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X and Y are two positive numbers. Is X > Y ? (1) (X+Y)/2 > Y (2)

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X and Y are two positive numbers. Is X > Y ? (1) (X+Y)/2 > Y (2)  [#permalink]

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New post 13 Aug 2018, 21:54
6
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

62% (01:23) correct 38% (01:21) wrong based on 146 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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X and Y are two positive numbers. Is X > Y ? (1) (X+Y)/2 > Y (2)  [#permalink]

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New post 13 Aug 2018, 23:08
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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Re: X and Y are two positive numbers. Is X > Y ? (1) (X+Y)/2 > Y (2)  [#permalink]

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New post 25 Aug 2018, 18:01
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amanvermagmat wrote:
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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Re: X and Y are two positive numbers. Is X > Y ? (1) (X+Y)/2 > Y (2)  [#permalink]

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New post 18 Oct 2018, 19:13
Izzyjolly wrote:
amanvermagmat wrote:
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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Re: X and Y are two positive numbers. Is X > Y ? (1) (X+Y)/2 > Y (2)  [#permalink]

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New post 18 Oct 2018, 21:50
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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Re: X and Y are two positive numbers. Is X > Y ? (1) (X+Y)/2 > Y (2)  [#permalink]

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New post 19 Oct 2018, 00:59
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 wrote:
Izzyjolly wrote:
amanvermagmat wrote:
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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Re: X and Y are two positive numbers. Is X > Y ? (1) (X+Y)/2 > Y (2) &nbs [#permalink] 19 Oct 2018, 00:59
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