GMATPREP EXAM 1 (6)

Question

Does anyone have an efficient approach to this question?

thanks for all your help!

LM · Answer

Yeah, here you go....

r1:: A^2/B

r2:: A^2/2*B ( because concentration of B increase by 100%)

But we need to keep r2 same as r1,

we can clearly see the relation; r2 = 2*r1

That means we have to increase the concentration of A in such a way that it becomes 2 times of what it was. 

i.e., A^2 should be increased to 2*A^2 in order to make r2 same as r1.

A^2 -------------->2*A^2

=> A---------------> *A

which means A*(1+x%) =  *A

(1+x%) = (2)^1/2 

x% =   - 1 = 1.4 - 1 = 0.4

x/100 = 0.4

x = 40%.

mastergmat1 · Answer

Vlad77 · Answer

(A1^2)/B1=((A1*x)^2) /B1*2 
2A1^2 =((A1*x)^2)
sqrt(2)*A1=A1*x
x=sqrt(2)=1,41

So 40% increase

GMATBLACKBELT · Answer

This one took me a long time. No way on test day would have gotten this.

the rate is equal to A^2/B since it is directly proportional put A^2 in the top. the rate is inversly propotional to B so put B in the bottom.

now B is increased by 100% or just 2B. essentially the ratio has to stay the same so... 

let x be what A needs to be multiplied by in order to equate the same ratio as A^2/B

A^2/B=(A*x)^2/2B 

Get rid of B's. 2A^2=(A*x)^2 Now get rid of the A^2's. We are left with 2=x^2. So x= sqrt2. sqrt 2 = aprox 1.4 so an increase of 40%.

jamesrwrightiii · Answer

I got it just by guesstimating
The only two answers that make sense based on the premise are D and E
I thought E was too high and chose D

b14kumar · Answer

:-D

GMATPREP EXAM 1 (6)

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