philschmid/DMath huggingface dataset:
Category: Correspondence

Question (EN): If we multiply 4 by a number and subtract 23 again, what is a number if we get 33?
Question (KO): 4와 어떤 수를 곱하고 23만큼 다시 빼면 33이 된다면 어떤 수는 얼마입니까?

Answer: 14

Solution (Python):
var_a = 33
var_b = 23
var_c = var_a + var_b
var_d = 4
var_e = var_c / var_d
print(int(var_e))
Category: Arithmetic calculation

Question (EN): The milk that Suyeong's family drank today is 0.4 of the total amount. If the remaining milk is 0.69 liters (L), how many liters (L) of milk were there at the beginning?
Question (KO): 수영이네 가족이 오늘 마신 우유는 전체의 0.4입니다. 남은 우유가 0.69리터(L)라면 처음에 있던 우유는 몇 리터(L)입니까?

Answer: 1.15

Solution (Python):
var_a = 0.69
var_b = 1
var_c = 0.4
var_d = var_b - var_c
var_e = var_a / var_d
print('{:.2f}'.format(round(var_e+1e-10,2)))
Category: Arithmetic calculation

Question (EN): A dumpling factory makes 182.88 kilograms (kg) of dumplings per month. Find how many kilograms (kg) of dumplings are made in this factory in one year.
Question (KO): 어느 만두공장에서는 1달에 182.88킬로그램(㎏)의 만두를 만듭니다. 이 공장에서 1년 동안 만든 만두는 몇 킬로그램(㎏)인지 구하시오.

Answer: 2194.56

Solution (Python):
var_a = 182.88
var_b = 12
var_c = var_a * var_b
print('{:.2f}'.format(round(var_c+1e-10,2)))
Category: Possibility

Question (EN): Kahi bought 4 erasers with numbers on them from a stationery shop. Each eraser had 0, 2, 4, and 6 written on it. If she chooses three of them and makes numbers that are greater than 100 and less than 1000, what is the sum of the total numbers that she can make?
Question (KO): 가희는 숫자가 적힌 지우개 4개를 문구점에서 구입했다. 지우개에는 각각 0, 2, 4, 6이 적혀 있었는데, 세 개를 골라서 100보다 크고 1000보다 작은 수를 만든다고 할 때 만들 수 있는 숫자들의 총 합은 얼마인가?

Answer: 7728

Solution (Python):
var_a = 0
var_b = 2
var_c = 4
var_d = 6
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.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]
list_b = [float(i) for i in list_b]
var_f = sum(list_b)
print(int(var_f))
Category: Correspondence

Question (EN): 31+6A2=683 is true. What is A?
Question (KO): 31+6A2=683이 성립합니다. A는 무엇인가요?

Answer: 5

Solution (Python):
var_a = '31+6A2=683'
var_b = 'A'
ans_dict = dict()
var_a = var_a.replace('×','*')
var_a = var_a.replace('x','*')
var_a = var_a.replace('÷','/')
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] = 1
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]))
    term_list = []
    op_list = []
    temp_c = ''
    for tc in temp:
        if tc not in '+-*/=><().':
            temp_c += tc
        else:
            op_list.append(tc)
            term_list.append(temp_c)
            temp_c = ''
    term_list.append(temp_c)
    new_eq = ''
    for i in range(len(op_list)):
        new_eq += str(int(term_list[i]))+op_list[i]
    new_eq += str(int(term_list[-1]))
    if len(new_eq) == len(var_a):
        new_eq=new_eq.replace('=', '==')
        new_eq=new_eq.replace('>==', '>=')
        new_eq=new_eq.replace('<==', '<=')
        eval_result = False
        try:
            eval_result = eval(new_eq)
        except:
            pass
        if eval_result:
            for i, (k, _) in enumerate(ans_dict.items()):
                ans_dict[k] = int(c[i])
var_c = ans_dict[var_b]
print(int(var_c))
Category: Arithmetic calculation

Question (EN): The number of questions that Min-joo and Sang-hyeon got right in the quiz was 43, and Min-joo got 5 more questions than Sang-hyeon. How many quiz questions did Sanghyun get right?
Question (KO): 민주와 상현이가 퀴즈에서 맞힌 문제의 수는 43문제이고, 민주가 상현이보다 5문제를 더 맞췄습니다. 상현이가 맞힌 퀴즈는 몇 문제입니까?

Answer: 19

Solution (Python):
var_a = 43
var_b = 5
var_c = var_a - var_b
var_d = 2
var_e = var_c / var_d
print(int(var_e))
Category: Possibility

Question (EN): You want to create a four-digit number by drawing four out of 1, 7, 0, 3, 5, and 6 and using them only once. Find the difference between the second smallest number and the first smallest number that can be made.
Question (KO): 1, 7, 0, 3, 5, 6 중에서 4개를 뽑아 한 번씩만 사용하여 네 자리 수를 만들려고 합니다. 만들 수 있는 두 번째로 작은 수와 첫 번째로 작은 수의 차를 구하시오.

Answer: 1

Solution (Python):
var_a = 1
var_b = 7
var_c = 0
var_d = 3
var_e = 5
var_f = 6
list_a= []
if "/" in str(var_f):
    var_f = eval(str(var_f))
list_a.append(var_f)
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_g = 4
list_b = [str(i) for i in list_a]
list_b = list(itertools.permutations(list_b, var_g))
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_h = 2
list_c=list_b.copy()
list_c.sort()
var_i = list_c[var_h-1]
var_j = 1
list_d=list_b.copy()
list_d.sort()
var_k = list_d[var_j-1]
var_l = var_i - var_k
var_m = abs(var_l)
print(int(var_m))
Category: Possibility

Question (EN): Three students and two teachers are standing in line. Find the number of cases if students must go next to each other.
Question (KO): 학생들 3명과 선생님 2명이 줄을 서서 가고 있다. 학생들은 반드시 서로 이웃하여 가는 경우의 수를 구하시오.

Answer: 36

Solution (Python):
var_a = 2
var_b = 1
var_c = var_a + var_b
var_d = 2
var_e = 1
var_f = var_d + var_e
var_g = 1
var_c = int(var_c)
var_f = int(var_f)
for i, elem in enumerate(range(var_f)):
    var_g = var_g * (var_c-i)
var_h = 3
var_i = 3
var_j = 1
var_h = int(var_h)
var_i = int(var_i)
for i, elem in enumerate(range(var_i)):
    var_j = var_j * (var_h-i)
var_k = var_g * var_j
print(int(var_k))
Category: Possibility

Question (EN): Find the number of two-digit natural numbers that can be made by drawing two of the number cards 0, 1, 7, and 9.
Question (KO): 숫자 카드 0, 1, 7, 9 중 2장을 뽑아 만들 수 있는 두 자리의 자연수의 개수를 구하시오.

Answer: 9

Solution (Python):
var_a = 0
var_b = 1
var_c = 7
var_d = 9
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 = len(list_b)
print(int(var_f))
Category: Correspondence

Question (EN): A number is greater than 24 but less than 28. Find out which number it is among 20, 23, 26, and 29.
Question (KO): 어떤 수는 24보다 크지만 28보다는 작습니다. 20, 23, 26, 29 중 어떤 수를 구하시오.

Answer: 26

Solution (Python):
var_a = 20
var_b = 23
var_c = 26
var_d = 29
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 = 24
list_b = []
for i in list_a:
    if i > var_e:
        list_b.append(i)
var_f = 28
list_c = []
for i in list_b:
    if i < var_f:
        list_c.append(i)
var_g = 1
var_h = list_c[var_g-1]
print(int(var_h))
Category: Comparison

Question (EN): Yoongi collected 4 points, Yuna collected 5 points, and Jungkook collected 6 minus 3 points. Who has the smallest number?
Question (KO): 윤기는 4, 유나는 5, 정국에는 6에서 3을 뺀 수를 모았습니다. 누가 가장 작은 수를 가지고 있나요?

Answer: Jungkook

Solution (Python):
var_a = 'Yoongi'
var_b = 'Yuna'
var_c = 'Jungkook'
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 = 4
var_e = 5
var_f = 6
var_g = 3
var_h = var_f - var_g
list_b= []
if "/" in str(var_h):
    var_h = eval(str(var_h))
list_b.append(var_h)
if "/" in str(var_e):
    var_e = eval(str(var_e))
list_b.append(var_e)
if "/" in str(var_d):
    var_d = eval(str(var_d))
list_b.append(var_d)
list_b.reverse()
var_i = 1
list_c=list_b.copy()
list_c.sort()
var_j = list_c[var_i-1]
var_k = list_b.index(var_j)+1
var_l = list_a[var_k-1]
print(var_l)
Category: Comparison

Question (EN): Yoongi collected 4 points, and Jungkook collected 6 multiplied by 3 points. Whose number is smaller?
Question (KO): 윤기는 4, 정국이는 6에서 3을 곱한 수를 모았습니다. 누구의 수가 더 작나요?

Answer: Yoongi

Solution (Python):
var_a = 'Yoongi'
var_b = 'Jungkook'
list_a= []
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_c = 4
var_d = 6
var_e = 3
var_f = var_d * var_e
list_b= []
if "/" in str(var_f):
    var_f = eval(str(var_f))
list_b.append(var_f)
if "/" in str(var_c):
    var_c = eval(str(var_c))
list_b.append(var_c)
list_b.reverse()
var_g = 1
list_c=list_b.copy()
list_c.sort()
var_h = list_c[var_g-1]
var_i = list_b.index(var_h)+1
var_j = list_a[var_i-1]
print(var_j)
Category: Correspondence

Question (EN): When a number is divided by 5 and 6, it is divisible, and when divided by 7, the remainder is 6. Find the smallest two-digit number in any number.
Question (KO): 어떤 수를 5와 6으로 나누면 나누어떨어지고 7로 나누면 나머지가 6이라고 합니다. 어떤 수 중에서 가장 작은 두 자리 수를 구하시오.

Answer: 90

Solution (Python):
var_a = 10
var_b = 99
var_c = 1
list_a = [i for i in range(var_a, var_b + 1, var_c)]
var_d = 5
var_e = 6
var_f = var_e * var_d / math.gcd(int(var_e), int(var_d))
list_b = []
var_f = int(var_f)
for i in list_a:
    i = int(i)
    if i % var_f == 0:
        list_b.append(i)
var_g = 7
var_h = 6
list_c = [] 
var_g = int(var_g)
var_h = int(var_h)
if var_h < 0:
    var_h = var_h + var_g
for i in list_b:
    i = int(i)
    if i%var_g == var_h:
        list_c.append(i)
var_i = 1
list_d=list_c.copy()
list_d.sort()
var_j = list_d[var_i-1]
print(int(var_j))
Category: Comparison

Question (EN): Rabbits and ducks are playing in the garden. If the sum of the legs of all animals is 48 and there are 9 rabbits, how many ducks are there?
Question (KO): 토끼와 오리가 정원에서 뛰어놀고 있습니다. 모든 동물의 다리 수의 합은 48개이고, 토끼가 9마리 있다고 할 때 오리는 몇 마리가 있습니까?

Answer: 6

Solution (Python):
var_a = 48
var_b = 4
var_c = 9
var_d = var_b * var_c
var_e = var_a - var_d
var_f = 2
var_g = var_e / var_f
print(int(var_g))
Category: Correspondence

Question (EN): A is a single digit number. If A567 is rounded down in the tens place to get 2560, What is A?
Question (KO): A는 한 자리 숫자입니다. A567을 십의 자리에서 버림하여 2560을 얻었습니다. A를 구하시오.

Answer: 2

Solution (Python):
var_a = 'A567'
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 = 2560
list_b = []
for i in list_a:
    if i >= var_b:
        list_b.append(i)
var_c = 2560
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 = 'A567'
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: Arithmetic calculation

Question (EN): How many divisors of 24 are there?
Question (KO): 24의 약수의 개수는 몇 개입니까?

Answer: 8

Solution (Python):
var_a = 24
list_a = []
num_sqrt = int(math.sqrt(var_a))
for i in range(1, num_sqrt+1):
    if var_a % i == 0:
        list_a.append(i)
        list_a.append(int(var_a/i))
list_a = sorted(set(list_a))
var_b = len(list_a)
print(int(var_b))
Category: Possibility

Question (EN): We want to create the largest number by using a given number once. Given the numbers 1, 4, and 5, what number should we make?
Question (KO): 주어진 숫자를 한 번씩 사용하여 가장 큰 수를 만들고자 합니다. 숫자 1, 4, 5가 있을 때, 어떤 수를 만들어야 합니까?

Answer: 541

Solution (Python):
var_a = 1
var_b = 4
var_c = 5
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 = len(list_a)
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 = 1
list_c=list_b.copy()
list_c.sort()
var_f = list_c[-var_e]
print(int(var_f))
Category: Comparison

Question (EN): The least common multiple of any two numbers is 21. Write the common multiples of these two numbers in order, starting with the smallest number. What is the sum of the numbers between the 8th and the 12th from the left?
Question (KO): 어떤 두 수의 최소공배수가 21입니다. 이 두 수의 공배수를 작은 수부터 차례대로 썼습니다. 왼쪽에서 8번째 수와 12번째 수 사이에 있는 수들의 합은 얼마입니까?

Answer: 630

Solution (Python):
var_a = 21
var_b = 8
var_c = 1
var_d = var_b + var_c
var_e = var_a * var_d
var_f = 21
var_g = 12
var_h = -1
var_i = var_g + var_h
var_j = var_f * var_i
var_k = 21
list_a = [i for i in range(var_e, var_j + 1, var_k)]
list_a = [float(i) for i in list_a]
var_l = sum(list_a)
print(int(var_l))
Category: Arithmetic calculation

Question (EN): If you pour 1/6 of the milk, pour the remaining 2/5 of the milk, and then pour the remaining 2/3 of the milk, you have 120 milliliters (mL) left, how many milliliters (mL) of the original milk were there?
Question (KO): 우유를 1/6만큼 따르고, 남은 우유의 2/5를 따르고, 다시 남은 우유의 2/3을 따랐더니 120밀리리터(㎖)가 남았다면 처음에 있던 우유는 몇 밀리리터(㎖)입니까?

Answer: 720

Solution (Python):
var_a = 120
var_b = 1
var_c = 0.16666666666666666
var_d = var_b - var_c
var_e = var_a / var_d
var_f = 1
var_g = 0.4
var_h = var_f - var_g
var_i = var_e / var_h
var_j = 1
var_k = 0.6666666666666666
var_l = var_j - var_k
var_m = var_i / var_l
print(int(eval('{:.2f}'.format(round(var_m+1e-10,2)))))
Category: Geometry

Question (EN): You have a square piece of paper with a side of 14 centimeters (cm). How many rectangular pieces of paper that are 8 centimeters (cm) long and 2 centimeters (cm) wide can be cut out of this piece of paper?
Question (KO): 한 변의 길이가 14센티미터(㎝)인 정사각형 모양의 종이가 있습니다. 이 종이로 가로가 8센티미터(㎝), 세로가 2센티미터(㎝)인 직사각형 모양의 종이 조각을 최대 몇 개나 오려낼 수 있습니까?

Answer: 12

Solution (Python):
var_a = 14
var_b = 2
var_c = var_a ** var_b
var_d = 8
var_e = 2
var_f = var_d * var_e
var_g = var_c // var_f
print(int(var_g))
Category: Arithmetic calculation

Question (EN): The two players teamed up to participate in the competition. Gaga has to cross the river on a boat, and Nana has to ride a bicycle through a tunnel. Boats move at 3 kilometers (km) per hour, and bicycles move at 4 kilometers (km) per hour. If the tunnel is 2 kilometers (km) longer than the river, and the total time of two players is 4 hours, how long does Gaga cross the river?
Question (KO): 두 선수가 함께 팀을 이뤄 대회에 참가했다. 가가 선수는 보트를 타고 강을 건너야 하고, 나나 선수는 자전거를 타고 터널을 지나가야 한다. 보트는 시속 3킬로미터(㎞)로 움직이고, 자전거는 시속 4킬로미터(㎞)로 탈 수 있다. 터널이 강 보다 2킬로미터(㎞) 더 길었고 두 선수가 걸린 시간의 합이 4시간일 때, 가가 선수가 강을 건넌 시간은?

Answer: 2

Solution (Python):
var_a = 4
var_b = 2
var_c = 4
var_d = var_b / var_c
var_e = var_a - var_d
var_f = 1
var_g = 3
var_h = var_f / var_g
var_i = 1
var_j = 4
var_k = var_i / var_j
var_l = var_h + var_k
var_m = var_e / var_l
var_n = 3
var_o = var_m / var_n
print(int(eval('{:.2f}'.format(round(var_o+1e-10,2)))))
Category: Geometry

Question (EN): Hoseok made an equilateral triangle with a side length of 19 centimeters (cm) with the wire he had, and there was 15 centimeters (cm) of wire left. Find the length in centimeters (cm) of the wire that Hoseok had at first.
Question (KO): 호석이는 가지고 있는 철사로 한 변의 길이가 19센티미터(㎝)인 정삼각형 1개를 만들었더니 철사가 15센티미터(㎝) 남았습니다. 호석이가 처음에 가지고 있던 철사의 길이는 몇 센티미터(㎝)인지 구하시오.

Answer: 72

Solution (Python):
var_a = 19
var_b = 3
var_c = var_a * var_b
var_d = 15
var_e = var_c + var_d
print(int(var_e))
Category: Geometry

Question (EN): If you cut a rectangular piece of paper 34 centimeters (cm) wide into squares with a side of 4 centimeters (cm), there are 2 centimeters (cm) are left over and there are 24 squares in total. What is the minimum perimeter of the rectangle in centimeters (cm) before cutting?
Question (KO): 가로가 34센티미터(㎝)인 직사각형 모양의 종이를 한 변이 4센티미터(㎝)인 정사각형 모양으로 잘랐더니 가로는 2센티미터(㎝)가 남고 정사각형은 모두 24개가 되었습니다. 자르기 전 직사각형의 둘레는 최소 몇 센티미터(㎝)입니까?

Answer: 92

Solution (Python):
var_a = 24
var_b = 34
var_c = 4
var_d = var_b // var_c
var_e = var_a / var_d
var_f = 4
var_g = var_e * var_f
var_h = 34
var_i = var_g + var_h
var_j = 2
var_k = var_i * var_j
print(int(var_k))
Category: Correspondence

Question (EN): When you multiply 0.46 by a certain number twice, you get 0.72. What number is that?
Question (KO): 0.46 에 어떤 수를 두 번 곱했더니 0.72 가 되었습니다. 어떤 수는 얼마입니까?

Answer: 3.4

Solution (Python):
var_a = 0.72
var_b = 0.46
var_c = var_a / var_b
var_d = 0.46
var_e = var_c / var_d
print('{:.2f}'.format(round(var_e+1e-10,2)))
Category: Arithmetic calculation

Question (EN): You sold 15 bundles of ballpoint pens in 5 days. If there are 40 ballpoint pens in each bundle, how many did you sell on average?
Question (KO): 5일동안 볼펜 15묶음을 팔았다. 각 묶음에 볼펜이 40개 있다면, 평균 몇 개를 팔았나요?

Answer: 120

Solution (Python):
var_a = 15
var_b = 40
var_c = var_a * var_b
var_d = 5
var_e = var_c // var_d
print(int(var_e))
Category: Comparison

Question (EN): There are seven number cards with the numbers 1.7, 1/5, 1/5, 1, 3/5, 3/8, and 1.4 respectively. If you arrange these cards in order from the card with the largest number, what number will be written on the fourth card?
Question (KO): 각각 1.7, 1/5, 1/5, 1, 3/5, 3/8, 1.4가 적힌 일곱 장의 숫자 카드가 있습니다. 이 카드들을 숫자의 크기가 큰 카드부터 차례로 배열할 때 네 번째에 놓이는 카드에는 어떤 숫자가 쓰여있겠습니까?

Answer: 0.6

Solution (Python):
var_a = 1.7
var_b = 0.2
var_c = 0.2
var_d = 1
var_e = 3
var_f = 5
var_g = var_e / var_f
var_h = 3
var_i = 8
var_j = var_h / var_i
var_k = 1.4
list_a= []
if "/" in str(var_k):
    var_k = eval(str(var_k))
list_a.append(var_k)
if "/" in str(var_j):
    var_j = eval(str(var_j))
list_a.append(var_j)
if "/" in str(var_g):
    var_g = eval(str(var_g))
list_a.append(var_g)
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_l = 4
list_b=list_a.copy()
list_b.sort()
var_m = list_b[-var_l]
print('{:.2f}'.format(round(var_m+1e-10,2)))
Category: Arithmetic calculation

Question (EN): 252 fifth-grade students and 8 teachers at Yeji's school are going on a field trip. If the cost of renting a 41-seater bus is 300,000 won and the highway toll per bus is 7,500 won, how much does it cost to rent the bus and pay the toll?
Question (KO): 예지네 학교 5학년 학생 252명과 선생님 8명이 현장 학습을 가려고 합니다. 41인승 버스 한 대를 빌리는 값은 300000원이고 버스 1대당 고속도로 통행료는 7500원이라고 할 때 버스를 빌리고, 통행료를 내는 데 모두 얼마가 필요합니까?

Answer: 2152500

Solution (Python):
var_a = 252
var_b = 8
var_c = var_a + var_b
var_d = 41
var_e = var_c / var_d
var_f = 1
var_g=int(((var_e+9*10**(var_f-2))//(10**(var_f-1)))*10**(var_f-1))

var_h = 300000
var_i = 7500
var_j = var_h + var_i
var_k = var_g * var_j
print(int(var_k))
Category: Comparison

Question (EN): What is the largest number less than 0.5 among 1/4, 0.7, 2/3, 0.4, and 2/5?
Question (KO): 1/4, 0.7, 2/3, 0.4, 2/5 중 0.5보다 작은 수 중 가장 큰 수는 무엇입니까?

Answer: 0.4

Solution (Python):
var_a = 0.25
var_b = 0.7
var_c = 0.6666666666666666
var_d = 0.4
var_e = 0.4
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 = 0.5
list_b = []
for i in list_a:
    if i < var_f:
        list_b.append(i)
var_g = 1
list_c=list_b.copy()
list_c.sort()
var_h = list_c[-var_g]
print('{:.2f}'.format(round(var_h+1e-10,2)))
Category: Geometry

Question (EN): What is the total volume of the stacked dice if they are stacked 7 horizontally, 5 vertically, and 3 layers high, each measuring 1 centimeters (cm) in width, length, and height?
Question (KO): 가로, 세로, 높이가 1센티미터(㎝)인 주사위를 가로로 7줄, 세로로 5줄, 높이를 3층으로 쌓아 올렸다면 쌓인 주사위들의 전체 부피는 얼마입니까?

Answer: 105

Solution (Python):
var_a = 1
var_b = 7
var_c = var_a * var_b
var_d = 1
var_e = 5
var_f = var_d * var_e
var_g = var_c * var_f
var_h = 1
var_i = 3
var_j = var_h * var_i
var_k = var_g * var_j
print(int(var_k))
Category: Geometry

Question (EN): There is a cube with a edge length of 12 centimeters (cm). If you cut a cube from the middle of each edge with one edge being 4 centimeters (cm) long, what square centimeter (cm2) is the surface area of the remaining part?
Question (KO): 한 모서리의 길이가 12㎝인 정육면체가 있습니다. 각 모서리의 중간에서 한 모서리의 길이가 4㎝인 정육면체를 잘랐을 때, 남은 부분의 겉넓이는 몇 제곱센티미터입니까?

Answer: 1248

Solution (Python):
var_a = 12
var_b = 12
var_c = var_a * var_b
var_d = 6
var_e = var_c * var_d
var_f = 4
var_g = 4
var_h = var_f * var_g
var_i = 2
var_j = var_h * var_i
var_k = 12
var_l = var_j * var_k
var_m = var_e + var_l
print(int(var_m))
Category: Possibility

Question (EN): Among the six number cards 8, 3, 6, 5, 0, and 7, A is the largest two-digit number, and B is the smallest two-digit number. Find the value of (A-B)×6.
Question (KO): 여섯 장의 숫자 카드 8, 3, 6, 5, 0, 7 중 두 장을 뽑아 만든 두 자리 수 중에서 가장 큰 수를 A, 가장 작은 수를 B라고 할 때, (A-B)×6의 값을 구하시오.

Answer: 342

Solution (Python):
var_a = 8
var_b = 3
var_c = 6
var_d = 5
var_e = 0
var_f = 7
list_a= []
if "/" in str(var_f):
    var_f = eval(str(var_f))
list_a.append(var_f)
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_g = 2
list_b = [str(i) for i in list_a]
list_b = list(itertools.permutations(list_b, var_g))
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_h = 1
list_c=list_b.copy()
list_c.sort()
var_i = list_c[-var_h]
var_j = 1
list_d=list_b.copy()
list_d.sort()
var_k = list_d[var_j-1]
var_l = var_i - var_k
var_m = 6
var_n = var_l * var_m
print(int(var_n))