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mixture problem

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Junior | Next Rank: 30 Posts
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by galit_d2d » Sun Apr 20, 2008 2:25 pm
hi, this problem is also from the math review material I downloaded here.

the question is :
how many ounces of a solution i.e. 30% salt must be added to a 50-ounce solution i.e. 10% salt so that the resulting sol is 20% salt .

the solution(which I don't understand...)
x= the added amount
500+30x=(x+50)*20
500 +30x =20x +1000
10x=500
x=50...

help plz...
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by amitansu » Mon Apr 21, 2008 12:27 am
galit_d2d,

Here is the simplified soln :

according to the q: let there be "x" ounces of soln. which contains 30% of salt and when that is added to existing 50 ounces of salt carrying 10% of salt , the result becomes some amount of mixture that would contain a 20% of salt. Hope this explanation clears the understanding of this prob.

Then : putting it into equation :

the new solution is (50+x);
Now if x is the amt. of ounces that we assume then according to q it has 30% of salt so that makes .3x of salt .
Again x would be added to 50 unces which contains 10% of salt and that makes 5 ounces of salt.
So finally when 30% of x anf 10% of 50 ounces sum up they become : 5+.3x.

Now 5+.3x=.2(50+x) (since the salt % is 20 of the final mixture)

solve it you would get 50 as the answer.

Though this explanation is lenghty but i feel this is pretty self explanetory.

thanks,
Amit
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by galit_d2d » Mon Apr 21, 2008 12:47 am
Amit, thanx for the explanation, I understand this now:)
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