VeritasPrepDennis wrote:
appleid wrote:
As of the example, let x and y be the quantities of 10% and 3%. I'd like to show you how I am thinking of your methods:
I am 100% clear about method 1 and 2 so focusing on ration thinking with WA. The situation we have now:
2/7 distance from x
5/7 distance from y
I understand the ratio of your simple example, its easy for me. Whereas, when it comes to finding ratio from reciprocal multiplication of x and y, the ratio I am getting is
2x/7= 5y/7
2x/7 * 7/5y = 5y/7 * 7/2x
2x/5y = 5y/2x #stuck!
how would I think of this in a numerical ratio of x to y??
another way to think about it
2/7 distance from x (like 3/5 of apples)
5/7 distance from y (like 2/5 of pears)
ratio of x to y is 2:5 (like ratio of apples to pears is 3:2)
This is certainly not right according to my method 2, there's more of x (10%) than y. So the ratio should be x:y is 5:2.
It would be helpful if you please explain
"Re-engineering" process, that'd be enough for now, visuals can be the last resort.
Thanks a lot. I completely understood the first two method by both of your help and now trying to get into ration thinking from weighted average. Excuse me if I am pushing it too much. I hope it will help me to connect WA, Fractions, Ratios better in long run.
appleid -
You have got this 95% correct - but have unnecessarily complicated it at the end.
You are perfect up to this point : 2x/7 = 5y/7 and then make a minor algebra mistake.
If 2x/7 = 5y/7, you can multiply both sides by 7 and get 2x = 5y
If 2x = 5y, then y=2/5x and x=5/2y. Looking at it this way, you can see that you need MORE of x (the 10% solution) to make the 8% solution
Also, you can clearly see how the ratio of 5:2 now is represented by the fractions (ie, that you will use 2.5 times more x than y, or only 40% the amount of x for y)
In the end, if you get one method stick with it, although being flexible helps on the exam. Let me know if this makes sense.
Its all clear now, I was thinking of taking reciprocals and somewhat lost myself in working through fractions. Thanks both of you. I am compiling the
steps in method 3:
3.
The difference between 10% and 3% is 7%; and, the WA is 8%
Here, x is (10-8)/7 = 2/7 away from 8%, to right in number line (if I see visually)
and y is (3-8)/7 = -5/7 away from 8%, to the left in number line.
As the differentials are distance from WA, you can omit negative sign for calculation and make them equal *
The fractions are:
2/7 distance from x = \(\frac{2}{7}x\)
5/7 distance from y = \(\frac{5}{7}y\)
As both distances come and meet at WA 8%* so,
\(\frac{2}{7}x = \frac{5}{7}y\)
\(2x = 5y\) #Multiply by 7 (Not taking reciprocals)
\(x = \frac{5y}{2}\)
\(\frac{x}{y} = \frac{5}{2}\)
So the ratio from the fraction is \(x:y = 5:2\)
Hope its perfect now. Please confirm me if everything is alright.
Another question (let it be last one..haha) to VeritasPrepDennis,
I tried to apply the same method on the example you gave me on 3 apple (a) and 2 pears(p):
\(\frac{3}{5}a = \frac{2}{5}p\)
\(3a = 2p\)
\(a = \frac{2p}{3}\)
\(\frac{a}{p} = \frac{2}{3}\)
So the ratio from the fraction is \(a:p = 2:3\)
Where am I getting it wrong? Is it because it's a simple fraction-ratio math compared to weighted average math?