X and Y are two positive numbers. Is X > Y ?

(1) (X+Y)/2 > Y

(2) X is Z% less than the sum (X+Y), where Z is a positive number and Z < 40.

If \(X=Y\) then \(X=Y=\frac{(X+Y)}{2}=(X+Y)∗(1−\frac{50}{100})\)
(1) \(\frac{(X+Y)}{2}>Y\) --> X < Y -> Sufficient.
(2) \(X= (X+Y)*(1-\frac{Z}{100})\).
Since \(Z<40\), \((X+Y)*(1-\frac{Z}{100}) > (X+Y)∗(1−\frac{50}{100})\) --> \(X>(X+Y)∗(1−\frac{50}{100})\), thus X > Y -> Sufficient.
Answer D.
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Hi,

It’s a pretty straight forward question, where it requires basic word to math conversion knowledge.

Given: X and Y are two positive numbers.

Question: Is X > Y ?

Statement I is sufficient:

(X+Y)/2 > Y

X + Y > 2Y

So, X > Y.

It is sufficient.

Statement II is sufficient:

X is Z% less than the sum (X+Y), where Z is a positive number and Z < 40.

Converting to Math,

X = (X+Y) – Z%(X+Y)

X = (X+Y) – Z%(X+Y)

X = (X+Y)*(1-Z/100)

Given,

Z < 40

Lets assume for a moment, Z = 40

X = (X+Y) * (1-40/100)

X = (X+Y) * 3/5

Solving it we get,

5X = 3X + 3Y

2X = 3Y

X = 3/2 * Y

So, X > Y when we take Z = 40, taking Z further lesser value will have X > Y only.

So statement II is also sufficient.

So the answer is D.

Hope it is clear.
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