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Director  V
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If x, y and b are non-zero integers such that x>y, is bx^2 –by^2>0?  [#permalink]

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Question Stats: 29% (01:53) correct 71% (02:17) wrong based on 56 sessions

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If x, y and b are non-zero integers such that x>y, is bx^2 –by^2>0?

I. b>0
II. b^3(|x|−|y|)>0
Math Expert V
Joined: 02 Aug 2009
Posts: 8004
Re: If x, y and b are non-zero integers such that x>y, is bx^2 –by^2>0?  [#permalink]

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If x, y and b are non-zero integers such that x>y, is bx^2 –by^2>0?
$$bx^2 –by^2>0.....b(x^2-y^2)>0$$. This is possible in two cases
#Both b and $$x^2-y^2$$ are positive, that is b>0 and $$x^2-y^2>0.....|x|-|y|>0$$
#Both b and $$x^2-y^2$$ are negative, that is b<0 and $$x^2-y^2<0.....|x|-|y|<0$$
Let us check the two conditions for this

I. b>0

II. b^3(|x|−|y|)>0
$$b^3$$ will have same sign as b, so this tells us that $$b^3(|x|−|y|)>0$$ means the same as $$b(|x|−|y|)>0$$.
This further tells us that both b and |x|-|y| have the same sign.
Answer is always YES, irrespective of the exact values of b, x and y.
Suff

B
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Re: If x, y and b are non-zero integers such that x>y, is bx^2 –by^2>0?  [#permalink]

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chetan2u wrote:
If x, y and b are non-zero integers such that x>y, is bx^2 –by^2>0?
$$bx^2 –by^2>0.....b(x^2-y^2)>0$$. This is possible in two cases
#Both b and $$x^2-y^2$$ are positive, that is b>0 and $$x^2-y^2>0.....|x|-|y|>0$$
#Both b and $$x^2-y^2$$ are negative, that is b<0 and $$x^2-y^2<0.....|x|-|y|<0$$
Let us check the two conditions for this

I. b>0

II. b^3(|x|−|y|)>0
$$b^3$$ will have same sign as b, so this tells us that $$b^3(|x|−|y|)>0$$ means the same as $$b(|x|−|y|)>0$$.
This further tells us that both b and |x|-|y| have the same sign.
Answer is always YES, irrespective of the exact values of b, x and y.
Suff

B

Sir but it is clearly given that x>y then why isn't the option b>0 is sufficient.

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Math Expert V
Joined: 02 Aug 2009
Posts: 8004
Re: If x, y and b are non-zero integers such that x>y, is bx^2 –by^2>0?  [#permalink]

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2
Shreshtha55 wrote:
chetan2u wrote:
If x, y and b are non-zero integers such that x>y, is bx^2 –by^2>0?
$$bx^2 –by^2>0.....b(x^2-y^2)>0$$. This is possible in two cases
#Both b and $$x^2-y^2$$ are positive, that is b>0 and $$x^2-y^2>0.....|x|-|y|>0$$
#Both b and $$x^2-y^2$$ are negative, that is b<0 and $$x^2-y^2<0.....|x|-|y|<0$$
Let us check the two conditions for this

I. b>0

II. b^3(|x|−|y|)>0
$$b^3$$ will have same sign as b, so this tells us that $$b^3(|x|−|y|)>0$$ means the same as $$b(|x|−|y|)>0$$.
This further tells us that both b and |x|-|y| have the same sign.
Answer is always YES, irrespective of the exact values of b, x and y.
Suff

B

Sir but it is clearly given that x>y then why isn't the option b>0 is sufficient.

Posted from my mobile device

b>0 requires |x|-|y|>0...
say b = 2, x = -7 and y = 3 will give |x|-|y|=|-7|-|6|=1>0
But say b = 2, x = 3 and y = -7 will give |x|-|y|=|3|-|7|=3-7=-4, which is NOT >0..
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Re: If x, y and b are non-zero integers such that x>y, is bx^2 –by^2>0?  [#permalink]

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kiran120680 wrote:
If x, y and b are non-zero integers such that x>y, is bx^2 –by^2>0?

I. b>0
II. b^3(|x|−|y|)>0

#1
b>0 , no relation or value of x & y given insufficient

#2
b^3(|x|−|y|)>0
this would only be true when b is +ve and irrelevant whether x ,y are + or -ve or opposite sign because they are both in modulus so always +ve
therefore bx^2 –by^2>0
shall always be true
IMO B
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Joined: 11 Aug 2017
Posts: 59
Re: If x, y and b are non-zero integers such that x>y, is bx^2 –by^2>0?  [#permalink]

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chetan2u wrote:
If x, y and b are non-zero integers such that x>y, is bx^2 –by^2>0?
$$bx^2 –by^2>0.....b(x^2-y^2)>0$$. This is possible in two cases
#Both b and $$x^2-y^2$$ are positive, that is b>0 and $$x^2-y^2>0.....|x|-|y|>0$$
#Both b and $$x^2-y^2$$ are negative, that is b<0 and $$x^2-y^2<0.....|x|-|y|<0$$
Let us check the two conditions for this

I. b>0

II. b^3(|x|−|y|)>0
$$b^3$$ will have same sign as b, so this tells us that $$b^3(|x|−|y|)>0$$ means the same as $$b(|x|−|y|)>0$$.
This further tells us that both b and |x|-|y| have the same sign.
Answer is always YES, irrespective of the exact values of b, x and y.
Suff

B

Dear chetan I still struggle to understand the 3rd n 4th line of your explanation. hoe did you get x^2-y^2>0.....|x|-|y|>0 and x^2-y^2<0.....|x|-|y|<0
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Posts: 1687
Location: India
Concentration: International Business, Operations
Schools: INSEAD Jan '19
GPA: 3.01
WE: Engineering (Real Estate)
If x, y and b are non-zero integers such that x>y, is bx^2 –by^2>0?  [#permalink]

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1
Mohammad Ali Khan wrote:
chetan2u wrote:
If x, y and b are non-zero integers such that x>y, is bx^2 –by^2>0?
$$bx^2 –by^2>0.....b(x^2-y^2)>0$$. This is possible in two cases
#Both b and $$x^2-y^2$$ are positive, that is b>0 and $$x^2-y^2>0.....|x|-|y|>0$$
#Both b and $$x^2-y^2$$ are negative, that is b<0 and $$x^2-y^2<0.....|x|-|y|<0$$
Let us check the two conditions for this

I. b>0

II. b^3(|x|−|y|)>0
$$b^3$$ will have same sign as b, so this tells us that $$b^3(|x|−|y|)>0$$ means the same as $$b(|x|−|y|)>0$$.
This further tells us that both b and |x|-|y| have the same sign.
Answer is always YES, irrespective of the exact values of b, x and y.
Suff

B

Dear chetan I still struggle to understand the 3rd n 4th line of your explanation. hoe did you get x^2-y^2>0.....|x|-|y|>0 and x^2-y^2<0.....|x|-|y|<0

You can also try easy numbers :

If b^3 > 0, this means b>0

now we are left with |x| - |y| > 0

Take x = 2 and y = -10

Given x > y

|2| - |-10| > 0

2 - 10 > 0

-8 > 0 (But this is not true) Reject these values

Take x = 2 and y = 1

|2| - |1| > 0

1 > 0 True

b > 0 and |x| - |y| > 0

Hence b^3 * |x| - |y| > 0

Actual values does not matter here but signs of the values matter.

x^2 - y^2 can be considered |x| - |y| > 0 because x^2 and |x| will never be negative. Same goes for values of "y"
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"Do not watch clock; Do what it does. KEEP GOING." If x, y and b are non-zero integers such that x>y, is bx^2 –by^2>0?   [#permalink] 31 Mar 2019, 06:19
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