Question 1138226: Someone please help me!!! The question is:
Error in the scale. Alex is using a scale that is known to have a constant error. A can of soup and a can of tuna are placed on this scale, and it reads 24 ounces. Now four identical cans of soup and three identical cans of tuna are placed on an accurate scale, and a weight of 80 ounces is recorded. If two cans of tuna weigh 18 ounces on the bad scale, then what is the amount of error in the scale and what is the correct weight of each type of can?
I know how to do the math I just have no idea what the three equations would be.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the bad scale has a constant error.
i take this to mean it is constantly a certain number of ounces over or under what the real weight should be.
on the bad scale:
2 cans of tuna weigh 18 ounces.
1 can of soup and 1 can of tuna weigh 24 ounces.
if 2 cans of tuna weigh 18 ounces, then 1 can of tuna weighs 9 ounces on the bad scale.
this means that 1 can of soup weighs 15 ounces on the bad scale.
on the good scale:
4 cans of soup and 3 cans of tuna weigh 80 ounces.
if you assume 1 can of tune weighs 9 ounces and you assume 1 can of soup weighs 15 ounces, then 4 cans of soup and 3 cans of tuna would weight 4 * 15 + 3 * 9 = 87 ounces.
that means the bad scale says they weigh 87 ounces when the good scale says they weight 80 ounces.
that means the bad scale is reading 7 ounces too high.
since the error is constant, divide the total error by the number of cans being weighed to get a constant error per can of 1 ounce too high.
that means that the actual weight of a can of tuna is 8 ounces and the actual weight of a can of soup is 14 ounces.
4 cans of soup at 14 ounces each and 3 cans of tuna at 8 ounces each would then weight a total of 80 ounces, which would be correct.
if your were to solve this algebraically, you might do the following:
on the bad sale, you have:
2 cans of tuna = 18.
1 can of soup and 1 can of tuna = 24.
let t equal the true weight of a can of tuna and let s equal the true weight of a can of soup.
let a equal the error.
a can be positive or negative.
then:
t + a equals the bad weight of a can of tuna.
s + a equals the bad weight of a can of soup.
the same letter for the error is used because the error is constant across everything that's being weighed by the bad scale.
on the bad scale, you get:
2 cans of tuna weigh 18 ounces.
1 can of soup and 1 can of tuna weigh 24 ounces.
becomes:
2 * (t + a) = 18
(s + a) + (t + a) = 24
simplify the first equatio to get 2t + 2a = 18.
divide both sides of that equation by 2 to get t + a = 9.
in the second equaiton, replace t + a with 9 to get s + a + 9 = 24.
solve for s + a to get s + a = 15.
you have:
t + a = 9
s + a = 25.
on the good scale, you have:
4s + 3t = 80
on the bad scale, this would becomes 4 * (s + a) + 3 * (5 + a) = 4 * 15 + 3 * 9 = 87.
therefore you have:
4s + 3t = 80 on the good scale.
4(s+a) + 3(t + a) = 87 on the bad scale.
simplify the second equation and keep the first equation as is to get:
4s + 3t = 80
4s + 3t + 7a = 87
subtract the first equation from the second to get 7a = 7
solve for a to get a = 1.
that tells you that the constant error is equal to 1 ounce over per can.
when a = 1, then t + a = 9 results in t = 8 and s + a = 15 results in s = 14.
when t = 8 and s = 14, 4s +3t = 80, as it should.
your solution should be that the amount of error in the scale should be 1 ounce over per can and the correct weight per can of tuna is 8 ounces and the correct weight per can of soup is 14 ounces if i understood this problem correctly.
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