OA: C

Let the number be \(XY\) , where \(X\)= Digit at tens place and \(Y\) = Digit at unit's place

Number formed by reversing the digits \(= YX\)

As per Question

\(\frac{XY}{YX}=\frac{4}{7}\)

\(\frac{10X+Y}{10Y+X}=\frac{4}{7}\)

\(70X+7Y = 40Y+4X\)

\(66X=33Y; \quad 2X=Y\)

Taking \(X=1,\quad Y\) would be \(2\)

Taking \(X=2,\quad Y\) would be \(4\)

Taking \(X=3,\quad Y\) would be \(6\)

Taking \(X=4, \quad Y\) would be \(8\)

~~Taking X=5,Y would be 10~~( Not possible as Y is single digit)

Possible value of \(XY : 12,24,36,48\)

Possible value of \(YX : 21,42,63,84\)

Their Sum\(= (12+24+36+48) + (21+42+63+84) = 12(1+2+3+4)+21(1+2+3+4)= 12*10+21*10 =10(12+21)=10*33=330\)

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