What is the value of a ?

Question

(1) a ^2 = b^2
(2) b − a = −12

Bunuel · Answer

What is the value of a ?

(1) a ^2 = b^2. Clearly insufficient. But from this statement  b^2-a^2=0  -->  (b-a)(b+a)=0  -->  b-a=0  or  b+a=0 .

(2) b − a = −12. Also clearly insufficient.

(1)+(2) Since from (2)  b-a=-12
eq{0} , then from (1)  b+a=0 . So, we have two equations:  b-a=-12  and  b+a=0 . Solving gives  a=6  and  b=-6 . Sufficient.

Answer: C.

atalpanditgmat · Answer

Thanks Bunuel....your explanation is much more easier than Kaplan's...

Zarrolou · Answer

What is the value of a ?

1 .a^2  =  b^2 
 |a|=|b|  so or  a=b  or  a=-b , we have no value of b. Not sufficient

2.b - a = -12
Clearly not sufficient.

1+2
 a-b=12 , so from statement 1 we can deduce that a=-b ( it cannot be a=b because a-b=b-b=0 and not 12).
 -b-b=12,b=-6  and  a=6 . C

mau5 · Answer

From F.S 1, we have (a-b)(a+b) = 0. Insufficient.

From F.S 2, we have b-a = -12. Insufficient.

Taking both together, we know that (a-b) is not equal to zero. Thus, (a+b) = 0, and we can get the value for a. Sufficient.

C.

GmatDaddy · Answer

Statement 1:  a^2 = b^2  
This implies, a=+b or -b
Insufficient.

Statement 2: b - a = -12 
Clearly inssuficient.

Statement 1 and 2 Together:

Case 1: a=+b
b-b=-12
0=-12
Absurd Result.

Case 2: a=-b
b+b=-12
b=-6

Hence, C is our answer.

fskilnik · Answer

To prove that each statement alone is insufficient, we present a BIFURCATION (algebraic, in this case):

? = a

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

Take\,\,\left\{ \begin{gathered}
 {a^2} = {b^2} = 0\,\,\,\,\, \Rightarrow \,\,\,\,? = 0 \hfill \
 {a^2} = {b^2} = 1\,\,\,\,\, \Rightarrow \,\,\,\,? 
e 0\,\,\,\,\,\,\,\,\left( {a = \pm \,1} ight) \hfill \ 
\end{gathered} ight.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{	ext{INSUF}}{	ext{.}}

\left( 2 ight)\,\,a\, - b = 12

Take\,\,\left\{ \begin{gathered}
 b = 0\,\,\,\,\, \Rightarrow \,\,\,\,? = 12 \hfill \
 b = 1\,\,\,\, \Rightarrow \,\,\,\,? 
e 12\,\,\,\,\,\,\,\,\left( {a = \,13} ight) \hfill \ 
\end{gathered} ight.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{	ext{INSUF}}{	ext{.}}

\left( {1 + 2} ight)\,\,\,\,\,\left\{ \begin{gathered}
 {a^2} - {b^2} = 0\,\,\,\,\, \hfill \
 a - b = 12 \hfill \ 
\end{gathered} ight. \Rightarrow \,\,\,\,\,\,\left\{ \begin{gathered}
 \left( {a + b} ight)\left( {a - b} ight) = 0 \hfill \
 a - b = 12 \hfill \ 
\end{gathered} ight.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\{ \begin{gathered}
 a + b = 0 \hfill \
 a - b = 12 \hfill \ 
\end{gathered} ight.\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( + ight)} \,\,\,\,\,2a = 12\,\,\,\,\,\, \Rightarrow \,\,\,\,\,a\,\,{	ext{unique}} \Rightarrow \,\,\,\,\,\,{	ext{SUF}}{	ext{.}}\,

The above follows the notations and rationale taught in the GMATH method.

MHIKER · Answer

(1)  a ^2 = b^2

a ^2 - b^2=0

(a-b)(a+b)=0  
 a-b=0 
 a=b

OR, 
 a+b=0 
 a=-b

(2) b − a = −12; Insufficient.

Considering both:

a=b, so, b-(-b)=12, 2b=12, b=6; Sufficient.

The answer is C

What is the value of a ?

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