How do make r the subject in the following fomula:- A= P(1+R/100)5

You should take more care in writing your questions. Because of how you have written it, I cannot tell if you are asking me to rearrange A = P(1 + (R/100)) x 5 or if you are asking me to rearrange A = P((1+R)/100) x 5. The 5 at the end could even be interpreted as being a power, rather than a coefficient, but I will assume that it is not. I will answer these two different interpretations of your question respectively and would encourage you to be more clear in future. The key to answering this question is that whatever you do to one side you must also do to the other. i) A = P(1 + (R/100)) x 5 =) P(1+(R/100)) = A/5 [divide both sides by 5] =) 1+(R/100) = A/5P [divide both sides by P] =) R/100 = A/5P - 1 [subtract 1 from both sides] =) R = 100(A/5P - 1) [multiply both sides by 100] At this point, we simplify: R = 100(A/5P - 1) = 100A/5P - 100 = 20A/P - 100 = 20(A/P - 5) ii)A = P((1+R)/100)x5 =) P((1+R)/100) = A/5 =) (1+R)/100 = A/5P =) 1+R = 100(A/5P) =) R = 100(A/5P) - 1 Simplifying: R = 100(A/5P) - 1 = (100A/5P) - 1 = (20A/P) - 1
By: Wyess.

Date of comment: Fri, Dec 24th 2010