# What formula can solve a problem where you want to find a and b, knowing c? [closed]

You have a ratio -- a / b = c

Condition1
c has to be within the range .80-.90

Condition2
You do not know a
OR You know that a is within the range 20-30

Condition3
You know b is within the range 25-35

The goal is to find out what combinations of a and b will outcome to 80-90

What formula is best to solve this kind of problem? Please also explain the formula in plain english.

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## closed as not a real question by Andres Caicedo, LVK, Chris Eagle, Old John, WilliamAug 31 '12 at 4:34

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

So what is it; you don't know anything about $a$, or you know $20\leq a\leq 30$? –  Arturo Magidin Apr 18 '12 at 4:26
If $a$ is between 20 and 30, and $b$ is between 25 and 35, then $a/b$ is between $20/35$ and $30/25$, so it's nowhere near where you want $c$ to be. –  Gerry Myerson Apr 18 '12 at 4:29
So... everything is perfectly clear, but you don't know what $\leq$ means, and you call the decimal point a "dot". And apparently it wasn't so clear that it was free of errors... Ooooookay.... –  Arturo Magidin Apr 18 '12 at 4:31
It's amazing you can type, with that enormous chip on your shoulder. –  Gerry Myerson Apr 18 '12 at 4:37
@Arturo, if I thought you thought I was referring to you, I'd be mortified, so I'm going to assume you know the reference was not to you. –  Gerry Myerson Apr 18 '12 at 6:43

If $.80 \le c \le .90$ (the $\le$ is less than or equal to-this shows the range of $c$) and $20 \le a \le 30$ we have $b=\frac ac$. $b$ is increased as $a$ increases and $c$ decreases, so the greatest $b$ can be is $\frac {30}{.80}=37.5$. The least $b$ can be is $\frac {20}{.90}\approx 22.2$. Is this what you were looking for? The pair 1 and 3 can be solved similarly.