philschmid/DMath huggingface dataset:
Category: Arithmetic calculation

Question (EN): What is the largest three-digit multiple of 6?
Question (KO): 6의 배수 중 가장 큰 세 자리 수는?

Answer: 996

Solution (Python):
var_a = 100
var_b = 999
var_c = 1
list_a = [i for i in range(var_a, var_b + 1, var_c)]
var_d = 6
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
list_c=list_b.copy()
list_c.sort()
var_f = list_c[-var_e]
print(int(var_f))
Category: Geometry

Question (EN): Minho has a triangular ruler with a base of 15 centimeters (cm) and a height of 20 centimeters (cm). When a triangle is drawn along the surface of this triangular ruler, find the area of the drawn triangle in square centimeters (cm2).
Question (KO): 민호는 밑변이 15센티미터(㎝)이고 높이가 20센티미터(㎝)인 삼각형 모양의 삼각자를 갖고 있습니다. 이 삼각자의 겉을 따라서 삼각형을 그렸을 때, 그려진 삼각형의 넓이는 몇 제곱센티미터(㎠)인지 구하시오.

Answer: 150

Solution (Python):
var_a = 15
var_b = 20
var_c = var_a * var_b
var_d = 2
var_e = var_c / var_d
print(int(var_e))
Category: Comparison

Question (EN): We compared the swimming speeds of five fish A, B, C, D, and E. A is faster than B but slower than C. D is faster than E but slower than B. Which of these fish is the slowest?
Question (KO): 5마리의 물고기 (가), (나), (다), (라), (마)가 헤엄치는 속도를 비교해 보았습니다. (가)는 (나)보다 빠르지만 (다)보다는 느립니다. (라)는 (바)보다는 빠르지만 (나)보다는 느립니다. 이 중에서 가장 느린 물고기는 무엇입니까?

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: Geometry

Question (EN): When the small square-shaped notebooks are arraged without gaps to form a large square, there are 20 notebooks left. When I tried to increase the width and height by one row each, 9 notebooks were short. How many notebooks are there in total?
Question (KO): 작은 정사각형 모양의 공책을 빈틈없이 늘어놓아 큰 정사각형을 만들었더니 20개의 공책이 남았습니다. 가로 한 변과 세로 한 변을 1열씩 늘리려고 하였더니 9개의 공책이 부족하였습니다. 공책의 개수는 총 몇 개입니까?

Answer: 216

Solution (Python):
var_a = 20
var_b = 9
var_c = var_a + var_b
var_d = 1
var_e = var_c - var_d
var_f = 2
var_g = var_e / var_f
var_h = 2
var_i = var_g ** var_h
var_j = 20
var_k = var_i + var_j
print(int(var_k))
Category: Correspondence

Question (EN): A47BC is a five-digit number and is a multiple of 4, 5, or 9. How many possible numbers can be A in A47BC?
Question (KO): A47BC는 다섯 자리 수이고 4, 5, 9의 배수입니다. A47BC에서 A가 될 수 있는 숫자는 모두 몇 개입니까?

Answer: 5

Solution (Python):
var_a = 'A47BC'
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 = 4
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 = 5
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 = 9
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 = 'A47BC'
var_f = 'A'
var_e = str(var_e)
var_f = str(var_f)
unk_idx = var_e.index(var_f)
list_e = []
for elem in list_d:
    elem = str(elem)
    list_e.append(int(elem[unk_idx]))
list_e = list(set(list_e))
var_g = len(list_e)
print(int(var_g))
Category: Geometry

Question (EN): What is the area of a square with a perimeter of 48 centimeters (cm)?
Question (KO): 둘레가 48센티미터(㎝)인 정사각형의 넓이는 얼마인가?

Answer: 144

Solution (Python):
var_a = 48
var_b = 4
var_c = var_a / var_b
var_d = 2
var_e = var_c ** var_d
print(int(var_e))
Category: Arithmetic calculation

Question (EN): There are 82 blue and red pens in total. Find the number of blue pens when there are 6 more blue pens than red pens.
Question (KO): 파란 펜과 빨간 펜이 합쳐서 82개 있다. 파란 펜이 빨간 펜보다 6개 많을 때 파란 펜은 몇 개인지 구하시오.

Answer: 44

Solution (Python):
var_a = 82
var_b = 6
var_c = var_a - var_b
var_d = 2
var_e = var_c / var_d
var_f = 6
var_g = var_e + var_f
print(int(var_g))
Category: Geometry

Question (EN): There are 27 cube dice, and each side is 3 centimeters (cm) long. When these dice are stacked to form a cube, what is the surface area of the cube?
Question (KO): 27개의 정육면체 주사위가 있고, 이 주사위들의 한 모서리의 길이는 3센티미터(㎝)입니다. 이 주사위들을 쌓아 정육면체를 만들었을 때, 쌓아 만든 정육면체의 겉넓이는?

Answer: 486

Solution (Python):
var_a = 3
var_b = 27
var_c = 0.3333333333333333
var_d = var_b ** var_c
var_e = var_a * var_d
var_f = 2
var_g = var_e ** var_f
var_h = 6
var_i = var_g * var_h
print(int(var_i))
Category: Geometry

Question (EN): What is the sum of the measures of the six interior angles of a hexagon?
Question (KO): 육각형의 여섯 내각의 크기의 합은 몇 도입니까?

Answer: 720

Solution (Python):
var_a = 6
var_b = 2
var_c = var_a - var_b
var_d = 180
var_e = var_c * var_d
print(int(var_e))
Category: Comparison

Question (EN): The red bus took 4 hours to drive 200 kilometers (km), and the yellow bus took 3 hours to drive 180 kilometers (km). Which bus is faster?
Question (KO): 빨간색 버스는 200킬로미터(㎞)를 가는데 4시간이 걸렸고, 노란색 버스는 180킬로미터(㎞)를 가는 데 3시간이 걸렸습니다. 어느 버스가 더 빠른가요?

Answer: yellow

Solution (Python):
var_a = 'red'
var_b = 'yellow'
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 = 200
var_d = 4
var_e = var_c / var_d
var_f = 180
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)
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): There are numbers when divided by 8, the remainder is 5, and when divided by 3, the remainder is 2. When these numbers are counted from the smallest to largest, find the 20th number.
Question (KO): 8로 나누었을 떄 나머지가 5이면서 3으로 나눈 나머지가 2인 수가 있다. 이러한 수를 작은 수부터 세었을 때 20째 수를 구하여라

Answer: 461

Solution (Python):
var_a = 0
var_b = 9999
var_c = 1
list_a = [i for i in range(var_a, var_b + 1, var_c)]
var_d = 3
var_e = 2
list_b = [] 
var_d = int(var_d)
var_e = int(var_e)
if var_e < 0:
    var_e = var_e + var_d
for i in list_a:
    i = int(i)
    if i%var_d == var_e:
        list_b.append(i)
var_f = 8
var_g = 5
list_c = [] 
var_f = int(var_f)
var_g = int(var_g)
if var_g < 0:
    var_g = var_g + var_f
for i in list_b:
    i = int(i)
    if i%var_f == var_g:
        list_c.append(i)
var_h = 20
list_d=list_c.copy()
list_d.sort()
var_i = list_d[var_h-1]
print(int(var_i))
Category: Correspondence

Question (EN): Subtracting 30 from a number is a multiple of 70. Find the smallest value of any possible number. However, a number is greater than 280.
Question (KO): 어떤 수에서 30을 빼면 70의 배수이다. 가능한 어떤 수 중 가장 작은 값을 구하시오. 단, 어떤 수는 280 이상이다.

Answer: 310

Solution (Python):
var_a = 280
var_b = 30
var_c = var_a + var_b
print(int(var_c))
Category: Possibility

Question (EN): When using the given number cards 7, 5, and 2 all once to make a number greater than 530 and less than 710, write the number that can be in the tens place.
Question (KO): 주어진 숫자 카드 7, 5, 2를 한 번씩 모두 사용하여 530보다 크고 710보다 작은 수를 만들 때, 십의 자리에 올 수 있는 숫자를 쓰시오.

Answer: 7

Solution (Python):
var_a = 7
var_b = 5
var_c = 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 = 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 = 530
list_c = []
for i in list_b:
    if i > var_e:
        list_c.append(i)
var_f = 710
list_d = []
for i in list_c:
    if i < var_f:
        list_d.append(i)
var_g = 1
var_h = list_d[var_g-1]
var_i = 100
var_j = var_h % var_i
var_k = 10
var_l = var_j // var_k
print(int(var_l))
Category: Possibility

Question (EN): Figure out the smallest sum of numbers that are made out of the number cards 1, 4, 9, and 5 using only once.
Question (KO): 숫자 카드 1, 4, 9, 5를 한 번씩 모두 사용하여 합이 가장 작은 값을 구하시오.

Answer: 19

Solution (Python):
var_a = 1
var_b = 4
var_c = var_a + var_b
var_d = 9
var_e = var_c + var_d
var_f = 5
var_g = var_e + var_f
print(int(var_g))
Category: Arithmetic calculation

Question (EN): There were 1.6 liters (L) of chocolate milk. If you drank 0.8 liters (L) yesterday and 0.3 liters (L) today, how many liters (L) of chocolate milk are left?
Question (KO): 초코 우유가 1.6리터(L) 있었습니다. 어제 0.8리터(L) 마시고 오늘 0.3리터(L) 마셨다면, 남은 초코 우유는 몇 리터(L)입니까?

Answer: 0.5

Solution (Python):
var_a = 1.6
var_b = 0.8
var_c = var_a - var_b
var_d = 0.3
var_e = var_c - var_d
print('{:.2f}'.format(round(var_e+1e-10,2)))
Category: Comparison

Question (EN): Choose the smallest number among 5, 9, 10, and 2.
Question (KO): 5, 9, 10, 2 중에서 가장 작은 숫자를 선택하십시오.

Answer: 2

Solution (Python):
var_a = 5
var_b = 9
var_c = 10
var_d = 2
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 = 1
list_b=list_a.copy()
list_b.sort()
var_f = list_b[var_e-1]
print(int(var_f))
Category: Geometry

Question (EN): There are some pieces of paper tape of 18 centimeters (cm) each, and 12 pieces of it are overlapped to form a round ring with a circumference of 210 centimeters (cm). If the length of the overlapped part is the same, write down how much is it in millimeters (mm).

Question (KO): 길이 18센티미터(㎝)인 종이 테이프 12장을 겹쳐 둘레 210센티미터(㎝)의 둥근 고리를 만들었습니다. 겹친 부분의 길이는 같을 때, 얼마인지 밀리미터(㎜)로 나타내시오.

Answer: 5

Solution (Python):
var_a = 18
var_b = 12
var_c = var_a * var_b
var_d = 210
var_e = var_c - var_d
var_f = 12
var_g = var_e / var_f
var_h = 10
var_i = var_g * var_h
print(int(var_i))
Category: Geometry

Question (EN): There is a cube with each edge 3 centimeters (cm) long. If one digs out cubes with its edge 1 centimeters (cm) long from each center of the faces of the cube, what is the volume in cubic centimeters (cm3) of the remaining part?
Question (KO): 한 모서리의 길이가 3㎝인 정육면체가 있습니다. 각 면의 한가운데서 한 모서리의 길이가 1㎝인 정육면체를 파내었을 때, 남은 부분의 부피는 몇 세제곱센티미터입니까?

Answer: 21

Solution (Python):
var_a = 3
var_b = 3
var_c = var_a ** var_b
var_d = 1
var_e = 3
var_f = var_d ** var_e
var_g = 6
var_h = var_f * var_g
var_i = var_c - var_h
print(int(var_i))
Category: Comparison

Question (EN): Thirteen students competed in jumping rope. After the match, there were 4 people who did fewer jump ropes than Namjoon. What rank is Namjoon?
Question (KO): 13명의 학생이 줄넘기 시합을 했습니다. 시합이 끝나고, 남준이 보다 개수가 적은 사람은 4명이었습니다. 남준이는 몇 등일까요?

Answer: 9

Solution (Python):
var_a = 13
var_b = 4
var_c = var_a - var_b
print(int(var_c))
Category: Arithmetic calculation

Question (EN): 2 times the sum of cows and pigs is 32. If there are 7 cows, how many pigs are there?
Question (KO): 소와 돼지의 합의 2배가 32이다. 소가 7마리이면 돼지는 몇마리인가?

Answer: 9

Solution (Python):
var_a = 32
var_b = 2
var_c = var_a / var_b
var_d = 7
var_e = var_c - var_d
print(int(var_e))
Category: Correspondence

Question (EN): The six-digit number A2468B is divisible by 55. How many possible numbers can be A2468B?
Question (KO): 여섯 자리 수 A2468B은 55로 나누어떨어지는 수이다. A2468B이 될 수 있는 수는 모두 몇 개인가?

Answer: 2

Solution (Python):
var_a = 'A2468B'
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 = 55
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 = len(list_b)
print(int(var_c))
Category: Arithmetic calculation

Question (EN): Among the students in Yuna's class, there are 28 students who like Korean, 27 students who like math, and 22 students who like both Korean and math. If there is no student who does not like both Korean and math, how many students are there in Yuna's class?
Question (KO): 유나네 반 학생 중에서 국어를 좋아하는 학생은 28명, 수학을 좋아하는 학생은 27명, 국어와 수학을 모두 좋아하는 학생은 22명입니다. 국어와 수학을 모두 좋아하지 않는 학생이 한 명도 없을 때, 유나네 반 학생은 모두 몇 명입니까?

Answer: 33

Solution (Python):
var_a = 28
var_b = 27
var_c = var_a + var_b
var_d = 22
var_e = var_c - var_d
print(int(var_e))
Category: Possibility

Question (EN): The seven digits 0, 1, 2, 3, 4, 5, and 6 are used once to form a seven-digit number. If this number is a multiple of 55, find the smallest number.
Question (KO): 일곱 개의 숫자 0, 1, 2, 3, 4, 5, 6을 한 번씩 사용하여 일곱 자리 수를 만들었다. 이 수가 55의 배수일 때, 가장 작은 수를 구하여라.

Answer: 1042635

Solution (Python):
var_a = 0
var_b = 1
var_c = 2
var_d = 3
var_e = 4
var_f = 5
var_g = 6
list_a= []
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_h = 7
list_b = [str(i) for i in list_a]
list_b = list(itertools.permutations(list_b, var_h))
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_i = 55
list_c = []
var_i = int(var_i)
for i in list_b:
    i = int(i)
    if i % var_i == 0:
        list_c.append(i)
var_j = 1
list_d=list_c.copy()
list_d.sort()
var_k = list_d[var_j-1]
print(int(var_k))
Category: Correspondence

Question (EN): If you round down 9A7 to the tens place, you get 930. Find the value of A.
Question (KO): 9A7을 버림하여 십의 자리까지 나타내면 930이다. A의 값을 구하시오.

Answer: 3

Solution (Python):
var_a = '9A7'
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 = 930
list_b = []
for i in list_a:
    if i >= var_b:
        list_b.append(i)
var_c = 930
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 = '9A7'
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: Comparison

Question (EN): You want to drink 1 liter (L) of juice and 3 liters (L) of juice equally divided into cups of the same shape and size. If 1 liter (L) of juice is divided equally into 2 A cups and 3 liters (L) of juice is divided equally into 4 B cups, which cup has more juice, Cup A or Cup B?
Question (KO): 주스 1리터(L)와 주스 3리터(L)를 모양과 크기가 같은 컵에 각각 똑같이 나누어 마시려고 합니다. 주스 1리터(L)를 컵 가 2개에, 주스 3리터(L)를 컵 나 4개에 똑같이 나누어 담을 때, 컵 가와 컵 나 중 어느 컵에 주스가 더 많을까요?

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 = 1
var_d = 2
var_e = var_c * var_d
var_f = 3
var_g = 4
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): When 12 is divided by A, the quotient is B and the remainder is C. A, B, and C are natural numbers. If the quotient and the remainder are equal, what is the largest possible A?
Question (KO): 12를 A로 나눴더니 몫이 B가 되었고, 나머지는 C가 되었습니다. A, B, C는 자연수 입니다. 몫과 나머지가 같을 때, 가능한 A 중 가장 큰 수는 무엇인가요?

Answer: 11

Solution (Python):
var_a = 12
var_b = 1
var_c = var_a - var_b
print(int(var_c))
Category: Geometry

Question (EN): If you multiply the width of a square by 0.8 and the length by 5 to create a new rectangle, its area increases by 15.18 square centimeters (cm2) than the area of the original square. What is the area of the first square in square centimeters (cm2)?
Question (KO): 어떤 정사각형의 가로를 0.8배, 세로를 5배로 하여 새로운 직사각형을 만들었더니 그 넓이가 처음 정사각형의 넓이보다 15.18제곱센티미터(㎠)만큼 늘었습니다. 처음 정사각형의 넓이는 몇 제곱센티미터(㎠)입니까?

Answer: 5.06

Solution (Python):
var_a = 15.18
var_b = 0.8
var_c = 5
var_d = var_b * var_c
var_e = 1
var_f = var_d - var_e
var_g = var_a / var_f
print('{:.2f}'.format(round(var_g+1e-10,2)))
Category: Correspondence

Question (EN): You had to multiply a number by 5 and then multiply by 10, but mistakenly multiplied 15 instead to get 45. Find the correctly calculated value.
Question (KO): 어떤 수에 5를 곱한 다음 10을 곱해야 할 것을 잘못하여 15를 곱하였더니 45가 되었습니다. 바르게 계산한 결과를 구하시오.

Answer: 150

Solution (Python):
var_a = 45
var_b = 15
var_c = var_a / var_b
var_d = 5
var_e = var_c * var_d
var_f = 10
var_g = var_e * var_f
print(int(var_g))
Category: Comparison

Question (EN): How many of 0.8, 1/2, 0.9, and 1/3 are less than or equal to 0.4?
Question (KO): 0.8, 1/2, 0.9, 1/3 중에서 0.4보다 작거나 같은 수는 몇 개입니까?

Answer: 1

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 = 0.4
list_b = []
for i in list_a:
    if i <= var_e:
        list_b.append(i)
var_f = len(list_b)
print(int(var_f))
Category: Comparison

Question (EN): A total of 108 coins consists of 10-won coins, 50-won coins, and 100-won coins. The amount of the 50-won coins is 1/3 of the amount of the 10-won coins, and the amount of the 100-won coins is 1/2 of that of the 50-won coins. What kind of won-coins do you have the least?
Question (KO): 10원짜리 동전, 50원짜리 동전, 100원짜리 동전이 모두 108개 있습니다. 50원짜리 동전은 10원짜리 동전의 1/3개 있고, 100원짜리 동전은 50원짜리 동전의 1/2개 있습니다. 몇 원짜리 동전이 가장 적게 있습니까?

Answer: 100-won

Solution (Python):
var_a = '10-won'
var_b = '50-won'
var_c = '100-won'
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
var_e = 1
var_f = 3
var_g = var_e / var_f
var_h = 1
var_i = 3
var_j = var_h / var_i
var_k = 2
var_l = var_j / var_k
list_b= []
if "/" in str(var_l):
    var_l = eval(str(var_l))
list_b.append(var_l)
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_m = 1
list_c=list_b.copy()
list_c.sort()
var_n = list_c[var_m-1]
var_o = list_b.index(var_n)+1
var_p = list_a[var_o-1]
print(var_p)
Category: Geometry

Question (EN): If you draw two diagonals from a square, you get B polygons having A sides and C right-angled polygons having A sides. What is A×(B+C)?
Question (KO): 사각형이 있을 때 두 개의 대각선을 그을경우 A각형이 B개, 직각 A각형이 C개 만들어진다. A×(B+C)는?

Answer: 24

Solution (Python):
var_a = 3
var_b = 8
var_c = var_a * var_b
print(int(var_c))
Category: Arithmetic calculation

Question (EN): There are 10.4 kilograms (kg) of flour in a large barrel. If you want to divide this flour into several bowls, how many bowls will you need to divide 0.8 kg (kg) into each bowl?
Question (KO): 큰 통 속에 밀가루가 10.4킬로그램(㎏) 있습니다. 이 밀가루를 여러 개의 그릇에 나누어 담으려고 할 때, 그릇 한 개에 0.8킬로그램(㎏) 씩 나누어 담으려면 그릇은 모두 몇 개가 필요할까요?

Answer: 13

Solution (Python):
var_a = 10.4
var_b = 0.8
var_c = var_a / var_b
var_d = 1
var_e=int(((var_c+9*10**(var_d-2))//(10**(var_d-1)))*10**(var_d-1))

print(int(var_e))
Category: Correspondence

Question (EN): When you divide a number by 4, you get 12. What is the result of dividing the number by 3?
Question (KO): 어떤 수를 4로 나누었더니 결과가 12가 나왔습니다. 3으로 어떤 수를 나눈 결과는 무엇입니까?

Answer: 16

Solution (Python):
var_a = 12
var_b = 4
var_c = var_a * var_b
var_d = 3
var_e = var_c / var_d
print(int(var_e))