Proportion Question

fangtray wrote:The rate of a certain chemical reaction is directly porportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. if the concentration of chemical B is increased by 100%, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged?

Rate is directly proportional to ((CA)^2) / (CB), i.e

Rate of Chemical Reaction= K * ((CA)^2) / (CB), where K is a constant and CA, CB are concentration of chemical A and B respectively.

Rate of Chemical Reaction One = K * ((CA)^2) / (CB)
Rate of Chemical Reaction two = K * ((p*CA)^2)/(2CB), where p = Percentage of CA
Since Rate of Chemical Reaction One = Rate of Chemical Reaction two
K * ((CA)^2) / (CB) = K * ((p*CA)^2)/(2CB)
Implies, p^2 = 2
p = 1.414.

So it is a 41.4% Increase ~ 40% Increase

p.s: In a hurry hence the crappy notes, let me know if you need any further help!

I posted a solution here:

https://www.beatthegmat.com/the-rate-of- ... 60321.html