The sum of the digits of a two-digit number is 11. If the digits are reversed, the new number is 27 more than the original number. What is the original number?
47
Let the tens digit be x and units digit be y, with \(x+y=11\).
Reversed number minus original: \((10y+x)-(10x+y) = 27 \Rightarrow 9y-9x=27 \Rightarrow y-x=3\).
Solving \(x+y=11\) and \(y-x=3\): \(y=7,\ x=4\).
Original number: \(10\times4+7 = 47\).
Hence, the original number is 47.
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