Is 4^(x+3)/64 1 ?

Question

Is  \frac{4^{x+3}}{64} > 1  ?

(1)  4^{x+3} > 0

(2)  4^x > 1

Source : McGraw Hill Education

Archit3110 · Answer

Express
 \frac{4^{x+3}}{64} > 1 
we get
4^x>1
#1
insufficient
#2
4^x>1
sufficient
IMO B

Rohitmariner · Answer

#1) 4^x+3>0 here x by putting X=-1,-2,-3 and putting it in the equation proves it wrong so insufficient
#2) 4^x>1 here X can be0,1..n
Putting in original proves it right
So B

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KarthikaD · Answer

#1) 4^x+3>0 here x by putting X=-1,-2,-3 and putting it in the equation proves it wrong so insufficient
#2) 4^x>1 here X can be0,1..n
Putting in original proves it right
So B

DIDN'T get this well. Can someone help me to understand this?

SaquibHGMATWhiz · Answer

Solution:

Pre-Analysis:    
  We are asked if  \frac{4^{x+3}}{64} > 1  or not  
 This can be simplified a little
 ⇒4^{x+3}>64 
 ⇒4^x 	imes 4^3>64 
 ⇒4^x>\frac{64}{64} 
 ⇒4^x>1 
 ⇒4^x>4^0 
 ⇒x>0  
 Thus, the question is asking us if  4^x>1  or if  x>0  or not 
 This is a yes-no question

Statement 1:    4^{x+3} > 0 
  This statement doesn't tell us if  x > 0  or not  
  4^{x+3} > 0   will satisfy if x is negative like -1 and also when x is positive like +1  
 Thus,  statement 1 alone is not sufficient and we can eliminate options A and D

Statement 2:    4^x > 1 
  This is exactly in line with our pre-analysis 
 Thus, statement 2 alone is sufficient

Hence the right answer is    Option B

Is 4^(x+3)/64 > 1 ?

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