A hardware supplier manufactures three kinds of clamps, types A, B, and C. Production restrictions force it to make 10 more ty
1 answer:
Answer:
A = 120
B = 40
C = 70
Solution:
As per the question:
Manufacturer forced to make 10 more type C clamps than the total of A and b:
10 + A + B = C (1)
Also, 3 times as many type B as type A clamps are:
A = 3B (2)
The total no. of clamps produced per day:
A + B + C = 330 (3)
The no. of each type manufactured per day:
Now, from eqn (1), and (3):
A + B + 10 + A + B = 330
2A + 2B = 320
A + B = 160 (4)
Now, from eqn (2) and (4):
3B + B = 160
B = 40
Since, A = 3B
A =
A = 120
Put the values of A and C in eqn (3):
120 + 40 + C = 330
C = 70
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Answer:
Explanation:
The code does not fail on the first step since 1900 divided by 4 is actually 475 and has no remainder, meaning that it should return True. The code won't work because the if statements need to be nested in a different format. The correct algorithm would be the following, which can also be seen in the picture attached below that if we input 1900 it would output is not a leap year as it fails on the division by 400 which gives a remainder of 0.75
input_year = int(input())
if input_year % 4 == 0:
if input_year % 100 == 0:
if input_year % 400 == 0:
print(input_year, "is a leap year")
else:
print(input_year, "- not a leap year")
else:
print(input_year, "is a leap year")
else:
print(input_year, "- not a leap year")
