Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 19 Aug 2016
Posts: 153
Location: India
GPA: 3.82

If −1 < x < 0 and 0 < y < 1
[#permalink]
Show Tags
25 Apr 2017, 06:09
Question Stats:
51% (01:16) correct 49% (00:56) wrong based on 271 sessions
HideShow timer Statistics
If −1 < x < 0 and 0 < y < 1, which of the following has the least value? A. x/y B. (x/y)^2 C. x^2/y D. x^3/y E. x^2/y^3 I know the answer. I just have one confusion as to how a negative sign to the power 3 could be greater than the negative sign to the power 1. Just as in this question it is implied that x < x^3. If you do it simply isn't 3 > (3)^3. Kindly someone help clarify my confusion. Thanks!
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Consider giving me Kudos if you find my posts useful, challenging and helpful!



Math Expert
Joined: 02 Aug 2009
Posts: 6796

Re: If −1 < x < 0 and 0 < y < 1
[#permalink]
Show Tags
25 Apr 2017, 06:22
ashikaverma13 wrote: If −1 < x < 0 and 0 < y < 1, which of the following has the least value?
A. x/y
B. (x/y)^2
C. x^2/y
D. x^3/y
E. x^2/y^3
I know the answer. I just have one confusion as to how a negative sign to the power 3 could be greater than the negative sign to the power 1. Just as in this question it is implied that x < x^3. If you do it simply isn't 3 > (3)^3.
Kindly someone help clarify my confusion. Thanks! Hi.. It will be true whenever the value is between 0 and 1... Say the value is 1/2... (1/2)^3=1/8.. And 1/8 >1/2
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
GMAT online Tutor



Current Student
Joined: 12 Aug 2015
Posts: 2648

Re: If −1 < x < 0 and 0 < y < 1
[#permalink]
Show Tags
25 Apr 2017, 06:42
chetan2u wrote: ashikaverma13 wrote: If −1 < x < 0 and 0 < y < 1, which of the following has the least value?
A. x/y
B. (x/y)^2
C. x^2/y
D. x^3/y
E. x^2/y^3
I know the answer. I just have one confusion as to how a negative sign to the power 3 could be greater than the negative sign to the power 1. Just as in this question it is implied that x < x^3. If you do it simply isn't 3 > (3)^3.
Kindly someone help clarify my confusion. Thanks! Hi.. It will be true whenever the value is between 0 and 1... Say the value is 1/2... (1/2)^3=1/8.. And 1/8 >1/2 Hi chetan2u I am getting an E here.
Here is why =>
We can all agree that B,C,D are out as B and D are positive and in C x^3<x .
Let us compare A and E now.
Let us assume that A<E Hence> x/y<x^2/y^3 y^2/x<1 Which is always true.
What am i missing here ?
_________________
MBA Financing: INDIAN PUBLIC BANKS vs PRODIGY FINANCE! Getting into HOLLYWOOD with an MBA! The MOST AFFORDABLE MBA programs!STONECOLD's BRUTAL Mock Tests for GMATQuant(700+)AVERAGE GRE Scores At The Top Business Schools!



Math Expert
Joined: 02 Aug 2009
Posts: 6796

Re: If −1 < x < 0 and 0 < y < 1
[#permalink]
Show Tags
25 Apr 2017, 07:00
stonecold wrote: chetan2u wrote: ashikaverma13 wrote: If −1 < x < 0 and 0 < y < 1, which of the following has the least value?
A. x/y
B. (x/y)^2
C. x^2/y
D. x^3/y
E. x^2/y^3
I know the answer. I just have one confusion as to how a negative sign to the power 3 could be greater than the negative sign to the power 1. Just as in this question it is implied that x < x^3. If you do it simply isn't 3 > (3)^3.
Kindly someone help clarify my confusion. Thanks! Hi.. It will be true whenever the value is between 0 and 1... Say the value is 1/2... (1/2)^3=1/8.. And 1/8 >1/2 Hi chetan2u I am getting an E here.
Here is why =>
We can all agree that B,C,D are out as B and D are positive and in C x^3<x .
Let us compare A and E now.
Let us assume that A<E Hence> x/y<x^2/y^3 y^2/x<1 Which is always true.
What am i missing here ? Hi.. never cross multiply negative values.. x is NEGATIVE, y is positive...... So x/y is /+ or a negative quantity.. Also x^2 will be positive and y^3 is also POSITIVE.. x^2/y^3 means */+*+*+ , this will be positive.. So A is NEGATIVE and E is positive.. So A<E.. Hope it helps
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
GMAT online Tutor



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8282
Location: Pune, India

Re: If −1 < x < 0 and 0 < y < 1
[#permalink]
Show Tags
25 Apr 2017, 08:58
ashikaverma13 wrote: If −1 < x < 0 and 0 < y < 1, which of the following has the least value?
A. x/y
B. (x/y)^2
C. x^2/y
D. x^3/y
E. x^2/y^3
I know the answer. I just have one confusion as to how a negative sign to the power 3 could be greater than the negative sign to the power 1. Just as in this question it is implied that x < x^3. If you do it simply isn't 3 > (3)^3.
Kindly someone help clarify my confusion. Thanks! Numbers on the number line behave differently between these ranges: n < 1 1 < n < 0 0 < n < 1 n > 1 Take one number from each range and find the relation between n, n^2 and n^3. The way I would approach this question: x is negative so the smallest number would have negative x. Ignore options (B), (C) and (E). Out of (A) and (D), both will be negative and both have y in the denominator. The absolute value of x will be more than the absolute value of x^3 so x will be more negative than x^3. So x/y will be smaller than x^3/y. Answer (A)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



PS Forum Moderator
Joined: 25 Feb 2013
Posts: 1212
Location: India
GPA: 3.82

If −1 < x < 0 and 0 < y < 1
[#permalink]
Show Tags
25 Apr 2017, 09:25
ashikaverma13 wrote: If −1 < x < 0 and 0 < y < 1, which of the following has the least value?
A. x/y
B. (x/y)^2
C. x^2/y
D. x^3/y
E. x^2/y^3
I know the answer. I just have one confusion as to how a negative sign to the power 3 could be greater than the negative sign to the power 1. Just as in this question it is implied that x < x^3. If you do it simply isn't 3 > (3)^3.
Kindly someone help clarify my confusion. Thanks! As per the inequality, x is negative and y is positive. So to get the least value we will have to choose the operation where x is negative. Options  B, C & E have x^2, which is positive hence they are out. Option A & D are the contenders. Easiest and shortest way to solve could be substitution  let x = 0.5 & y =0.5. Now evaluate option A & D Option A: x/y = 0.5/0.5 = 1 Option D: x^3/y = (0.5)^3/0.5 = 0.125/0.5 =  0.25 Clearly Option A is least as 1<0.25 Just FYI  as x is negative and has to be a decimal number greater than 1, so higher powers will of x will move it towards 0. therefore higher powers of x will be greater than x itself, for example (0.5)^11 will be 0.00049 which is greater than 0.5



Manager
Joined: 19 Aug 2016
Posts: 153
Location: India
GPA: 3.82

Re: If −1 < x < 0 and 0 < y < 1
[#permalink]
Show Tags
25 Apr 2017, 11:07
chetan2u wrote: ashikaverma13 wrote: If −1 < x < 0 and 0 < y < 1, which of the following has the least value?
A. x/y
B. (x/y)^2
C. x^2/y
D. x^3/y
E. x^2/y^3
I know the answer. I just have one confusion as to how a negative sign to the power 3 could be greater than the negative sign to the power 1. Just as in this question it is implied that x < x^3. If you do it simply isn't 3 > (3)^3.
Kindly someone help clarify my confusion. Thanks! Hi.. It will be true whenever the value is between 0 and 1... Say the value is 1/2... (1/2)^3=1/8.. And 1/8 >1/2 Thanks for your response. I guess the same rule applies here that applies for numbers between 0 and 1 i.e. if x is 0<x<1 then x^2<x. But conversely for negative number it'll go like x^3<x<x^2. Kindly correct me if I am wrong.
_________________
Consider giving me Kudos if you find my posts useful, challenging and helpful!



Manager
Joined: 19 Aug 2016
Posts: 153
Location: India
GPA: 3.82

Re: If −1 < x < 0 and 0 < y < 1
[#permalink]
Show Tags
26 Apr 2017, 11:26
VeritasPrepKarishma wrote: The absolute value of x will be more than the absolute value of x^3 so x will be more negative than x^3. So x/y will be smaller than x^3/y. Answer (A)
Could you elaborate more on the above? Thanks!
_________________
Consider giving me Kudos if you find my posts useful, challenging and helpful!



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8282
Location: Pune, India

Re: If −1 < x < 0 and 0 < y < 1
[#permalink]
Show Tags
27 Apr 2017, 05:35
ashikaverma13 wrote: VeritasPrepKarishma wrote: The absolute value of x will be more than the absolute value of x^3 so x will be more negative than x^3. So x/y will be smaller than x^3/y. Answer (A)
Could you elaborate more on the above? Thanks! This is related to number properties. When x > 1 \(x = 3\) \(x^3 = 27\) (greater than x) When 0 < x < 1 \(x = \frac{1}{3}\) \(x^3 = \frac{1}{27}\) (smaller than x) When 1 < x < 0 \(x = \frac{1}{3}\) \(x^3 = \frac{1}{27}\) (absolute value of x^3 is smaller than absolute value of x. Since both are negative, x^3 > x) When x < 1 \(x = 3\) \(x^3 = 27\) (absolute value of x^3 is greater than absolute value of x. Since both are negative, x^3 < x) Coming back to our question, when 1 < x < 0 \(x^3 > x\) Dividing both sides by y (which is positive so the inequality sign stays the same), \(\frac{x^3}{y} > \frac{x}{y}\)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Senior Manager
Joined: 24 Oct 2016
Posts: 285
Location: India
Concentration: Finance, International Business
GPA: 3.96
WE: Human Resources (Retail Banking)

Re: If −1 < x < 0 and 0 < y < 1
[#permalink]
Show Tags
27 Apr 2017, 22:15
VeritasPrepKarishma wrote: ashikaverma13 wrote: VeritasPrepKarishma wrote: The absolute value of x will be more than the absolute value of x^3 so x will be more negative than x^3. So x/y will be smaller than x^3/y. Answer (A)
Could you elaborate more on the above? Thanks! This is related to number properties. When x > 1 \(x = 3\) \(x^3 = 27\) (greater than x) When 0 < x < 1 \(x = \frac{1}{3}\) \(x^3 = \frac{1}{27}\) (smaller than x) When 1 < x < 0 \(x = \frac{1}{3}\) \(x^3 = \frac{1}{27}\) (absolute value of x^3 is smaller than absolute value of x. Since both are negative, x^3 > x) When x < 1 \(x = 3\) \(x^3 = 27\) (absolute value of x^3 is greater than absolute value of x. Since both are negative, x^3 < x) Coming back to our question, when 1 < x < 0 \(x^3 > x\) Dividing both sides by y (which is positive so the inequality sign stays the same), \(\frac{x^3}{y} > \frac{x}{y}\) hi, karishma , although your explanation is quite clear , but how i usually decide which is grater or less, such as say x=1/3 and x^3= 1/27 then x>x^3 now for the ve one x=1/3 then x^3= 1/27 so signs change 1/3 < 1/27 is my way of understanding is correct ? please comment . many thanks.



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 3516
Location: United States (CA)

Re: If −1 < x < 0 and 0 < y < 1
[#permalink]
Show Tags
28 Apr 2017, 15:08
ashikaverma13 wrote: If −1 < x < 0 and 0 < y < 1, which of the following has the least value?
A. x/y
B. (x/y)^2
C. x^2/y
D. x^3/y
E. x^2/y^3 Notice that x is negative and y is positive and that all of the answer choices involve powers or division. Thus, to determine the expression that has the least value, we want the expression to be negative. We see that choices B, C, and E are all positive, so we can eliminate these choices. Choices A and D are both negative. To determine the one with the least value, let’s use numbers for x and y. Let’s say x = ½ and y = ½: A. x/y = (½)/(½) = 1 D. x^3/y = (½)^3/(½) = (⅛)/(½) = (⅛) x 2 = ¼ We see that choice A has the least value. Answer: A
_________________
Scott WoodburyStewart
Founder and CEO
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



Director
Joined: 12 Nov 2016
Posts: 759
Location: United States
GPA: 2.66

Re: If −1 < x < 0 and 0 < y < 1
[#permalink]
Show Tags
09 May 2017, 19:52
ashikaverma13 wrote: If −1 < x < 0 and 0 < y < 1, which of the following has the least value?
A. x/y
B. (x/y)^2
C. x^2/y
D. x^3/y
E. x^2/y^3
I know the answer. I just have one confusion as to how a negative sign to the power 3 could be greater than the negative sign to the power 1. Just as in this question it is implied that x < x^3. If you do it simply isn't 3 > (3)^3.
Kindly someone help clarify my confusion. Thanks! The most efficient way to solve this problem is possibly through both plugging in and testing values and process of elimination (made possible through knowing properties such as (x)^2 is positive) B, C and E all result in positive numbers because the exponent is even so the only choices are A & D If we use (.5) and (.5) .5/.5= 1 (.5)^3/ (.5) (.0125)/(.5) = .025 Hence "A" is correct



NonHuman User
Joined: 09 Sep 2013
Posts: 8105

Re: If −1 < x < 0 and 0 < y < 1
[#permalink]
Show Tags
25 Jun 2018, 11:32
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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: If −1 < x < 0 and 0 < y < 1 &nbs
[#permalink]
25 Jun 2018, 11:32






