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amanvermagmat
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X and Y are two positive numbers. Is X > Y ? (1) (X+Y)/2 > Y (2) 13 Aug 2018, 22:54
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D
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Difficulty:

45% (medium)

Question Stats:

67% (01:52) correct 33% (01:42) wrong based on 257 sessions

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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.
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D
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rahulkashyap
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X and Y are two positive numbers. Is X > Y ? (1) (X+Y)/2 > Y (2) 14 Aug 2018, 00:08
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option a straight away gives the ans

but for option b, on simplifying, we get that x/y>1.5
so yes

ans is d
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Izzyjolly
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X and Y are two positive numbers. Is X > Y ? (1) (X+Y)/2 > Y (2) 25 Aug 2018, 19:01
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amanvermagmat
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.
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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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X and Y are two positive numbers. Is X > Y ? (1) (X+Y)/2 > Y (2) 18 Oct 2018, 20:13
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amanvermagmat
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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I am still not clear on how you got 50/100 from Z<40, can you please elaborate on this?

Thanks!
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X and Y are two positive numbers. Is X > Y ? (1) (X+Y)/2 > Y (2) 18 Oct 2018, 22:50
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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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X and Y are two positive numbers. Is X > Y ? (1) (X+Y)/2 > Y (2) 19 Oct 2018, 01:59
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Hi khosld64,

\(Z<40\) --> \(\frac{Z}{100} < \frac{40}{100} < \frac{50}{100}\) --> \(1-\frac{Z}{100} > 1−\frac{50}{100}\)

Multiple both sides to (X+Y)

Thus \((X+Y)*(1-\frac{Z}{100}) > (X+Y)∗(1−\frac{50}{100})\)

Hope it's clear.

khosld64
Izzyjolly
amanvermagmat
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.

I am still not clear on how you got 50/100 from Z<40, can you please elaborate on this?

Thanks!
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X and Y are two positive numbers. Is X > Y ? (1) (X+Y)/2 > Y (2) 21 Jun 2024, 17:43
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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.­

We need to find whether X > Y
statement 1 ,

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

Statement 2 ,

x= (x+y)(1- z/100)

If z = 40 , x= 1.5y
Hence x > y

If z=1 , x= 99y
then also x > y

Hence statement 2 is also sufficient.

D is the answer.

GMATinsight , any better method than this if exists..
 
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