Math Question
Started by Jab1113
why did you make the pulp additional 1*x? Do you assume it is 100%?
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michigan family full of wolverines or spartans? lol
This is a lot easier if you just take a step back and think about it logically and use high school algebra.
You're starting with 30 gallons of 20% solution so it has already 6 gallons of pulp:
30 gallons * 20% = 6 gallons
You want to know how much more pulp (x) you need to add so the percentage equals 50%:
(6+x) / (30+x) = 50%
Solve for x
You're starting with 30 gallons of 20% solution so it has already 6 gallons of pulp:
30 gallons * 20% = 6 gallons
You want to know how much more pulp (x) you need to add so the percentage equals 50%:
(6+x) / (30+x) = 50%
Solve for x
Anyone who has the Math destroyer(2010 ed)
look at TEST 1 #8....They used a simliar formulae but did not have 1*x
how are these 2 questions different?
look at TEST 1 #8....They used a simliar formulae but did not have 1*x
how are these 2 questions different?
i'm guessing you mean something like this...?
(0.20)(30) = (0.10)(30+x)
that would tell you how much water you would need to add to 30 gal. of 20% pulp to dilute it down to 10% pulp. it's m1v1 = m2v2.
the reason I wrote it out the way I did in the original problem, was to keep things consistent with (m)(v) terms:
[original] + [added] = [total]
(0.20)(30) + (1.00)(x) = (0.50)(30+x)
if you're adding something pure, obviously (1.00)(x) is just x.
if you're adding water, obviously (0.00)(x) is 0 and that term goes away and you get m1v1 = m2v2.
if you're mixing in x gal. of another concentration, you have the versatility to use (0.75)(x), etc.
my point was that the (m)(v) thing is more versatile than just m1v1 = m2v2. sure, you can do the same thing step-wise (a la LetsGo) but I don't see how that's any easier than being able to set up a one-line equation once you understand how (m)(v) terms can be additive.
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