If ab bc^2

Question

If ab − bc^2 < 0, is a < 0?

(1) c > 0
(2) b < 0

(ab —bc^2) < 0 → ab < bc^2
If b > 0 → a < c^2
If b < 0 → a > c^2
Since number c^2 must be either positive or zero, then we can revise this second situation to say "If b is negative, a is positive," or "If b is negative, answer to question is No."
(Note: b cannot be zero, since that would cause ab = bc^2 = 0.)

(1): c > 0. Suppose that c = 2. Then ab – b(2^2) < 0, and ab < 4b. Look at inequality ab < 4b. With no restrictions on either a or b (except that b is not zero), we should think that there is a good chance that a could be either positive or negative and that this statement is Insufficient.
If b = 3, then ab < 4b becomes 3a < 4(3), or 3a < 12. Dividing both sides of inequality 3a < 12 by positive number 3 keeps direction of inequality sign same, and we then have that a < 4. inequality a < 4 can be true for a positive value for a like 1, since 1 < 4. Indeed, if a = 1, b = 3, and c = 2, then ab —bc^2 < 0 is true,
[1(3) – 3(2^2) < 0 because 1(3) – 3(2^2) = -9]. > 0 is true, and since a < 0, answer to question would be No.
However, inequality a < 4 can also be true for a negative value for a like –1, since –1 < 4. This would cause answer to question to be Yes
Insufficient
(2): If b < 0 → a > c2
Always No
Sufficient

Hi, I want to know if we have simpler approach to Statement (1), please.

Show Answer

Official Answer

B

Bunuel · Answer

If ab − bc^2 < 0, is a < 0?

(1) c > 0. If \(c=1\) and \(b=1\), then we get \(a<1\), so \(a\) may or may not be less than 0. Not sufficient.

(2) b < 0 --> reduce by negative \(b\) and flip the sign: \(a-c^2>0\) --> \(a>c^2\). Since the square of a number is always non-negative, then we have that \(a>c^2\geq{0}\) --> \(a>0\). Sufficient.

Answer: B.

Hope it's clear.

MathRevolution · Answer

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. Visit https://www.mathrevolution.com/gmat/lesson for details. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary. ab - bc^2 < 0 ⇔ b(a-c^2) < 0 ⇔ b<0, a>c^2 or b>0, a c^2 ≥ 0 . Thus a > 0 . The answer is 'no'. Condition 1) a = 2, b = 1, c = 1 and a = 2, b = 1, c = 1 are possible solutions. If a = -2, b = 1, c = 1 , then we have ab - bc^2 = (-2)\cdot(-1) - (-1) \cdot 1^2 = 3 > 0, a<0 and the answer is 'yes'. If a = 2, b = 1, c = 1 , then we have ab - bc^2 = 2\cdot1 - 1 \cdot 1^2 = 1 > 0, a>0 and the answer is 'no'. Since condition 1) does not yield a unique solution, it is not sufficient. Therefore, B is the answer.

hiranmay · Answer

If ab − bc^2 < 0, is a < 0?

(1) c > 0
(2) b < 0

Solution:
Answer: B
ab − bc^2 < 0
=> b(a-c^2) <0
=> + - < 0 => b > 0 and a < c^2 => c^2>= 0, so "a" can be +ve/-ve
or - + < 0 => b < 0 and a> c^2 => c^2>=0, so a>0 i.e "a" is +ve

(1) c > 0 --> insuff: if c> 0, "a" can be +ve/-ve --> answer of "a<0?" can be yes/no
(2) b < 0 --> suff: if b < 0, a>0 i.e "a" is +ve --> answer of "a<0?" is no

Pratham_Patil24 · Answer

Hi  Bunuel ,

In these question types, pertaining to numbers and pure contexts especially, how do we know when to and most importantly which numbers to plug in? I am really struggling in this aspect of DS.

If ab bc^2 < 0, is a < 0?

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