Consider the equation \[\dfrac{k}{x} = 12\] where \[k\] is any number between \[20\] and \[65\] and \[x\] is a positive integer. What are the possible values of \[x\]?

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Hint: We will first write the given equation \[\dfrac{k}{x} = 12\] as \[k = 12x\]. As given in the question, \[k\] is any number between \[20\] and \[65\]. So, using this and substituting \[k = 12x\], we will write \[20 < 12x < 65\]. Then we will simplify it. Given, \[x\] is a positive integer, so we will further simplify it using the definition of positive integer to find the possible values of \[x\].

Complete step-by-step answer:
The given equation is \[\dfrac{k}{x} = 12\].
On cross multiplication, we can write the given equation as \[k = 12x\].
Now, as given in the question, \[k\] is any number between \[20\] and \[65\]. So, we can write,
\[ \Rightarrow 20 < k < 65\]
On putting \[k = 12x\], we get
\[ \Rightarrow 20 < 12x < 65\]
Now, as we know, the inequality remains the same on dividing by a positive number. So, on dividing by \[12\], we get
\[ \Rightarrow \dfrac{{20}}{{12}} < x < \dfrac{{65}}{{12}}\]
Cancelling the common terms, if any, from the numerator and the denominator, we get
\[ \Rightarrow \dfrac{5}{3} < x < \dfrac{{65}}{{12}}\]
On calculating, we get
\[ \Rightarrow 1.67 < x < 5.42\]
But, given in the question that \[x\] is a positive integer. As we know that the first integer that will come just after \[1.67\] is \[2\] and the first integer that will come just before \[5.42\] is \[5\].
So, we can rewrite it as,
\[ \Rightarrow 2 \leqslant x \leqslant 5\]
Therefore, the possible values of \[x\] are \[2\], \[3\], \[4\] and \[5\].

Note: An integer can never be a fraction, a decimal, or a percent. A positive integer are those numbers that are prefixed with a plus sign \[( + )\]. But, most of the time positive numbers are represented simply as numbers without the plus sign. Also, zero is a neutral integer because it can neither be a positive nor a negative integer i.e., zero has no \[ + ve\] sign or \[ - ve\] sign.