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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If y > x, is y > 64 (1) x has 7 factors (all even but one) and is div

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Math Expert V
Joined: 02 Sep 2009
Posts: 61302
If y > x, is y > 64 (1) x has 7 factors (all even but one) and is div  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 37% (02:37) correct 63% (02:35) wrong based on 52 sessions

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If y > x, is y > 64

(1) x has 7 factors (all even but one) and is divisible by only one prime number.
(2) |y - 64| > |x - 64|

Are You Up For the Challenge: 700 Level Questions

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Re: If y > x, is y > 64 (1) x has 7 factors (all even but one) and is div  [#permalink]

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Bunuel wrote:
If y > x, is y > 64?

(1) x has 7 factors (all even but one) and is divisible by only one prime number.
(2) |y - 64| > |x - 64|

Are You Up For the Challenge: 700 Level Questions

Given: y > x

Question : is y > 64?

Statement 1: x has 7 factors (all even but one) and is divisible by only one prime number.

i.e. $$x = 2^6 = 64$$ which is the only possible value of x

P.S. 2^6 has 7 factors {1, 2, 4, 8, 16, 32, 64} only one odd rest all even factors

and y > x therefore y > 64

SUFFICIENT

Statement 2: |y - 64| > |x - 64|

DIstance between y and 64 is Greater than distance between x and 64 and Y is to the right of x on number line because y > x therefore

y must be to the right of 64 and x must be somewhere between y and 64

SUFFICIENT

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VP  P
Joined: 24 Nov 2016
Posts: 1218
Location: United States
If y > x, is y > 64 (1) x has 7 factors (all even but one) and is div  [#permalink]

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Bunuel wrote:
If y > x, is y > 64

(1) x has 7 factors (all even but one) and is divisible by only one prime number.
(2) |y - 64| > |x - 64|

(1) x has 7 factors (all even but one) and is divisible by only one prime number. sufic

num.f(x)=7 (all even except one); x=prime^6=2^6=64; y>64.

(2) |y - 64| > |x - 64| sufic

y-64≥0…y≥64; x-64≥0…x≥64
---(-)--(-)---64----+--+---
x,y<64: -(y-64)>-(x-64)…-y+x>0…x>y=false;
x,y≥64: (y-64)>(x-64)…y-x>0…y>x=true; y>64.

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Director  P
Joined: 25 Jul 2018
Posts: 549
If y > x, is y > 64 (1) x has 7 factors (all even but one) and is div  [#permalink]

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If y > x, is y > 64 ?

(Statement1): x has 7 factors (all even but one) and is divisible by only one prime number.

$$a^{6}$$ —> (6+1)= 7 factors
—> a should be prime number and even number at the same time—> 2 could be prime and even
$$x= 2^{6} = 64$$ —> y >64 (yes)
Sufficient

(Statement2): |y - 64| > |x - 64|
—> square both sides
$$(y—64)^{2} > (x—64)^{2}$$

$$y^{2} —128y+ 64^{2} > x^{2} —128x+64^{2}$$
$$y^{2}—128y > x^{2} —128x$$
(y—x)( y+x) —128(y—x) > 0
(y—x)(y+x—128) >0

Given that y—x >0, then —> y+x—128 >0
—> y+x >128

Now, we have two inequalities:
y—x > 0
y+x >128
—> 2y > 128 —> y > 64 (Yes)
Sufficient

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Manager  G
Joined: 10 Dec 2017
Posts: 178
Location: India
Re: If y > x, is y > 64 (1) x has 7 factors (all even but one) and is div  [#permalink]

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Bunuel wrote:
If y > x, is y > 64

(1) x has 7 factors (all even but one) and is divisible by only one prime number.
(2) |y - 64| > |x - 64|

Are You Up For the Challenge: 700 Level Questions

From 1
X=2^6=total 7 factors ( 6+ 1)
X=64
since Y>X
Y>64
Sufficient
From 2
Y>X
so Y-64>X-64( we can add or subtract in inequality)
and lY-64l>lX-64l
for this to be true Y>64
Sufficient
D:)
Intern  B
Joined: 23 Dec 2019
Posts: 4
Re: If y > x, is y > 64 (1) x has 7 factors (all even but one) and is div  [#permalink]

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I didn't understand how Statement 2 is sufficient. Could someone post step by step answer
Manager  B
Joined: 22 Sep 2018
Posts: 66
If y > x, is y > 64 (1) x has 7 factors (all even but one) and is div  [#permalink]

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ayushsaxena04 wrote:
I didn't understand how Statement 2 is sufficient. Could someone post step by step answer

if using step process , here how it works

consider what is given in the question , y > x

consider both mods give negative value ie , . suppose y < 64 and X < 64, you get

-(y-64)>-(x-64)
-y+64 > -x+ 64

ie X > Y , but this contradicts with condition given in the question iteself, so in statement 2 consider Y>64 and solve you again get Y > X . so you can say Y>64. If y > x, is y > 64 (1) x has 7 factors (all even but one) and is div   [#permalink] 23 Jan 2020, 06:31
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