philschmid/DMath huggingface dataset:
Category: Comparison

Question (EN): The truck is loaded with two boxes, (a) and (b). Box (a) is smaller than box (b). Which of the two boxes is the larger one?
Question (KO): 트럭에 (가), (나) 2개의 상자를 실었습니다. (가) 상자는 (나) 상자보다 작습니다. 두 상자 중 큰 상자는 무엇입니까?

Answer: (b)

Solution (Python):
var_a = '(a)'
var_b = '(b)'
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 = '(a)'
var_d = '(b)'
var_e = '<'
list_b= []
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)
if "/" in str(var_c):
    var_c = eval(str(var_c))
list_b.append(var_c)
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_f = 1
var_g = list_c[var_f-1]
print(var_g)
Category: Possibility

Question (EN): Sehun wants to make a three-digit number by using the single-digit natural numbers 0, 2, and 9 once each. How many total can he make?
Question (KO): 세훈이는 한 자리 자연수 0,2,9를 한 번 씩 사용하여 세 자리 수를 만들고자 한다. 만들 수 있는 수는 총 몇 개인가?

Answer: 4

Solution (Python):
var_a = 0
var_b = 2
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 = 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 = len(list_b)
print(int(var_e))
Category: Geometry

Question (EN): There are regular hexagonal prisms and square pyramids with the same edges. The sum of the lengths of all the edges of the hexagonal prism is 81 centimeters (cm). What is the sum of the lengths of all the edges of the pyramid in centimeters (cm)?
Question (KO): 모서리의 길이가 같은 정육각기둥과 정사각뿔이 있습니다. 육각기둥의 모든 모서리의 길이의 합은 81센티미터(㎝)입니다. 사각뿔의 모든 모서리의 길이의 합은 몇 센티미터(㎝)입니까?

Answer: 36

Solution (Python):
var_a = 81
var_b = 6
var_c = 3
var_d = var_b * var_c
var_e = var_a / var_d
var_f = 4
var_g = 2
var_h = var_f * var_g
var_i = var_e * var_h
print(int(var_i))
Category: Comparison

Question (EN): While Yoongi, Jungkook, and Yuna have following numbers: 4, 6 minus 3, and 5, who has the smallest number?
Question (KO): 윤기는 4, 정국이는 6에서 3을 뺀 수를 모았고, 유나는 5를 모았습니다. 누가 가장 작은 수를 가지고 있나요?

Answer: Jungkook

Solution (Python):
var_a = 'Yoongi'
var_b = 'Jungkook'
var_c = 'Yuna'
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 = 6
var_f = 3
var_g = var_e - var_f
var_h = 5
list_b= []
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_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): There are numbers 0.8, 1/2, 0.9, and 1/3. Find the smallest number that is greater than or equal to 0.6.
Question (KO): 0.8, 1/2, 0.9, 1/3의 숫자가 있다. 이 중 0.6보다 크거나 같은 숫자들 중 가장 작거나 같은 숫자를 구하시오.

Answer: 0.8

Solution (Python):
var_a = 0.8
var_b = 0.5
var_c = 0.9
var_d = 0.3333333333333333
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 = 0.6
list_b = []
for i in list_a:
    if i >= var_e:
        list_b.append(i)
var_f = 1
list_c=list_b.copy()
list_c.sort()
var_g = list_c[var_f-1]
print('{:.2f}'.format(round(var_g+1e-10,2)))
Category: Correspondence

Question (EN): What number is A in 56-1A=39?
Question (KO): 56-1A=39일 때, A에 들어가야 하는 숫자는 무엇인가요?

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: Arithmetic calculation

Question (EN): There are two buckets of water. The left bucket holds 2800 litres (L) of water and the right bucket holds 1500 L of water. When the water in the left bucket was transferred to the right bucket, the left bucket contained 360 litres (L) more water than the right bucket. Find the number of litres (L) of water transferred to the right bucket.
Question (KO): 두 개의 물통이 있습니다. 왼쪽 물통에는 물이 2800리터(L), 오른쪽 물통에는 물이 1500리터(L) 있습니다. 왼쪽 물통에 있는 물을 오른쪽 물통에 옮겨 담았더니 왼쪽 물통에 들어있는 물은 오른쪽 물통보다 360리터(L)가 많았습니다. 오른쪽 물통으로 옮긴 물은 몇 리터(L)인지 구하세요

Answer: 470

Solution (Python):
var_a = 2800
var_b = 1500
var_c = var_a - var_b
var_d = 360
var_e = var_c - var_d
var_f = 2
var_g = var_e / var_f
print(int(var_g))
Category: Correspondence

Question (EN): 6A5+B03=748. How much is A?
Question (KO): 6A5+B03=748입니다. A는 얼마인가요?

Answer: 4

Solution (Python):
var_a = '6A5+B03=748'
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: Comparison

Question (EN): At Ha-gyeom's school, there are 221 students who are taller than 150 centimeters (cm) and 255 students who are not taller than 150 centimeters (cm). At Myeonghun's school, there are 275 students who wear glasses and 226 students who do not wear glasses. Which school has more students?
Question (KO): 하겸이네 학교에 키가 150센티미터(㎝)가 넘는 학생은 221명, 150센티미터(㎝)가 넘지 않는 학생은 255명입니다. 명훈이네 학교에 안경을 쓴 학생은 275명, 안경을 쓰지 않은 학생은 226명입니다. 누구네 학교의 학생이 더 많습니까?

Answer: Myeonghun

Solution (Python):
var_a = 'Ha-gyeom'
var_b = 'Myeonghun'
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 = 221
var_d = 255
var_e = var_c + var_d
var_f = 275
var_g = 226
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)
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: Correspondence

Question (EN): You want to multiply a number by 12. When you mistakenly multiplied by 14 to get 42, what is the correctly calculated value?
Question (KO): 어떤 수에 12를 곱하려고 합니다. 잘못하여 14를 곱했을 때의 답이 42라면 바르게 계산한 결과는 얼마일까요?

Answer: 36

Solution (Python):
var_a = 42
var_b = 14
var_c = var_a / var_b
var_d = 12
var_e = var_c * var_d
print(int(var_e))
Category: Comparison

Question (EN): Choose a number greater than 1.1 among 1.4, 9/10, 1.2, 0.5, and 13/10, and count the number of selected numbers.
Question (KO): 1.4, 9/10, 1.2, 0.5, 13/10 중에서 1.1보다 큰 숫자를 고르고, 선택된 숫자들의 개수를 구하세요

Answer: 3

Solution (Python):
var_a = 1.4
var_b = 0.9
var_c = 1.2
var_d = 0.5
var_e = 1.3
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 = 1.1
list_b = []
for i in list_a:
    if i > var_f:
        list_b.append(i)
var_g = len(list_b)
print(int(var_g))
Category: Possibility

Question (EN): Numbers are given: 2, 4, 5, 1, 3, 7, 6, and 0. Find the smallest number using all cards only once.
Question (KO): 2, 4, 5, 1, 3, 7, 6, 0 수 카드를 모두 한 번씩만 사용하여 가장 작은 수를 찾으시오

Answer: 10234567

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

Question (EN): 632-AB1=41 is valid. What is B?
Question (KO): 632-AB1=41이 성립합니다. B는 무엇인가요?

Answer: 9

Solution (Python):
var_a = '632-AB1=41'
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]
print(int(var_c))
Category: Possibility

Question (EN): Choose 3 of the number cards from 1 to 9 and find the number of combination of numbers that makes 19.
Question (KO): 1에서 9까지의 숫자 카드 중 3개를 선택하고 19가 되는 숫자 조합의 수를 찾으세요.

Answer: 30

Solution (Python):
var_a = 1
var_b = 9
var_c = 1
list_a = [i for i in range(var_a, var_b + 1, var_c)]
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]
list_c=[]
for i in list_b:
    var_e = 0
    i = int(i)
    while i//10 > 0:
        var_e = var_e + i%10
        i = i//10
    var_e = var_e + i%10
    list_c.append(var_e)
var_f = 19
var_g = 0
var_f = int(var_f)
for i in list_c:
    i = int(i)
    if i == var_f:
        var_g = var_g + 1
print(int(var_g))
Category: Arithmetic calculation

Question (EN): Which five-digit number is the smallest common multiple of 3 and 5?
Question (KO): 다섯 자리 수 중 3과 5의 공배수 중 가장 작은 수는?

Answer: 10005

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 = 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 = 5
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 = 1
list_d=list_c.copy()
list_d.sort()
var_g = list_d[var_f-1]
print(int(var_g))
Category: Possibility

Question (EN): 3 different boxes each contain 4 types of balls. Find the number of ways Yongwoo chooses the ball.
Question (KO): 서로 다른 상자 3개 안에 각각 네 종류의 공이 들어 있습니다. 용우가 공을 고르는 경우의 수를 구하시오.

Answer: 12

Solution (Python):
var_a = 3
var_b = 4
var_c = var_a * var_b
print(int(var_c))
Category: Possibility

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

Answer: 3426

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 = 3
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))
Category: Possibility

Question (EN): How many number of cases do you get a 4, 5, or 6 when you throw a die?
Question (KO): 한 개의 주사위를 던졌을 때, 4, 5, 6의 눈이 나오는 경우의 수는?

Answer: 3

Solution (Python):
var_a = 1
var_b = 6
var_c = 1
list_a = [i for i in range(var_a, var_b + 1, var_c)]
var_d = 4
var_e = 5
var_f = 6
list_b= []
if "/" in str(var_f):
    var_f = eval(str(var_f))
list_b.append(var_f)
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()
list_c = list(set(list_a) & set(list_b))
var_g = len(list_c)
print(int(var_g))
Category: Possibility

Question (EN): Find the largest four-digit number that can be formed by using the numbers 9, 4, 1, and 5 once each.
Question (KO): 숫자 9, 4, 1, 5를 한 번씩 사용하여 만들 수 있는 가장 큰 네 자리수를 구하시오.

Answer: 9541

Solution (Python):
var_a = 1
var_b = 4
var_c = 9
var_d = 5
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 = 4
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 = 1
list_c=list_b.copy()
list_c.sort()
var_g = list_c[-var_f]
print(int(var_g))
Category: Comparison

Question (EN): Several children are standing in a circle equally spaced apart. Based on one child, the 6th child and the 11th child are standing facing each other in a clockwise direction. How many children are there in total?
Question (KO): 여러 명의 아이들이 같은 간격으로 둥글게 서 있습니다. 한 아이를 기준으로 시계 방향으로 6번째 아이와 11번째 아이가 마주 보고 서 있습니다. 아이들은 모두 몇 명입니까?

Answer: 10

Solution (Python):
var_a = 11
var_b = 6
var_c = var_a - var_b
var_d = 2
var_e = var_c * var_d
print(int(var_e))
Category: Arithmetic calculation

Question (EN): When Haebin and 11 of his friends high-five each other once, how many high-fives are there?
Question (KO): 해빈이와 11명의 친구들이 서로 한 번씩 하이파이브를 하였을 때, 하이파이브는 모두 몇 번 이루어졌습니까?

Answer: 66

Solution (Python):
var_a = 11
var_b = 1
var_c = var_a + var_b
var_d = 11
var_e = var_c * var_d
var_f = 2
var_g = var_e / var_f
print(int(var_g))
Category: Arithmetic calculation

Question (EN): Minhyung went for a run with his friends. If it takes 20 seconds for Minhyung to run 150 meters (m), find how many meters (m) he can run in 1 minute.
Question (KO): 민형이가 친구들과 달리기를 했습니다. 민형이가 150미터(m)를 달리려면 20초의 시간이 걸릴 때, 1분동안은 몇 미터(m)를 달릴 수 있는지 구하세요.

Answer: 450

Solution (Python):
var_a = 1
var_b = 60
var_c = var_a * var_b
var_d = 20
var_e = var_c / var_d
var_f = 150
var_g = var_e * var_f
print(int(var_g))
Category: Comparison

Question (EN): Eunji was the 10th to arrive at school. If Taehyung arrived after Eunji and then Namjoon arrived, in which order did Namjoon arrive at school?
Question (KO): 은지는 10번째로 학교에 도착했습니다. 은지 다음에는 태형이가 도착했고 그 다음에는 남준이가 도착했다면 남준이는 몇 번째로 도착했을까요?

Answer: 12

Solution (Python):
var_a = 10
var_b = 1
var_c = var_a + var_b
var_d = 1
var_e = var_c + var_d
print(int(var_e))
Category: Arithmetic calculation

Question (EN): Jungkook owns 3 balls and Yoongi owns 2 balls. How many balls are the two players owning?
Question (KO): 정국이는 3개, 윤기는 2개의 공을 소유하고 있습니다. 두 명이 가지고 있는 공의 개수는 몇 개입니까?

Answer: 5

Solution (Python):
var_a = 3
var_b = 2
var_c = var_a + var_b
print(int(var_c))
Category: Geometry

Question (EN): A square was made by arranging 12 beads of the same size, 12 by 12, without gaps. How many marbles are on one side?
Question (KO): 가로 12개씩, 세로 12개씩 크기가 같은 구슬을 빈틈없이 늘어놓아 정사각형을 만들었습니다. 한 변에 놓인 구슬은 몇 개입니까?

Answer: 12

Solution (Python):
var_a = 12
print(int(var_a))
Category: Arithmetic calculation

Question (EN): On the road, cars stand equally spaced from the front to the back. If the distance between the first car and the last car is 242 meters (m), and the distance between the cars is 5.5 meters (m), how many cars are there in the street?
Question (KO): 도로에서 자동차들이 앞 뒤로 같은 간격으로 서 있다. 맨 앞 자동차와 맨 뒤 자동차의 거리가 242 미터(m)이고, 자동차 사이의 거리가 5.5미터(m)라면 거리에 있는 자동차는 모두 몇 대인가?

Answer: 45

Solution (Python):
var_a = 242
var_b = 5.5
var_c = var_a // var_b
var_d = 1
var_e = var_c + var_d
print(int(var_e))
Category: Possibility

Question (EN): From a fruit basket containing apple, peach, pear and melon, when a total of 3 fruits are taken out in order, allowing picking the same type of fruit, how many possible cases are there?
Question (KO): 사과, 복숭아, 배, 참외 과일 바구니에서 중복을 허락하여 총 3개의 과일을 순서대로 꺼낼 때, 가능한 경우의 수는 모두 몇 가지 입니까?

Answer: 64

Solution (Python):
var_a = 'apple'
var_b = 'peach'
var_c = 'pear'
var_d = 'melon'
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 = len(list_a)
var_f = 3
var_g = var_e ** var_f
print(int(var_g))
Category: Possibility

Question (EN): Seung-a had a group assignment with 5 friends and had to choose 1 leader, 1 secretary, and 2 presenters. Find the number of cases in which Seung-a becomes a clerk.
Question (KO): 승아는 5명의 친구들과 조별과제를 하게 되어 조장 1명, 서기 1명, 발표자 2명을 뽑게 되었다. 승아가 서기를 하게 되는 경우의 수를 구하시오.

Answer: 12

Solution (Python):
var_a = 5
var_b = 1
var_c = var_a - var_b
var_d = 1
var_e = 1
var_f = 1
var_g = var_e - var_f
var_h = var_d + var_g
var_i = 2
var_j = var_h + var_i
var_k = 1
var_c = int(var_c)
var_j = int(var_j)
for i, elem in enumerate(range(var_j)):
    var_k = var_k * (var_c-i)
var_l = 2
var_m = 2
var_n = 1
var_l = int(var_l)
var_m = int(var_m)
for i, elem in enumerate(range(var_m)):
    var_n = var_n * (var_l-i)
var_o = var_k / var_n
print(int(var_o))
Category: Arithmetic calculation

Question (EN): If the elements of set A are 1/2, 1/3, 1/5 and the elements of set B are 0.2, 0.4, 0.5, 0.7, find the number of elements in the intersection of set A and B.
Question (KO): 집합 A의 원소는 1/2, 1/3, 1/5이고, 집합 B의 원소는 0.2, 0.4, 0.5, 0.7일 때 A와 B의 교집합의 원소의 개수를 구하시오.

Answer: 2

Solution (Python):
var_a = 0.5
var_b = 0.3333333333333333
var_c = 0.2
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 = 0.2
var_e = 0.4
var_f = 0.5
var_g = 0.7
list_b= []
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)
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()
list_c = list(set(list_a) & set(list_b))
var_h = len(list_c)
print(int(var_h))