# Solving linear inequalities

Problem:

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?

Solution:

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 $$n = 0, 1, 2, \ldots?$$