The answer is not going to be B, as even statement II is also not sufficient..
Statement I is contained within statement II, so let us straight way check statement II..

Now x=1.2y..
The final price is
\(a(1+\frac{1.2y}{100})(1-\frac{y}{100})=a(1+\frac{1.2y}{100}-\frac{y}{100} +\frac{1.2y^2}{10000}=a(10000+2y-1.2y^2)/10000\)..
Now is this greater than initial a

If you take initial price as 100 and y a small %, say 10%..x=1.2y=12%..
So final price is 100*(1+0.12)(1-0.1)=100*(1.12)*0.9=112*0.9=100.8 , so greater than 100..

But let us take y slightly bigger % say 50% and x=1.2*50%=60%
So final price is 100*(1+0.6)(1-0.5)=100*1.6*0.5=160*0.5=80, which is less than 100..

So different answers..

E

Always be careful while solving % problems when everything is in variables.
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
4) Base while finding % increase and % decrease : https://gmatclub.com/forum/percentage-increase-decrease-what-should-be-the-denominator-287528.html


GMAT Expert

A price marked up by 25% and subsequently marked down 20% will result in original price.

i.e. If mark up= 1.2*mark down, final price = original price

Eg.
Original price: $100
Mark up 25%: 25
Price after mark up: 125
Mark down 20%: 125/5= 25
Price after mark down: 100

St.1 is Insufficient.
X= 25%, Y >20% No
X= 25%, Y= < 20% Yes

St.2 is sufficient as calculated above. Final price will be equal to original price.

Ans B

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