Minimums
LOUiS
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"In a major hotel, clients can use several elevators. An elevator takes
a maximum load of 1560 kg. It is assumed that the average mass of a man
is 75 kg and that of a woman is 60 kg. At one point, the elevator
contains twice as many women as men and there are at least 3 men. We
want to know the minimum load of the elevator".......The minimum load
of the elevator is zero......
75 x + 60y = 1560 or 78x + 60y = 1560 or 75x + 57y = 1560 etc......
75(x + 3) + 60(y - 3) = 1560 cannot solve it.
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Here is my understanding of the problem.
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Problem:
In a major hotel, clients can use several elevators. An elevator takes
a maximum load of 1560 kg. It is assumed that the average mass of a
man is 75 kg and that of a woman is 60 kg. At one point, the elevator
contains twice as many women as men and there are at least 3 men. We
want to know the minimum load of the elevator?
Solution:
Let x= number of men
Let y= number of women
x = 0 X cannot be negative because we are dealing with people.
y = 0 Y cannot be negative because we are dealing
with people.
75x + 60y = 1560 A man weights 75 kg, a woman weights 60 kg, and
combined they cannot exceed 1560 kg.
y = 2x There are twice as many women as men.
x = 3 There are at least 3 men.
Algebraically:
Since there are at least 3 men, we can assume that that would be the
minimum number of men on the elevator. To solve for the number of
women, we can simply substitute in the number of men into the y = 2x
equation.
y = 2x
y = 2(3)
y = 6
3 men and 6 women is the minimum number of people on this elevator.
You can find a detailed answer here:
http://www.thelouisguy.com/files/probelevator.pdf
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I hope this helps you understand the problem. The wording of the
problem is a little deceiving, yes, but a lot about word problems is
trying to decipher what they are looking for.