지 구하시오.

Answer: 34

Solution (Python):
var_a = 1
var_b = 1000
var_c = 1
list_a = [i for i in range(var_a, var_b + 1, var_c)]
var_d = 3
list_b = []
var_d = int(var_d)
for i in list_a:
    i = int(i)
    if i % var_d == 0:
        list_b.append(i)
var_e = 1
var_f = 3
list_c = [] 
var_e = int(var_e)
var_f = int(var_f)
for i in list_b:
    i = int(i)
    if (i//var_e)%10 == var_f:
        list_c.append(i)
var_g = len(list_c)
print(int(var_g))
Category: Comparison

Question (EN): Students stand in a line. There is no one in front of Taehyung and there are three people between Taehyung and Namjoon. If 8 people are standing behind Namjoon, how many students are in the line?
Question (KO): 학생들이 한 줄로 서 있습니다. 태형의 앞에는 아무도 없고 태형이와 남준이 사이에는 3명이 있습니다. 남준이 뒤에 8명이 서 있다면 줄을 서 있는 학생은 모두 몇 명입니까?

Answer: 13

Solution (Python):
var_a = 1
var_b = 3
var_c = var_a + var_b
var_d = 1
var_e = var_c + var_d
var_f = 8
var_g = var_e + var_f
print(int(var_g))
Category: Correspondence

Question (EN): A number is subtracted by 1.9, multiplied by 1.5, and added by 32. And finally, you divided it by 2.5 and you got 20. What is the number, including the decimal point?
Question (KO): 어떤 수를 1.9만큼 빼고 다시 1.5만큼 곱하고 32를 더했습니다. 마지막으로 어떤 수를 2.5로 나누었더니 20이 되었습니다. 소수점을 포함한 어떤 수는 얼마입니까?

Answer: 13.9

Solution (Python):
var_a = 20
var_b = 2.5
var_c = var_a * var_b
var_d = 32
var_e = var_c - var_d
var_f = 1.5
var_g = var_e / var_f
var_h = 1.9
var_i = var_g + var_h
print('{:.2f}'.format(round(var_i+1e-10,2)))
Category: Possibility

Question (EN): If you write down all the even numbers that are less than 100 and that are multiples of 3 on a piece of paper, how many times do you have to write down the figure '1'?
Question (KO): 100이하인 짝수이면서 3의 배수인 수를 모두 종이에 적으면, 숫자 1을 몇 번 적어야 하는지 구하시오.

Answer: 2

Solution (Python):
var_a = 1
var_b = 100
list_a = []
if var_a%2!=0:
    for i in range(var_a+1, var_b+1, 2):
        list_a.append(i)
else:
    for i in range(var_a, var_b+1, 2):
        list_a.append(i)
var_c = 3
list_b = []
var_c = int(var_c)
for i in list_a:
    i = int(i)
    if i % var_c == 0:
        list_b.append(i)
var_d=""
for i in list_b:
    i = str(i)
    var_d = var_d + i
list_c = []
var_d = int(var_d)
while var_d//10 > 0:
    list_c.append(var_d%10)
    var_d = var_d//10
list_c.append(var_d%10)
list_c = list_c[::-1]
var_e = 1
var_f = 0
var_e = int(var_e)
for i in list_c:
    i = int(i)
    if i == var_e:
        var_f = var_f + 1
print(int(var_f))
Category: Geometry

Question (EN): When the roof is drawn on drawing paper, it is the trapezoid whose length of the upper side is 6 centimeters (cm) shorter than the length of the lower side, and the height is 8 centimeters (cm). If the area of the roof is 72 square centimeters (cm2), find the length of the upper side of the roof.
Question (KO): 도화지에 지붕을 그렸을 때, 윗변의 길이가 아랫변의 길이보다 6 센티미터(㎝) 짧고, 높이가 8 센티미터(㎝)인 사다리꼴이라고 합니다. 지붕의 넓이가 72제곱센티미터(㎠)일 때, 이 지붕의 윗변의 길이를 구하시오.

Answer: 6

Solution (Python):
var_a = 72
var_b = 2
var_c = var_a * var_b
var_d = 8
var_e = var_c / var_d
var_f = 6
var_g = var_e - var_f
var_h = 2
var_i = var_g / var_h
print(int(var_i))
Category: Correspondence

Question (EN): Multiply a number by 5 to get 100. What is the result when the number is divided by 10?
Question (KO): 어떤 수에 5를 곱하면 100이 됩니다. 10으로 어떤 수를 나누었을 때 결과값은 얼마인가요?

Answer: 2

Solution (Python):
var_a = 100
var_b = 5
var_c = var_a / var_b
var_d = 10
var_e = var_c / var_d
print(int(var_e))
Category: Correspondence

Question (EN): The number of apples Minjin has is divided by 8. Seongyeong gave 8 to Minjin, but Bong-Geon took 30 of them. If the remaining number of apples is 12 when multiplied by 6, how many apples does Minjin have at the beginning?
Question (KO): 민진이가 가진 사과 개수를 8로 나누었습니다. 선경이가 8개를 주었는데 봉건이가 30개를 빼앗아갔습니다. 남은 사과 개수에 6을 곱했더니 12개였다면, 처음 민진이가 가진 사과의 개수는 몇 개입니까?

Answer: 192

Solution (Python):
var_a = 12
var_b = 6
var_c = var_a / var_b
var_d = 30
var_e = var_c + var_d
var_f = 8
var_g = var_e - var_f
var_h = 8
var_i = var_g * var_h
print(int(var_i))
Category: Possibility

Question (EN): You are given three numbers 1, 7, and 9. How many two-digit numbers can be formed using a given number once?
Question (KO): 3개의 숫자 1, 7, 9가 주어져 있습니다. 주어진 숫자를 한 번씩 사용하여 만들 수 있는 두 자리 수는 모두 몇 개입니까?

Answer: 6

Solution (Python):
var_a = 1
var_b = 7
var_c = 9
list_a= []
if "/" in str(var_c):
    var_c = eval(str(var_c))
list_a.append(var_c)
if "/" in str(var_b):
    var_b = eval(str(var_b))
list_a.append(var_b)
if "/" in str(var_a):
    var_a = eval(str(var_a))
list_a.append(var_a)
list_a.reverse()
var_d = 2
list_b = [str(i) for i in list_a]
list_b = list(itertools.permutations(list_b, var_d))
list_b = [''.join(num_list) for num_list in list_b]
list_b = [str_num for str_num in list_b if str_num[0] != '0']
list_b = [float(i) for i in list_b]
var_e = len(list_b)
print(int(var_e))
Category: Comparison

Question (EN): Hoseok kicked jegi 48 times. Jinsu kicked jegi 15 times each in the 1st and 2nd attempt. At least how many times should Jinsu kick jegi in the 3rd attempt, in order to kick more than Hoseok?
Question (KO): 호석이는 제기를 48번 찼습니다. 진수는 제기를 1번째와 2번째에 각각 15번씩 찼습니다. 진수가 호석이보다 제기를 더 많이 차려면 3번째에는 적어도 몇 번을 차야 합니까?

Answer: 19

Solution (Python):
var_a = 48
var_b = 15
var_c = 15
var_d = var_b + var_c
var_e = var_a - var_d
var_f = 1
var_g = var_e + var_f
print(int(var_g))