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If −1 < x < 0 and 0 < y < 1
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25 Apr 2017, 05:09
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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!
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Re: If −1 < x < 0 and 0 < y < 1
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25 Apr 2017, 05:22
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
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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
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Re: If −1 < x < 0 and 0 < y < 1
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25 Apr 2017, 05:42
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 ?
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Re: If −1 < x < 0 and 0 < y < 1
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25 Apr 2017, 06:00
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
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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
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Re: If −1 < x < 0 and 0 < y < 1
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25 Apr 2017, 07:58
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)
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If −1 < x < 0 and 0 < y < 1
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25 Apr 2017, 08:25
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



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Re: If −1 < x < 0 and 0 < y < 1
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25 Apr 2017, 10:07
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.
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Re: If −1 < x < 0 and 0 < y < 1
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26 Apr 2017, 10:26
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!
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Re: If −1 < x < 0 and 0 < y < 1
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27 Apr 2017, 04:35
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), \(\frac{x^3}{y} > \frac{x}{y}\)
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Re: If −1 < x < 0 and 0 < y < 1
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27 Apr 2017, 21:15
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.



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Re: If −1 < x < 0 and 0 < y < 1
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28 Apr 2017, 14:08
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
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Re: If −1 < x < 0 and 0 < y < 1
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09 May 2017, 18:52
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



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Re: If −1 < x < 0 and 0 < y < 1
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