Is a^2 > 3a – b^4?

(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

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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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.

Is a^2 > 3a – b^4?

(1) 3a – b^4 = -5.
a^2 is always greater or equal to 0, while (1) tells us that 3a - b^4 = -5 <0. So a^2 > 3a - b^4. SUFFICIENT.

(2) a > 5 and b > 0
a > 5 --> a^2 > 3a > 0
b > 0 --> b^4 > 0.
So a^2 > 3a - b^4. SUFFICIENT.

Answer: (D).

hey! I took a-6 and b=10
and it doesn't satisfy right ??

When b =10 and a=6;
36>18-10^4?
This does not satisfy right?

Is \(a^2 > 3a – b^4\)?
Is \(a^2 - 3a + b^4 > 0\)?
Is \(a^2 - 3a > 0\)?
Is \(a(a-3) > 0\)?

(1) \(3a – b^4 = -5\)

\(a^2 > -5\)
\(a^2\) can never be negative; SUFFICIENT.

(2) a > 5 and b > 0
If a > 5, we can answer that Is \(a(a-3) > 0.\)

SUFFICIENT.

Answer is D.

for statement 1, why cant a be 0?

Of course it can be 0. I meant non-negative, but wrote positive.

Too easy to be tagged with 700-level. This is basic.

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