Key concept: a.bc \(= a + b(\frac{1}{10}) + c(\frac{1}{100})\)
For example: \(3.25 = 3 + 2(\frac{1}{10}) + 5(\frac{1}{100})=3+0.2+0.05=3.25\)

Jorge wrote a check for $x.yz BUT recorded it as $x.zy. His bank statement showed a balance that was $0.54 greater than what his records showed
This tells us that: $x.zy - $x.yz = 0.54
Now apply the above concept to write: \([x + z(\frac{1}{10}) + y(\frac{1}{100})]-[x + y(\frac{1}{10}) + z(\frac{1}{100})]= 0.54\)

Subtract \(x\) from both sides to get: \([z(\frac{1}{10}) + y(\frac{1}{100})]- [y(\frac{1}{10}) + z(\frac{1}{100})]= 0.54\)

Multiply both sides by \(100\) to get: \([10z + y]- [10y + z]= 54\)

Simplify to get: \(9z - 9y= 54\)

Divide both sides by \(9\) to get: \(z - y= 6\)

Rearrange to get: \(z = y + 6\)

Since y is a DIGIT, the possible solutions are
y = 0 and z = 6
y = 1 and z = 7
y = 2 and z = 8
y = 3 and z = 9
And that's it!

Since z can be 6, 7, 8 or 9, the correct answer is E

Cheers,
Brent