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 August 15th, 2019, 02:07 PM #1 Newbie   Joined: Aug 2019 From: Terrace Posts: 2 Thanks: 0 Can someone solve X in this Does a formula exist for this or am I out in left field? Here goes: Where X + 13% of X = 200. Solve X Note - The percentage and the number 200 are arbitrary. Either of these could be infinite.
 August 15th, 2019, 02:18 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,529 Thanks: 1389 $X + p\% \text{ of } X = C$ $\left(1+\dfrac{p}{100}\right)X = C$ $X = \dfrac{C}{1+\frac{p}{100}}$ letting $p=13,~C = 200$ $X = \dfrac{200}{1+\frac{13}{100}} \approx 176.99$ Thanks from greg1313, topsquark and idontknow
 August 15th, 2019, 02:26 PM #3 Newbie   Joined: Aug 2019 From: Terrace Posts: 2 Thanks: 0 Alright then - I figured that there must be a formula as opposed to putting in random numbers to eventually obtain an outcome. Thank You!
August 16th, 2019, 12:46 PM   #4
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Quote:
 Originally Posted by floydy12 … The percentage and the number 200 are arbitrary. Either of these could be infinite.
Well, the number 200 can't be infinite, but I know what you mean. In cases where either the percent or the given constant is infinite, there is no solution for X.

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