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If A varies directly proportional to C and B also varies directly proportional to C, which one of the following is not correct ?

A. $$\left( {A + B} \right)\alpha \,C$$

B. $$\left( {A - B} \right)\alpha \,\frac{1}{C}$$

C. $$\sqrt {AB} \,\alpha \,C$$

D. $$\frac{A}{B} = {\text{constant}}$$

Answer: Option B

Solution(By Examveda Team)

$$\eqalign{ & {\text{A }}\alpha {\text{ C and B }}\alpha {\text{ C}} \cr & \Rightarrow {\text{A}} = {\text{kC and B}} = {\text{mC}} \cr & {\text{for some constants k and m}}{\text{.}} \cr & \therefore A + B = kC + mC = \left( {{\text{k}} + {\text{m}}} \right){\text{C}} \cr & \Rightarrow \left( {A + B} \right)\alpha {\text{ C}}. \cr & {\text{A}} - {\text{B}} = {\text{kC}} - {\text{mC}} = \left( {{\text{k}} - {\text{m}}} \right){\text{C}} \cr & \Rightarrow \left( {A - B} \right)\alpha {\text{ C}}. \cr & \sqrt {AB} = \sqrt {kC \times mC} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {{\text{km}}{{\text{C}}^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {{\text{km}}.} {\text{C}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {{\text{AB}}} {\text{ }}\alpha {\text{ C}} \cr & = \frac{{\text{A}}}{{\text{B}}} = \frac{{{\text{kC}}}}{{{\text{mC}}}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{\text{k}}}{{\text{m}}} = {\text{Constant}}{\text{.}} \cr} $$

This Question Belongs to Arithmetic Ability >> Ratio

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Comments ( 1 )

  1. HIRANMAY DAS
    HIRANMAY DAS :
    1 year ago

    Why A-B is not a constant?

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