philschmid/DMath huggingface dataset:
Category: Arithmetic calculation

Question (EN): The average score on the math test taken in May and April is 89. If the average score on the math test taken in April, May, and June is 88, what is the score on the math test taken in June?
Question (KO): 수학시험을 5월에 본것과 4월에 본 것의 평균은 89점이다. 4,5,6월에 본 평균이 88점이라면 6월에 본 시험은 몇 점인가?

Answer: 86

Solution (Python):
var_a = 88
var_b = 3
var_c = var_a * var_b
var_d = 89
var_e = var_c - var_d
var_f = 89
var_g = var_e - var_f
print(int(var_g))
Category: Comparison

Question (EN): There are bags weighing 1.2 kilograms (kg), 3.1 kilograms (kg), 2.4 kilograms (kg), 3.0 kilograms (kg), and 1.8 kilograms (kg), respectively. Of the bags that weigh more than 2 kilograms (kg), how many kilograms (kg) is the lightest?
Question (KO): 무게가 각각 1.2킬로그램(㎏), 3.1킬로그램(㎏), 2.4킬로그램(㎏), 3.0킬로그램(㎏), 1.8킬로그램(㎏)인 가방이 있습니다. 2킬로그램(㎏)이 넘는 가방 중 가장 가벼운 가방은 몇 킬로그램(㎏)입니까?

Answer: 2.4

Solution (Python):
var_a = 1.2
var_b = 3.1
var_c = 2.4
var_d = 3
var_e = 1.8
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 = 2
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-1]
print('{:.2f}'.format(round(var_h+1e-10,2)))
Category: Possibility

Question (EN): Find the sum of the numbers larger than 670 and the sum of the numbers smaller than 500 among three digits that can be made using 4, 6, and 7 only once.  
Question (KO): 4, 6, 7을 한 번씩만 사용하여 만들 수 있는 세 자리 수 중에서 670보다 큰 수의 합과 500보다 작은 수의 합의 차를 구하시오.

Answer: 1241

Solution (Python):
var_a = 4
var_b = 6
var_c = 7
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 = 3
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 = 670
list_c = []
for i in list_b:
    if i > var_e:
        list_c.append(i)
list_c = [float(i) for i in list_c]
var_f = sum(list_c)
var_g = 500
list_d = []
for i in list_b:
    if i < var_g:
        list_d.append(i)
list_d = [float(i) for i in list_d]
var_h = sum(list_d)
var_i = var_f - var_h
var_j = abs(var_i)
print(int(var_j))
Category: Arithmetic calculation

Question (EN): Jimin gave Yuna 25 candies, and there are 13 left over. How many candies does Jimin have at the start?
Question (KO): 지민이가 유나에게 사탕 25개를 주었더니 13개가 남았습니다. 처음에 지민이가 가진 사탕은 몇 개입니까?

Answer: 38

Solution (Python):
var_a = 13
var_b = 25
var_c = var_a + var_b
print(int(var_c))
Category: Arithmetic calculation

Question (EN): A rose's price is 500 won. If you are going to sell 200 roses parceling into 25 bundles, how much is a bundle?
Question (KO): 한 송이에 500원인 장미가 있다. 200송이의 장미를 25묶음으로 나누어 팔때 한 묶음 당 얼마인가?

Answer: 4000

Solution (Python):
var_a = 200
var_b = 25
var_c = var_a / var_b
var_d = 500
var_e = var_c * var_d
print(int(var_e))
Category: Geometry

Question (EN): When a large box in the shape of a cuboid measuring 6 centimeters (cm) wide, 4 centimeters (cm) long, and 1 centimeters (cm) high became larger into a volume of 30 centimeters (cm) wide, 20 centimeters (cm) long, and 5 centimeters (cm) high, find how many times it has grown.
Question (KO): 6센티미터(㎝), 세로 4센티미터(㎝), 높이 1센티미터(㎝) 인 직육면체 모양의 큰 상자가 가로 가로 30 센티미터(㎝) , 세로 20 센티미터(㎝) , 높이 5센티미터(㎝) 로 부피가 커졌을 때, 몇 배가 커진전지 구하시오.

Answer: 125

Solution (Python):
var_a = 30
var_b = 20
var_c = var_a * var_b
var_d = 5
var_e = var_c * var_d
var_f = 6
var_g = 4
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: Comparison

Question (EN): There is a flower garden. 3/8 of this flower garden is planted with snapweed, and 2/7 is planted with sunplant. Which part of the area is larger: snapweed or sunplant?
Question (KO): 어느 꽃밭이 있습니다. 이 꽃밭의 3/8에는 봉선화가 심어져 있으며, 2/7에는 채송화가 심어져 있습니다. 어느 꽃을 심은 부분이 더 넓을까요?

Answer: snapweed

Solution (Python):
var_a = 'sunplant'
var_b = 'snapweed'
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 = 0.2857142857142857
var_d = 0.375
list_b= []
if "/" in str(var_d):
    var_d = eval(str(var_d))
list_b.append(var_d)
if "/" in str(var_c):
    var_c = eval(str(var_c))
list_b.append(var_c)
list_b.reverse()
var_e = 1
list_c=list_b.copy()
list_c.sort()
var_f = list_c[-var_e]
var_g = list_b.index(var_f)+1
var_h = list_a[var_g-1]
print(var_h)
Category: Correspondence

Question (EN): Subtracting 9 from a certain number results 5. Find the certain number.
Question (KO): 처음 수에 9를 빼면 5입니다. 처음 수를 구하시오.

Answer: 14

Solution (Python):
var_a = 5
var_b = 9
var_c = var_a + var_b
print(int(var_c))
Category: Arithmetic calculation

Question (EN): All students in the (a) class are participating in at least one art or music class. There are 35 students in the art class and 32 students in the music class. If the number of students participating in both classes is 19, how many students are in the  (a) class?
Question (KO): (가) 반 학생들은 모두 미술, 음악 수업 중 최소 한 개 이상에 참여하고 있습니다. 미술 수업에 참여하는 학생이 35명이고 음악 수업에 참여하는 학생은 32명입니다. 두 수업 모두 참여하는 학생 수는 19명일 때, (가) 반의 학생 수는 몇 명입니까?

Answer: 48

Solution (Python):
var_a = 35
var_b = 32
var_c = var_a + var_b
var_d = 19
var_e = var_c - var_d
print(int(var_e))
Category: Comparison

Question (EN): A bowl with a volume of 1 liter (L) was estimated as 1100 milliliters (㎖) by Miae and 850 milliliters (㎖) by Hyori. Which of the two people gave the closest estimate of the volume of the bowl?
Question (KO): 실제 들이가 1리터(L)인 그릇을 미애는 1100밀리리터(㎖), 효리는 850밀리리터(㎖)로 어림하였습니다. 두 사람 중 그릇의 들이를 가장 가깝게 어림한 사람은 누구인가?

Answer: Miae

Solution (Python):
var_a = 'Miae'
var_b = 'Hyori'
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 = 1000
var_d = 1100
var_e = var_c - var_d
var_f = abs(var_e)
var_g = 1000
var_h = 850
var_i = var_g - var_h
var_j = abs(var_i)
list_b= []
if "/" in str(var_j):
    var_j = eval(str(var_j))
list_b.append(var_j)
if "/" in str(var_f):
    var_f = eval(str(var_f))
list_b.append(var_f)
list_b.reverse()
var_k = 1
list_c=list_b.copy()
list_c.sort()
var_l = list_c[var_k-1]
var_m = list_b.index(var_l)+1
var_n = list_a[var_m-1]
print(var_n)
Category: Correspondence

Question (EN): There are two different numbers A and B. Find the value of B-A from 29+AB=B6.
Question (KO): 서로 다른 두 수 A, B가 있습니다. 두 자리 수끼리의 덧셈식 29+AB=B6에서 B-A의 값을 구하시오.

Answer: 3

Solution (Python):
var_a = '29+AB=B6'
var_b = 'B'
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]
var_d = '29+AB=B6'
var_e = 'A'
ans_dict = dict()
var_d = var_d.replace('×','*')
var_d = var_d.replace('x','*')
var_d = var_d.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_d):
    if v in variable_candi:
        ans_dict[v] = 1
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]))
    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_d):
        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_f = ans_dict[var_e]
var_g = var_c - var_f
print(int(var_g))
Category: Arithmetic calculation

Question (EN): How many five-digit numbers are common multiples of 9, 6, and 2?
Question (KO): 다섯 자리 수 중 9, 6, 2의 공배수인 수는 모두 몇 개 입니까?

Answer: 5000

Solution (Python):
var_a = 10000
var_b = 99999
var_c = 1
list_a = [i for i in range(var_a, var_b + 1, var_c)]
var_d = 9
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 = 6
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 = 2
list_d = []
var_f = int(var_f)
for i in list_c:
    i = int(i)
    if i % var_f == 0:
        list_d.append(i)
var_g = len(list_d)
print(int(var_g))
Category: Possibility

Question (EN): The dice was rolled and the numbers 2, 3, and 4 came out. How many three-digit numbers made by using all these numbers once are greater than 340?
Question (KO): 주사위를 던져 나온 숫자가 2, 3, 4입니다. 나온 숫자를 한 번씩 모두 사용하여 만든 세 자리 수 중 340보다 큰 수는 모두 몇 개입니까?

Answer: 3

Solution (Python):
var_a = 2
var_b = 3
var_c = 4
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 = 3
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 = 340
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: Geometry

Question (EN): There is a tank in the shape of a cuboid which is 30 centimeters (cm) wide, 20 centimeters (cm) long, and 5 centimeters (cm) high. When pouring water into the tank with a cuboid-shaped bowl measuring 6 centimeters (cm) wide, 4 centimeters (cm) long, and 1 cm (cm high), find out how many times it takes to fill the tank.
Question (KO): 가로 30 센티미터(㎝) , 세로 20 센티미터(㎝) , 높이 5센티미터(㎝) 인 직육면체 모양의 수조가 있다. 가로 6센티미터(㎝), 세로 4센티미터(㎝), 높이 1센티미터(㎝) 인 직육면체 모양의 그릇으로 물을 퍼서 수조에 채울 떄, 몇 번을 퍼야 수조가 꽉 찰지 구하시오.

Answer: 125

Solution (Python):
var_a = 30
var_b = 20
var_c = var_a * var_b
var_d = 5
var_e = var_c * var_d
var_f = 6
var_g = 4
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 seven-digit number 25AB33C is said to be both a multiple of 8, and a multiple of 9, and a multiple of 11. get B.
Question (KO): 일곱자리 수 25AB33C는 8의 배수이고 9의 배수이면서, 11의 배수라고 한다. B를 구하시오.

Answer: 0

Solution (Python):
var_a = '25AB33C'
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 = 8
list_b = []
var_b = int(var_b)
for i in list_a:
    i = int(i)
    if i % var_b == 0:
        list_b.append(i)
var_c = 9
list_c = []
var_c = int(var_c)
for i in list_b:
    i = int(i)
    if i % var_c == 0:
        list_c.append(i)
var_d = 11
list_d = []
var_d = int(var_d)
for i in list_c:
    i = int(i)
    if i % var_d == 0:
        list_d.append(i)
var_e = '25AB33C'
var_f = 'B'
var_e = str(var_e)
var_f = str(var_f)
unk_idx = var_e.index(var_f)
var_g = 0
for elem in list_d:
    elem = str(elem)
    var_g = int(elem[unk_idx])
print(int(var_g))
Category: Arithmetic calculation

Question (EN): There are five numbers 10, 11, 12, 13, and 14. What is the difference between the smallest number and the second smallest number?
Question (KO): 5개의 수 10, 11, 12, 13, 14가 있습니다. 그 중에서 가장 작은 수와 2번째로 작은 수의 차는 얼마입니까?

Answer: -1

Solution (Python):
var_a = 10
var_b = 11
var_c = 12
var_d = 13
var_e = 14
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 = 1
list_b=list_a.copy()
list_b.sort()
var_g = list_b[var_f-1]
var_h = 2
list_c=list_a.copy()
list_c.sort()
var_i = list_c[var_h-1]
var_j = var_g - var_i
print(int(var_j))
Category: Arithmetic calculation

Question (EN): Siwoo folds 180 cranes in 30 minutes. Find how many cranes Siwoo made in 1 hour at this speed.
Question (KO): 시우는 30분 동안 180개의 학을 접는다. 시우가 이와 같은 속도로 1시간 동안 만든 학의 개수는 몇 개일지 구하시오.

Answer: 360

Solution (Python):
var_a = 1
var_b = 60
var_c = var_a * var_b
var_d = 30
var_e = var_c / var_d
var_f = 180
var_g = var_e * var_f
print(int(var_g))
Category: Comparison

Question (EN): If there are three numbers 0.8, 1/2, and 0.5, what is the sum of all numbers greater than 0.1?
Question (KO): 총 3개의 숫자 0.8, 1/2, 0.5이 있습니다. 이 중 0.1보다 큰 수를 모두 더하면 얼마인가요?

Answer: 1.8

Solution (Python):
var_a = 0.8
var_b = 0.5
var_c = 0.5
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 = 0.1
list_b = []
for i in list_a:
    if i > var_d:
        list_b.append(i)
list_b = [float(i) for i in list_b]
var_e = sum(list_b)
print('{:.2f}'.format(round(var_e+1e-10,2)))
Category: Geometry

Question (EN): If A and B are the numbers of vertices and the numbers of edges, what is the value of A+B for a pentagon?
Question (KO): 꼭짓점의 개수와 선분의 개수를 각각 A, B라 할 때, 오각형에 대한 A+B의 값은?

Answer: 10

Solution (Python):
var_a = 5
var_b = 5
var_c = var_a + var_b
print(int(var_c))
Category: Arithmetic calculation

Question (EN): There are three numbers 10, 11, and 12. What is the sum of the largest and smallest numbers?
Question (KO): 3개의 수 10, 11, 12가 있습니다. 그 중에서 가장 큰 수와 가장 작은 수의 합은 얼마입니까?

Answer: 22

Solution (Python):
var_a = 10
var_b = 11
var_c = 12
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 = 1
list_b=list_a.copy()
list_b.sort()
var_e = list_b[-var_d]
var_f = 1
list_c=list_a.copy()
list_c.sort()
var_g = list_c[var_f-1]
var_h = var_e + var_g
print(int(var_h))
Category: Geometry

Question (EN): A cuboid was created using three cubes, each of which has a length of 8 centimeters (cm). Find the surface area of this cuboid.
Question (KO): 한 모서리의 길이가 8센티미터(㎝)인 정육면체 3개를 모두 사용해서 1개의 직육면체를 만들었습니다. 이 직육면체의 겉넓이를 구하시오.

Answer: 896

Solution (Python):
var_a = 8
var_b = 8
var_c = var_a * var_b
var_d = 6
var_e = 3
var_f = var_d * var_e
var_g = 4
var_h = var_f - var_g
var_i = var_c * var_h
print(int(var_i))
Category: Geometry

Question (EN): Find the number of all diagonals of a figure with 11 vertices.
Question (KO): 꼭짓점이 11개인 도형의 모든 대각선의 개수를 구하시오.

Answer: 44

Solution (Python):
var_a = 11
var_b = 11
var_c = 3
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): Adding a particular number to 0.46 gives 0.72. What is that particular number?
Question (KO): 0.46 에 어떤 수를 더했더니 0.72 가 되었습니다. 어떤 수는 얼마입니까?

Answer: 0.26

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

Question (EN): 56-1A=39 is true. If A is one-digit number, how much is A?
Question (KO): 56-1A=39입니다. A는 얼마인가요?

Answer: 7

Solution (Python):
var_a = '56-1A=39'
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: Correspondence

Question (EN): 24 is a result of subtracting 63 from a particular number by mistake when the original calculation was to subtract 36 from that particular number. Find the result of the original calculation.
Question (KO): 24는 어떤 수에서 36을 빼야할 것을 잘못하여 63을 빼서 나온 결과입니다. 바르게 계산한 결과를 구하시오.

Answer: 51

Solution (Python):
var_a = 24
var_b = 63
var_c = var_a + var_b
var_d = 36
var_e = var_c - var_d
print(int(var_e))
Category: Arithmetic calculation

Question (EN): Hojun ate 1/3 of the bread yesterday and 2/5 of it today. What is the remaining part of the whole?
Question (KO): 호준이가 빵을 어제는 전체의 1/3을, 오늘은 전체의 2/5를 먹었습니다. 남은 부분은 전체의 얼마인지 구하시오.

Answer: 0.27

Solution (Python):
var_a = 1
var_b = 0.3333333333333333
var_c = var_a - var_b
var_d = 0.4
var_e = var_c - var_d
print('{:.2f}'.format(round(var_e+1e-10,2)))
Category: Possibility

Question (EN): Find the difference between the smallest 3-digit number and the second smallest 3-digit number using the number cards 1, 6, and 8 all once.
Question (KO): 숫자 카드 1, 6, 8을 한 번씩 모두 사용하여 가장 작은 세 자리수와 두 번째로 작은 세 자리수의 차를 구하시오.

Answer: 18

Solution (Python):
var_a = 1
var_b = 6
var_c = 8
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 = 3
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-1]
var_g = 2
list_d=list_b.copy()
list_d.sort()
var_h = list_d[var_g-1]
var_i = var_f - var_h
var_j = abs(var_i)
print(int(var_j))
Category: Comparison

Question (EN): Yoongi has 4 apples, Yuna has 5, and Jungkook has 6 divided by 3 apples. Who has the most apples?
Question (KO): 윤기는 4개, 유나는 5개, 정국이는 6 나누기 3만큼 사과를 가졌습니다. 누가 가진 사과가 가장 많습니까?

Answer: Yuna

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]
var_k = list_b.index(var_j)+1
var_l = list_a[var_k-1]
print(var_l)
Category: Arithmetic calculation

Question (EN): The fruit shop sold 3/10 of the total apples in the morning and 4/10 of the total in the afternoon. If the number of apples sold today is 140, how many apples were there in the fruit shop yesterday?
Question (KO): 과일가게에서 오전에 전체 사과의 3/10을 팔았고, 오후에 전체의 4/10을 팔았습니다. 오늘 판 사과의 수가 140개라면 어제 과일 가게에 있던 사과는 모두 몇 개인지 구하시오.

Answer: 200

Solution (Python):
var_a = 140
var_b = 0.3
var_c = 0.4
var_d = var_b + var_c
var_e = var_a / var_d
print(int(var_e))
Category: Arithmetic calculation

Question (EN): There are boxes numbered 1 to 22 in the warehouse, and each box is divided into 6 parts. If each section contains 5 notebooks, how many notebooks are there in this warehouse?
Question (KO): 창고에는 1번부터 22번 박스가 있고 각 박스마다 6부분으로 나뉘어져 있습니다. 각 부분마다 노트북이 5개씩 담겨져 있다면, 이 창고에는 모두 몇 개의 노트북이 있습니까?

Answer: 660

Solution (Python):
var_a = 22
var_b = 6
var_c = var_a * var_b
var_d = 5
var_e = var_c * var_d
print(int(var_e))
Category: Arithmetic calculation

Question (EN): How many four-digit numbers are divisible by 3?
Question (KO): 3으로 나누어 떨어질 수 있는 네 자리 수는 모두 몇 개 있습니까?

Answer: 3000

Solution (Python):
var_a = 1000
var_b = 9999
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 = len(list_b)
print(int(var_e))