GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
A tricky question
[#permalink]
13 Aug 2016, 12:13
13 Aug 2016, 12:13
Could any body please help me solve this question:
Attachments
File comment: Quantitative Comparison Question
QC1.jpg [ 528.51 KiB | Viewed 1925 times ]
Re: A tricky question
[#permalink]
15 Aug 2016, 08:43
15 Aug 2016, 08:43
Hi dear Sandy,I am new to this forum and this is my first post so I am not aware of the format for posting the question,please also tell me what format I need to follow for posting questions?
As far as source of this question is concerned,this question has been taken from the notes provided by US Educational Foundation GRE Academy and I do not know the name of the book.There are some other tricky questions as well that I would share later
Your solution is not clear to me,only two equations are obvious i.e.
-10 < Y^X < -1 ....... 1
-1 < X^Y < 0 ....... 2
x and y both are integers (given)
but we cannot assume that -1 < X < 0 as x is integer & no integer falls in this range;as you say that x is negative and y must be positive then suppose x = -2 & y =3
=> x^y = (-2)^3 = -8 but this does not fall in the range specified for x^y ,i.e. -1<x^y<0
=> y^x = (3)^-2 = 1/(3^2) = 1/6 = 0.167 but this does not fall in the range specified for y^x ,i.e. -10<y^x<-1
So based upon these calculations the solution does not seem clear ...
regards
m.yasir
As far as source of this question is concerned,this question has been taken from the notes provided by US Educational Foundation GRE Academy and I do not know the name of the book.There are some other tricky questions as well that I would share later
Your solution is not clear to me,only two equations are obvious i.e.
-10 < Y^X < -1 ....... 1
-1 < X^Y < 0 ....... 2
x and y both are integers (given)
but we cannot assume that -1 < X < 0 as x is integer & no integer falls in this range;as you say that x is negative and y must be positive then suppose x = -2 & y =3
=> x^y = (-2)^3 = -8 but this does not fall in the range specified for x^y ,i.e. -1<x^y<0
=> y^x = (3)^-2 = 1/(3^2) = 1/6 = 0.167 but this does not fall in the range specified for y^x ,i.e. -10<y^x<-1
So based upon these calculations the solution does not seem clear ...
regards
m.yasir
Re: A tricky question
[#permalink]
15 Aug 2016, 11:28
15 Aug 2016, 11:28
Expert Reply
yasir9909 wrote:
Hi dear Sandy,I am new to this forum and this is my first post so I am not aware of the format for posting the question,please also tell me what format I need to follow for posting questions?
As far as source of this question is concerned,this question has been taken from the notes provided by US Educational Foundation GRE Academy and I do not know the name of the book.There are some other tricky questions as well that I would share later
Your solution is not clear to me,only two equations are obvious i.e.
-10 < Y^X < -1 ....... 1
-1 < X^Y < 0 ....... 2
x and y both are integers (given)
but we cannot assume that -1 < X < 0 as x is integer & no integer falls in this range;as you say that x is negative and y must be positive then suppose x = -2 & y =3
=> x^y = (-2)^3 = -8 but this does not fall in the range specified for x^y ,i.e. -1<x^y<0
=> y^x = (3)^-2 = 1/(3^2) = 1/6 = 0.167 but this does not fall in the range specified for y^x ,i.e. -10<y^x<-1
So based upon these calculations the solution does not seem clear ...
regards
m.yasir
As far as source of this question is concerned,this question has been taken from the notes provided by US Educational Foundation GRE Academy and I do not know the name of the book.There are some other tricky questions as well that I would share later
Your solution is not clear to me,only two equations are obvious i.e.
-10 < Y^X < -1 ....... 1
-1 < X^Y < 0 ....... 2
x and y both are integers (given)
but we cannot assume that -1 < X < 0 as x is integer & no integer falls in this range;as you say that x is negative and y must be positive then suppose x = -2 & y =3
=> x^y = (-2)^3 = -8 but this does not fall in the range specified for x^y ,i.e. -1<x^y<0
=> y^x = (3)^-2 = 1/(3^2) = 1/6 = 0.167 but this does not fall in the range specified for y^x ,i.e. -10<y^x<-1
So based upon these calculations the solution does not seem clear ...
regards
m.yasir
Please ignore my solution its wrong!
Regards,
