If a and b are integers, is a < 0?

Condition 1: b^2 -a^2 < 0
simplifying this equation gives us (b-a)*(b+a)<0.
We can form different cases where a and b assume -ve and +ve values and find that equation is possible under multiple circumstances.
Thus this data is insufficient.

Condition 2: b^2-a^3 > 0

we cant form 2 different cases where both a and b are +ve and -ve and 2 cases where one of them is -ve and find that second condition holds for multiple cases. Thus insufficient.

Combining both the conditions, we find that a^3 is less than a^2. This is only possible when either a <0 or 0<a<1. since a and b both are integers, as given in the question. a<0 is the only possibility.

Hence C is answer.

1) This just means that \(a^2 > b^2.\) We still don't know anything about specific values for a or b. For \(a = 2, b = 1\) we have FALSE, but for \(a = -2, b = 1\) we have TRUE. Insufficient.

2) Similar to above, this just means that \(a^3 < b^2.\) This doesn't tell us anything about specific values of a or b. For \(a = 1, b = 2\) we have FALSE, but for \(a = -1, b = 2\) we have TRUE. Insufficient.

Together, we have that \(a^3 < b^2 < a^2\), or that \(a^3 < a^2\). Since a and b are both integers, the only possibility is that \(a<0.\)

Answer: C

If a & b are integers, is a<0
1. b^2 - a^2 < 0
2. b^2 - a^3 >0

In stat 1 since b^2 - a^2 < 0 means a>b but we cannot say anything about its sign. Not sufficient
In stat 2 again we cannot infer anything about a's sign. For ex- b=4, a=2 satisfies it and b=4 and a=-2 also satisfies it
Taking the two stat together we can say that since a>b and b^2 - a^3 >0 and since a and b are integers a should be less than 0.
C should be the correct choice.

1. b^2 - a^2 < 0

\(b^2 < a^2\)
b and a can be both positive and negative

not sufficient

2. b^2 - a^3 >0

\(b^2 > a^3\)

b = 1 and a =1/2 => a>0
b=10 and a=-50 => a<0

not sufficient

using both

\(b^2 > a^3\)
\(b^2 < a^2\)
is only possible when a is negative

sufficient
C

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\(a,b\,\,{\text{ints}}\)

\({\text{a}}\,\,\mathop {\text{ < }}\limits^? \,\,{\text{0}}\)


\(\left( 1 \right)\,\,\,{a^2} > {b^2}\,\,\,\left[ {\, \geqslant 0\,\,\,\, \Rightarrow \,\,\,{a^2} > 0\,\,\, \Rightarrow \,\,\,a \ne 0} \right]\,\,\,\,\,\,\left( * \right)\)

\(\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,\left( {a,b} \right) = \left( {1,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\\\
\,{\text{Take}}\,\,\left( {a,b} \right) = \left( { - 1,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\ \\
\end{gathered} \right.\)


\(\left( 2 \right)\,\,{b^2} > {a^3}\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,\left( {a,b} \right) = \left( {0,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\\\
\,{\text{Take}}\,\,\left( {a,b} \right) = \left( { - 1,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\ \\
\end{gathered} \right.\)


\(\left( {1 + 2} \right)\,\,\,{a^2} > {b^2} > {a^3}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \begin{gathered}\\
{a^2} > {a^3} \hfill \\\\
\left( * \right)\,\,a \ne 0 \hfill \\ \\
\end{gathered} \right.\,\,\,\,\,\mathop \Rightarrow \limits^{:\,\,{a^2}\,\, > \,\,0} \,\,\,1 > a\,\,\,\,\,\mathop \Rightarrow \limits^{a\,\,\operatorname{int} \, \ne \,\,0} \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.