philschmid/DMath huggingface dataset:
Category: Possibility

Question (EN): When all three natural numbers 1, 6, and 8 are used to form a natural number less than 1000, find the sum of the smallest number and the third smallest number.
Question (KO): 세 자연수 1, 6, 8을 한 개씩 모두 사용하여 1000보다 작은 자연수를 만들 때, 가장 작은 수와 세 번째로 작은 수의 합을 구하시오.

Answer: 786

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 = 3
list_d=list_b.copy()
list_d.sort()
var_h = list_d[var_g-1]
var_i = var_f + var_h
print(int(var_i))
Category: Arithmetic calculation

Question (EN): Yoongi caught 5 frogs. If he caught 2 more frogs, how many frogs did Yoongi catch?
Question (KO): 윤기는 개구리 5마리를 잡았습니다. 개구리 2마리를 더 잡았을 때, 윤기가 잡은 개구리는 몇 마리 인가요?

Answer: 7

Solution (Python):
var_a = 5
var_b = 2
var_c = var_a + var_b
print(int(var_c))
Category: Comparison

Question (EN): Gray cats are heavier than white cats and lighter than black cats. Yellow cats are the lightest. Which color is the heaviest among gray, white, black and yellow cats?
Question (KO): 회색 고양이는 흰색 고양이보다 무겁고 검은색 고양이보다 가볍습니다. 노란색 고양이는 가장 가볍습니다. 회색, 흰색, 검은색, 노란색 고양이 중에서 가장 무거운 고양이의 색깔은 무엇입니까?

Answer: black

Solution (Python):
var_a = 'gray'
var_b = 'white'
var_c = 'black'
var_d = 'yellow'
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 = 'gray'
var_f = 'white'
var_g = '>'
var_h = 'gray'
var_i = 'black'
var_j = '<'
var_k = 'yellow'
var_l = 'white'
var_m = '<'
var_n = 'yellow'
var_o = 'gray'
var_p = '<'
var_q = 'yellow'
var_r = 'black'
var_s = '<'
list_b= []
if "/" in str(var_s):
    var_s = eval(str(var_s))
list_b.append(var_s)
if "/" in str(var_r):
    var_r = eval(str(var_r))
list_b.append(var_r)
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)
if "/" in str(var_e):
    var_e = eval(str(var_e))
list_b.append(var_e)
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_t = 1
var_u = list_c[var_t-1]
print(var_u)
Category: Geometry

Question (EN): Five pieces of tape, every 2.7 meters (m) long, were stacked so that there was an overlap of 0.3 meters (m). If you divide this spliced tape into 6 equal parts, what is the length of each equal part?
Question (KO): 각 길이가 2.7미터(m)인 테이프 5장을 0.3미터(m)씩 겹치는 부분이 있도록 이었습니다. 이 이어 붙인 테이프를 6등분할 때, 각 등분의 길이는 얼마입니까?

Answer: 2.05

Solution (Python):
var_a = 2.7
var_b = 5
var_c = var_a * var_b
var_d = 0.3
var_e = 5
var_f = 1
var_g = var_e - var_f
var_h = var_d * var_g
var_i = var_c - var_h
var_j = 6
var_k = var_i / var_j
print('{:.2f}'.format(round(var_k+1e-10,2)))
Category: Geometry

Question (EN): Given that the area of a trapezoid is 56 square centimeters (cm), the top side is 5 centimeters (cm), and the bottom side is 9 centimeters (cm), what is the height of the trapezoid in centimeters (cm)?
Question (KO): 어떤 사다리꼴의 넓이가 56제곱센티미터(㎠), 윗변이 5센티미터(㎝), 아랫변이 9센티미터(㎝)이라면, 이 사다리꼴의 높이는 몇 센티미터(㎝)입니까?

Answer: 8

Solution (Python):
var_a = 56
var_b = 2
var_c = var_a * var_b
var_d = 5
var_e = 9
var_f = var_d + var_e
var_g = var_c / var_f
print(int(var_g))
Category: Correspondence

Question (EN): You mistakenly subtracted 13/5 from a number instead of adding 13/5 and got 9/7. How much is it when calculated correctly?
Question (KO): 어떤 수에 13/5을 더해야 할 것을 잘못하여 빼었더니 9/7가 되었습니다. 바르게 계산하면 얼마입니까?

Answer: 6.49

Solution (Python):
var_a = 1.2857142857142858
var_b = 2.6
var_c = var_a + var_b
var_d = 2.6
var_e = var_c + var_d
print('{:.2f}'.format(round(var_e+1e-10,2)))
Category: Geometry

Question (EN): There is one triangle. The length of the hypotenuse and base of this triangle is 15 centimeters (cm) and 9 centimeters (cm), respectively. Find the area.
Question (KO): 한 삼각형이 있다. 이 삼각형의 빗변의 길이과 밑변의 길이는 각각 15센티미터(㎝), 9센티미터(㎝)이다. 넓이를 구하라.

Answer: 54

Solution (Python):
var_a = 15
var_b = 2
var_c = var_a ** var_b
var_d = 9
var_e = 2
var_f = var_d ** var_e
var_g = var_c - var_f
var_h = 0.5
var_i = var_g ** var_h
var_j = 9
var_k = var_i * var_j
var_l = 2
var_m = var_k / var_l
print(int(var_m))
Category: Comparison

Question (EN): 20 students are waiting for the bus. If 11 people came after Yoongi, how many people came before Yoongi?
Question (KO): 20명의 학생들이 버스를 기다리고 있습니다. 윤기보다 늦게 온 사람은 11명이라면 윤기보다 먼저 온 사람은 몇 명입니까?

Answer: 8

Solution (Python):
var_a = 20
var_b = 11
var_c = var_a - var_b
var_d = 1
var_e = var_c - var_d
print(int(var_e))
Category: Arithmetic calculation

Question (EN): What is the smallest 5-digit number that is a multiple of 3 and a multiple of 4?
Question (KO): 3의 배수이면서 4의 배수인 수 중 가장 작은 다섯 자리 수는?

Answer: 10008

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 = 4
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: Geometry

Question (EN): There is a quadrangular pyramid whose edges have the same length. If the sum of the lengths of the edges is 14.8 meters (m), how many meters (m) is the length of one edge?
Question (KO): 모서리의 길이가 같은 사각뿔이 있습니다. 모서리의 길이의 합이 14.8미터(m) 일 때 한 모서리의 길이는 몇 미터(m) 입니까?

Answer: 1.85

Solution (Python):
var_a = 14.8
var_b = 4
var_c = 2
var_d = var_b * var_c
var_e = var_a / var_d
print('{:.2f}'.format(round(var_e+1e-10,2)))
Category: Comparison

Question (EN): Three people A, B, and C have books. The average number of books owned by the three people is 23 fewer than the average of the books owned by two people A and B, and 17 more than the average number of books owned by B and C. If B has 66 books, how many books does C have?
Question (KO): 가, 나, 다 세 사람이 책을 가지고 있습니다. 세 사람이 가지고 있는 책의 평균은, 가, 나 두 사람이 가진 책의 평균보다 23권 더 적고, 나, 다 두 사람이 가진 평균보다 17권 더 많습니다. 나가 가지고 있는 책이 66권일 때, 다가 가지고 있는 책은 몇 권입니까?

Answer: 44

Solution (Python):
var_a = 66
var_b = 17
var_c = 2
var_d = var_b * var_c
var_e = var_a - var_d
var_f = 3
var_g = var_e * var_f
var_h = 17
var_i = 23
var_j = var_h + var_i
var_k = 2
var_l = var_j * var_k
var_m = var_g + var_l
var_n = 66
var_o = 2
var_p = var_n * var_o
var_q = var_m - var_p
var_r = 2
var_s = 2
var_t = var_r + var_s
var_u = 3
var_v = var_t - var_u
var_w = var_q / var_v
print(int(var_w))
Category: Geometry

Question (EN): You are going to build a fence in the shape of an equilateral triangle with wooden sticks that are 8 meters (m) long. Find how many meters (m) each side of the fence should be.
Question (KO): 길이가 8미터(m)인 나무 막대로 정삼각형 모양의 울타리를 만들려고 합니다. 울타리의 한 변은 몇 미터(m)로 해야 할지 구하시오.

Answer: 2.67

Solution (Python):
var_a = 8
var_b = 3
var_c = var_a / var_b
print('{:.2f}'.format(round(var_c+1e-10,2)))
Category: Correspondence

Question (EN): Jongkook drinks milk every day. He drank 3 liters (L) and 7 milliliters (㎖) on the first day, a few milliliters (㎖) on the second day, and 840 milliliters (㎖) on the third day. As he drank 6 liters (L) and 30 milliliters (㎖) in total, how many milliliters (㎖) of milk did Jongkook drink on the second day?
Question (KO): 종국이는 매일 우유를 마신다. 첫 째날 3리터(L) 7밀리리터(㎖)를 마시고, 둘째 날 몇 미터(m)를 마시고, 셋째 날 840미터(m)를 마셔서 전체로 6리터(L) 30밀리리터(㎖)를 마셨다. 종국이가 둘째 날 마신 우유의 양은 몇 미터(m)인가?

Answer: 2183

Solution (Python):
var_a = 6
var_b = 1000
var_c = var_a * var_b
var_d = 30
var_e = var_c + var_d
var_f = 840
var_g = var_e - var_f
var_h = 3
var_i = 1000
var_j = var_h * var_i
var_k = 7
var_l = var_j + var_k
var_m = var_g - var_l
print(int(var_m))
Category: Arithmetic calculation

Question (EN): There is no child who cannot do either of A and B. If there are 38 children who can do A and 29 children who can do B and the total number of children is 48, how many children can do both?
Question (KO): (가)와 (나) 중에서 하나도 못 하는 어린이는 없습니다. (가)를 할 수 있는 어린이는 38명, (나)를 할 수 있는 어린이가 29명이고 전체 어린이는 48명일 때, 둘 모두를 할 수 있는 어린이는 몇 명입니까?

Answer: 19

Solution (Python):
var_a = 38
var_b = 29
var_c = var_a + var_b
var_d = 48
var_e = var_c - var_d
print(int(var_e))
Category: Geometry

Question (EN): If you put 10 pieces of square paper in each horizontal and vertical line without gaps, you get a large square. Find the number of pieces of paper placed around the perimeter.
Question (KO): 정사각형 종이조각을 가로와 세로 모두 10장씩 빈틈없이 놓았더니 큰 정사각형이 되었습니다. 둘레에 놓인 종이조각의 수를 구하시오.

Answer: 36

Solution (Python):
var_a = 10
var_b = 4
var_c = var_a * var_b
var_d = 4
var_e = var_c - var_d
print(int(var_e))
Category: Correspondence

Question (EN): A, B, C are different numbers from 0 to 9. Find A given the two-digit addition equation AB+BC=BCB.
Question (KO): A, B, C는 0에서 9까지의 서로 다른 숫자입니다. 두 자리 수끼리의 덧셈식 AB+BC=BCB에서 A를 구하시오.

Answer: 9

Solution (Python):
var_a = 'AB+BC=BCB'
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): On the first day of participating in the group activity, Miju met 11 new friends and shook hands. If Mijoo and her friends shook hands once, how many times did they all shake hands?
Question (KO): 미주는 모둠 활동에 참여한 첫째 날 11명의 친구들을 새로 만나서 악수를 하였습니다. 미주와 친구들이 서로 한 번씩 악수를 하였다면 악수는 모두 몇 번 하였습니까?

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): Find the sum of all numbers greater than or equal to 30, less than or equal to 50, and that are multiples of 6.
Question (KO): 30보다 같거나 크고 50보다 같거나 작으면서 6의 배수인 수의 총합을 구하시오.

Answer: 156

Solution (Python):
var_a = 30
var_b = 50
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)
list_b = [float(i) for i in list_b]
var_e = sum(list_b)
print(int(var_e))
Category: Correspondence

Question (EN): While painting, Hyungjun used half of the paint to paint the house and he used 5 more than 1/3 of the remaining paint to paint the fence. If there are 5 cans of paint left, how many cans of paint did Hyeongjun have to begin with?
Question (KO): 형준이는 페인트 칠을 하면서 전체의 절반을 집을 칠하는데 썼고, 담장을 칠하는데 남은 페인트의 1/3보다 5통 많게 썼습니다. 남은 페인트가 5통이라면 형준이는 처음 몇 통의 페인트를 가지고 있었습니까?

Answer: 30

Solution (Python):
var_a = 5
var_b = 5
var_c = var_a + var_b
var_d = 1
var_e = 0.3333333333333333
var_f = var_d - var_e
var_g = var_c / var_f
var_h = 1
var_i = 0.5
var_j = var_h - var_i
var_k = var_g / var_j
print(int(eval('{:.2f}'.format(round(var_k+1e-10,2)))))
Category: Possibility

Question (EN): Five animals are on the way: Rat, Cow, Tiger, Rabbit, Dragon and Snake. The mouse with the best eye sight should come first, and the snake with the longest tail should come last. Find the number of all cases in which five animals form a line.
Question (KO): 쥐, 소, 호랑이, 토끼, 용, 뱀 다섯 동물이 길을 가고 있습니다. 눈이 제일 좋은 쥐는 맨 앞에 와야 하고, 꼬리가 제일 긴 뱀은 맨 뒤에 와야 합니다. 다섯 동물이 줄을 서는 모든 경우의 수를 구하시오.

Answer: 24

Solution (Python):
var_a = 'Rat'
var_b = 'Cow'
var_c = 'Tiger'
var_d = 'Rabbit'
var_e = 'Dragon'
var_f = 'Snake'
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 = 'Rat'
list_b= []
if "/" in str(var_g):
    var_g = eval(str(var_g))
list_b.append(var_g)
list_b.reverse()
list_c = list(set(list_a) - set(list_b))
var_h = 'Snake'
list_d= []
if "/" in str(var_h):
    var_h = eval(str(var_h))
list_d.append(var_h)
list_d.reverse()
list_e = list(set(list_c) - set(list_d))
var_i = len(list_e)
var_j = len(list_e)
var_k = 1
var_i = int(var_i)
var_j = int(var_j)
for i, elem in enumerate(range(var_j)):
    var_k = var_k * (var_i-i)
print(int(var_k))
Category: Comparison

Question (EN): Among five numbers 1.4, 9/10, 1.2, 0.5, and 13/10, what is the sum of the numbers greater than 1.1?
Question (KO): 5가지 숫자 1.4, 9/10, 1.2, 0.5, 13/10 중에서 1.1보다 큰 수들의 합을 구하면 얼마입니까?

Answer: 3.9

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)
list_b = [float(i) for i in list_b]
var_g = sum(list_b)
print('{:.2f}'.format(round(var_g+1e-10,2)))
Category: Arithmetic calculation

Question (EN): It is said that a factory produces 230 TVs on average. Find how many TVs are produced in 15 days.
Question (KO): 어느 공장에서 평균적으로 230개의 티비를 생산한다고 한다. 보름동안 몇 개의 티비를 만드는지 구하시오.

Answer: 3450

Solution (Python):
var_a = 230
var_b = 15
var_c = var_a * var_b
print(int(var_c))
Category: Arithmetic calculation

Question (EN): There are five numbers 10, 11, 12, 13, and 14. What is the product of the 3rd largest number and the 2nd largest number?
Question (KO): 5개의 수 10, 11, 12, 13, 14가 있습니다. 그 중에서 3번째로 큰 수와 2번째로 큰 수의 곱은 얼마입니까?

Answer: 156

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 = 3
list_b=list_a.copy()
list_b.sort()
var_g = list_b[-var_f]
var_h = 2
list_c=list_a.copy()
list_c.sort()
var_i = list_c[-var_h]
var_j = var_g * var_i
print(int(var_j))
Category: Possibility

Question (EN): Divide a picture into four sections, and try to color each one with red, blue, green, and yellow. How many times are there where you paint them all different colors?
Question (KO): 하나의 그림을 네 구역으로 나누고, 각자 빨강, 파랑, 초록, 노랑으로 칠하려고 한다. 모두 다른 색으로 칠하는 경우는 몇 가지인가?

Answer: 24

Solution (Python):
var_a = 4
var_b = 4
var_c = 1
var_a = int(var_a)
var_b = int(var_b)
for i, elem in enumerate(range(var_b)):
    var_c = var_c * (var_a-i)
print(int(var_c))
Category: Geometry

Question (EN): I made an equilateral triangle with a wire 63 cm (cm) long, and there was neither excess nor shortage of wire. How many centimeters (cm) is the length of one side of an equilateral triangle?
Question (KO): 길이가 63㎝인 철사로 정삼각형을 만들었더니 철사가 남지도 모자라지도 않았습니다. 정삼각형의 한 변의 길이는 몇 센티미터(㎝)입니까?

Answer: 21

Solution (Python):
var_a = 63
var_b = 3
var_c = var_a / var_b
print(int(var_c))
Category: Arithmetic calculation

Question (EN): When there are numbers from 100 to 200, what is sum of odd numbers in the range?
Question (KO): 100부터 200까지의 수 가 있습니다. 이들 중 홀수들의 합을 구하시오

Answer: 7500

Solution (Python):
var_a = 100
var_b = 200
list_a = []
if var_a%2==0:
    for i in range(var_a+1, var_b+1, 2):
        list_a.append(i)
else:
    for i in range(var_a, var_b+1, 2):
        list_a.append(i)
list_a = [float(i) for i in list_a]
var_c = sum(list_a)
print(int(var_c))
Category: Arithmetic calculation

Question (EN): There are 45 ducks and 28 geese on the farm. Also, hens are 13 less than ducks, and roosters are 9 more than geese. How many chickens are there in the farm in total?
Question (KO): 농장에 오리 45마리, 거위 28마리가 있습니다. 또 암탉은 오리보다 13마리 적고, 수탉은 거위보다 9마리 많습니다. 농장에 있는 닭은 모두 몇 마리입니까?

Answer: 69

Solution (Python):
var_a = 45
var_b = 13
var_c = var_a - var_b
var_d = 28
var_e = 9
var_f = var_d + var_e
var_g = var_c + var_f
print(int(var_g))
Category: Comparison

Question (EN): 20 students stood in a line. Jungkook stood third in line, and Yoongi stood right behind Jungkook. How many students are standing behind Yoongi?
Question (KO): 20명의 학생들이 한줄로 섰습니다. 정국이는 3등으로 줄을 섰고, 윤기는 정국이 바로 뒤에 섰습니다. 윤기 뒤에 서 있는 학생은 몇 명입니까?

Answer: 16

Solution (Python):
var_a = 20
var_b = 3
var_c = var_a - var_b
var_d = 1
var_e = var_c - var_d
print(int(var_e))
Category: Possibility

Question (EN): Joohan is about to choose a hat and pair of shoes to wear tomorrow. Joohan has 4 hats and 3 pairs of shoes. In how many ways can Juhan choose one hat and one pair of shoes?
Question (KO): 주한이가 내일 쓸 모자 하나와 신을 신발 하나를 고르려고 합니다. 주한이가 가진 모자는 4개이고 주한이가 가진 신발은 3개입니다. 주한이가 모자와 신발을 각각 1종류씩 고르는 경우의 수는 모두 몇가지 인가요?

Answer: 12

Solution (Python):
var_a = 4
var_b = 1
var_c = 1
var_a = int(var_a)
var_b = int(var_b)
for i, elem in enumerate(range(var_b)):
    var_c = var_c * (var_a-i)
for i, elem in enumerate(range(var_b)):
    var_c = var_c / (i+1)
var_d = 3
var_e = 1
var_f = 1
var_d = int(var_d)
var_e = int(var_e)
for i, elem in enumerate(range(var_e)):
    var_f = var_f * (var_d-i)
for i, elem in enumerate(range(var_e)):
    var_f = var_f / (i+1)
var_g = var_c * var_f
print(int(var_g))
Category: Arithmetic calculation

Question (EN): The price of clay is 17.25 won per gram (g). If Seungjun buys one 1000 gram (g) clay and one 10 gram (g) clay, how much should he pay?
Question (KO): 찰흙 1그램(g) 당 17.25원의 가격이라고 합니다. 승준이가 1000그램(g)짜리 찰흙 1개와 10그램(g)짜리 찰흙 1개를 살 때, 얼마를 지불해야합니까?

Answer: 17422.5

Solution (Python):
var_a = 17.25
var_b = 1000
var_c = var_a * var_b
var_d = 17.25
var_e = 10
var_f = var_d * var_e
var_g = var_c + var_f
print('{:.2f}'.format(round(var_g+1e-10,2)))
Category: Possibility

Question (EN): We want to select 4 out of 10 distinguishable balls. How many ways are there to choose a ball?
Question (KO): 구별이 가능한 10개의 공 중에서 4개를 선택하려고 합니다. 공을 선택할 수 있는 경우는 모두 몇 가지입니까?

Answer: 210

Solution (Python):
var_a = 10
var_b = 4
var_c = 1
var_a = int(var_a)
var_b = int(var_b)
for i, elem in enumerate(range(var_b)):
    var_c = var_c * (var_a-i)
for i, elem in enumerate(range(var_b)):
    var_c = var_c / (i+1)
print(int(var_c))
Category: Geometry

Question (EN): A fishbowl containing an object called (A) has been filled with water to a height of 16 centimeters (cm). If the length of one corner of the fishbowl is 20 centimeters (cm) and if (A) is removed, the height is 9 centimeters (cm), then what is the volume of the (A)?
Question (KO): (가) 물건 1개가 들어있는 어항에 물이 16센티미터(㎝) 높이만큼 담겨있었습니다. 어항의 한 모서리의 길이가 20센티미터(㎝)이고, (가) 물건을 뺐더니 높이가 9센티미터(㎝)가 되었다면, (가) 물건의 부피는 얼마인가?

Answer: 2800

Solution (Python):
var_a = 20
var_b = 20
var_c = var_a * var_b
var_d = 16
var_e = 9
var_f = var_d - var_e
var_g = var_c * var_f
print(int(var_g))
Category: Comparison

Question (EN): 60 presses printed 40680 sheets of paper in print shop A and 55 presses printed 37510 sheets of paper in print shop B. Which printshop printed more sheets per press on average?
Question (KO): (가) 인쇄소에선 60대의 인쇄기로 40680장의 종이를 인쇄했고, (나) 인쇄소에선 55대의 인쇄기로 37510장의 종이를 인쇄했습니다. 인쇄기 하나 당 평균적으로 인쇄한 종이의 장수가 더 많은 인쇄소는 어디입니까?

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 = 40680
var_d = 60
var_e = var_c / var_d
var_f = 37510
var_g = 55
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: Arithmetic calculation

Question (EN): Eunji bought a 350 won snack at the mart with the money she had and received 800 won from her mother as an allowance. I bought a pencil with half of the remaining money, and I had 550 won left. How much money did Eunji have at first?
Question (KO): 은지는 가지고 있던 돈으로 마트에서 350원짜리 과자를 사고 어머니께 용돈으로 800원을 받았습니다. 남은 돈의 반으로 연필을 샀더니 550원이 남았습니다. 처음에 은지가 가지고 있던 돈은 얼마입니까?

Answer: 650

Solution (Python):
var_a = 550
var_b = 2
var_c = var_a * var_b
var_d = 800
var_e = var_c - var_d
var_f = 350
var_g = var_e + var_f
print(int(var_g))