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

Question

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|

Anshkarha · Accepted Answer

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

Hence, the answer is D

Princ · Answer

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^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

MikeScarn · Answer

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. Answer: D PS - I got this wrong at first

700 level IMO.

Mo2men · Answer

Hi

What is the source if the question. It is really good.

Shobhit7 · Answer

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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TheNightKing · Answer

Hello,
I am kind of confused with what the question is saying.
Given: x and y are non zero numbers less than 1 which means (at least as per me)  x<=-1  and  y<=-1 . Then how can statement 1 says  y>0 ? What am I missing?

Sorry may be a stupid question.
Thank you!

Anshkarha · Answer

0.5 is a non zero number less than 1

Posted from my mobile device

Kinshook · Answer

Given: x and y are non zero numbers less than 1 Asked: Is (y^4 - x^4) > (y^3 - x^5) ? Q: y^4 - y^3 > x^4 - x^5? Q: -y^3(1-y) > x^4(1-x)? Since x and y are non-zero numbers <1: 1-y>0; 1-x>0; x^4>0 Q: -y^3> x^4(1-x)/(1-y)>0? Q: y<0? (1) y > 0 y is not <0 SUFFICIENT (2) y > |x| y>|x|>0 SUFFICIENT IMO D

Kinshook · Answer

a70

TheNightKing

x & y are non zero numbers less than 1 means
x < 1; y<1 ;  x 
eq 0 ; y 
eq 0

TheNightKing · Answer

Yes, I understood. I was missing the range of 0 to 1.
Thank you!

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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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