eaakbari wrote:There is an interesting formula I use that works with mixtures and weighted averages.
% of higher ratio - % of mean : % of mean - %of lower = lower :higher
Here we have 1 part containing 100% sand
1 part containing 30 % sand
what we want is 50 % sand
substitute in formula
100 - 50 : 50 - 30 = lower:higher
5:2 = lower :higher
That means the ratio of 30%mixture to pure sand is 5:2
The total weight has to be 10 kgs
5x+2x=7x
7x=10 ; x=10/7
Hence 5 x = 50/7
Hence we can find how many kgs would a 10 kg misture contain of the 30%misture
= 50/7
Therefore amount removed = 10 -50/7 = 20/7
which is the answer
indeed, it's a good formula.
in cases like this one, however, you can save some more time by having a "birds eye view".
look at the percentages. one is 30 and the other one is 70. You want to make them equal. or 50-50.
Or you have to remove that 20% of clay.
Now, we know that 70% clay comes from 100% of mixture. Therefore, 20% will come from (20 * 100)/70.
or 200/7 % of the mixture OR 2/7 of the mixture.
Hence, the answer will be 2/7th of 10 = 20/7 . in fact it is even possible to do it orally
