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GMAT Prep percentage

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Master | Next Rank: 500 Posts
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by moneyman » Thu Nov 15, 2007 6:11 am
The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversly proportional to the concentration of Chemical B present. If the concentration of chemical B is increased by 100%, which of the follwing is the closest to the % change in the concentration of chemical A required to keep the reaction rate unchanged??

100% decrease

50% dec

40% dec

40% inc

50% inc
Maxx
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by samirpandeyit62 » Thu Nov 15, 2007 10:18 pm
Posted: Thu Nov 15, 2007 7:11 am Post subject: GMAT Prep percentage

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The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversly proportional to the concentration of Chemical B present. If the concentration of chemical B is increased by 100%, which of the follwing is the closest to the % change in the concentration of chemical A required to keep the reaction rate unchanged??

100% decrease

50% dec

40% dec

40% inc

50% inc


first of all if B increase then rate will decrease (inv prop) so to keep the rate unchanged Av should inc

given

R = xA ^2

R = y/ B ( x,y = constanats , A= conc of A , B = conc o-f B)

so when B become 2B

Rd (dec rate) = y/ 2B i.e reduces by 1/2

so R = 2 * y/2B

= 2*xA^2

= x * (1.4)^2

so A should be inc to 1.4 A

i.e by 40%

D
Regards
Samir
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by moneyman » Fri Nov 16, 2007 3:15 am
Hey Samir if u dont mind can you pls explain this part ??

Rd (dec rate) = y/ 2B i.e reduces by 1/2

so R = 2 * y/2B

= 2*xA^2

= x * (1.4)^2
Maxx
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by samirpandeyit62 » Sat Nov 17, 2007 12:06 am
I've used Rd to signify the reduced rate R

i.r Rd = 1/2 R = 1/2 *y/B

so to bring Rd back to R we need to multiply it by 2

2Rd = R

2 * y/2B = R

so 2 * xA^2 =R

so x*(rt(2)*A)^2 =R

rt(2) =1.4 approx

so rt(2)*A = 1.4 A

i.e 40% more than A

hope this is clear.
Regards
Samir
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