If a and b are integers, is a

Question

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

(1) b^2 – a^2 < 0
(2) b^2 – a^3 > 0

raarun · Answer

The question ask us to check where a is a negative integer.

Stmnt 1:
b^2 – a^2 < 0
b^2<a^2
e.g 4 < a^2 so a can be both positive or negative
Insufficient

Stmnt 2:
b^2 – a^3 > 0
b^2 > a^3
e.g when a= 1 b=4 the answer is no
e.g when a=1 b=-1 the answer is yes
Insufficient

Stmnt 1+2
Adding the inequalities we get
b^2-a^2 < b^2-a^3
Editing my answer .I did not change sign when i multiplied it by -1
~~a^2 > a^3~~
a^3 >a^2
~~This states that 'a' can never be negative.The answer to the question is NO~~
So a can only be negative.However the answer remains same

Sufficient

Ans C

varundixitmro2512 · Answer

I will go for E

wse cant just add equalities without knowig sign of variables

prashant212 · Answer

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.

GSBae · 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

Abhi077 · Answer

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

dvishal387 · Answer

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.

CounterSniper · Answer

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

Bunuel · Answer

Merging topics. Please search before posting and name topics properly. Check rules here: https://gmatclub.com/forum/rules-for-po ... 33935.html Thank you.

fskilnik · Answer

a,b\,\,{	ext{ints}}

{	ext{a}}\,\,\mathop {	ext{ < }}\limits^? \,\,{	ext{0}}

\left( 1 ight)\,\,\,{a^2} > {b^2}\,\,\,\left \,\,\,\,\,\,\left( * ight)

\left\{ \begin{gathered}
 \,{	ext{Take}}\,\,\left( {a,b} ight) = \left( {1,0} ight)\,\,\,\, \Rightarrow \,\,\,\left\langle {{	ext{NO}}} ightangle \,\, \hfill \
 \,{	ext{Take}}\,\,\left( {a,b} ight) = \left( { - 1,0} ight)\,\,\,\, \Rightarrow \,\,\,\left\langle {{	ext{YES}}} ightangle \,\, \hfill \ 
\end{gathered} ight.

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

\left( {1 + 2} ight)\,\,\,{a^2} > {b^2} > {a^3}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \begin{gathered}
 {a^2} > {a^3} \hfill \
 \left( * ight)\,\,a 
e 0 \hfill \ 
\end{gathered} ight.\,\,\,\,\,\mathop \Rightarrow \limits^{:\,\,{a^2}\,\, > \,\,0} \,\,\,1 > a\,\,\,\,\,\mathop \Rightarrow \limits^{a\,\,\operatorname{int} \, 
e \,\,0} \,\,\,\,\,\left\langle {{	ext{YES}}} ightangle

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

Regards, 
Fabio.

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

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