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If c is a nonzero constant such that x*f(x) = c, variables x

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If c is a nonzero constant such that x*f(x) = c, variables x

by fskilnik@GMATH » Wed Mar 13, 2019 7:39 am

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GMATH practice exercise (Quant Class 20)

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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

Answer: [spoiler]_____(B)__[/spoiler]
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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by fskilnik@GMATH » Wed Mar 13, 2019 10:48 am
fskilnik@GMATH wrote:GMATH practice exercise (Quant Class 20)

Image

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
Image

$$? = c = {1 \over 2} \cdot b\,\,\,\,\,\,\left[ {\left( {{1 \over 2};\,b} \right)\,\,\, \in \,\,\,{\rm{graph}}\left( f \right)} \right]$$
$${\rm{line}}\,{\rm{:}}\,\,\,\left\{ \matrix{
\,\left( { - 2;0} \right)\,\,\, \in \,\,\,{\rm{line}} \hfill \cr
\,{\rm{slope}} = {{1 - 0} \over {0 - \left( { - 2} \right)}} = {1 \over 2} \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{1 \over 2} = {{y - 0} \over {x - \left( { - 2} \right)}}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,y = {1 \over 2}\left( {x + 2} \right)\,\,\,\,\,\left( * \right)$$
$$\left( {{1 \over 2};\,b} \right)\,\,\, \in \,\,\,{\rm{line}}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,b = {1 \over 2}\left( {{1 \over 2} + 2} \right) = {1 \over 4} + 1 = 1.25$$
$$? = {{1.25} \over 2} = 0.625\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left( {\rm{B}} \right)$$

We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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