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GMATH practice exercise (Quant Class 20)
If c is a nonzero constant such that x*f(x) = c, variables x and y=f(x) are said to be inversely proportional. If the dotted curve shown above, defined by the points (x,y) where x>0 and y=f(x), represents a relationship between two inversely proportional variables, what is the value of the proportionality constant c?
(A) 0.5
(B) 0.625
(C) 0.8
(D) 1.25
(E) 1.5
If c is a nonzero constant such that x*f(x) = c, variables x and y=f(x) are said to be inversely proportional. If the dotted curve shown above, defined by the points (x,y) where x>0 and y=f(x), represents a relationship between two inversely proportional variables, what is the value of the proportionality constant c?
(A) 0.5
(B) 0.625
(C) 0.8
(D) 1.25
(E) 1.5
13 Mar 2019, 09:34
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Compute the equation of the line. Since there are two points,
\(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1}{2}\)
Line equation: \(y = \frac{1}{2}x + 1\)
At \((0.5, 0)\), \(y = \frac{5}{4}\)
We know that \(x*f(x) = c\). Hence, \(\frac{1}{2}*\frac{5}{4} = \frac{5}{8} = 0.625\)
\(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1}{2}\)
Line equation: \(y = \frac{1}{2}x + 1\)
At \((0.5, 0)\), \(y = \frac{5}{4}\)
We know that \(x*f(x) = c\). Hence, \(\frac{1}{2}*\frac{5}{4} = \frac{5}{8} = 0.625\)
General Discussion
13 Mar 2019, 11:59
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fskilnik
GMATH practice exercise (Quant Class 20)
If c is a nonzero constant such that x*f(x) = c, variables x and y=f(x) are said to be inversely proportional. If the dotted curve shown above, defined by the points (x,y) where x>0 and y=f(x), represents a relationship between two inversely proportional variables, what is the value of the proportionality constant c?
(A) 0.5
(B) 0.625
(C) 0.8
(D) 1.25
(E) 1.5
If c is a nonzero constant such that x*f(x) = c, variables x and y=f(x) are said to be inversely proportional. If the dotted curve shown above, defined by the points (x,y) where x>0 and y=f(x), represents a relationship between two inversely proportional variables, what is the value of the proportionality constant c?
(A) 0.5
(B) 0.625
(C) 0.8
(D) 1.25
(E) 1.5
\(? = c = {1 \over 2} \cdot b\,\,\,\,\,\,\left[ {\left( {{1 \over 2};\,b} \right)\,\,\, \in \,\,\,{\rm{graph}}\left( f \right)} \right]\)
\({\rm{line}}\,{\rm{:}}\,\,\,\left\{ \matrix{\\
\,\left( { - 2;0} \right)\,\,\, \in \,\,\,{\rm{line}} \hfill \cr \\
\,{\rm{slope}} = {{1 - 0} \over {0 - \left( { - 2} \right)}} = {1 \over 2} \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{1 \over 2} = {{y - 0} \over {x - \left( { - 2} \right)}}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,y = {1 \over 2}\left( {x + 2} \right)\,\,\,\,\,\left( * \right)\)
\(\left( {{1 \over 2};\,b} \right)\,\,\, \in \,\,\,{\rm{line}}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,b = {1 \over 2}\left( {{1 \over 2} + 2} \right) = {1 \over 4} + 1 = 1.25\)
\(? = {{1.25} \over 2} = 0.625\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left( {\rm{B}} \right)\)
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
