philschmid/DMath huggingface dataset:
Category: Possibility

Question (EN): You want to create an eight-digit number by using the four numbers 4, 0, 2, and 6 twice. Find the sum of the largest and smallest possible numbers.
Question (KO): 4개의 숫자 4, 0, 2, 6을 두 번씩 사용하여 여덟 자리 수를 만들려고 합니다. 만들 수 있는 가장 큰 수와 가장 작은 수의 합을 구하시오.

Answer: 86466666

Solution (Python):
var_a = 4
var_b = 4
var_c = 0
var_d = 0
var_e = 2
var_f = 2
var_g = 6
var_h = 6
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 = 8
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]
var_l = 1
list_d=list_b.copy()
list_d.sort()
var_m = list_d[var_l-1]
var_n = var_k + var_m
print(int(var_n))
Category: Possibility

Question (EN): I'm trying to find a two-digit number that is divisible by 3 and 4 and has a remainder of 4 when divided by 5. How many two-digit numbers can you make?
Question (KO): 3과 4로 나누어떨어지고 5로 나누면 나머지가 4인 두 자리 수를 찾으려고 합니다. 만들 수 있는 두 자리 수는 모두 몇 개입니까?

Answer: 2

Solution (Python):
var_a = 10
var_b = 99
var_c = 1
list_a = [i for i in range(var_a, var_b + 1, var_c)]
var_d = 3
var_e = 4
var_f = var_e * var_d / math.gcd(int(var_e), int(var_d))
list_b = []
var_f = int(var_f)
for i in list_a:
    i = int(i)
    if i % var_f == 0:
        list_b.append(i)
var_g = 5
var_h = 4
list_c = [] 
var_g = int(var_g)
var_h = int(var_h)
if var_h < 0:
    var_h = var_h + var_g
for i in list_b:
    i = int(i)
    if i%var_g == var_h:
        list_c.append(i)
var_i = len(list_c)
print(int(var_i))
Category: Correspondence

Question (EN): The teacher distributes chocolate to students evenly. There will be the following leftovers when each 6, 8, and 10 are distributed with an equal amount of candies: 4, 6, and 8. What is the minimum number of chocolates the teacher has?
Question (KO): 선생님이 학생들에게 초콜릿을 나누어줍니다. 이때 6명에게 똑같이 나누어 주면 4개가 남고 8명에게 똑같이 나누어 주면 6개가 남고 10명에게 똑같이 나누어 주면 8개가 남는다고 한다. 이떄 선생님이 가지고 있는 초콜릿의 최소 개수는 몇 개입니까?

Answer: 118

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 = 6
var_e = 4
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 = 6
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 = 10
var_i = 8
list_d = [] 
var_h = int(var_h)
var_i = int(var_i)
if var_i < 0:
    var_i = var_i + var_h
for i in list_c:
    i = int(i)
    if i%var_h == var_i:
        list_d.append(i)
var_j = 1
list_e=list_d.copy()
list_e.sort()
var_k = list_e[var_j-1]
print(int(var_k))
Category: Geometry

Question (EN): How many vertices are there in a regular octahedron?
Question (KO): 정팔면체에서 꼭짓점은 모두 몇 개 일까요?

Answer: 6

Solution (Python):
var_a = 6
print(int(var_a))
Category: Arithmetic calculation

Question (EN): There are four numbers 10, 11, 12, and 13. What is the remainder of the largest number divided by the next largest number?
Question (KO): 4개의 수 10, 11, 12, 13이 있습니다. 그 중에서 가장 큰 수와 그 다음으로 큰 수로 나눈 나머지는 얼마입니까?

Answer: 1

Solution (Python):
var_a = 10
var_b = 11
var_c = 12
var_d = 13
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]
var_g = 2
list_c=list_a.copy()
list_c.sort()
var_h = list_c[-var_g]
var_i = var_f % var_h
print(int(var_i))
Category: Correspondence

Question (EN): Subtracting some number from 913 gives 514. How much is 514 minus some number?
Question (KO): 913에서 어떤 수를 빼었더니 514가 나왔습니다. 514에서 어떤 수를 빼면 얼마입니까?

Answer: 115

Solution (Python):
var_a = 514
var_b = 913
var_c = 514
var_d = var_b - var_c
var_e = var_a - var_d
print(int(var_e))
Category: Correspondence

Question (EN): A number is multiples of 5, and when divided by 5, the quotient is 25. Find the value of subtracting 17 from the number and then divided by 6.
Question (KO): 어떤 수는 5의 배수이고 5로 나누었더니 몫이 25 였습니다. 어떤 수에서 17을 뺀 값을 6으로 나눈 값을 구하시오.

Answer: 18

Solution (Python):
var_a = 25
var_b = 5
var_c = var_a * var_b
var_d = 17
var_e = var_c - var_d
var_f = 6
var_g = var_e / var_f
print(int(var_g))
Category: Arithmetic calculation

Question (EN): You are going to put 235 bracelets in a bag of 10 each and sell them for 3,000 won. What is the maximum amount you can get for selling a bagged bracelet?
Question (KO): 팔찌 235개를 한 봉지에 10개씩 담아 3000원에 팔려고 합니다. 봉지에 담은 팔찌를 팔 때 최대 얼마를 받을 수 있습니까?

Answer: 69000

Solution (Python):
var_a = 235
var_b = 10
var_c = var_a // var_b
var_d = 3000
var_e = var_c * var_d
print(int(var_e))
Category: Arithmetic calculation

Question (EN): There are four numbers 10, 11, 12, and 13. What is the remainder of the largest number divided by the next largest number?
Question (KO): 4개의 수 10, 11, 12, 13이 있습니다. 그 중에서 가장 큰 수와 가장 작은 수의 곱은 얼마입니까?

Answer: 130

Solution (Python):
var_a = 10
var_b = 11
var_c = 12
var_d = 13
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]
var_g = 1
list_c=list_a.copy()
list_c.sort()
var_h = list_c[var_g-1]
var_i = var_f * var_h
print(int(var_i))
Category: Correspondence

Question (EN): I mistakenly added what should have subtracted 2.95 from a number, and I got 9.28. Find the correct calculated value.
Question (KO): 어떤 수에서 2.95를 빼야 할 것을 잘못하여 더했더니 9.28이 되었습니다. 바르게 계산한 값을 구해 보시오.

Answer: 3.38

Solution (Python):
var_a = 9.28
var_b = 2.95
var_c = var_a - var_b
var_d = 2.95
var_e = var_c - var_d
print('{:.2f}'.format(round(var_e+1e-10,2)))
Category: Arithmetic calculation

Question (EN): When you put the snail on the table and measured the distance it moved for an hour, it was 4.25 centimeters (cm). After a few hours, the snail moved 29.75 centimeters (cm) from the position where the distance was measured earlier. How many hours have passed since the snail was placed on the table?
Question (KO): 달팽이 책상에 놓고 1시간 동안 움직인 거리를 재봤더니 4.25센티미터(㎝)였다. 몇 시간 이후 달팽이가 아까 거리를 쟀던 위치에서 부터 29.75센티미터(㎝)만큼 움직였다. 달팽이를 책상에 놓은지 몇 시간이 지났는지 구하시오.

Answer: 8

Solution (Python):
var_a = 29.75
var_b = 4.25
var_c = var_a + var_b
var_d = 4.25
var_e = var_c / var_d
print(int(var_e))
Category: Correspondence

Question (EN): When you subtract 2A from A5, you get B7. Find the sum of A and B
Question (KO): A5-2A=B7이 됩니다. A와 B의 합을 구하시오

Answer: 13

Solution (Python):
var_a = 'A5-2A=B7'
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]
var_d = 'A5-2A=B7'
var_e = 'B'
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: Correspondence

Question (EN): A particular number multiplied by 4 results in 5/2. What number do you get when you divide that particular number by 5/24?
Question (KO): 4를 어떤 수에 곱하면 5/2가 된다. 어떤 수를 5/24로 나눈다면 얼마가 되는가?

Answer: 3

Solution (Python):
var_a = 2.5
var_b = 4
var_c = var_a / var_b
var_d = 0.20833333333333334
var_e = var_c / var_d
print(int(var_e))
Category: Arithmetic calculation

Question (EN): A water tank has 13/4 liters (L) of water. You remove 12/5 liters (L) of water from the tank and then add back 11/6 liters (L) of water. Calculate the remaining amount of water in the tank in liters (L).
Question (KO): 물탱크에 13/4리터(L) 만큼의 물이 있다. 이 물탱크에서12/5리터(L) 만큼 퍼내고 11/6리터(L)만큼 다시 집어 넣었다. 물탱크에 남은 물은 몇 리터(L)인지 구하시오.

Answer: 2.68

Solution (Python):
var_a = 3.25
var_b = 2.4
var_c = var_a - var_b
var_d = 1.8333333333333333
var_e = var_c + var_d
print('{:.2f}'.format(round(var_e+1e-10,2)))
Category: Possibility

Question (EN): There are four types of milk sold at a store: chocolate milk, strawberry milk, banana milk, and coffee milk. When selecting milk from this store, find the number of ways to select 2 cartons of milk, allowing duplicatio.
Question (KO): 어느 상점에서 판매하는 우유는 초코우유, 딸기우유, 바나나우유, 커피우유 4종류이다. 이 상점에서 우유를 선택할 때, 중복을 허용하여 우유 2팩을 선택하는 방법의 수를 구하시오.

Answer: 10

Solution (Python):
var_a = 'chocolate'
var_b = 'strawberry'
var_c = 'banana'
var_d = 'coffee'
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 = 2
var_g = var_e + var_f
var_h = 1
var_i = var_g - var_h
var_j = 2
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)
for i, elem in enumerate(range(var_j)):
    var_k = var_k / (i+1)
print(int(var_k))
Category: Correspondence

Question (EN): The equation A82+3B=216 holds. What number should go in A?
Question (KO): 등식 A82+3B=216이 성립한다고 합니다. A에 들어가야 하는 숫자는 무엇인가요?

Answer: 1

Solution (Python):
var_a = 'A82+3B=216'
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: Possibility

Question (EN): There are 5 marbles: red, blue, yellow, purple, and black. You want to choose two of these and give one to your younger brother and the other to your older brother as a gift. Find the number of cases in this situation.
Question (KO): 구슬이 빨강, 파랑, 노랑, 보라, 검정 5개가 있다. 이 중 2개를 골라서 하나를 동생에게, 다른 하나는 형에게 선물로 주려고 한다.이 상황에서의 경우의 수를 구하여라.

Answer: 20

Solution (Python):
var_a = 5
var_b = 2
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: Arithmetic calculation

Question (EN): One bottle of grape juice contains 1 liter (L) and 200 milliliters (㎖). Three friends, Taeyeon, Jimin, and Jisoo, poured grape juice into a glass to share, and a friend complained about the difference in the amount of juice. If after the 200 milliliters (ml) of juice in Taeyeon's glass was poured into Jimin's glass and 100 milliliters (ml) into Jisoo's glass it became fair, how many milliliters (ml) of juice was originally in Jisoo's glass?
Question (KO): 포도 주스 1병에는 1리터(L) 200밀리리터(㎖)가 들어있습니다. 태연, 지민, 지수 세 명의 친구가 포도 주스를 나눠 마시기 위해 잔에 따랐더니 주스의 양이 달라 불평하는 친구가 생겼습니다. 태연이 잔에 든 주스를 지민이 잔으로 200밀리리터(㎖), 지수 잔으로 100밀리리터(㎖) 옮겨 따른 후에야 공평하게 따라졌다면, 지수 잔에는 원래 몇 밀리리터(㎖)의 주스가 있었습니까?

Answer: 300

Solution (Python):
var_a = 1
var_b = 1000
var_c = var_a * var_b
var_d = 200
var_e = var_c + var_d
var_f = 200
var_g = var_e - var_f
var_h = 100
var_i = var_g - var_h
var_j = 3
var_k = var_i / var_j
print(int(var_k))
Category: Arithmetic calculation

Question (EN): There is a two-digit natural number whose tens place is 3. Let A and B be the quotient of this number by 10 and the remainder of division by 10, respectively. If B multiplied by 10 plus A is 9 less than A multiplied by 10 plus B, what is the first number?
Question (KO): 십의 자리 수가 3인 두 자리 자연수가 있다. 이 숫자를 10으로 나눈 몫과, 10으로 나눈 나머지를 각각 A, B라고 하자. B에 10을 곱하고, A를 더한 수는 A에 10을 곱하고 B를 더한 수보다 9만큼 작다고 할 때, 처음 수는 무엇인지 구하라.

Answer: 32

Solution (Python):
var_a = 3
var_b = 10
var_c = var_a * var_b
var_d = 9
var_e = var_c - var_d
var_f = 3
var_g = var_e - var_f
var_h = 10
var_i = 1
var_j = var_h - var_i
var_k = var_g / var_j
var_l = 3
var_m = 10
var_n = var_l * var_m
var_o = var_k + var_n
print(int(var_o))
Category: Correspondence

Question (EN): The four-digit number 14AB is divisible by 9. If A and B are different numbers from 0 to 9, how many numbers can be 14AB?
Question (KO): 네 자리 수 14AB는 9로 나누어떨어지는 수입니다. A, B는 0부터 9까지의 서로 다른 수일 때 14AB가 될 수 있는 수는 모두 몇 개입니까?

Answer: 11

Solution (Python):
var_a = '14AB'
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 = 9
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: Correspondence

Question (EN): When you add 30 to a number, you get 55. What is the result of subtracting the number by 23?
Question (KO): 30을 어떤 수에 더했더니 55가 되었습니다. 23을 어떤 수에 뺀 결과를 구하시오.

Answer: 2

Solution (Python):
var_a = 55
var_b = 30
var_c = var_a - var_b
var_d = 23
var_e = var_c - var_d
print(int(var_e))
Category: Possibility

Question (EN): Find the number of cases in which A or B is at the forefront when three of the four students are selected from A, B, C, and D and placed in a line.
Question (KO): A, B, C, D 네 명의 학생 중에서 3명을 뽑아 일렬로 세울 때, A 또는 B가 맨 앞에 서게 되는 경우의 수를 구하시오.

Answer: 12

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 = len(list_a)
var_d = 4
var_e = 1
var_f = var_d - var_e
var_g = 3
var_h = 1
var_i = var_g - var_h
var_j = 1
var_f = int(var_f)
var_i = int(var_i)
for i, elem in enumerate(range(var_i)):
    var_j = var_j * (var_f-i)
for i, elem in enumerate(range(var_i)):
    var_j = var_j / (i+1)
var_k = var_c * var_j
var_l = 3
var_m = 1
var_n = var_l - var_m
var_o = 3
var_p = 1
var_q = var_o - var_p
var_r = 1
var_n = int(var_n)
var_q = int(var_q)
for i, elem in enumerate(range(var_q)):
    var_r = var_r * (var_n-i)
var_s = var_k * var_r
print(int(var_s))
Category: Arithmetic calculation

Question (EN): Taeyeon and Yura solved the problem book. Taeyeon solved 16 questions a day for 7 days and Yura solved 25 questions a day for 6 days. How many problems did Taeyeon and Yura solve?
Question (KO): 태연이와 유라가 문제집을 풀었습니다. 태연이는 하루에 16문제씩 7일을 풀었고 유라는 하루에 25문제씩 6일을 풀었습니다. 태연이와 유라가 푼 문제는 모두 몇 문제인가요?

Answer: 262

Solution (Python):
var_a = 16
var_b = 7
var_c = var_a * var_b
var_d = 25
var_e = 6
var_f = var_d * var_e
var_g = var_c + var_f
print(int(var_g))
Category: Arithmetic calculation

Question (EN): You want to place the desks at 3 meter (m) intervals on a 114 meter (m) playground. How many desks will you need in total if you place them at the beginning and at the end of the playground?
Question (KO): 책상을 114미터(m)인 운동장 위에 3미터(m) 간격으로 세우려고 합니다. 운동장 처음과 끝에 반드시 세워져 있다고 할 때 총 몇개의 책상이 필요합니까?

Answer: 39

Solution (Python):
var_a = 114
var_b = 3
var_c = var_a / var_b
var_d = 1
var_e = var_c + var_d
print(int(var_e))
Category: Arithmetic calculation

Question (EN): Use the three number cards 3, 4, and 7 to create a three-digit number. In here, find the largest value among the numbers with 7 at the front. (However, you cannot use number cards repeatedly.)
Question (KO): 세 장의 숫자카드 3, 4, 7을 사용하여 세자리 수를 만든다. 이때 7이 가장 앞에 오는 수들 중 가장 큰 값을 구하시오.(단, 숫자카드를 중복 사용할 수 없다.)

Answer: 743

Solution (Python):
var_a = 3
var_b = 4
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 = 100
var_f = 7
list_c = [] 
var_e = int(var_e)
var_f = int(var_f)
for i in list_b:
    i = int(i)
    if (i//var_e)%10 == var_f:
        list_c.append(i)
var_g = 1
list_d=list_c.copy()
list_d.sort()
var_h = list_d[-var_g]
print(int(var_h))
Category: Arithmetic calculation

Question (EN): Four friends want to eat food and split the money equally. If you ate 5 fish-shaped bread which is 200 won each and 7 hotteoks which is 800 won each, how much would each person pay?
Question (KO): 4명의 친구들이 음식을 먹고 돈을 똑같이 나누어 내려고 합니다. 한 개에 200원인 붕어빵 5개와 하나에 800원인 호떡 7개를 먹었다면 한 사람이 얼마를 내야합니까?

Answer: 1650

Solution (Python):
var_a = 200
var_b = 5
var_c = var_a * var_b
var_d = 800
var_e = 7
var_f = var_d * var_e
var_g = var_c + var_f
var_h = 4
var_i = var_g / var_h
print(int(var_i))
Category: Possibility

Question (EN): Two of the numbers 3, 5, and 9 were drawn and used only once to create the largest two-digit number. Find the number twice as large as the two-digit number you created.
Question (KO): 3, 5, 9의 숫자 중에서 2장을 뽑아 한 번씩만 사용하여 가장 큰 두 자리 수를 만들었습니다. 만든 두 자리 수보다 2배 큰 수를 구하시오.

Answer: 190

Solution (Python):
var_a = 3
var_b = 5
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 = 2
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]
var_g = 2
var_h = var_f * var_g
print(int(var_h))
Category: Arithmetic calculation

Question (EN): Yuna threaded 380 beads in an hour, and Eunji threaded 325 beads in an hour. When the two of them threaded beads at a constant speed for 8 hours, how many more beads did Yuna thread than Eunji?
Question (KO): 유나는 1시간에 380개의 구슬을 꿰었고, 은지는 1시간에 325개의 구슬을 꿰었습니다. 두 사람이 같은 빠르기로 구슬을 8시간 동안 꿰었을 때, 유나는 은지보다 구슬을 몇 개 더 많이 꿰었습니까?

Answer: 440

Solution (Python):
var_a = 380
var_b = 325
var_c = var_a - var_b
var_d = 8
var_e = var_c * var_d
print(int(var_e))
Category: Comparison

Question (EN): At a constant speed, a bus travels a distance of 17.28 kilometers (km) in 16 minutes, and a car travels a distance of 8.52 kilometers (km) in 6 minutes. If a bus and a car start at the same time from the same place, which one will go farther after 15 minutes?
Question (KO): 일정한 빠르기로 버스는 17.28킬로미터(㎞) 의 거리를 16 분 동안에 가고, 승용차는 8.52킬로미터(㎞) 의 거리를 6 분 동안에 갑니다. 버스와 승용차가 같은 곳에서 동시에 출발한다 면 15 분 후에는 어느 것이 더 멀리 가겠습니까?

Answer: car

Solution (Python):
var_a = 'bus'
var_b = 'car'
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 = 17.28
var_d = 16
var_e = var_c / var_d
var_f = 15
var_g = var_e * var_f
var_h = 8.52
var_i = 6
var_j = var_h / var_i
var_k = 15
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)
list_b.reverse()
var_m = 1
list_c=list_b.copy()
list_c.sort()
var_n = list_c[-var_m]
var_o = list_b.index(var_n)+1
var_p = list_a[var_o-1]
print(var_p)