SOLUTION: A water gallon weighs 5 kg when it is filled with water. When the gallon is half empty, it weighs 2.6 kg. How much does the gallon weigh when it is empty?

Algebra ->  Test -> SOLUTION: A water gallon weighs 5 kg when it is filled with water. When the gallon is half empty, it weighs 2.6 kg. How much does the gallon weigh when it is empty?      Log On


   



Question 1210263: A water gallon weighs 5 kg when it is filled with water. When the gallon is half empty, it weighs 2.6 kg. How much does the gallon weigh when it is empty?
Found 4 solutions by greenestamps, ikleyn, josgarithmetic, timofer:
Answer by greenestamps(13200) About Me  (Show Source):
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The container with the gallon of water weighs 5 kg.

When half of the water is emptied, the container and the remaining water weigh 2.6 kg. So the weight of the half gallon of water that was emptied is (5-2.6) = 2.4 kg.

The weight of the half gallon of water that remains in the container is also 2.4 kg.

Since the combined weight of the container and the remaining half gallon of water is 2.6 kg, the weight of the container alone is (2.6-2.4) = 0.2 kg.

ANSWER: 0.2 kg


Answer by ikleyn(52781) About Me  (Show Source):
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.


In the formulation of this problem the word/(the term)  " gallon "  is used by  MISTAKE.

The correct formulation of the problem should use the term  " a container "  instead of  " gallon ".


For your information and education: one gallon of water weights  3.875  kilograms.


Math problem,  when formulated correctly,  should not create trouble/disturbance in the mind of a reader.



Answer by josgarithmetic(39617) About Me  (Show Source):
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"A water gallon weighs 5 kg when it is filled with water. "

That is completely wrong.

Answer by timofer(104) About Me  (Show Source):
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You wrote that wrong. Say it like this:
A water container "weighs" 5 kg when filled full with water. When the container is half empty, it holds 2.6 kg of water.


x, amount filled
y, how much kilograms in container

Two points are (0.5, 2.6) and (1, 5).

Slope would be m=%285-2.6%29%2F%281-0.5%29=4.8.

Point-Slope equation format could be used.
y-5=4.8%28x-1%29
y-5=4.8x-4.8
y=4.8x-4.8%2B5
y=4.8x%2B0.2
So if container were empty, x=0, so y=0.2.
The container is 0.2 kg when empty.