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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 $p=13,~C = 200$ $X = \dfrac{200}{1+\frac{13}{100}} \approx 176.99$ 
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 
Member Joined: Jun 2019 From: AZ, Seattle, San Diego Posts: 30 Thanks: 21  

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