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Seven times a two digit number is equal to 4 times the number obtained by reversing the order of the digits and the sum of the digit is 3 find the number.

Let first digit is x & second digit is y. then the number is (10x + y) x + y = 3 -------------------------(1) (given) 7(10x + y) = 4(10y + x) (given) 2x = y put the value of y in equation (1) 3x = 3 x = 1 y = 2 hence number is (10*1 + 2) = 12

seven times a two digit number is equal to 4 times the number
obtained by reversing the order of the digits and the sum of the digit
is 3 find the number. Solution : let the digit at units place = x and the digit at tens place = y therefore original number = 10y +x reversed Number = 10x+y According to question , 7 (10y +x) = 4 (10x+y) ............(given) 70y + 7x = 40x +4y 70y - 4y = 40x - 7x 66y = 33x 2y = x x =2y ...............(1)

Sum of digits of two digit number = 3 ........(given) x + y = 3 Substitute x in above equation , we get 2y + y = 3 3y =3 y= 3/3 y=1 substitute y in 1 , we get x=2*1 x=2

Hence original number = 10y+x = 10*1 + 2 = 10 + 2 = 12