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3(x 7) 9 [#permalink]
1
\(3(x-7) \geq 9 \text{ ---(1)}\)
\(0.25y-3 \leq 1 \text{ ---(2)}\)

Lets deal with (1) first

\(3(x-7) \geq 9\)

\((x-7) \geq \frac{9}{3}\)

\(x-7 \geq 3\)

\(x \geq 3+7\)

\(x\geq 10\)


Lets now handle (2)

\(0.25y-3 \leq 1\)

\(0.25y \leq 1+3\)

\(0.25y \leq 4\)

\(y \leq \frac{4}{0.25}\)

\(y\leq 16\)

So we now have two ranges.

\(x\) ranges from 10 to infinity

\(y\) ranges from 16 to minus infinity

\(x\) and \(y\) can therefore take any of the following values \(10,11,12,13,14,15,16\)

If \(x = 16\) and \(y= 10\), Quantity A is greater, if it is the other way round, Quantity B is greater. If both \(x\) and \(y\) equal any of the above numbers, both Quantities are Equal. Therefore, there is no discernible relationship between the two quantities.

The correct answer is Choice D.
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3(x 7) 9 [#permalink]
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