Is y between 1 and 2, exclusive?
1). y^2 is less than y => 0<y<1, sufficient
2). y^2 is between 1 and 2, exclusive => -1.4x<y<-1 or 1<y<1.4x, Insufficient
Is it A
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Is y between 1 and 2, exclusive?
Source: Beat The GMAT — Data Sufficiency |
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neelgandham
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Amiable Scholar
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1) y^2 < y => y (y-1) < 0 => y< 0 or y <1 => y <1 Sufficient
2) 1<=y^2 <=2 => {y:-1.4 <= y <= -1} U {y: 1<=y<=1.4} not sufficient
2) 1<=y^2 <=2 => {y:-1.4 <= y <= -1} U {y: 1<=y<=1.4} not sufficient
Last edited by Amiable Scholar on Fri Oct 28, 2011 8:01 am, edited 1 time in total.
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GmatMathPro
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How do you figure?Amiable Scholar wrote: (btw no lower boundary here)
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neelgandham
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y^2 > y for all y<0, eg y=-1, y^2=1 ; y=-1/2, Y^2 =1/4 ....Amiable Scholar wrote:(btw no lower boundary here
y^2 >= y for all y>0, eg y=1,2,3 ; y^2 =1,4,9
So, it does !
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Amiable Scholar
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NeelGandham and GmatMathPro.
My bad ,
really was out of touch from the quant problems for 1 year
so missed some basic fundamentals in simple problem earlier.
x(x-1) < 0
first need to find out the roots
0 and 1 here .. (considering equality) ..
so need to check the sign of expression in following intervals
(-∞ , 0) (0,1) (1,∞)
for (-∞ , 0) x is -ve, (x-1) is -ve so x(x-1) is positive => x(x-1) > 0 for -∞ < x < 0
for (0 , 1) x is +ve, (x-1) is -ve so x(x-1) is negative => x(x-1) < 0 for 0 < x < 1
for (1 , ∞) x is +ve, (x-1) is +ve so x(x-1) is positive => x(x-1) > 0 for 1 < x < ∞
I am going to update my post. and thanks for pointing out.
My bad ,
really was out of touch from the quant problems for 1 year
so missed some basic fundamentals in simple problem earlier.
x(x-1) < 0
first need to find out the roots
0 and 1 here .. (considering equality) ..
so need to check the sign of expression in following intervals
(-∞ , 0) (0,1) (1,∞)
for (-∞ , 0) x is -ve, (x-1) is -ve so x(x-1) is positive => x(x-1) > 0 for -∞ < x < 0
for (0 , 1) x is +ve, (x-1) is -ve so x(x-1) is negative => x(x-1) < 0 for 0 < x < 1
for (1 , ∞) x is +ve, (x-1) is +ve so x(x-1) is positive => x(x-1) > 0 for 1 < x < ∞
I am going to update my post. and thanks for pointing out.
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