Carcass wrote:
Jorge's bank statement showed a balance that was $0.54 greater than what his records showed. He discovered that he had written a check for $x.yz and had recorded it as $x.zy, where each of x, y, and z represents a digit from 0 though 9. Which of the following could be the value of z ?
A. 2
B. 3
C. 4
D. 5
E. 6
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 showedThis 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