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Math Expert V
Joined: 02 Sep 2009
Posts: 62467
Is x > 3? (1) 5^y > 25^8 and y = x^2 (2) 2^(15x) > 8^(4x)*8  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 52% (01:54) correct 48% (02:09) wrong based on 60 sessions

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Is $$x > 3$$?

(1) $$5^y > 25^8$$ and $$y=x^2$$
(2) $$2^{15x} > 8^{4x}*8$$

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Re: Is x > 3? (1) 5^y > 25^8 and y = x^2 (2) 2^(15x) > 8^(4x)*8  [#permalink]

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1
1) y=16 Therefore, x= 4 or -4. Insufficuient
2) 15x>12x+3, x>1, Insufficuient
Combining. x=4.
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Re: Is x > 3? (1) 5^y > 25^8 and y = x^2 (2) 2^(15x) > 8^(4x)*8  [#permalink]

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1
(1) $$5^y > 25^8$$ and $$y = x^2$$
--> $$5^y > 5^{16}$$
--> $$y > 16$$
--> $$x^2 > 16$$
--> $$x < -4$$ or $$x > 4$$ --> Insufficient

(2) $$2^{15x} > 8^{4x}∗8$$
-->$$2^{15x} > 2^{12x}∗2^3$$
-->$$2^{15x} > 2^{12x + 3}$$
--> $$15x > 12x + 3$$
--> $$3x > 3$$
--> $$x > 1$$
--> Possible values of $$x = {1.5, 2, 3, 5, . . . }$$ --> Insufficient

Combining (1) & (2),
Since $$x > 1$$, $$x < -4$$ or $$x > 4$$
--> $$x > 4$$ only
--> $$x > 3$$ always--> Sufficient
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Is x > 3? (1) 5^y > 25^8 and y = x^2 (2) 2^(15x) > 8^(4x)*8  [#permalink]

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#1
5^y>25^8 and y=x^2
5^y>5^16 and y=x^2
y has to be atleast 17 and value can be +/-4
so x>3 sufficient
#2
2^15x>8^4x∗8
2^15x>2^12x+3
15x>12x+3
x>1
insufficient
from 1 &2
sufficient
IMO C

Is x>3?

(1) 5^y>25^8 and y=x^2
(2) 2^15x>8^4x∗8

Originally posted by Archit3110 on 18 Nov 2019, 01:39.
Last edited by Archit3110 on 19 Nov 2019, 00:02, edited 1 time in total.
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Re: Is x > 3? (1) 5^y > 25^8 and y = x^2 (2) 2^(15x) > 8^(4x)*8  [#permalink]

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1
Quote:
Is x>3?

(1) 5^y>25^8 and y=x^2
(2) 2^{15x}>8^4x∗8

(1) 5^y>25^8 and y=x^2 insufic

$$5^y>{25}^8…5^y>5^{16}…y>16$$
$$y=x^2:x^2>16…|x|>4…x>4,x<-4$$

(2) 2^{15x}>8^4x∗8 insufic
$$2^{15x}>8^{4x}∗8…2^{15x-3}>2^{12x}…15x-3>12x…3x>3…x>1$$

(1&2) sufic
$$x>1:|x|>4…x>4$$

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Re: Is x > 3? (1) 5^y > 25^8 and y = x^2 (2) 2^(15x) > 8^(4x)*8  [#permalink]

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1
Is x>3?

(1) $$5^y>25^8 and y=x^2$$
$$5^{x^2}>(5^2)^8$$
$$x^2 > 16$$

Let $$x^2 = 25$$
=> x = +/- 5

INSUFFICIENT.

(2) $$2^{15x}>8^{4x}∗8$$
$$2^{15x}>8^{4x + 1}$$
$$2^{15x}>2^{3*(4x + 1)}$$
15x > 3*(4x + 1)
x > 1

INSUFFICIENT.

Together 1 and 2.
x = 5, 6, 7 ....
Hence x > 3 always

SUFFICIENT.

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Re: Is x > 3? (1) 5^y > 25^8 and y = x^2 (2) 2^(15x) > 8^(4x)*8  [#permalink]

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1
IMO C

From 1 - x>+-4
From 2- x>1
Combining both, x>3..
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Re: Is x > 3? (1) 5^y > 25^8 and y = x^2 (2) 2^(15x) > 8^(4x)*8  [#permalink]

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1
(1)This gives y>16

Then x^2>16
x can be any value except -4 to 4
INSUFFICIENT

(2)this gives x>1....so INSUFFICIENT

combining both we can get.....x>4.

So SUFFICIENT

OA:C

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Re: Is x > 3? (1) 5^y > 25^8 and y = x^2 (2) 2^(15x) > 8^(4x)*8  [#permalink]

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1

1. 5^y > 25^8 and y=x^2
5^y > (5^2)^8
5^y > 5^16
hence y>16. But y=x^2
hence x^2 > 16
(x-4)(x+4) > 0
x<-4 or x>4
This is insufficient because when x=-5, x<3, but when x=5, then x>3.

2. 2^15x > 8^4x * 8
2^15x > (8)^(4x+1)
2^15x > (2)^(3[4x+1])
This implies 15x>12x+3 hence x>1
This is insufficient because when x=2, then x<3. but when x=4, then x>3.

1+2
x<-4 or x>4 and x>1.
This means we are able to conclude that x>4 since that is the range that satisfies both statements.
Within the range of x>4, x>3. Hence both statements taken together is sufficient.

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Re: Is x > 3? (1) 5^y > 25^8 and y = x^2 (2) 2^(15x) > 8^(4x)*8  [#permalink]

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Is x>3?

(1) 5^y > 25^8 and y=x^2
(2) 2^(15x) > 8^(4x)*8

(1) 5^ y > 5^2^8 so, y > 16 . now, x^2 > 16. so, x > 4 or x < -4. Not sufficient.
(2) 15 x > 12x + 3 or, 3x > 3 or, x > 1 . not sufficient
Together, x is greater than 3. sufficient.
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Re: Is x > 3? (1) 5^y > 25^8 and y = x^2 (2) 2^(15x) > 8^(4x)*8  [#permalink]

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1
Is x >3?

(Statement1):
y >16 and $$y = x^{2}$$

(x —4)(x+4) >0
Clearly insufficient

(Statement2): 15x > 12x+ 3
3x > 3
x >1
—> if x=4, then Yes
—> If x=2, then No
Insufficient

Taken together 1&2,
x >1
(x—4)(x+ 4) >0

—> x >4
Sufficient

Posted from my mobile device Re: Is x > 3? (1) 5^y > 25^8 and y = x^2 (2) 2^(15x) > 8^(4x)*8   [#permalink] 18 Nov 2019, 14:05
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# Is x > 3? (1) 5^y > 25^8 and y = x^2 (2) 2^(15x) > 8^(4x)*8  