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stringlengths 20
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stringlengths 1
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Correspondence | You need to add 45 to a number, but mistakenly added 54, and got 78. Find the result of the correct calculation. | 69 | 78 54 [OP_SUB] 45 [OP_ADD] | var_a = 78
var_b = 54
var_c = var_a - var_b
var_d = 45
var_e = var_c + var_d
print(int(var_e)) |
Geometry | There is a rectangle whose width is 0.875 times its length. If the length is 24 centimeters (cm), find the area in square centimeters (cm2). | 504 | 24 0.875 [OP_MUL] 24 [OP_MUL] | var_a = 24
var_b = 0.875
var_c = var_a * var_b
var_d = 24
var_e = var_c * var_d
print(int(var_e)) |
Arithmetic calculation | Find the sum of the even numbers from 1 to 10 that are less than or equal to 6. | 12 | 1 10 [OP_LIST_EVEN] 6 [OP_LIST_LESS_EQUAL] [OP_LIST_SUM] | var_a = 1
var_b = 10
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)
var_c = 6
list_b = []
for i in list_a:
if i <= var_c:
list_b.append(i)
list_b = [float(i) for i in list_b]
var_d = sum(list_b)
print(int(var_d)) |
Arithmetic calculation | How many four-digit numbers are divisible by the four numbers 2, 3, 8, and 9? | 125 | 1000 9999 1 [OP_LIST_ARANGE] 2 [OP_LIST_DIVISIBLE] 3 [OP_LIST_DIVISIBLE] 8 [OP_LIST_DIVISIBLE] 9 [OP_LIST_DIVISIBLE] [OP_LIST_LEN] | var_a = 1000
var_b = 9999
var_c = 1
list_a = [i for i in range(var_a, var_b + 1, var_c)]
var_d = 2
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 = 3
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 = 8
list_d = []
var_f = int(var_f)
for i in list_c:
i = int(i)
if i % var_f == 0:
list_d.append(i)
var_g = 9
list_e = []
var_g = int(var_g)
for i in list_d:
i = int(i)
if i % var_g == 0:
list_e.append(i)
var_h = len(list_e)
print(int(var_h)) |
Correspondence | Add a number to 52, multiply by 3, subtract 60, and divide by 8, and get 15.Find the number. | 8 | 15 8 [OP_MUL] 60 [OP_ADD] 3 [OP_DIV] 52 [OP_SUB] | var_a = 15
var_b = 8
var_c = var_a * var_b
var_d = 60
var_e = var_c + var_d
var_f = 3
var_g = var_e / var_f
var_h = 52
var_i = var_g - var_h
print(int(var_i)) |
Arithmetic calculation | I bought 3 notebooks at a stationery store for 750 won. How much should I pay if I buy 9 copies of the same notebook? | 2250 | 750 3 [OP_DIV] 9 [OP_MUL] | var_a = 750
var_b = 3
var_c = var_a / var_b
var_d = 9
var_e = var_c * var_d
print(int(var_e)) |
Geometry | The sum of the lengths of all the edges of a regular prism is 256 centimeters (cm). The length of the base is 4 times the width, and the height is 3 times the width. What is the length of this rectangular prism in centimeters (cm)? | 32 | 256 4 [OP_DIV] 4 3 [OP_ADD] 1 [OP_ADD] [OP_DIV] 4 [OP_MUL] | var_a = 256
var_b = 4
var_c = var_a / var_b
var_d = 4
var_e = 3
var_f = var_d + var_e
var_g = 1
var_h = var_f + var_g
var_i = var_c / var_h
var_j = 4
var_k = var_i * var_j
print(int(var_k)) |
Geometry | There is a fishbowl in the shape of a cuboid. When the length of each corner is 4 centimeters (cm), 6 centimeters (cm), and 15 centimeters (cm), find the volume of the fish tank. | 360 | 15 6 [OP_MUL] 4 [OP_MUL] | var_a = 15
var_b = 6
var_c = var_a * var_b
var_d = 4
var_e = var_c * var_d
print(int(var_e)) |
Correspondence | There is a six-digit number A15B94 that can be made with the single-digit numbers A and B. When this number is divisible by 99, find the value that can be B. | 3 | A15B94 [OP_GEN_POSSIBLE_LIST] 99 [OP_LIST_DIVISIBLE] A15B94 B [OP_LIST_FIND_UNK] | var_a = 'A15B94'
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 = 99
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 = 'A15B94'
var_d = 'B'
var_c = str(var_c)
var_d = str(var_d)
unk_idx = var_c.index(var_d)
var_e = 0
for elem in list_b:
elem = str(elem)
var_e = int(elem[unk_idx])
print(int(var_e)) |
Correspondence | When a number is multiplied by 9 and then divided by 3, it is 27. Find the number. | 9 | 27 3 [OP_MUL] 9 [OP_DIV] | var_a = 27
var_b = 3
var_c = var_a * var_b
var_d = 9
var_e = var_c / var_d
print(int(var_e)) |
Arithmetic calculation | Toys come in 6 boxes, 8 in each box. How many boxes are needed to pack 4 toys in a box? | 12 | 8 6 [OP_MUL] 4 [OP_FDIV] | var_a = 8
var_b = 6
var_c = var_a * var_b
var_d = 4
var_e = var_c // var_d
print(int(var_e)) |
Possibility | When two of the number cards 4, 5, and 6 are used once to form the smallest two-digit number, what is the number of unused cards? | 6 | [OP_LIST_SOL] 4 5 6 [OP_LIST_EOL] 2 [OP_LIST_GET_PERM] 1 [OP_LIST_MIN] [OP_LIST_POP] [OP_NUM2LIST] [OP_SET_DIFFERENCE] 1 [OP_LIST_GET] | var_a = 4
var_b = 5
var_c = 6
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-1]
list_d = []
var_f = int(var_f)
while var_f//10 > 0:
list_d.append(var_f%10)
var_f = var_f//10
list_d.append(var_f%10)
list_d = list_d[::-1]
list_e = list(set(list_a) - set(list_d))
var_g = 1
var_h = list_e[var_g-1]
print(int(var_h)) |
Comparison | The following is the long jump record of four people : Kyungsoo 2.3 meters (m), Younghee 9/10 meters (m), Jinju 1.8 meters (m), Chanho 2.5 meters (m). Who among these following people jumped the second furthest? | Kyungsoo | [OP_LIST_SOL] Kyungsoo Younghee Jinju Chanho [OP_LIST_EOL] [OP_LIST_SOL] 2.3 9/10 1.8 2.5 [OP_LIST_EOL] 2 [OP_LIST_MAX] [OP_LIST_INDEX] [OP_LIST_POP] [OP_LIST_GET] | var_a = 'Kyungsoo'
var_b = 'Younghee'
var_c = 'Jinju'
var_d = 'Chanho'
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 = 2.3
var_f = 0.9
var_g = 1.8
var_h = 2.5
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_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()
var_i = 2
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) |
Correspondence | When A is divided by B, the quotient is C and the remainder is 4. A, B, and C are natural numbers. If B and C are equal, find the smallest possible number of A. | 29 | 4 1 [OP_ADD] 2 [OP_POW] 4 [OP_ADD] | var_a = 4
var_b = 1
var_c = var_a + var_b
var_d = 2
var_e = var_c ** var_d
var_f = 4
var_g = var_e + var_f
print(int(var_g)) |
Correspondence | Minsu plans to go on a bicycle trip for 28 days. If he must travel a total distance of 82.04 kilometers (km) and the same distance a day, find how many kilometers (km) are required to take in one day. | 2.93 | 1 28 82.04 [OP_DIV] [OP_DIV] | var_a = 1
var_b = 28
var_c = 82.04
var_d = var_b / var_c
var_e = var_a / var_d
print('{:.2f}'.format(round(var_e+1e-10,2))) |
Geometry | There is an auditorium in the shape of a square with a side length of 14.25 meters (m). Find the area in square meters (m2) of this auditorium. | 203.06 | 14.25 2 [OP_POW] | var_a = 14.25
var_b = 2
var_c = var_a ** var_b
print('{:.2f}'.format(round(var_c+1e-10,2))) |
Arithmetic calculation | You want to plant street trees on one side of a road that is 1500 meters (m) long at intervals of 25 meters (m). How many street trees do you need? (Note: trees are planted at the beginning and end of the road as well.) | 61 | 1500 25 [OP_DIV] 1 [OP_ADD] | var_a = 1500
var_b = 25
var_c = var_a / var_b
var_d = 1
var_e = var_c + var_d
print(int(var_e)) |
Arithmetic calculation | Jihoon read 3 books this month, Taeyeon read 2 books, Dongyeop 5 books, Hanhae 3 books, and Kibum 1 book. How many of them have read 3 books this month? | 3 | [OP_LIST_SOL] 3 2 5 3 1 [OP_LIST_EOL] 3 [OP_LIST_MORE_EQUAL] [OP_LIST_LEN] | var_a = 3
var_b = 2
var_c = 5
var_d = 3
var_e = 1
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 = []
for i in list_a:
if i >= var_f:
list_b.append(i)
var_g = len(list_b)
print(int(var_g)) |
Arithmetic calculation | There are 18 absent students are absent in Jungkook's school. If there are 848 male students and 49 fewer female students than male students, how many students came to school today? | 1629 | 848 848 49 [OP_SUB] [OP_ADD] 18 [OP_SUB] | var_a = 848
var_b = 848
var_c = 49
var_d = var_b - var_c
var_e = var_a + var_d
var_f = 18
var_g = var_e - var_f
print(int(var_g)) |
Possibility | We want to create a three-digit natural number using the numbers 1, 2, and 3. When you can use the same number, find the sum of all numbers that can be made. | 5994 | [OP_LIST_SOL] 1 2 3 [OP_LIST_EOL] 3 [OP_LIST_GET_PRODUCT] [OP_LIST_SUM] | var_a = 1
var_b = 2
var_c = 3
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.product(list_b, repeat=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_b = [float(i) for i in list_b]
var_e = sum(list_b)
print(int(var_e)) |
Correspondence | There are 45 chocolates 15 bags. 5 Find how many chocolates are in 5 bags. | 15 | 45 15 [OP_DIV] 5 [OP_MUL] | var_a = 45
var_b = 15
var_c = var_a / var_b
var_d = 5
var_e = var_c * var_d
print(int(var_e)) |
Arithmetic calculation | If the sum of two consecutive even numbers is 46, find the larger of the two even numbers. | 24 | 46 2 [OP_DIV] 1 [OP_ADD] | var_a = 46
var_b = 2
var_c = var_a / var_b
var_d = 1
var_e = var_c + var_d
print(int(var_e)) |
Correspondence | You want to write an even number less than or equal to 100. How many times do you write the number 0 in all? | 26 | 1 200 [OP_LIST_EVEN] [OP_LIST2NUM] [OP_NUM2LIST] 0 [OP_LIST_FIND_NUM] | var_a = 1
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)
var_c=""
for i in list_a:
i = str(i)
var_c = var_c + i
list_b = []
var_c = int(var_c)
while var_c//10 > 0:
list_b.append(var_c%10)
var_c = var_c//10
list_b.append(var_c%10)
list_b = list_b[::-1]
var_d = 0
var_e = 0
var_d = int(var_d)
for i in list_b:
i = int(i)
if i == var_d:
var_e = var_e + 1
print(int(var_e)) |
Comparison | There are three bowls A, B, and C containing food in a refrigerator. Bowl C has more food than bowl B, and bowl B has more food than bowl A. When the bowl with the least amount of food is emptied first, which bowl is emptied first? | A | [OP_LIST_SOL] A B C [OP_LIST_EOL] [OP_LIST_SOL] B A > C B > [OP_LIST_EOL] [OP_LIST_COND_MAX_MIN] [OP_LIST_LEN] [OP_LIST_GET] | var_a = 'A'
var_b = 'B'
var_c = 'C'
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 = 'B'
var_e = 'A'
var_f = '>'
var_g = 'C'
var_h = 'B'
var_i = '>'
list_b= []
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)
if "/" in str(var_d):
var_d = eval(str(var_d))
list_b.append(var_d)
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_j = len(list_c)
var_k = list_c[var_j-1]
print(var_k) |
Arithmetic calculation | If each person wants to eat 2/5 pizzas, how many pizzas you should buy to share it with 10 people? | 4 | 2/5 10 [OP_MUL] | var_a = 0.4
var_b = 10
var_c = var_a * var_b
print(int(var_c)) |
Possibility | Jungkook wants to buy different fruits and give a present to Jimin and Yoongi. If the fruit shop sells apples, peaches, pears, and melons, how many number of ways are there for him to give a gift? | 12 | [OP_LIST_SOL] apples peaches pears melons [OP_LIST_EOL] [OP_LIST_LEN] 2 [OP_PERM] | var_a = 'apples'
var_b = 'peaches'
var_c = 'pears'
var_d = 'melons'
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 = 1
var_e = int(var_e)
var_f = int(var_f)
for i, elem in enumerate(range(var_f)):
var_g = var_g * (var_e-i)
print(int(var_g)) |
Geometry | You cannot draw a straight line between certain three points. How many semi-straight line can you make with these three points? | 6 | 3 2 [OP_PERM] | var_a = 3
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)) |
Possibility | Find the sum of the largest and smallest of the two-digit numbers that can be formed by drawing two different numbers from 3, 5, 7, and 8. | 122 | [OP_LIST_SOL] 3 5 7 8 [OP_LIST_EOL] 2 [OP_LIST_GET_PERM] 1 [OP_LIST_MAX] 1 [OP_LIST_MIN] [OP_ADD] | var_a = 3
var_b = 5
var_c = 7
var_d = 8
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 = 2
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]
var_h = 1
list_d=list_b.copy()
list_d.sort()
var_i = list_d[var_h-1]
var_j = var_g + var_i
print(int(var_j)) |
Arithmetic calculation | How many boxes are needed to store 27 notebooks in 9 boxes? | 3 | 27 9 [OP_FDIV] | var_a = 27
var_b = 9
var_c = var_a // var_b
print(int(var_c)) |
Correspondence | There are three digit numbers where the tens digit is 3 and the sum of the ones digit and the hundreds digit is 5. Each digit in this three-digit number is different, and when you add 124 to this number, all digits are equal. Find a three digit number. | 431 | [OP_LIST_SOL] [OP_LIST_SOL] A 3 B + 124 = C C C [OP_LIST_EOL] [OP_LIST2NUM] A [OP_DIGIT_UNK_SOLVER] 3 [OP_LIST_SOL] A 3 B + 124 = C C C [OP_LIST_EOL] [OP_LIST2NUM] B [OP_DIGIT_UNK_SOLVER] [OP_LIST_EOL] [OP_LIST2NUM] | var_a = 'A'
var_b = 3
var_c = 'B'
var_d = '+'
var_e = 124
var_f = '='
var_g = 'C'
var_h = 'C'
var_i = 'C'
list_a= []
if "/" in str(var_i):
var_i = eval(str(var_i))
list_a.append(var_i)
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_j=""
for i in list_a:
i = str(i)
var_j = var_j + i
var_k = 'A'
ans_dict = dict()
var_j = var_j.replace('×','*')
var_j = var_j.replace('x','*')
var_j = var_j.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_j):
if v in variable_candi:
ans_dict[v] = 1
candi = list(itertools.product('0123456789', repeat=len(ans_dict)))
for c in candi:
temp = var_j
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_j):
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_l = ans_dict[var_k]
var_m = 3
var_n = 'A'
var_o = 3
var_p = 'B'
var_q = '+'
var_r = 124
var_s = '='
var_t = 'C'
var_u = 'C'
var_v = 'C'
list_b= []
if "/" in str(var_v):
var_v = eval(str(var_v))
list_b.append(var_v)
if "/" in str(var_u):
var_u = eval(str(var_u))
list_b.append(var_u)
if "/" in str(var_t):
var_t = eval(str(var_t))
list_b.append(var_t)
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)
list_b.reverse()
var_w=""
for i in list_b:
i = str(i)
var_w = var_w + i
var_x = 'B'
ans_dict = dict()
var_w = var_w.replace('×','*')
var_w = var_w.replace('x','*')
var_w = var_w.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_w):
if v in variable_candi:
ans_dict[v] = 1
candi = list(itertools.product('0123456789', repeat=len(ans_dict)))
for c in candi:
temp = var_w
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_w):
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_y = ans_dict[var_x]
list_c= []
if "/" in str(var_y):
var_y = eval(str(var_y))
list_c.append(var_y)
if "/" in str(var_m):
var_m = eval(str(var_m))
list_c.append(var_m)
if "/" in str(var_l):
var_l = eval(str(var_l))
list_c.append(var_l)
list_c.reverse()
var_z=""
for i in list_c:
i = str(i)
var_z = var_z + i
print(int(var_z)) |
Possibility | You want to create a three-digit number by picking three different numbers from 0, 1, 2, 3, and 5. How many three-digit numbers can you make? | 48 | [OP_LIST_SOL] 0 1 2 3 5 [OP_LIST_EOL] 3 [OP_LIST_GET_PERM] [OP_LIST_LEN] | var_a = 0
var_b = 1
var_c = 2
var_d = 3
var_e = 5
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 = [str(i) for i in list_a]
list_b = list(itertools.permutations(list_b, var_f))
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_g = len(list_b)
print(int(var_g)) |
Arithmetic calculation | Gyuri had 1.6 kg (kg) of plums. She gave 0.8 kg (kg) to Sungmin and 0.3 kg (kg) to Dongju. How many remaining kilograms (kg) of plums does Gyuri have? | 0.5 | 1.6 0.8 [OP_SUB] 0.3 [OP_SUB] | 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))) |
Arithmetic calculation | There are total 240 marbles of yellow and blue marbles. How many yellow marbles are there if there are 2 fewer blue marbles than yellow marbles? | 121 | 240 2 [OP_SUB] 2 [OP_DIV] 2 [OP_ADD] | var_a = 240
var_b = 2
var_c = var_a - var_b
var_d = 2
var_e = var_c / var_d
var_f = 2
var_g = var_e + var_f
print(int(var_g)) |
Arithmetic calculation | It takes 30 minutes to collect 6/11 liters (l) of sap from a tree. How many minutes does it take to collect 8 liters (l) of sap from the same tree? | 440 | 30 6/11 [OP_DIV] 8 [OP_MUL] | var_a = 30
var_b = 0.5454545454545454
var_c = var_a / var_b
var_d = 8
var_e = var_c * var_d
print(int(eval('{:.2f}'.format(round(var_e+1e-10,2))))) |
Arithmetic calculation | Jungkook can travel 2.7 kilometers (km) per hour. How many hours will it take Jimin to travel 4.86 kilometers (km) at the same speed? | 1.8 | 1 2.7 [OP_DIV] 4.86 [OP_MUL] | var_a = 1
var_b = 2.7
var_c = var_a / var_b
var_d = 4.86
var_e = var_c * var_d
print('{:.2f}'.format(round(var_e+1e-10,2))) |
Arithmetic calculation | There are 45 apples. There are 21 more apples than pears. If there are 18 more tangerines than pears, how many tangerines are there? | 42 | 45 21 [OP_SUB] 18 [OP_ADD] | var_a = 45
var_b = 21
var_c = var_a - var_b
var_d = 18
var_e = var_c + var_d
print(int(var_e)) |
Arithmetic calculation | You want to cut a ribbon into 0.82 meter (m) each in order to make ribbon loops. Make a loop with 53.73 meters (m) of ribbon and find how many millimeters (mm) of this ribbon are left. | 430 | 53.73 0.82 [OP_MOD] 1000 [OP_MUL] | var_a = 53.73
var_b = 0.82
var_c = var_a % var_b
var_d = 1000
var_e = var_c * var_d
print(int(eval('{:.2f}'.format(round(var_e+1e-10,2))))) |
Possibility | You want to create a three-digit number using five different single-digit numbers. How many ways are there to have different digits? | 60 | 1 5 1 [OP_LIST_ARANGE] 3 [OP_LIST_GET_PERM] [OP_LIST_LEN] | var_a = 1
var_b = 5
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]
var_e = len(list_b)
print(int(var_e)) |
Geometry | Wrapping paper was attached to the outer surface of a cuboid measuring 6 centimeters (cm) wide, 5 centimeters (cm) long, and 4 centimeters (cm) high. When this cuboid is cut into 120 cubes with a volume of 1 cubic centimeter (cm3), find the number of cubes without wrapping paper attached. | 24 | 6 2 [OP_SUB] 5 2 [OP_SUB] [OP_MUL] 4 2 [OP_SUB] [OP_MUL] | var_a = 6
var_b = 2
var_c = var_a - var_b
var_d = 5
var_e = 2
var_f = var_d - var_e
var_g = var_c * var_f
var_h = 4
var_i = 2
var_j = var_h - var_i
var_k = var_g * var_j
print(int(var_k)) |
Possibility | There are 4 different pairs of pants and 2 skirts. You want to wrap 3 of these in 3 different wrapping paper. How many cases can be made? | 720 | 4 2 [OP_ADD] 3 [OP_PERM] 3 3 [OP_PERM] [OP_MUL] | var_a = 4
var_b = 2
var_c = var_a + var_b
var_d = 3
var_e = 1
var_c = int(var_c)
var_d = int(var_d)
for i, elem in enumerate(range(var_d)):
var_e = var_e * (var_c-i)
var_f = 3
var_g = 3
var_h = 1
var_f = int(var_f)
var_g = int(var_g)
for i, elem in enumerate(range(var_g)):
var_h = var_h * (var_f-i)
var_i = var_e * var_h
print(int(var_i)) |
Comparison | (A) is shorter than (B), and (B) is taller than (C). Also, (A) is shorter than (D) and the (C) is taller than (D). Find who is the shortest. | (A) | [OP_LIST_SOL] (A) (B) (C) (D) [OP_LIST_EOL] [OP_LIST_SOL] [OP_LIST_EOL] [OP_LIST_COND_MAX_MIN] [OP_LIST_LEN] [OP_LIST_GET] | var_a = '(A)'
var_b = '(B)'
var_c = '(C)'
var_d = '(D)'
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()
list_b= []
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_e = len(list_c)
var_f = list_c[var_e-1]
print(var_f) |
Correspondence | When 11 is divided by 4, the remainder is A and the quotient is 2. Find the value of A. | 3 | 11 4 2 [OP_MUL] [OP_SUB] | var_a = 11
var_b = 4
var_c = 2
var_d = var_b * var_c
var_e = var_a - var_d
print(int(var_e)) |
Arithmetic calculation | Hyerin folded 16 cranes a day for 7 days, and Taeyeong folded 25 cranes a day for 6 days. Find how many cranes Hyerin and Taeyeong folded. | 262 | 16 7 [OP_MUL] 25 6 [OP_MUL] [OP_ADD] | 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)) |
Correspondence | A three-digit number has an 8 in the hundreds, an A in the tens, and 2 in the ones. Rounding this to the nearest hundred makes it 800. Find the sum of the numbers that can be A. | 35 | [OP_LIST_SOL] 8 A 2 [OP_LIST_EOL] [OP_LIST2NUM] [OP_GEN_POSSIBLE_LIST] 850 [OP_LIST_MORE_EQUAL] 8A2 A [OP_LIST_FIND_UNK] [OP_LIST_SUM] | var_a = 8
var_b = 'A'
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=""
for i in list_a:
i = str(i)
var_d = var_d + i
ans_dict = dict()
var_d = str(var_d)
list_b = []
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] = 0
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]))
if len(var_d) == len(str(int(temp))):
new_elem = int(temp)
list_b.append(new_elem)
var_e = 850
list_c = []
for i in list_b:
if i >= var_e:
list_c.append(i)
var_f = '8A2'
var_g = 'A'
var_f = str(var_f)
var_g = str(var_g)
unk_idx = var_f.index(var_g)
list_d = []
for elem in list_c:
elem = str(elem)
list_d.append(int(elem[unk_idx]))
list_d = list(set(list_d))
list_d = [float(i) for i in list_d]
var_h = sum(list_d)
print(int(var_h)) |
Geometry | What is the volume in cubic meters (m3) of a cuboid whose width is 80 centimeters (cm), its length is 75 centimeters (cm), and its height is 120 centimeters (cm), and its length, width, and height are twice each? | 5.76 | 80 100 [OP_DIV] 2 [OP_MUL] 75 100 [OP_DIV] 2 [OP_MUL] 120 100 [OP_DIV] 2 [OP_MUL] [OP_MUL] [OP_MUL] | var_a = 80
var_b = 100
var_c = var_a / var_b
var_d = 2
var_e = var_c * var_d
var_f = 75
var_g = 100
var_h = var_f / var_g
var_i = 2
var_j = var_h * var_i
var_k = 120
var_l = 100
var_m = var_k / var_l
var_n = 2
var_o = var_m * var_n
var_p = var_j * var_o
var_q = var_e * var_p
print('{:.2f}'.format(round(var_q+1e-10,2))) |
Geometry | 5 people each have 8 tickets in the shape of a square which is 30 centimeters (cm) wide and 30 centimeters (cm) long. Find how many square meters (m2) of space the tickets occupy when they are spread out on the floor without overlapping each other. | 3.6 | 30 100 [OP_DIV] 2 [OP_POW] 8 [OP_MUL] 5 [OP_MUL] | var_a = 30
var_b = 100
var_c = var_a / var_b
var_d = 2
var_e = var_c ** var_d
var_f = 8
var_g = var_e * var_f
var_h = 5
var_i = var_g * var_h
print('{:.2f}'.format(round(var_i+1e-10,2))) |
Arithmetic calculation | Ten trees are planted on one side of the road at intervals of 10 meters (m). If trees are planted only at the beginning of the road, find the length in meters (m) of the road. | 90 | 10 1 [OP_SUB] 10 [OP_MUL] | var_a = 10
var_b = 1
var_c = var_a - var_b
var_d = 10
var_e = var_c * var_d
print(int(var_e)) |
Correspondence | A and B are single-digit whole numbers. What is the sum of all A's and B's such that satisfy both 4/23<A/23<32/43 and 0.23<1/B<0.6? | 8 | 4/23<A/23<32/43 A [OP_DIGIT_UNK_SOLVER] [OP_LIST_LEN] 0.23<1/B<0.6 B [OP_DIGIT_UNK_SOLVER] [OP_LIST_LEN] [OP_ADD] | var_a = '4/23<A/23<32/43'
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] = []
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].append(int(c[i]))
list_a = list(set(ans_dict[var_b]))
var_c = len(list_a)
var_d = '0.23<1/B<0.6'
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] = []
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].append(int(c[i]))
list_b = list(set(ans_dict[var_e]))
var_f = len(list_b)
var_g = var_c + var_f
print(int(var_g)) |
Correspondence | When three-digit number ABC is divided by 17, the quotient and the remainder is 28 and 9. Find the sum of A, B, and C. | 17 | 17 28 [OP_MUL] 9 [OP_ADD] [OP_NUM2LIST] [OP_LIST_SUM] | var_a = 17
var_b = 28
var_c = var_a * var_b
var_d = 9
var_e = var_c + var_d
list_a = []
var_e = int(var_e)
while var_e//10 > 0:
list_a.append(var_e%10)
var_e = var_e//10
list_a.append(var_e%10)
list_a = list_a[::-1]
list_a = [float(i) for i in list_a]
var_f = sum(list_a)
print(int(var_f)) |
Arithmetic calculation | There are five numbers 3.4, 7/2, 1.7, 27/10, and 2.9. Write the smallest number including the decimal point. | 1.7 | [OP_LIST_SOL] 3.4 7/2 1.7 27/10 2.9 [OP_LIST_EOL] 1 [OP_LIST_MIN] | var_a = 3.4
var_b = 3.5
var_c = 1.7
var_d = 2.7
var_e = 2.9
list_a= []
if "/" in str(var_e):
var_e = eval(str(var_e))
list_a.append(var_e)
if "/" in str(var_d):
var_d = eval(str(var_d))
list_a.append(var_d)
if "/" in str(var_c):
var_c = eval(str(var_c))
list_a.append(var_c)
if "/" in str(var_b):
var_b = eval(str(var_b))
list_a.append(var_b)
if "/" in str(var_a):
var_a = eval(str(var_a))
list_a.append(var_a)
list_a.reverse()
var_f = 1
list_b=list_a.copy()
list_b.sort()
var_g = list_b[var_f-1]
print('{:.2f}'.format(round(var_g+1e-10,2))) |
Comparison | As some students are lined up, there are 6 people in front of Eunji and 5 people behind her. Find the total number of students in the line. | 12 | 6 1 [OP_ADD] 5 [OP_ADD] | var_a = 6
var_b = 1
var_c = var_a + var_b
var_d = 5
var_e = var_c + var_d
print(int(var_e)) |
Geometry | Calculate the area covered by the paint when a cuboid with a width of 3 centimeters (cm), a length of 4 centimeters (cm), and a height of 5 centimeters (cm) is put in a paint bucket and removed. | 94 | 3 4 [OP_MUL] 3 5 [OP_MUL] 4 5 [OP_MUL] [OP_ADD] [OP_ADD] 2 [OP_MUL] | var_a = 3
var_b = 4
var_c = var_a * var_b
var_d = 3
var_e = 5
var_f = var_d * var_e
var_g = 4
var_h = 5
var_i = var_g * var_h
var_j = var_f + var_i
var_k = var_c + var_j
var_l = 2
var_m = var_k * var_l
print(int(var_m)) |
Possibility | Among the three-digit natural numbers formed by three different digits picked from the following ten digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9, find the number of odd numbers less than or equal to 300. | 72 | [OP_LIST_SOL] 0 1 2 3 4 5 6 7 8 9 [OP_LIST_EOL] 3 [OP_LIST_GET_PERM] 300 [OP_LIST_LESS_EQUAL] 2 [OP_LIST_DIVISIBLE] [OP_SET_DIFFERENCE] [OP_LIST_LEN] | var_a = 0
var_b = 1
var_c = 2
var_d = 3
var_e = 4
var_f = 5
var_g = 6
var_h = 7
var_i = 8
var_j = 9
list_a= []
if "/" in str(var_j):
var_j = eval(str(var_j))
list_a.append(var_j)
if "/" in str(var_i):
var_i = eval(str(var_i))
list_a.append(var_i)
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_k = 3
list_b = [str(i) for i in list_a]
list_b = list(itertools.permutations(list_b, var_k))
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_l = 300
list_c = []
for i in list_b:
if i <= var_l:
list_c.append(i)
var_m = 2
list_d = []
var_m = int(var_m)
for i in list_c:
i = int(i)
if i % var_m == 0:
list_d.append(i)
list_e = list(set(list_c) - set(list_d))
var_n = len(list_e)
print(int(var_n)) |
Geometry | We want to make a circle with a radius of 3 centimeters (cm) in a square with a side length of 30 centimeters (cm) with no overlapping of circles. How many circles can you make in all? | 25 | 30 3 2 [OP_MUL] [OP_FDIV] 2 [OP_POW] | var_a = 30
var_b = 3
var_c = 2
var_d = var_b * var_c
var_e = var_a // var_d
var_f = 2
var_g = var_e ** var_f
print(int(var_g)) |
Correspondence | 25A7 rounded up to the nearest tens is 2570. Find A. | 6 | 25A7 [OP_GEN_POSSIBLE_LIST] 2570 [OP_LIST_LESS_EQUAL] 2570 10 [OP_SUB] [OP_LIST_MORE] 25A7 A [OP_LIST_FIND_UNK] | var_a = '25A7'
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 = 2570
list_b = []
for i in list_a:
if i <= var_b:
list_b.append(i)
var_c = 2570
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 = '25A7'
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)) |
Correspondence | Yoongi divides some number by 4 to get 12. What does he get when the number divides by 3? | 16 | 12 4 [OP_MUL] 3 [OP_DIV] | 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)) |
Possibility | I wrote the numbers from 1000 to 2000 sequentially. Find the sum of all the digits written. | 14502 | 1000 2000 1 [OP_LIST_ARANGE] [OP_LIST_NUM2SUM] [OP_LIST_SUM] | var_a = 1000
var_b = 2000
var_c = 1
list_a = [i for i in range(var_a, var_b + 1, var_c)]
list_b=[]
for i in list_a:
var_d = 0
i = int(i)
while i//10 > 0:
var_d = var_d + i%10
i = i//10
var_d = var_d + i%10
list_b.append(var_d)
list_b = [float(i) for i in list_b]
var_e = sum(list_b)
print(int(var_e)) |
Correspondence | Sang-woo decided to share some of his 12 notebooks and 34 pencils with his friends. Sang-woo gave away several notebooks and three times as many pencils to his friends. If the total number of pencils and notebooks left is 30, find how many notebooks were distributed. | 4 | 12 34 [OP_ADD] 30 [OP_SUB] 1 3 [OP_ADD] [OP_DIV] | var_a = 12
var_b = 34
var_c = var_a + var_b
var_d = 30
var_e = var_c - var_d
var_f = 1
var_g = 3
var_h = var_f + var_g
var_i = var_e / var_h
print(int(var_i)) |
Arithmetic calculation | The length of the wooden block in Yoojung has is 30 centimeters (cm) longer than 31 meters (m). Write the length of the wooden block that Yoojung has in centimeters (cm). | 3130 | 31 100 [OP_MUL] 30 [OP_ADD] | var_a = 31
var_b = 100
var_c = var_a * var_b
var_d = 30
var_e = var_c + var_d
print(int(var_e)) |
Arithmetic calculation | There are five numbers 10, 11, 12, 13, and 14. What is the value of the second largest number divided by the smallest number? | 1.3 | [OP_LIST_SOL] 10 11 12 13 14 [OP_LIST_EOL] 2 [OP_LIST_MAX] 1 [OP_LIST_MIN] [OP_DIV] | 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 = 2
list_b=list_a.copy()
list_b.sort()
var_g = list_b[-var_f]
var_h = 1
list_c=list_a.copy()
list_c.sort()
var_i = list_c[var_h-1]
var_j = var_g / var_i
print('{:.2f}'.format(round(var_j+1e-10,2))) |
Correspondence | A3-41=52, where A3 is two-digit number. What is A? | 9 | A3-41=52 A [OP_DIGIT_UNK_SOLVER] | var_a = 'A3-41=52'
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)) |
Arithmetic calculation | Kyungho likes to ride a bike. The guard rode a bicycle at 3 kilometers (km) per hour from home and went for a walk, and returned home from city hall at 4 kilometers (km) per hour. When he returned, he chose a road that was 2 kilometers (km) longer than when he went, and when he returned home, 4 hours had passed since before he left. Find the time taken to return. | 2 | 4 2 4 [OP_DIV] [OP_SUB] 1 3 [OP_DIV] 1 4 [OP_DIV] [OP_ADD] [OP_DIV] 2 [OP_ADD] 4 [OP_DIV] | var_a = 4
var_b = 2
var_c = 4
var_d = var_b / var_c
var_e = var_a - var_d
var_f = 1
var_g = 3
var_h = var_f / var_g
var_i = 1
var_j = 4
var_k = var_i / var_j
var_l = var_h + var_k
var_m = var_e / var_l
var_n = 2
var_o = var_m + var_n
var_p = 4
var_q = var_o / var_p
print(int(var_q)) |
Geometry | What is the surface area in square centimeters (cm2) of a cube with one edge of 20 centimeters (cm)? | 2400 | 20 2 [OP_POW] 6 [OP_MUL] | var_a = 20
var_b = 2
var_c = var_a ** var_b
var_d = 6
var_e = var_c * var_d
print(int(var_e)) |
Geometry | There is a cuboid with a height of 13 centimeters (cm) and a base area of 14 square centimeters (cm2). What is the volume of this cuboid? | 182 | 14 13 [OP_MUL] | var_a = 14
var_b = 13
var_c = var_a * var_b
print(int(var_c)) |
Arithmetic calculation | In a factory, (a) machines make 5 product (c) per minute, and (b) machines make 8 product (c) per minute. If the two machines startd to make product (c) at the same time and (b) machine made 40 product (c), how many fewer product (c) did (a) machines make than the (b) machines? | 15 | 40 5 40 8 [OP_DIV] [OP_MUL] [OP_SUB] | var_a = 40
var_b = 5
var_c = 40
var_d = 8
var_e = var_c / var_d
var_f = var_b * var_e
var_g = var_a - var_f
print(int(var_g)) |
Possibility | There are balls with the numbers 1 through 9 written on them. Take out 3 of them and find the total number of ways where the numbers on the ball can add up to 19. | 30 | 1 9 1 [OP_LIST_ARANGE] 3 [OP_LIST_GET_PERM] [OP_LIST_NUM2SUM] 19 [OP_LIST_FIND_NUM] | 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)) |
Correspondence | The natural numbers A and B satisfy 15 = 3 × A = 5 × B. Find A. | 5 | 15=3×A=5×B A [OP_NUM_UNK_SOLVER] | var_a = '15=3×A=5×B'
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] = 0
candidate_num = [i for i in range(51)]
candi = list(itertools.product(candidate_num, 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)):
if term_list[i] == '':
new_eq += str(term_list[i])+op_list[i]
else:
new_eq += str(int(term_list[i]))+op_list[i]
new_eq += str(int(term_list[-1]))
new_eq=new_eq.replace('=', '==')
new_eq=new_eq.replace('>==', '>=')
new_eq=new_eq.replace('<==', '<=')
eval_result = False
try:
if '=' in new_eq and '>' not in new_eq and '<' not in new_eq:
new_eq=new_eq.replace('==','=')
new_eq=new_eq.replace('>','')
new_eq=new_eq.replace('<','')
new_eq=new_eq.split('=')
for i in range(len(new_eq)-1):
eval_result = math.isclose(eval(new_eq[i]), eval(new_eq[i+1]))
if not eval_result:
break
else:
eval_result = eval(new_eq)
except:
eval_result = False
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)) |
Correspondence | You are going to divide a number 20. When you mistakenly multiply the number by 10, the answer is 50. What is the correct calculation result? | 4 | 20 50 10 [OP_DIV] [OP_DIV] | var_a = 20
var_b = 50
var_c = 10
var_d = var_b / var_c
var_e = var_a / var_d
print(int(var_e)) |
Comparison | Horses, cows, pigs, sheep, rabbits, and squirrels entered the fence in that order. Which animal came in the fence at 4th place? | sheep | [OP_LIST_SOL] Horses cows pigs sheep rabbits squirrels [OP_LIST_EOL] 4 [OP_LIST_GET] | var_a = 'Horses'
var_b = 'cows'
var_c = 'pigs'
var_d = 'sheep'
var_e = 'rabbits'
var_f = 'squirrels'
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 = 4
var_h = list_a[var_g-1]
print(var_h) |
Geometry | Minsu is trying to make a square with once used wire to make a regular hexagon whose side is 4 centimeters (cm) long. Find the area of the square. | 36 | 4 6 [OP_MUL] 4 [OP_DIV] 2 [OP_POW] | var_a = 4
var_b = 6
var_c = var_a * var_b
var_d = 4
var_e = var_c / var_d
var_f = 2
var_g = var_e ** var_f
print(int(var_g)) |
Geometry | Beverages were arranged in a square shape with 10 cans on each row and column. How many beverage cans are there at the perimeter? | 36 | 10 4 [OP_MUL] 4 [OP_SUB] | 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)) |
Possibility | Toss the coin twice, then roll the dice once. In how many ways can two tosses of a coin come up the same side and the dice roll is a multiple of 3? | 4 | 2 1 6 1 [OP_LIST_ARANGE] 3 [OP_LIST_DIVISIBLE] [OP_LIST_LEN] [OP_MUL] | var_a = 2
var_b = 1
var_c = 6
var_d = 1
list_a = [i for i in range(var_b, var_c + 1, var_d)]
var_e = 3
list_b = []
var_e = int(var_e)
for i in list_a:
i = int(i)
if i % var_e == 0:
list_b.append(i)
var_f = len(list_b)
var_g = var_a * var_f
print(int(var_g)) |
Correspondence | A is the number of 23 groups of 10 and 1 of 100 groups. B is the number that can be counted by jumping from 172 to 105 four times. Find the sum of A and B. | 922 | 100 1 [OP_MUL] 10 23 [OP_MUL] [OP_ADD] 172 105 4 [OP_MUL] [OP_ADD] [OP_ADD] | var_a = 100
var_b = 1
var_c = var_a * var_b
var_d = 10
var_e = 23
var_f = var_d * var_e
var_g = var_c + var_f
var_h = 172
var_i = 105
var_j = 4
var_k = var_i * var_j
var_l = var_h + var_k
var_m = var_g + var_l
print(int(var_m)) |
Geometry | Find the number of sides and vertices of the pentagon and find the sum of them all. | 10 | 5 5 [OP_ADD] | var_a = 5
var_b = 5
var_c = var_a + var_b
print(int(var_c)) |
Comparison | Figure A is a rectangle that is 15 centimeters (cm) wide and 8 centimeters (cm) long. Figure B is a rectangle with a width of 10 centimeters (cm) and a length of 13 centimeters (cm). Which of the two shapes has greater area? | B | [OP_LIST_SOL] A B [OP_LIST_EOL] [OP_LIST_SOL] 15 8 [OP_MUL] 10 13 [OP_MUL] [OP_LIST_EOL] 1 [OP_LIST_MAX] [OP_LIST_INDEX] [OP_LIST_POP] [OP_LIST_GET] | 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 = 15
var_d = 8
var_e = var_c * var_d
var_f = 10
var_g = 13
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) |
Comparison | What is the sum of the numbers greater than or equal to 0.4 among 0.8, 1/2, and 0.3? | 1.3 | [OP_LIST_SOL] 0.8 1/2 0.3 [OP_LIST_EOL] 0.4 [OP_LIST_MORE_EQUAL] [OP_LIST_SUM] | var_a = 0.8
var_b = 0.5
var_c = 0.3
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 = 0.4
list_b = []
for i in list_a:
if i >= var_d:
list_b.append(i)
list_b = [float(i) for i in list_b]
var_e = sum(list_b)
print('{:.2f}'.format(round(var_e+1e-10,2))) |
Geometry | An octahedron has a side length of 2 centimeters (cm). What is the surface area of this regular octahedron in square centimeters (cm2)? | 10.39 | 3 3 1/2 [OP_POW] [OP_MUL] 2 [OP_DIV] 2 2 [OP_POW] [OP_MUL] | var_a = 3
var_b = 3
var_c = 0.5
var_d = var_b ** var_c
var_e = var_a * var_d
var_f = 2
var_g = var_e / var_f
var_h = 2
var_i = 2
var_j = var_h ** var_i
var_k = var_g * var_j
print('{:.2f}'.format(round(var_k+1e-10,2))) |
Arithmetic calculation | There are three numbers 100, -1, and 2. What is the difference between the largest number and the smallest number? | 101 | [OP_LIST_SOL] 100 -1 2 [OP_LIST_EOL] 1 [OP_LIST_MAX] 1 [OP_LIST_MIN] [OP_SUB] | var_a = 100
var_b = -1
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 = 1
list_b=list_a.copy()
list_b.sort()
var_e = list_b[-var_d]
var_f = 1
list_c=list_a.copy()
list_c.sort()
var_g = list_c[var_f-1]
var_h = var_e - var_g
print(int(var_h)) |
Possibility | Among the people of A, B, C, and D, two people are selected, one to be the president and the other to be the vice president. Find the number of possible cases. | 12 | [OP_LIST_SOL] A B C D [OP_LIST_EOL] [OP_LIST_LEN] 2 [OP_PERM] | var_a = 'A'
var_b = 'B'
var_c = 'C'
var_d = 'D'
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 = 1
var_e = int(var_e)
var_f = int(var_f)
for i, elem in enumerate(range(var_f)):
var_g = var_g * (var_e-i)
print(int(var_g)) |
Geometry | What is the area of a square when one diagonal is 12 meters (m) long? | 72 | 12 2 [OP_POW] 2 [OP_DIV] | var_a = 12
var_b = 2
var_c = var_a ** var_b
var_d = 2
var_e = var_c / var_d
print(int(var_e)) |
Arithmetic calculation | Among the people who visited the cinema during the week, 1518 were women, and among them, 536 were office workers. There were 525 more males than females, and among them, 1257 were non-workers. How many office workers visited the cinema in total? | 1322 | 1518 525 [OP_ADD] 1257 [OP_SUB] 536 [OP_ADD] | var_a = 1518
var_b = 525
var_c = var_a + var_b
var_d = 1257
var_e = var_c - var_d
var_f = 536
var_g = var_e + var_f
print(int(var_g)) |
Arithmetic calculation | There are a total of 24 students with and without after-school activities. Of these, 50 candies were distributed equally only to students doing after-school activities, leaving 2 candies. When students who do after-school activities are twice as many students who do not do after-school activities, how many candies do students who do after-school activities get? | 6 | 50 2 [OP_SUB] 24 2 1 [OP_ADD] [OP_FDIV] [OP_FDIV] | var_a = 50
var_b = 2
var_c = var_a - var_b
var_d = 24
var_e = 2
var_f = 1
var_g = var_e + var_f
var_h = var_d // var_g
var_i = var_c // var_h
print(int(var_i)) |
Geometry | How many diagonals can be drawn in a tetragon? | 2 | 4 4 3 [OP_SUB] [OP_MUL] 2 [OP_DIV] | var_a = 4
var_b = 4
var_c = 3
var_d = var_b - var_c
var_e = var_a * var_d
var_f = 2
var_g = var_e / var_f
print(int(var_g)) |
Correspondence | Yechan was supposed to divide the tape by 2.5, but he divided it into 2 and 2/5 by mistake, and it became 15/32 centimeters (cm). Find the length in centimeters (cm) of the tape that Yechan originally intended to use. | 0.45 | 15/32 2 2/5 [OP_ADD] [OP_MUL] 2.5 [OP_DIV] | var_a = 0.46875
var_b = 2
var_c = 0.4
var_d = var_b + var_c
var_e = var_a * var_d
var_f = 2.5
var_g = var_e / var_f
print('{:.2f}'.format(round(var_g+1e-10,2))) |
Arithmetic calculation | Two 275 centimeters (cm) long tapes are overlapped and attached, resulting in a total length of 512 centimeters (cm). Find the length of the overlapped part in centimeters (cm). | 19 | 275 275 [OP_ADD] 512 [OP_SUB] 2 [OP_DIV] | var_a = 275
var_b = 275
var_c = var_a + var_b
var_d = 512
var_e = var_c - var_d
var_f = 2
var_g = var_e / var_f
print(int(var_g)) |
Possibility | Nine digits from 1 to 9 are used once to form a nine-digit number. If this number is a multiple of 55, find the largest 9-digit number. | 987642315 | 1 9 1 [OP_LIST_ARANGE] 9 [OP_LIST_GET_PERM] 55 [OP_LIST_DIVISIBLE] 1 [OP_LIST_MAX] | 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 = 9
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 = 55
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]
print(int(var_g)) |
Arithmetic calculation | If you have 100 sheets of paper to make 3 notebooks of 30 sheets, how many sheets are left? | 10 | 100 30 3 [OP_MUL] [OP_SUB] | var_a = 100
var_b = 30
var_c = 3
var_d = var_b * var_c
var_e = var_a - var_d
print(int(var_e)) |
Possibility | Find the number of ways to put 5 people in a line. | 120 | 5 5 [OP_PERM] | var_a = 5
var_b = 5
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)) |
Comparison | Jungkook has 0.8 and Yoongi has 1/2 number cards. How many people have number cards less than or equal to 0.3? | 0 | [OP_LIST_SOL] 0.8 1/2 [OP_LIST_EOL] 0.3 [OP_LIST_LESS_EQUAL] [OP_LIST_LEN] | var_a = 0.8
var_b = 0.5
list_a= []
if "/" in str(var_b):
var_b = eval(str(var_b))
list_a.append(var_b)
if "/" in str(var_a):
var_a = eval(str(var_a))
list_a.append(var_a)
list_a.reverse()
var_c = 0.3
list_b = []
for i in list_a:
if i <= var_c:
list_b.append(i)
var_d = len(list_b)
print(int(var_d)) |
Arithmetic calculation | If the minimum weight that a scale with numbers written at 5 kilograms (kg) intervals can display is 0 kilograms (kg) and the maximum weight is 199 kilograms (kg), find how many times 0 is written on this scale. | 21 | 0 199 5 [OP_LIST_ARANGE] [OP_LIST2NUM] [OP_NUM2LIST] 0 [OP_LIST_FIND_NUM] | var_a = 0
var_b = 199
var_c = 5
list_a = [i for i in range(var_a, var_b + 1, var_c)]
var_d=""
for i in list_a:
i = str(i)
var_d = var_d + i
list_b = []
var_d = int(var_d)
while var_d//10 > 0:
list_b.append(var_d%10)
var_d = var_d//10
list_b.append(var_d%10)
list_b = list_b[::-1]
var_e = 0
var_f = 0
var_e = int(var_e)
for i in list_b:
i = int(i)
if i == var_e:
var_f = var_f + 1
print(int(var_f)) |
Correspondence | Adding some number to 37 gives 52. What number is it? | 15 | 52 37 [OP_SUB] | var_a = 52
var_b = 37
var_c = var_a - var_b
print(int(var_c)) |
Comparison | Seokgi read 1/4 of the children's book, Shinyoung read 1/3 of it, and Woong read 1/5 of it. Find the person who has read the most of the children's book. | Shinyoung | [OP_LIST_SOL] Shinyoung Seokgi Woong [OP_LIST_EOL] [OP_LIST_SOL] 1/3 1/4 1/5 [OP_LIST_EOL] 1 [OP_LIST_MAX] [OP_LIST_INDEX] [OP_LIST_POP] [OP_LIST_GET] | var_a = 'Shinyoung'
var_b = 'Seokgi'
var_c = 'Woong'
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 = 0.3333333333333333
var_e = 0.25
var_f = 0.2
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()
var_g = 1
list_c=list_b.copy()
list_c.sort()
var_h = list_c[-var_g]
var_i = list_b.index(var_h)+1
var_j = list_a[var_i-1]
print(var_j) |
Arithmetic calculation | There are 3 books in Daehyun's bag, 2 books in Donggil's bag, 5 books in Minjoo's bag, 3 books in Haeun's bag, and 1 book in Soyeong's bag. How many of them have 3 books? | 2 | [OP_LIST_SOL] 3 2 5 3 1 [OP_LIST_EOL] 3 [OP_LIST_FIND_NUM] | var_a = 3
var_b = 2
var_c = 5
var_d = 3
var_e = 1
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
var_g = 0
var_f = int(var_f)
for i in list_a:
i = int(i)
if i == var_f:
var_g = var_g + 1
print(int(var_g)) |
Arithmetic calculation | Among Korean, Mathematics, Science, and English, the average score before taking the English test was 92 points, and the average score after taking the English test turned to 94 points. What is your English score? | 100 | 92 2 [OP_ADD] 4 [OP_MUL] 92 3 [OP_MUL] [OP_SUB] | var_a = 92
var_b = 2
var_c = var_a + var_b
var_d = 4
var_e = var_c * var_d
var_f = 92
var_g = 3
var_h = var_f * var_g
var_i = var_e - var_h
print(int(var_i)) |
Arithmetic calculation | There is a hole through which 15 liters (l) of water is drained in 3 minutes at a constant rate. How many minutes does it take for 140 liters (l) of water to drain? | 28 | 140 15 3 [OP_DIV] [OP_DIV] | var_a = 140
var_b = 15
var_c = 3
var_d = var_b / var_c
var_e = var_a / var_d
print(int(var_e)) |
Correspondence | Find A from the 4-digit equation A2B2+1C1D=3333. | 2 | A2B2+1C1D=3333 A [OP_DIGIT_UNK_SOLVER] | var_a = 'A2B2+1C1D=3333'
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)) |
Comparison | At school, students take turns giving presentations. Eunjeong will be the 6th speaker, followed by 7 other students. How many students are giving presentations? | 13 | 6 7 [OP_ADD] | var_a = 6
var_b = 7
var_c = var_a + var_b
print(int(var_c)) |
Comparison | Namjoon gathered 3 and 6 together. Jimin gathered 7 and 1 together. Whose sum of numbers is greater? | Namjoon | [OP_LIST_SOL] Namjoon Jimin [OP_LIST_EOL] [OP_LIST_SOL] [OP_LIST_SOL] 3 6 [OP_LIST_EOL] [OP_LIST_SUM] [OP_LIST_SOL] 7 1 [OP_LIST_EOL] [OP_LIST_SUM] [OP_LIST_EOL] 1 [OP_LIST_MAX] [OP_LIST_INDEX] [OP_LIST_POP] [OP_LIST_POP] [OP_LIST_POP] [OP_LIST_GET] | var_a = 'Namjoon'
var_b = 'Jimin'
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 = 3
var_d = 6
list_b= []
if "/" in str(var_d):
var_d = eval(str(var_d))
list_b.append(var_d)
if "/" in str(var_c):
var_c = eval(str(var_c))
list_b.append(var_c)
list_b.reverse()
list_b = [float(i) for i in list_b]
var_e = sum(list_b)
var_f = 7
var_g = 1
list_c= []
if "/" in str(var_g):
var_g = eval(str(var_g))
list_c.append(var_g)
if "/" in str(var_f):
var_f = eval(str(var_f))
list_c.append(var_f)
list_c.reverse()
list_c = [float(i) for i in list_c]
var_h = sum(list_c)
list_d= []
if "/" in str(var_h):
var_h = eval(str(var_h))
list_d.append(var_h)
if "/" in str(var_e):
var_e = eval(str(var_e))
list_d.append(var_e)
list_d.reverse()
var_i = 1
list_e=list_d.copy()
list_e.sort()
var_j = list_e[-var_i]
var_k = list_d.index(var_j)+1
var_l = list_a[var_k-1]
print(var_l) |
Geometry | Find the length in centimeters (cm) of the base of a triangle with a height of 8 centimeters (cm) and an area of 24 square centimeters (cm2). | 6 | 24 2 [OP_MUL] 8 [OP_DIV] | var_a = 24
var_b = 2
var_c = var_a * var_b
var_d = 8
var_e = var_c / var_d
print(int(var_e)) |
Arithmetic calculation | There are five numbers 44, 16, 2, 77, and 241. Find the difference between the sum and the average of the five numbers given. | 304 | [OP_LIST_SOL] 44 16 2 77 241 [OP_LIST_EOL] [OP_LIST_SUM] [OP_LIST_MEAN] [OP_SUB] | var_a = 44
var_b = 16
var_c = 2
var_d = 77
var_e = 241
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()
list_a = [float(i) for i in list_a]
var_f = sum(list_a)
list_a = [float(i) for i in list_a]
var_g = sum(list_a)/len(list_a)
var_h = var_f - var_g
print(int(var_h)) |
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