GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Mar 2019, 12:34 ### GMAT 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

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. ### Request Expert Reply # NEW!!! Tough and tricky exponents and roots questions

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 53738
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

Kconfused wrote:
Bunuel wrote:
1. If $$357^x*117^y=a$$, where $$x$$ and $$y$$ are positive integers, what is the units digit of $$a$$?
(1) $$100<y^2<x^2<169$$
(2) $$x^2-y^2=23$$

(1) $$100<y^2<x^2<169$$ --> since both $$x$$ and $$y$$ are positive integers then $$x^2$$ and $$y^2$$ are perfect squares --> there are only two perfect squares in the given range 121=11^2 and 144=12^2 --> $$y=11$$ and $$x=12$$. Sufficient.(As cyclicity of units digit of $$7$$ in integer power is $$4$$, therefore the units digit of $$7^{23}$$ is the same as the units digit of $$7^3$$, so 3).

(2) $$x^2-y^2=23$$ --> $$(x-y)(x+y)=23=prime$$ --> since both $$x$$ and $$y$$ are positive integers then: $$x-y=1$$ and $$x+y=23$$ --> $$y=11$$ and $$x=12$$. Sufficient.

Hello Bunuel, for the second statement, you seem to indicate that since the difference of the squares of two positive integers is a prime number, the two integers are consecutive on the number scale. Does this always hold?
Thank you!

Yes, positive integer solutions of x^2-y^2=prime must be consecutive, because of x-y=1 (x=y+1).
_________________
Senior Manager  Joined: 29 Oct 2008
Posts: 348
Location: United States
Concentration: Marketing, Technology
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

Bunuel wrote:
8. What is the value of $$x^2+y^3$$?
(1) $$x^6+y^9=0$$
(2) $$27^{x^2}=\frac{3}{3^{3y^2+1}}$$

(1) $$x^6+y^9=0$$ --> $$(x^2)^3=(-y3)^3$$ ---> $$x^2=-y^3$$ --> $$x^2+y^3=0$$. Sufficient.

(2) $$27^{x^2}=\frac{3}{3^{3y^2+1}}$$ --> $$3^{3x^2}=\frac{3}{3^{3y^2}*3}$$ --> $$3^{3x^2}*3^{3y^2}=1$$ --> $$3^{3x^2+3y^2}=1$$ --> $$3x^2+3y^2=0$$ (the power of 3 must be zero in order this equation to hold true) --> $$x^2+y^2=0$$ the sum of two non-negative values is zero --> both $$x$$ and $$y$$ must be zero --> $$x=y=0$$ --> $$x^2+y^3=0$$. Sufficient.

Hi Bunuel, I have a minor doubt. Using the corollary a^3+b^3=(a+b)(a^2-a*b+b^2) and plugging in a=x^2 and b=y^3 in statement 1,
we can say that (x^2)^3+(y^3)^3=(x^2+y^3)(x^4-x^2y^3+y^6)=0
Since any of expressions (x^2+y^3) or (x^4-x^2y^3+y^6) can be zero. It seems that statement 1 is insufficient by this logic.
Kindly comment.
Math Expert V
Joined: 02 Sep 2009
Posts: 53738
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

joshnsit wrote:
Bunuel wrote:
8. What is the value of $$x^2+y^3$$?
(1) $$x^6+y^9=0$$
(2) $$27^{x^2}=\frac{3}{3^{3y^2+1}}$$

(1) $$x^6+y^9=0$$ --> $$(x^2)^3=(-y3)^3$$ ---> $$x^2=-y^3$$ --> $$x^2+y^3=0$$. Sufficient.

(2) $$27^{x^2}=\frac{3}{3^{3y^2+1}}$$ --> $$3^{3x^2}=\frac{3}{3^{3y^2}*3}$$ --> $$3^{3x^2}*3^{3y^2}=1$$ --> $$3^{3x^2+3y^2}=1$$ --> $$3x^2+3y^2=0$$ (the power of 3 must be zero in order this equation to hold true) --> $$x^2+y^2=0$$ the sum of two non-negative values is zero --> both $$x$$ and $$y$$ must be zero --> $$x=y=0$$ --> $$x^2+y^3=0$$. Sufficient.

Hi Bunuel, I have a minor doubt. Using the corollary a^3+b^3=(a+b)(a^2-a*b+b^2) and plugging in a=x^2 and b=y^3 in statement 1,
we can say that (x^2)^3+(y^3)^3=(x^2+y^3)(x^4-x^2y^3+y^6)=0
Since any of expressions (x^2+y^3) or (x^4-x^2y^3+y^6) can be zero. It seems that statement 1 is insufficient by this logic.
Kindly comment.

The point is that only real roots of x^4 - x^2y^3 + y^6 = 0 is x = y = 0, for which x^2 + y^3 = 0.
_________________
Manager  Joined: 25 Mar 2014
Posts: 133
Location: India
Concentration: Operations, Finance
GMAT Date: 05-10-2015
GPA: 3.51
WE: Programming (Computer Software)
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

Bunuel wrote:
8. What is the value of $$x^2+y^3$$?
(1) $$x^6+y^9=0$$
(2) $$27^{x^2}=\frac{3}{3^{3y^2+1}}$$

(1) $$x^6+y^9=0$$ --> $$(x^2)^3=(-y3)^3$$ ---> $$x^2=-y^3$$ --> $$x^2+y^3=0$$. Sufficient.

(2) $$27^{x^2}=\frac{3}{3^{3y^2+1}}$$ --> $$3^{3x^2}=\frac{3}{3^{3y^2}*3}$$ --> $$3^{3x^2}*3^{3y^2}=1$$ --> $$3^{3x^2+3y^2}=1$$ --> $$3x^2+3y^2=0$$ (the power of 3 must be zero in order this equation to hold true) --> $$x^2+y^2=0$$ the sum of two non-negative values is zero --> both $$x$$ and $$y$$ must be zero --> $$x=y=0$$ --> $$x^2+y^3=0$$. Sufficient.

Hi Bunuel,
I have a confusion. The problem statement with option no. 1 can be written as follows:
if a^3 + b^3 = 0 (1st option) then what is the value of a + b ? (where a = x^2 and b = y^3).
Now, we know that (a^3 + b^3) = (a+b) * (a^2 - a*b + b^2). (a^3 + b^3) has 2 factors.
And according to the explanation, since, (a^3 + b^3) =0 => a^3 = -(b^3) => a = -b => a + b = 0. So, whenever a^3 + b^3 is 0, a+b will also be 0 !!!!
Can these steps be true always? Please reply... i am really confused.
_________________

Please give Kudos to the post if you liked.

Math Expert V
Joined: 02 Sep 2009
Posts: 53738
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

aniteshgmat1101 wrote:
Bunuel wrote:
8. What is the value of $$x^2+y^3$$?
(1) $$x^6+y^9=0$$
(2) $$27^{x^2}=\frac{3}{3^{3y^2+1}}$$

(1) $$x^6+y^9=0$$ --> $$(x^2)^3=(-y3)^3$$ ---> $$x^2=-y^3$$ --> $$x^2+y^3=0$$. Sufficient.

(2) $$27^{x^2}=\frac{3}{3^{3y^2+1}}$$ --> $$3^{3x^2}=\frac{3}{3^{3y^2}*3}$$ --> $$3^{3x^2}*3^{3y^2}=1$$ --> $$3^{3x^2+3y^2}=1$$ --> $$3x^2+3y^2=0$$ (the power of 3 must be zero in order this equation to hold true) --> $$x^2+y^2=0$$ the sum of two non-negative values is zero --> both $$x$$ and $$y$$ must be zero --> $$x=y=0$$ --> $$x^2+y^3=0$$. Sufficient.

Hi Bunuel,
I have a confusion. The problem statement with option no. 1 can be written as follows:
if a^3 + b^3 = 0 (1st option) then what is the value of a + b ? (where a = x^2 and b = y^3).
Now, we know that (a^3 + b^3) = (a+b) * (a^2 - a*b + b^2). (a^3 + b^3) has 2 factors.
And according to the explanation, since, (a^3 + b^3) =0 => a^3 = -(b^3) => a = -b => a + b = 0. So, whenever a^3 + b^3 is 0, a+b will also be 0 !!!!
Can these steps be true always? Please reply... i am really confused.

Yes, if a^3 + b^3 = 0, then a + b = 0.
_________________
Manager  B
Joined: 16 Jan 2013
Posts: 80
GMAT 1: 490 Q41 V18 GMAT 2: 610 Q45 V28 GPA: 2.75
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

Bunuel wrote:
4. If $$xyz\neq{0}$$ is $$(x^{-4})*(\sqrt{y})*(z^{-2})<0$$?
(1) $$\sqrt{y}>\sqrt{x^2}$$
(2) $$y^3>\frac{1}{z{^4}}$$

$$xyz\neq{0}$$ means that neither of unknown is equal to zero. Next, $$(x^{-4})*(\sqrt{y})*(z^{-2})=\frac{\sqrt{y}}{x^4*z^2}$$, so the question becomes: is $$\frac{\sqrt{y}}{x^4*z^2}<0$$? Since $$x^4$$ and $$z^2$$ are positive numbers then the question boils down whether $$\sqrt{y}<0$$, which is the same as whether $$y<0$$ (recall that odd roots have the same sign as the base of the root, for example: $$\sqrt{125} =5$$ and $$\sqrt{-64} =-4$$).

(1) $$\sqrt{y}>\sqrt{x^2}$$ --> as even root from positive number ($$x^2$$ in our case) is positive then $$\sqrt{y}>\sqrt{x^2}>0$$, (or which is the same $$y>0$$). Therefore answer to the original question is NO. Sufficient.

(2) $$y^3>\frac{1}{z{^4}}$$ --> the same here as $$\frac{1}{z{^4}}>0$$ then $$y^3>\frac{1}{z{^4}}>0$$, (or which is the same $$y>0$$). Therefore answer to the original question is NO. Sufficient.

Hello sir,
It was not mentioned in the question stem whether x, y and z are integers. In this question, if x, y and z were non-integers, wouldn't the procedure be different? Would you please explain?

Thanks! :D
_________________

CEO  S
Joined: 20 Mar 2014
Posts: 2624
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44 GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

1
Bunuel wrote:
4. If $$xyz\neq{0}$$ is $$(x^{-4})*(\sqrt{y})*(z^{-2})<0$$?
(1) $$\sqrt{y}>\sqrt{x^2}$$
(2) $$y^3>\frac{1}{z{^4}}$$

$$xyz\neq{0}$$ means that neither of unknown is equal to zero. Next, $$(x^{-4})*(\sqrt{y})*(z^{-2})=\frac{\sqrt{y}}{x^4*z^2}$$, so the question becomes: is $$\frac{\sqrt{y}}{x^4*z^2}<0$$? Since $$x^4$$ and $$z^2$$ are positive numbers then the question boils down whether $$\sqrt{y}<0$$, which is the same as whether $$y<0$$ (recall that odd roots have the same sign as the base of the root, for example: $$\sqrt{125} =5$$ and $$\sqrt{-64} =-4$$).

(1) $$\sqrt{y}>\sqrt{x^2}$$ --> as even root from positive number ($$x^2$$ in our case) is positive then $$\sqrt{y}>\sqrt{x^2}>0$$, (or which is the same $$y>0$$). Therefore answer to the original question is NO. Sufficient.

(2) $$y^3>\frac{1}{z{^4}}$$ --> the same here as $$\frac{1}{z{^4}}>0$$ then $$y^3>\frac{1}{z{^4}}>0$$, (or which is the same $$y>0$$). Therefore answer to the original question is NO. Sufficient.

Hello sir,
It was not mentioned in the question stem whether x, y and z are integers. In this question, if x, y and z were non-integers, wouldn't the procedure be different? Would you please explain?

Thanks! :D

This question has a trick in that it will be true for both integers and non-integers alike:

1. x^2 > 0 no matter x=integer or not. Similarly, $$\sqrt{x^2}$$ will be positive for x=integer or non-integer. Try it out. Let x = 1 , $$\sqrt{1^2} = 1 > 0$$ and if x= 0.5, $$\sqrt{0.5^2} = 0.707 >0.$$

Thus it does not matter whether you take x = integer or a fraction.

Coming back to the question,

Per statement 1, as mentioned above, $$\sqrt{1^2} = 1 > 0$$ ---> $$\sqrt{y} > 0$$ -----> y >0. Thus the statement $$(x^{-4})*(\sqrt{y})*(z^{-2})<0$$ will be false. Thus this statement is sufficient.

Per statement 2, z^4 > 0 whether z =1 or z= 0.5 ----> 1/(z^4) >0 ----> y^3 >0---> y>0 (satisfies for both, y =1 or y = 0.25). Thus the given statement $$(x^{-4})*(\sqrt{y})*(z^{-2})<0$$ will be false. Thus this statement is sufficient.

As both the given statements are sufficient, the correct answer is D.

whenever you are in doubt, always test the given statements/conditions with different values (integers, fractions , etc). This way you will get a better picture.
Manager  B
Joined: 16 Jan 2013
Posts: 80
GMAT 1: 490 Q41 V18 GMAT 2: 610 Q45 V28 GPA: 2.75
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

Bunuel wrote:
8. What is the value of $$x^2+y^3$$?
(1) $$x^6+y^9=0$$
(2) $$27^{x^2}=\frac{3}{3^{3y^2+1}}$$

(1) $$x^6+y^9=0$$ --> $$(x^2)^3=(-y3)^3$$ ---> $$x^2=-y^3$$ --> $$x^2+y^3=0$$. Sufficient.

(2) $$27^{x^2}=\frac{3}{3^{3y^2+1}}$$ --> $$3^{3x^2}=\frac{3}{3^{3y^2}*3}$$ --> $$3^{3x^2}*3^{3y^2}=1$$ --> $$3^{3x^2+3y^2}=1$$ --> $$3x^2+3y^2=0$$ (the power of 3 must be zero in order this equation to hold true) --> $$x^2+y^2=0$$ the sum of two non-negative values is zero --> both $$x$$ and $$y$$ must be zero --> $$x=y=0$$ --> $$x^2+y^3=0$$. Sufficient.

Hello Bunuel,
Nice explanations. I can't figure out why exponents over exponents act different sometimes. When to multiply and when to add them? Here, why isn't (x^2)^3 = x^8 and (-y^3)^3 = -y^27, please explain? As we know, 3^3^3 = 27.

Could you please explicate?
_________________

Math Expert V
Joined: 02 Sep 2009
Posts: 53738
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

1
Bunuel wrote:
8. What is the value of $$x^2+y^3$$?
(1) $$x^6+y^9=0$$
(2) $$27^{x^2}=\frac{3}{3^{3y^2+1}}$$

(1) $$x^6+y^9=0$$ --> $$(x^2)^3=(-y3)^3$$ ---> $$x^2=-y^3$$ --> $$x^2+y^3=0$$. Sufficient.

(2) $$27^{x^2}=\frac{3}{3^{3y^2+1}}$$ --> $$3^{3x^2}=\frac{3}{3^{3y^2}*3}$$ --> $$3^{3x^2}*3^{3y^2}=1$$ --> $$3^{3x^2+3y^2}=1$$ --> $$3x^2+3y^2=0$$ (the power of 3 must be zero in order this equation to hold true) --> $$x^2+y^2=0$$ the sum of two non-negative values is zero --> both $$x$$ and $$y$$ must be zero --> $$x=y=0$$ --> $$x^2+y^3=0$$. Sufficient.

Hello Bunuel,
Nice explanations. I can't figure out why exponents over exponents act different sometimes. When to multiply and when to add them? Here, why isn't (x^2)^3 = x^8 and (-y^3)^3 = -y^27, please explain? As we know, 3^3^3 = 27.

Could you please explicate?

Please read the second post of this topic: new-tough-and-tricky-exponents-and-roots-questions-125967.html#p1027927

$$(a^m)^n=a^{mn}$$

$$a^m^n=a^{(m^n)}$$ and not $$(a^m)^n$$ (if exponentiation is indicated by stacked symbols, the rule is to work from the top down)
_________________
Retired Moderator B
Joined: 05 Jul 2006
Posts: 1713
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

Bunuel wrote:
10. If $$x$$ and $$y$$ are non-negative integers and $$x+y>0$$ is $$(x+y)^{xy}$$ an even integer?
(1) $$2^{x-y}=\sqrt[(x+y)]{16}$$
(2) $$2^x+3^y=\sqrt[(x+y)]{25}$$

(1) $$2^{x-y}=\sqrt[(x+y)]{16}$$ --> $$2^{x-y}=16^{\frac{1}{x+y}}=2^{\frac{4}{x+y}}$$ --> equate the powers: $$x-y=\frac{4}{x+y}$$ --> $$(x-y)(x+y)=4$$.

Since both $$x$$ and $$y$$ are integers (and $$x+y>0$$) then $$x-y=2$$ and $$x+y=2$$ --> $$x=2$$ and $$y=0$$ --> $$(x+y)^{xy}=2^0=1=odd$$, so the answer to the question is No. Sufficient. (Note that $$x-y=1$$ and $$x+y=4$$ --> $$x=2.5$$ and $$y=1.5$$ is not a valid scenario (solution) as both unknowns must be integers)

(2) $$2^x+3^y=\sqrt[(x+y)]{25}$$ --> obviously $$\sqrt[(x+y)]{25}$$ must be an integer (since $$2^x+3^y=integer$$) and as $$x+y=integer$$ then the only solution is $$\sqrt[(x+y)]{25}=\sqrt{25}=5$$ --> $$x+y=2$$. So, $$2^x+3^y=5$$ --> two scenarios are possible:
A. $$x=2$$ and $$y=0$$ (notice that $$x+y=2$$ holds true) --> $$2^x+3^y=2^2+3^0=5$$, and in this case: $$(x+y)^{xy}=2^0=1=odd$$;
B. $$x=1$$ and $$y=1$$ (notice that $$x+y=2$$ holds true) --> $$2^x+3^y=2^1+3^1=5$$, and in this case: $$(x+y)^{xy}=2^1=2=even$$.

Two different answers. Not sufficient.

if y = 0 then statement 1 become 2^x = 2^2x , i.e. x = 2x , i.e. x = 0 not 2 , what am i doing wrong here if you plz??
Math Expert V
Joined: 02 Sep 2009
Posts: 53738
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

yezz wrote:
Bunuel wrote:
10. If $$x$$ and $$y$$ are non-negative integers and $$x+y>0$$ is $$(x+y)^{xy}$$ an even integer?
(1) $$2^{x-y}=\sqrt[(x+y)]{16}$$
(2) $$2^x+3^y=\sqrt[(x+y)]{25}$$

(1) $$2^{x-y}=\sqrt[(x+y)]{16}$$ --> $$2^{x-y}=16^{\frac{1}{x+y}}=2^{\frac{4}{x+y}}$$ --> equate the powers: $$x-y=\frac{4}{x+y}$$ --> $$(x-y)(x+y)=4$$.

Since both $$x$$ and $$y$$ are integers (and $$x+y>0$$) then $$x-y=2$$ and $$x+y=2$$ --> $$x=2$$ and $$y=0$$ --> $$(x+y)^{xy}=2^0=1=odd$$, so the answer to the question is No. Sufficient. (Note that $$x-y=1$$ and $$x+y=4$$ --> $$x=2.5$$ and $$y=1.5$$ is not a valid scenario (solution) as both unknowns must be integers)

(2) $$2^x+3^y=\sqrt[(x+y)]{25}$$ --> obviously $$\sqrt[(x+y)]{25}$$ must be an integer (since $$2^x+3^y=integer$$) and as $$x+y=integer$$ then the only solution is $$\sqrt[(x+y)]{25}=\sqrt{25}=5$$ --> $$x+y=2$$. So, $$2^x+3^y=5$$ --> two scenarios are possible:
A. $$x=2$$ and $$y=0$$ (notice that $$x+y=2$$ holds true) --> $$2^x+3^y=2^2+3^0=5$$, and in this case: $$(x+y)^{xy}=2^0=1=odd$$;
B. $$x=1$$ and $$y=1$$ (notice that $$x+y=2$$ holds true) --> $$2^x+3^y=2^1+3^1=5$$, and in this case: $$(x+y)^{xy}=2^1=2=even$$.

Two different answers. Not sufficient.

if y = 0 then statement 1 become 2^x = 2^2x , i.e. x = 2x , i.e. x = 0 not 2 , what am i doing wrong here if you plz??

If y=0, then we'll have $$2^{x}=\sqrt[x]{16}$$:

$$2^x=2^{\frac{4}{x}}$$;

$$x=\frac{4}{x}$$;

$$x^2=4$$

$$x=2$$ or $$x=-2$$ (discard since x is non-negative).
_________________
Intern  Joined: 31 Oct 2017
Posts: 4
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

Bunuel wrote:
4. If $$xyz\neq{0}$$ is $$(x^{-4})*(\sqrt{y})*(z^{-2})<0$$?
(1) $$\sqrt{y}>\sqrt{x^2}$$
(2) $$y^3>\frac{1}{z{^4}}$$

$$xyz\neq{0}$$ means that neither of unknown is equal to zero. Next, $$(x^{-4})*(\sqrt{y})*(z^{-2})=\frac{\sqrt{y}}{x^4*z^2}$$, so the question becomes: is $$\frac{\sqrt{y}}{x^4*z^2}<0$$? Since $$x^4$$ and $$z^2$$ are positive numbers then the question boils down whether $$\sqrt{y}<0$$, which is the same as whether $$y<0$$ (recall that odd roots have the same sign as the base of the root, for example: $$\sqrt{125} =5$$ and $$\sqrt{-64} =-4$$).

(1) $$\sqrt{y}>\sqrt{x^2}$$ --> as even root from positive number ($$x^2$$ in our case) is positive then $$\sqrt{y}>\sqrt{x^2}>0$$, (or which is the same $$y>0$$). Therefore answer to the original question is NO. Sufficient.

(2) $$y^3>\frac{1}{z{^4}}$$ --> the same here as $$\frac{1}{z{^4}}>0$$ then $$y^3>\frac{1}{z{^4}}>0$$, (or which is the same $$y>0$$). Therefore answer to the original question is NO. Sufficient.

I could not understand the highlighted part. I also don't understand when to consider both positive and negative solutions and when to consider only positive solution when dealing with roots. Please explain.
Math Expert V
Joined: 02 Sep 2009
Posts: 53738
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

1
scrumptious829 wrote:
Bunuel wrote:
4. If $$xyz\neq{0}$$ is $$(x^{-4})*(\sqrt{y})*(z^{-2})<0$$?
(1) $$\sqrt{y}>\sqrt{x^2}$$
(2) $$y^3>\frac{1}{z{^4}}$$

$$xyz\neq{0}$$ means that neither of unknown is equal to zero. Next, $$(x^{-4})*(\sqrt{y})*(z^{-2})=\frac{\sqrt{y}}{x^4*z^2}$$, so the question becomes: is $$\frac{\sqrt{y}}{x^4*z^2}<0$$? Since $$x^4$$ and $$z^2$$ are positive numbers then the question boils down whether $$\sqrt{y}<0$$, which is the same as whether $$y<0$$ (recall that odd roots have the same sign as the base of the root, for example: $$\sqrt{125} =5$$ and $$\sqrt{-64} =-4$$).

(1) $$\sqrt{y}>\sqrt{x^2}$$ --> as even root from positive number ($$x^2$$ in our case) is positive then $$\sqrt{y}>\sqrt{x^2}>0$$, (or which is the same $$y>0$$). Therefore answer to the original question is NO. Sufficient.

(2) $$y^3>\frac{1}{z{^4}}$$ --> the same here as $$\frac{1}{z{^4}}>0$$ then $$y^3>\frac{1}{z{^4}}>0$$, (or which is the same $$y>0$$). Therefore answer to the original question is NO. Sufficient.

I could not understand the highlighted part. I also don't understand when to consider both positive and negative solutions and when to consider only positive solution when dealing with roots. Please explain.

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

$$\sqrt{9} = 3$$, NOT +3 or -3;
$$\sqrt{16} = 2$$, NOT +2 or -2;

Notice that in contrast, the equation $$x^2 = 9$$ has TWO solutions, +3 and -3. Because $$x^2 = 9$$ means that $$x =-\sqrt{9}=-3$$ or $$x=\sqrt{9}=3$$.
_________________
Manager  G
Joined: 02 Jun 2015
Posts: 191
Location: Ghana
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

Bunuel wrote:
4. If $$xyz\neq{0}$$ is $$(x^{-4})*(\sqrt{y})*(z^{-2})<0$$?
(1) $$\sqrt{y}>\sqrt{x^2}$$
(2) $$y^3>\frac{1}{z{^4}}$$

$$xyz\neq{0}$$ means that neither of unknown is equal to zero. Next, $$(x^{-4})*(\sqrt{y})*(z^{-2})=\frac{\sqrt{y}}{x^4*z^2}$$, so the question becomes: is $$\frac{\sqrt{y}}{x^4*z^2}<0$$? Since $$x^4$$ and $$z^2$$ are positive numbers then the question boils down whether $$\sqrt{y}<0$$, which is the same as whether $$y<0$$ (recall that odd roots have the same sign as the base of the root, for example: $$\sqrt{125} =5$$ and $$\sqrt{-64} =-4$$).

(1) $$\sqrt{y}>\sqrt{x^2}$$ --> as even root from positive number ($$x^2$$ in our case) is positive then $$\sqrt{y}>\sqrt{x^2}>0$$, (or which is the same $$y>0$$). Therefore answer to the original question is NO. Sufficient.

(2) $$y^3>\frac{1}{z{^4}}$$ --> the same here as $$\frac{1}{z{^4}}>0$$ then $$y^3>\frac{1}{z{^4}}>0$$, (or which is the same $$y>0$$). Therefore answer to the original question is NO. Sufficient.

Hi Bunuel,

Can you kindly help me understand why 4th root of x^2 > 0 and not 4th root of x^2 = 4th root of |x| similar to square of x^2 = square root of |x| as shown in the step 2 of your solution to the question @ https://gmatclub.com/forum/what-is-x-2- ... 99157.html?

Thank you
_________________

Kindly press kudos if you find my post helpful

Math Expert V
Joined: 02 Sep 2009
Posts: 53738
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

1
duahsolo wrote:
Bunuel wrote:
4. If $$xyz\neq{0}$$ is $$(x^{-4})*(\sqrt{y})*(z^{-2})<0$$?
(1) $$\sqrt{y}>\sqrt{x^2}$$
(2) $$y^3>\frac{1}{z{^4}}$$

$$xyz\neq{0}$$ means that neither of unknown is equal to zero. Next, $$(x^{-4})*(\sqrt{y})*(z^{-2})=\frac{\sqrt{y}}{x^4*z^2}$$, so the question becomes: is $$\frac{\sqrt{y}}{x^4*z^2}<0$$? Since $$x^4$$ and $$z^2$$ are positive numbers then the question boils down whether $$\sqrt{y}<0$$, which is the same as whether $$y<0$$ (recall that odd roots have the same sign as the base of the root, for example: $$\sqrt{125} =5$$ and $$\sqrt{-64} =-4$$).

(1) $$\sqrt{y}>\sqrt{x^2}$$ --> as even root from positive number ($$x^2$$ in our case) is positive then $$\sqrt{y}>\sqrt{x^2}>0$$, (or which is the same $$y>0$$). Therefore answer to the original question is NO. Sufficient.

(2) $$y^3>\frac{1}{z{^4}}$$ --> the same here as $$\frac{1}{z{^4}}>0$$ then $$y^3>\frac{1}{z{^4}}>0$$, (or which is the same $$y>0$$). Therefore answer to the original question is NO. Sufficient.

Hi Bunuel,

Can you kindly help me understand why 4th root of x^2 > 0 and not 4th root of x^2 = 4th root of |x| similar to square of x^2 = square root of |x| as shown in the step 2 of your solution to the question @ https://gmatclub.com/forum/what-is-x-2- ... 99157.html?

Thank you

Even roots from a non-negative number is non-negative, so 0 or positive: $$\sqrt[even]{something}\geq 0$$, not matter what that "something" is (provided it's positive or 0 because even roots from negative numbers are not defined for the GMAT) )

Next, $$\sqrt{x^2}=\sqrt{\sqrt{x^2}}=\sqrt{|x|}$$, which must be positive, so (1) basically says that $$\sqrt{y}>0$$, or that y > 0.

Finally, if it were
_________________
Non-Human User Joined: 09 Sep 2013
Posts: 10153
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________ Re: NEW!!! Tough and tricky exponents and roots questions   [#permalink] 18 Feb 2019, 04:11

Go to page   Previous    1   2   3   [ 56 posts ]

Display posts from previous: Sort by

# NEW!!! Tough and tricky exponents and roots questions

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.  