Question 1207014: 1.
A mixture A contains milk and water in the ratio 7: 2 and this mixture A is
mixed in the ratio 1: 3 with another mixture of milk and water in which ratio of
milk and water is 4:5 to form mixture B. Mixture A and B are mixed in the ratio
1: 2 to form mixture C and this mixture C is mixed in the ratio 6: 5 with
another mixture of milk and water in which ratio of milk and water is 3: 2 to
form mixture D. Mixture B and D are mixed in the ratio 22: 17 to form another
mixture E.
(a)Mixture B is sold after mixing 92 litres of pure milk to it and total profit
percent earned after selling the mixture is 75 percent, then what is the amount
of milk in the mixture B initially?
Found 3 solutions by mccravyedwin, Edwin McCravy, ikleyn:
Answer by mccravyedwin(407)   (Show Source):
You can put this solution on YOUR website!
The secret to solving a problem involving
"mixing an a:b solution in the ratio c:d with an e:f solution",
is:
1. The fraction of milk in an x:y milk to water solution is always x/(x+y),
and we only have to work with the fraction of milk.
2. Since c:d is a numerical ratio, we can let c and d units both be c and d
units of ANY size. It doesn't matter what size unit it is, as long as it is
the same size unit for both.
3. When we mix c units of an
a/(a+b) milk solution
with d units of an
e/(e+f) milk solution. That gives us a
[c*a/(a+b)+d*e/(e+f)]/(c+d)
milk solution. That will always give us the numerical fraction of milk in the
mixture.
Now, let me see you or one of the other tutors take that hint and solve the
problem. I'm sure that one of them will. If they don't, I'll solve it.
Edwin
Answer by Edwin McCravy(20056)  (Show Source):
You can put this solution on YOUR website!
I didn't read your problem all the way through. Now I see it was just a joke
problem.
(a)Mixture B is sold after mixing 92 litres of pure milk to it and total profit
percent earned after selling the mixture is 75 percent, then what is the amount
of milk in the mixture B initially?
You're not a math student, just a comedian.
Answer by ikleyn(52786)  (Show Source):
You can put this solution on YOUR website! .
Mathematically, this problem is illiterately posed and cannot be solved.
To say that this is a joke is too great an honor for this nonsense.
Just straight-up rubbish.
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