Solving linear inequalities


Someone said that the percentage of women employees in the office is more than 60% but less than 65%. What is the fewest number of employees for which this could be?


Suppose there are \(n\) employees. Then the number of women \(w,\) is more than \((60/100)n,\) and less than \((65/100)n.\) That is,

$$\dfrac{60n}{100} \lt w \lt \dfrac{65n}{100}$$

or equivalently,

$$12n \lt 20w \lt 13n.$$

Checking \(n = 0, 1, 2, \ldots\) we find that \(n = 8\) is the smallest solution. \(12n = 12 \cdot 8 = 96,\) and \(13n = 13 \cdot 8 = 104.\) Hence \(w\) must be \(5.\)

Going further:

What if the percentage of women is between m and n? Can we find \(n\) directly, instead of searching \(n = 0, 1, 2, \ldots?\)