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Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
18 Feb 2019, 14:42
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
73% (02:21) correct
27%
(02:38)
wrong
based on 153
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GMATH practice exercise (Quant Class 2)
\(\left\{ \matrix{\\
\,2\left( {x + y} \right) - A = 0 \hfill \cr \\
\,3x + By - 6 = 0 \hfill \cr} \right.\)
If \(A\) and \(B\) are two fixed constants such that the system of equations given above has more than one ordered pair \((x,y)\) in its solution set, what is the value of \(AB\) ?
(A) 10
(B) 12
(C) 14
(D) 16
(E) 18
\(\left\{ \matrix{\\
\,2\left( {x + y} \right) - A = 0 \hfill \cr \\
\,3x + By - 6 = 0 \hfill \cr} \right.\)
If \(A\) and \(B\) are two fixed constants such that the system of equations given above has more than one ordered pair \((x,y)\) in its solution set, what is the value of \(AB\) ?
(A) 10
(B) 12
(C) 14
(D) 16
(E) 18
18 Feb 2019, 15:50
Kudos
Bookmarks
fskilnik
GMATH practice exercise (Quant Class 2)
\(\left\{ \matrix{\\
\,2\left( {x + y} \right) - A = 0 \hfill \cr \\
\,3x + By - 6 = 0 \hfill \cr} \right.\)
If \(A\) and \(B\) are two fixed constants such that the system of equations given above has more than one ordered pair \((x,y)\) in its solution set, what is the value of \(AB\) ?
(A) 10
(B) 12
(C) 14
(D) 16
(E) 18
\(\left\{ \matrix{\\
\,2\left( {x + y} \right) - A = 0 \hfill \cr \\
\,3x + By - 6 = 0 \hfill \cr} \right.\)
If \(A\) and \(B\) are two fixed constants such that the system of equations given above has more than one ordered pair \((x,y)\) in its solution set, what is the value of \(AB\) ?
(A) 10
(B) 12
(C) 14
(D) 16
(E) 18
The system will have an infinite number of solutions for (x, y) if the two equations are THE SAME:
2(x+y) - A = 0 --> 2x + 2y - A = 0 --> 6x + 6y - 3A = 0
3x + By - 6 = 0 --------------------------> 6x +2By - 12 = 0
The equations in blue will be the same if 3A=12 and 2B=6, with the result that A=4, B=3, and A*B=12.
.
18 Feb 2019, 18:08
Kudos
Bookmarks
fskilnik
GMATH practice exercise (Quant Class 2)
\(\left\{ \matrix{\\
\,2\left( {x + y} \right) - A = 0 \hfill \cr \\
\,3x + By - 6 = 0 \hfill \cr} \right.\)
If \(A\) and \(B\) are two fixed constants such that the system of equations given above has more than one ordered pair \((x,y)\) in its solution set, what is the value of \(AB\) ?
(A) 10
(B) 12
(C) 14
(D) 16
(E) 18
Mitch´s nice solution is based on the fact that (A and B already fixed,) the solution sets of ordered pairs (x,y) that satisfy each equation alone represent (each one) a line in the rectangular coordinate system. \(\left\{ \matrix{\\
\,2\left( {x + y} \right) - A = 0 \hfill \cr \\
\,3x + By - 6 = 0 \hfill \cr} \right.\)
If \(A\) and \(B\) are two fixed constants such that the system of equations given above has more than one ordered pair \((x,y)\) in its solution set, what is the value of \(AB\) ?
(A) 10
(B) 12
(C) 14
(D) 16
(E) 18
We have two lines in the plane, hence they are concurrent (just one point (x,y) in common), parallel and distinct (no point (x,y) in common) or parallel and coincident (every point (x,y) in common).
From the question stem, we are looking for the last scenario, and Mitch´s solution is validated.
Let me present an alternate solution, using not only algebraic operations, but only algebraic arguments/insights. (We offer this problem in our SECOND class... no Analytic Geometry yet!)
\(? = A \cdot B\,\,\,\,\left( {{\rm{system}}\,\,{\rm{with}}\,\,{\rm{more}}\,\,{\rm{than}}\,\,{\rm{one}}\,\,{\rm{solution}}} \right)\)
\(\left\{ \matrix{\\
\,2\left( {x + y} \right) - A = 0 \hfill \cr \\
\,3x + By - 6 = 0 \hfill \cr} \right.\,\,\,\,\, \cong \,\,\,\,\,\left\{ \matrix{\\
\,2x + 2y = A\,\,\,\left( { \cdot \,3} \right) \hfill \cr \\
\,3x + By = 6\,\,\,\left( { \cdot \,2} \right) \hfill \cr} \right.\,\,\,\,\, \cong \,\,\,\,\,\left\{ \matrix{\\
\,6x + 6y = 3A\,\,\left( * \right) \hfill \cr \\
\,6x + 2By = 12 \hfill \cr} \right.\,\,\,\,\,\mathop \Rightarrow \limits^{\left( - \right)} \,\,\,\,\,y\left( {6 - 2B} \right) = 3\left( {A - 4} \right)\,\,\,\,\left( {**} \right)\)
\(6 - 2B \ne 0\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,y = {{3\left( {A - 4} \right)} \over {6 - 2B}}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,x = {1 \over 6}\left\{ {3A - 6\left[ {{{3\left( {A - 4} \right)} \over {6 - 2B}}} \right]} \right\}\,\,\,\,\mathop \Rightarrow \limits^{A,B\,\,{\rm{given}}} \,\,\,\,\left( {x,y} \right)\,\,\,{\rm{unique}}\,\,{\rm{solution}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{impossible}}\)
\(6 - 2B = 0\,\,\,\, \Rightarrow \,\,\,\,B = 3\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,A = 4\,\,\,\, \Rightarrow \,\,\,\,\,? = AB = 12\)
The correct answer is (B).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
