Is the value of a^2 - b^2 greater than the value of (3a + 3b)(2a - 2b) ?
(1) b < a
(2) a < -1
[DS][a2-b2] HSPA posts
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- bubbliiiiiiii
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Question:
(a^2 - b^2) > 6 (a^2 - b^2)
This is possible when a^2 - b^2 is negative.
a^2 - b^2 < 0
Is a^2 < b^2 ?
A: b<a consider a=1; b=2 and a=-3; b=2 INSUFFICIENT
B: a<-1 nothing known about B. INSUFFICIENT
Consider a and b,
a = [-.5, 0, 1, 2, 3 .. ] i.e., A can be both negative and positive and b will always be postive.
This is something very similar to what we had in option A.
Thus E.
(a^2 - b^2) > 6 (a^2 - b^2)
This is possible when a^2 - b^2 is negative.
a^2 - b^2 < 0
Is a^2 < b^2 ?
A: b<a consider a=1; b=2 and a=-3; b=2 INSUFFICIENT
B: a<-1 nothing known about B. INSUFFICIENT
Consider a and b,
a = [-.5, 0, 1, 2, 3 .. ] i.e., A can be both negative and positive and b will always be postive.
This is something very similar to what we had in option A.
Thus E.
Regards,
Pranay
Pranay
- manpsingh87
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3(a+b)*2(a-b)=6(a^2-b^2);HSPA wrote:Is the value of a^2 - b^2 greater than the value of (3a + 3b)(2a - 2b) ?
(1) b < a
(2) a < -1
1) b<a; a>0; b<0; let a=2 b=-3 a^2-b^2=4-9=-5 ; 6(a^2-b^2)=-30;
a>0, b>0 let a=4,b=2 a^2-b^2=16-4=12; 6(a^2-b^2)=6*12=72;
a<0, b<0 a=-1, b=-2 a^2-b^2=1-4=-3, 6(a^2-b^2)=-18;
therefore 1 alone is not sufficient to answer the question.
2) a<-1, as nothing is mentioned about b, therefore 2 alone is not sufficient to answer the question.
consider 1 and 2 together we have, b<a, and a<-1 now consider a=-2, b=-3 a^2-b^2= 4-9=-5; 6(a^2-b^2)= -30, which is less than a^2-b^2 hence C
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