BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

GMAT Prep percentage

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 468
Joined: Sat Mar 03, 2007 10:17 pm
Thanked: 5 times

GMAT Prep percentage

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
Top
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 460
Joined: Sun Mar 25, 2007 7:42 am
Thanked: 27 times

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

--------------------------------------------------------------------------------

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
Top
Master | Next Rank: 500 Posts
Posts: 468
Joined: Sat Mar 03, 2007 10:17 pm
Thanked: 5 times

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
Top
Master | Next Rank: 500 Posts
Posts: 460
Joined: Sun Mar 25, 2007 7:42 am
Thanked: 27 times

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
Top
Post new topic Post Reply
• Page 1 of 1