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For DS problems, we shouls always get the relationships of the variables in the equation in order to know what conditions are needed to be able to answer the question

A=profits first year
B=profits second year
C=profits third year

1ºequation: B=(1+x)A
2ºequation: C=(1-y)B
Question to answer C>A? in other words (1-y)(1+x)A>A. Crossing off A in both sides of the equation: (1-y)(1+x)>1

Therefore we need to know if (1-y)*(1+x)>1,

but what is the relationship between x and y in this equation? It does not seem we know the relationship very well to be able to answer. Let's multiply the brackets (1-y)*(1+x)=1-y+x-x*y

Ok, so now we know that we need to know both the x+(-)y reationship and the x*y one

Option (1): 1-x<1 -y, which is -x<-y, which is x>y. This option tells me the x+(-)y relationship, but what about the multiplication? NOT SUFFICIENT

Option (2): xy/100<x-y. We inmediately see we have the x+(-)y reationship and the x*y one. No need to calculate. If we know the relationship between both, we can already conclude if C>A or not. Therefore: SUFFICIENT

Option B
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Bunuel
The profits of a new company were x percent higher in the second year than the first year and y percent lower in the third year than the second year. Were the profits higher in the third year than the first year?

(1) 1 - x < 1 - y
(2) xy/100 < x - y

Video Explanation



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Bunuel
The profits of a new company were x percent higher in the second year than the first year and y percent lower in the third year than the second year. Were the profits higher in the third year than the first year?

(1) 1 - x < 1 - y
(2) xy/100 < x - y

The profits of a new company were x percent higher in the second year than the first year and y percent lower in the third year than the second year. Were the profits higher in the third year than the first year?
Q. (100+x)(100-y)/10000 -1 = (x-y)/100-xy/10000 = (x-y) -xy/100> 0 ?

(1) 1 - x < 1 - y
x - y > 0; x > y; In (x-y)-xy/100, x-y>0 but -xy/100<0
NOT SUFFICIENT

(2) xy/100 < x - y
x-y - xy/100>0
SUFFICIENT

IMO B
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