Question : Let XYZ be a 3-digit number and the difference between XYZ and ZYX is equal to PQR. Is (P + R) equal to Q ?
Statement-I : P = 3
Statement-II : R = 6
Which one of the following is correct in respect of the above Question and the Statements ?
Let the 3-digit number XYZ be represented algebraically as: \(XYZ = 100X + 10Y + Z\) Let the 3-digit number ZYX be represented algebraically as: \(ZYX = 100Z + 10Y + X\) For XYZ and ZYX to be 3-digit numbers, \(X \neq 0\) and \(Z \neq 0\). Assuming \(XYZ > ZYX\), we must have \(X > Z\). Thus, \(X > Z \ge 1\). The difference \(X - Z\) is a positive integer between 1 and 8.
The difference is calculated as:
\((XYZ - ZYX) = (100X + 10Y + Z) - (100Z + 10Y + X)\) \(= 99X - 99Z\) \(= 99(X - Z)\)This difference equals PQR, represented as \(100P + 10Q + R\). Therefore, \(100P + 10Q + R = 99(X - Z)\).
Let's examine the structure of \(99(X - Z)\):
In all cases, the tens digit \(Q\) is 9, and the sum of the hundreds digit \(P\) and the units digit \(R\) is also 9 (\(P+R=9\)). Therefore, \(P + R = Q\) is always true.
The question "Is (P + R) equal to Q?" can be answered affirmatively using only the general properties of the number system, specifically the result \(XYZ - ZYX = 99(X - Z)\), which leads to \(P+R=Q\). The specific values \(P=3\) (Statement-I) and \(R=6\) (Statement-II) are not required to arrive at this conclusion.
Hence, the question can be answered even without using both statements.
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