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Is a^2 > 3a – b^4?

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Is a^2 > 3a – b^4? [#permalink]

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Re: Is a^2 > 3a – b^4? [#permalink]

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New post 02 Oct 2015, 01:13
Bunuel wrote:
Is a^2 > 3a – b^4?

(1) 3a – b^4 = -5
(2) a > 5 and b > 0

Kudos for a correct solution.



(1) 3a – b^4 = -5
as a^2 will lways be positive... it will be greater than 3a – b^4, which is -5... suff
(2) a > 5 and b > 0
bothsides have a positive a, but the RHS has 'a' multiplied by 3 and then some positive value is subtracted .. so a^2 will always be greater than RHS where 'a' is greater than or equal to 3... so suff
ans D
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Re: Is a^2 > 3a – b^4? [#permalink]

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New post 02 Oct 2015, 09:56
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Bunuel wrote:
Is a^2 > 3a – b^4?

(1) 3a – b^4 = -5
(2) a > 5 and b > 0

Kudos for a correct solution.


Question:

Is \(a^2 > 3a - b^4?\)

Solution:

Is \(a^2 > 3a - b^4\)?

→ question becomes: Is \(a^2 + b^4 -3a > 0\)?


1) \(3a – b^4 = -5 → b^4 – 3a = 5.\) Substituting in inequality above, we obtain:

Is \(a^2 + 5 > 0?\) Since \(a^2\) is always positive and 5 is a positive number, their sum will always be greater than zero. Hence, statement I is sufficient.

2) a>5 and b>0.

Let a = 5. →\(a^2 = 25.\)
Let \(b=1 → b^4 = 1 → 3a = 15\)

→ Is 25 > 15-1 ? → Is 25 > 14? Answer is yes. This is an extreme case as a is always greater than 5. Integer a could be 5.001 for example.

With any value of a and b chosen that satisfy above condition in statement 2, inequality always hold. Hence, statement 2 is sufficient.

Answer: D.

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Re: Is a^2 > 3a – b^4? [#permalink]

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New post 02 Oct 2015, 10:02
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Bunuel wrote:
Is a^2 > 3a – b^4?

(1) 3a – b^4 = -5
(2) a > 5 and b > 0

Kudos for a correct solution.



(1) After applying statement 1, the stem can be written as Is \(a^2>-5\). a^2 is always non-negative and therefore is always greater than -5. This statement is sufficient.

(2) The stem can be written as Is \(a^2-3a>-b^4?\) --> Is \(a*(a-3)>-b^4?\). The left hand side of the equality will always be positive because it is given that a>5. The right hand side will always be non-positive because it is the negative of a non-negative number. Since a positive number is always greater than a non-positive number, this statement is always sufficient.

Correct answer choice is D

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Re: Is a^2 > 3a – b^4? [#permalink]

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New post 02 Oct 2015, 12:12
Bunuel wrote:
Is a^2 > 3a – b^4?

(1) 3a – b^4 = -5
(2) a > 5 and b > 0

Kudos for a correct solution.


Statement (1) makes right side of inequality -ve. Left side of the inequality will always be +ve or zero in case a=0.

hence sufficient.

Statement (2) a^2 will have values 36,49,64....etc.(for a=6,7,8........)
3a-b^4 will have values 15-somevalue,21-somevalue,24-somevalue (for a-6,7,8.....) , no impact of value of b as b^4 is always +ve
Hence Sufficient.

Ans: D
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Is a^2 > 3a – b^4? [#permalink]

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New post 04 Oct 2015, 21:19
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Here is the official GMATPrep question that the above question is based on:

Is 2x - 3y < x^2 ?

(1) 2x - 3y = -2
(2) x > 2 and y > 0

Discussion here: is-2x-3y-x-100596.html

Dabral

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Re: Is a^2 > 3a – b^4? [#permalink]

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New post 06 Oct 2015, 07:06
Bunuel wrote:
Is a^2 > 3a – b^4?

(1) 3a – b^4 = -5
(2) a > 5 and b > 0

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

Question Type: Yes/No This question asks whether a^2 > 3a – b^4.

Statement 1: 3a – b^4 = -5. This statement may not appear sufficient at first. You cannot manipulate it algebraically to mirror the question stem as might be the first impulse. However, the important thing is that “3a – b^4 equals a negative number.” Remember that a^2 cannot be less than zero, as a squared number cannot be negative. The answer, then, is “yes,” as you can absolutely conclude that a^2 will be greater than -5. The correct answer is either A or D.

Statement 2: a > 5 and b > 0. Again, this statement may not appear to be sufficient. It does not give specific values for a or b. However, if you “Just Do It” and plug in the numbers then you will see that it is sufficient. A good strategy for these “greater than” statements is to use the actual numbers given. Although the statement says a > 5 and b > 0 you can just use 5 and 0 in the inequality. The inequality becomes “25 > 15 – 0?” The answer is clearly “yes”: 25 > 15. And as you increase both “a” and “b” the result becomes a stronger “yes.” For example, if “a = 6 and b = 2” the inequality is “36 > 18 – 16?” Conceptually it looks like this, so long as a is greater than 3 then a^2 will be greater than 3a. Whatever number “b” is can only take away from the 3a it cannot add to it. In fact, Statement 2 gives more information than strictly necessary; “a > 3” would be sufficient.

The correct answer is D.
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Re: Is a^2 > 3a – b^4?   [#permalink] 30 Sep 2017, 06:02
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