If \(x\) and \(y\) are directly proportional and when \(x=13\), \(y=39\) , which of the following is not a possible pair of corresponding values of \(x\) and \(y\)?（ ）
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If \(x\) and \(y\) are directly proportional and when \(x=13\), \(y=39\) , which of the following is not a possible pair of corresponding values of \(x\) and \(y\)?（ ）
Answer
C
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Solution
Given, \(x\) and \(y\) are directly proportional
i.e. \(x\propto y\)
\(\Rightarrow\) \(x=ky\) where \(k\) is constant.
With \(x=13\) and \(y=39\), we have
\(13=39k\)
\(\Rightarrow\) \(k=\frac{13}{39}\)\(\Rightarrow\) \(k=\frac{1}{3}\).
So, \(\frac{x}{y}=\frac{1}{3}\).
Checking all the options as:
a.) \(x=1\) and \(y=3\), checking with \(k=\frac{1}{3}\)
i.e. \(x=\frac{1}{3}\times 3=1\).
b.) \(x=17\) and \(y=51\), checking with \(k=\frac{1}{3}\)
i.e. \(x=\frac{1}{3}\times 51=17\)
c.) \(x=30\) and \(y=10\), checking with \(k=\frac{1}{3}\)
i.e. \(x=\frac{1}{3}\times 10\neq 30\)
d.) \(x=6\) and \(y=18\), checking with \(k=\frac{1}{3}\)
i.e. \(x=\frac{1}{3}\times 18=6\).
Since option \((C)\) only does not follow the proportionality.
Hence, \((C)\) is the correct option.
i.e. \(x\propto y\)
\(\Rightarrow\) \(x=ky\) where \(k\) is constant.
With \(x=13\) and \(y=39\), we have
\(13=39k\)
\(\Rightarrow\) \(k=\frac{13}{39}\)\(\Rightarrow\) \(k=\frac{1}{3}\).
So, \(\frac{x}{y}=\frac{1}{3}\).
Checking all the options as:
a.) \(x=1\) and \(y=3\), checking with \(k=\frac{1}{3}\)
i.e. \(x=\frac{1}{3}\times 3=1\).
b.) \(x=17\) and \(y=51\), checking with \(k=\frac{1}{3}\)
i.e. \(x=\frac{1}{3}\times 51=17\)
c.) \(x=30\) and \(y=10\), checking with \(k=\frac{1}{3}\)
i.e. \(x=\frac{1}{3}\times 10\neq 30\)
d.) \(x=6\) and \(y=18\), checking with \(k=\frac{1}{3}\)
i.e. \(x=\frac{1}{3}\times 18=6\).
Since option \((C)\) only does not follow the proportionality.
Hence, \((C)\) is the correct option.
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