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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 $$y$$?（ ）
A. $$1$$ and $$3$$
B. $$17$$ and $$51$$
C. $$30$$ and $$10$$
D. $$6$$ and $$18$$
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## QuestionMathsClass 8

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 $$y$$?（ ）
A. $$1$$ and $$3$$
B. $$17$$ and $$51$$
C. $$30$$ and $$10$$
D. $$6$$ and $$18$$

C
4.6
4.6

## 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.