Let the no. be 10x + y.No. formed by the interchange of digits = 10y + xWe have y - x = 2 .........(i)y + x = 14 .........(ii)Solving (i) and (ii), we get x = 6, and y = 8∴ the no. is 68.
Let the price of a pen be 'x' and that of pencil be 'y' Then 87x + 29y = 783Multiplying eqn (i) by 30/29 we get (87x) x 30/29 + (29y) 30/29 = 783 x 30/29∴ 90x + 30y = 810
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