SOLUTION: Hi 3 identical basins, A, B, C were filled with water. Basin A had a mass of 3.1kg when it was half full. Basin B had a mass of 2.2kg when it was 1/5 full. What fraction of bas

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Question 1101815: Hi
3 identical basins, A, B, C were filled with water. Basin A had a mass of 3.1kg when it was half full.
Basin B had a mass of 2.2kg when it was 1/5 full. What fraction of basin C was filled with water when it hsd a mass of 2.8kg.
thanks
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39615) About Me  (Show Source):
You can put this solution on YOUR website!


x, mass of water+basin
y, fraction of basin filled
points (x,y) 

BASIN   MASS Total   FRACTION FILLED
A        3.1            0.5 or half
B        2.2            0.2 or a fifth
C        2.8            ? need to find

slope 1%2F3;

Using point-slope form and point ( 3.1, 1/2 ),
y-1%2F2=%281%2F3%29%28x-3.1%29
-
y=x%2F3-8%2F15

Question asked means, find y when x is 2.8
Just substitute and evaluate y.
-
y=2.8%2F3-8%2F15
y=%282.8%2A5-8%29%2F15
y=6%2F15
highlight%28y=2%2F5%29



NOT DONE - MAY CONTAIN MISTAKE
(Delay in finishing due to temporary confusion when substituting for x=2.8, which should have been no trouble)

Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


The other tutor started on a solution but did not finish it. While her method seems a bit unusual to me, it is valid; I assume she just did not see how to finish it.

I will finish her solution and then show my very different solution.

She got to the point where she had the equation
y+=+%28x%2F3%29-+%288%2F15%29
where y is the fraction of the basin that is filled and x is the total mass of the basin and water.

To continue solving the problem by this process, we can solve the equation for x when y=0 to give us the mass of the basin alone, and then solve the equation for x when y=1 to give us the total mass of the basin and the water.

mass of basin alone; y = 0:
0+=+%28x%2F3%29+-+%288%2F15%29
%28x%2F3%29+=+%288%2F15%29
x+=+24%2F15+=+1.6

mass of basin and water when basin is full; y = 1:
1+=+%28x%2F3%29+-+%288%2F15%29
23%2F15+=+x%2F3
x+=+23%2F5+=+4.6

So the basin empty has a mass of 1.6kg; full it has a mass of 4.6kg; so the mass of the water is 3kg.

So if basin C had a mass of 2.8kg, the mass of water it contained was 1.2kg; the fraction that basin C is full is 1.2%2F3+=+2%2F5

Now for a very different solution....

This shows you how very different the amount of work required to solve a problem can be, depending on how you define the variables in the problem.

And the lesson from that is that you should never be afraid to try solving a problem by a different method than everybody else -- you might find a better method.
That's how progress is made in the world of science and mathematics.

So here is my setup for the problem:

Let x be the mass of each basin
Let y be the mass of water in a full basin

Basin A has a mass of 3.1kg when half full: x+%2B+.5y+=+3.1
Basin B has a mass of 2.2kg when one-fifth full: x+%2B+.2y+=+2.2

Subtract the two equations:
.3y+=+.9
y+=+3

Substitute into either equation:
x+%2B+.5%283%29+=+3.1
x+=+1.6

Each basin has a mass of 1.6kg; the mass of water in a full basin is 3kg.

Now we are at the same point we were in the other solution; we finish as before, finding that basin C is 2/5 full if it has a mass of 2.8kg.