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# Is the value of a^2 - b^2 > greater than the value of

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Is the value of a^2 - b^2 > greater than the value of  [#permalink]

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29 Sep 2012, 11:21
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Difficulty:

55% (hard)

Question Stats:

65% (02:06) correct 35% (02:10) wrong based on 232 sessions

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Is the value of a^2 - b^2 greater than the value of (3a + 3b) (2a - 2b) ?

(1) b < a

(2) a < -1

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Re: Is the value of a^2 - b^2 > greater than the value of  [#permalink]

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29 Sep 2012, 12:04
1
carcass wrote:
Is the value of a^2 - b^2 greater than the value of (3a + 3b) (2a - 2b) ?

1) b < a

2) a < -1

A tricky one

The question "Is $$a^2-b^2>3(a+b)2(a-b)=6(a^2-b^2)$$?" translates to "Is $$a^2-b^2<0$$?" or "Is $$|a|<|b|$$"?

(1) We can have $$b=-7<3=a$$ or $$b=3<7=a$$.
Not sufficient.

(2) Obviously not sufficient, no information about $$b$$.

(1) and (2) together: now $$b<a<-1$$, so necessarily $$|a|<|b|$$.
Sufficient.

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Re: Is the value of a^2 - b^2 > greater than the value of  [#permalink]

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29 Sep 2012, 14:04
Hi evajager

I elaborated in this way

1) b < a the value depend on the absolute value of b (when we have a value $$n^2$$ we really do not know )

1) + 2) we know that b < a < - 1 $$a^2 - b^2$$ must be begative ------> 6 * negative = a big negative number ------> 6 ( $$a^2 - b^2$$ ) < $$a^2 - b^2$$

Can you explain me $$a^2 - b^2 < 0$$ ????

Thanks
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Re: Is the value of a^2 - b^2 > greater than the value of  [#permalink]

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29 Sep 2012, 14:15
carcass wrote:
Hi evajager

I elaborated in this way

1) b < a the value depend on the absolute value of b (when we have a value $$n^2$$ we really do not know )

1) + 2) we know that b < a < - 1 $$a^2 - b^2$$ must be begative ------> 6 * negative = a big negative number ------> 6 ( $$a^2 - b^2$$ ) < $$a^2 - b^2$$

Can you explain me $$a^2 - b^2 < 0$$ ????

Thanks

$$a^2-b^2>3(a+b)2(a-b)=6(a^2-b^2)$$ or $$0>5(a^2-b^2)$$ which means $$0>a^2-b^2$$.
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Re: Is the value of a^2 - b^2 > greater than the value of  [#permalink]

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29 Sep 2012, 14:19
ohhhhhhhhhhhh

got it. somtimes you do not see what is obvious.

Thanks
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Re: Is the value of a^2 - b^2 > greater than the value of  [#permalink]

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07 Oct 2015, 12:29
carcass wrote:
Is the value of a^2 - b^2 greater than the value of (3a + 3b) (2a - 2b) ?

(1) b < a

(2) a < -1

Is $$a^2-b^2 > 6* (a^2 - b^2)$$ ? --> Is $$5*(a^2-b^2) < 0$$ ? --> Is $$5*(a-b) * (a+b) < 0$$ ?

(1) Statement 1 can be written as b-a<0, which implies a-b>0. No information is given about a+b. Hence insufficient.

(2) Statement 2 is insufficient because no information is given about b

(Combined) a and b are both negative, hence we know a+b is also negative. The answer to the question stem is Yes. Answer choice C is correct.
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Re: Is the value of a^2 - b^2 > greater than the value of  [#permalink]

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26 Feb 2018, 02:20
The question can be rewritten as is (a+b)(a-b)> 6(a+b)(a-b) ??

st 1) b<a they can be any number with different signs not sufficient
st 2) a<-1 we don't know anything about B not sufficient

together, we know that a<-1 and b<a so b should be also negative
(a+b) should result a negative number whereas (a-b) will be positive
if we plug in the signs to our original question:
(-)(+)> 6(-)(+) ?? yes, a negative number will be greater than that negative number multiplied by 6.
so the answer should be C
Re: Is the value of a^2 - b^2 > greater than the value of &nbs [#permalink] 26 Feb 2018, 02:20
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