One approach is to look for a pattern

1 day: cost = \(z\) dollars

2 days: cost = \(z + \frac{z}{6}\) dollars

3 days: cost = \(z + \frac{z}{6}+ \frac{z}{6}\) dollars

4 days: cost = \(z + \frac{z}{6}+\frac{z}{6}+\frac{z}{6}\) dollars

Notice that the number of times we add z/6 is 1 less than the number of vacation days
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y days: cost = \(z + (y-1)(\frac{z}{6})\) dollars

Check answer choices....\(z + (y-1)(\frac{z}{6})\) is not an option. So, we must simplify this expression.

Take: \(z + (y-1)(\frac{z}{6})\)

Rewrite \(z\) as \(\frac{6z}{6}\) to get: \(\frac{6z}{6} + \frac{(y-1)z}{6}\)

Simplify: \(\frac{6z}{6} + \frac{yz-z}{6}\)

Simplify: \(\frac{6z+(yz-z)}{6}\)

Simplify: \(\frac{5z+yz}{6}\)

Answer: D

Cheers,
Brent
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let y=2
so total cost ; z+z/6 ; 7z/6
substitute value of y =2 in given answer options
IMO D: gives 7z/6

$ z - First day
$ z/6 - For each additional day
z+z/6(y-1) = 5z/6 + zy/6 = 1/6(5z+zy) --- D is the correct ans.

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