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If x and y are non zero numbers less than 1 is (y^4 - x^4)>(y^3 - x^5)

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Joined: 21 May 2016
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If x and y are non zero numbers less than 1 is (y^4 - x^4)>(y^3 - x^5)  [#permalink]

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Updated on: 14 Aug 2018, 04:16
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Difficulty:

95% (hard)

Question Stats:

27% (02:34) correct 73% (02:23) wrong based on 133 sessions

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If x and y are non zero numbers less than 1, is $$(y^4 - x^4) > (y^3 - x^5)$$ ?

(1) y > 0
(2) y > |x|

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Originally posted by a70 on 03 Aug 2018, 11:15.
Last edited by a70 on 14 Aug 2018, 04:16, edited 2 times in total.
Intern
Joined: 22 Jul 2018
Posts: 5
Re: If x and y are non zero numbers less than 1 is (y^4 - x^4)>(y^3 - x^5)  [#permalink]

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04 Aug 2018, 08:51
3
3
Rearrange the terms
y^4 -y^3 > -x^5 + x^4
factor, so the question now is:
is y^3(y - 1) > - (x^4) (x - 1)?

1) y>0
so y is between 0 and 1
This means the LHS is negative (because (y-1) is negative, and y^3 is positive)
RHS is always positive (because (x-1) is always negative since x is less than 1, -(x^4) is always negative, so their product is always positive)
If LHS is always negative and RHS is always positive, this means that RHS>LHS, so statement 1 is sufficient to determine that the inequality is false.

2) y> |x|
This means y is positive i.e. y>0
The statement is therefore equivalent to statement 1, and we can use the same logic to determine that the inequality is false

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Re: If x and y are non zero numbers less than 1 is (y^4 - x^4)>(y^3 - x^5)  [#permalink]

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03 Aug 2018, 11:59
2
ankit7055 wrote:
If x and y are non zero numbers less than 1, is (y^4 - x^4) > (y^3 - x^5) ?

(1) y > 0
(2) y > |x|

1 > x
1 > y

(1) y > 0

make y= $$\frac{1}{2}$$

1 > y > 0

($$.5^4$$ - $$x^4$$) > ($$.5^3$$ - $$x^5$$) ---> ($$\frac{1}{16}$$ - $$x^4$$) > ($$\frac{1}{8}$$ - $$x^5$$)

Make x=-1 ---> ($$\frac{1}{16}$$ - 1) > ($$\frac{1}{8}$$ + 1) Answer is NO

make x=$$\frac{1}{2}$$ ---> ($$\frac{1}{16}$$ - $$\frac{1}{16}$$) > ($$\frac{1}{8}$$ - $$\frac{1}{32}$$) Answer is NO

Sufficient

(2) y > |x|

Same applies from statement 1.

NO. Sufficient.

PS - I got this wrong at first 700 level IMO.
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Re: If x and y are non zero numbers less than 1 is (y^4 - x^4)>(y^3 - x^5)  [#permalink]

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03 Aug 2018, 18:47
ankit7055 wrote:
If x and y are non zero numbers less than 1, is (y^4 - x^4) > (y^3 - x^5) ?

(1) y > 0
(2) y > |x|

Hi

What is the source if the question. It is really good.
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Joined: 22 Feb 2018
Posts: 426
Re: If x and y are non zero numbers less than 1 is (y^4 - x^4)>(y^3 - x^5)  [#permalink]

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04 Aug 2018, 10:21
3
1
ankit7055 wrote:
If x and y are non zero numbers less than 1, is (y^4 - x^4) > (y^3 - x^5) ?

(1) y > 0
(2) y > |x|

OA:D

Is $$(y^4 - x^4) > (y^3 - x^5)$$?
Is $$y^3(y-1)>x^4(1-x)$$?

Statement 1 : $$y > 0$$
$$0<y<1$$
L.H.S $$y^3(y-1)$$ would always be negative

For $$x$$, there can be 2 cases,
1) $$0<x<1$$
2) $$x<0$$
R.H.S will be always positive, as $$x^4$$ is +ve, $$(1-x)$$ would give +ve value in both of the cases
So There is definite answer to Is $$(y^4 - x^4) > (y^3 - x^5)$$? , That is No
Statement 1 alone is sufficient

Statement 2 : $$y > |x|$$
it means $$y$$ is positive, i,e $$y>0$$ same as statement 1
So There is definite answer to Is $$(y^4 - x^4) > (y^3 - x^5)$$? , That is No
Statement 2 alone is sufficient
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Re: If x and y are non zero numbers less than 1 is (y^4 - x^4)>(y^3 - x^5)  [#permalink]

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04 Feb 2019, 09:37
Is (y4−x4)>(y3−x5)?

Rearrange:
Is y3(y−1)>x4(1−x)?

x4 is always positive (even power of x)

(1-x) is also always positive (plug in any value of x less than 1)

Therefore, x4(1-x) is always positive.

(y-1) is always negative, as y is less than 1.

St.1 says that y is positive.
Therefore, y(y-1) is negative and hence smaller than positive term x4(1-x)

St.2 says that y is greater than an absolute expression. Hence, Statement 2 provides the same information as st.1

Hence, Ans D

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Re: If x and y are non zero numbers less than 1 is (y^4 - x^4)>(y^3 - x^5)   [#permalink] 04 Feb 2019, 09:37
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