Is x 3? (1) 5^y 25^8 and y = x^2 (2) 2^(15x) 8^(4x)*8

Question

Competition Mode Question

Is  x > 3 ?

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

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vg18 · Answer

1) y=16 Therefore, x= 4 or -4. Insufficuient
2) 15x>12x+3, x>1, Insufficuient
Combining. x=4.
IMO C

CareerGeek · Answer

(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

Archit3110 · Answer

#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

exc4libur · Answer

(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

Ans (C)

unraveled · Answer

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.

IMO Answer C.

pandajee · Answer

IMO C

From 1 - x>+-4
From 2- x>1
Combining both, x>3..

madgmat2019 · Answer

(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

Posted from my mobile device

eakabuah · Answer

We are asked if x>3.

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.

The answer is option C.

minustark · Answer

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.
C is the answer.

lacktutor · Answer

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 
The answer is C

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bumpbot · Answer

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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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