# The digit at the tens place … ## CBSE, JEE, NEET, NDA

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The digit at the tens place of a two digit number is four times that in the units place. If the digits are reversed, the new number will be 54 less than the original number. Find the original number.

Preeti Dabral 1 month, 1 week ago

Let the digits at unit place be y. Let the digits at ten's place be x.
Number = 10x + y
Sum of digits of the two digits number is four times that in the unit's place. => x + y = 4y => x = 4y - y => x = 3y ......(i).
If the digits are reversed the new number will be 54 less that the original number.
Number obtained by reversing the digits = 10y + x
Number obtained by reversing the digits = 10x + y - 54 => 10y + x = 10x + y - 54 => 54 = 10x + y - 10y - x => 54 = 9x - 9y => 54 = 9(x - y) => 54/9 = x - y => 6 = x - y .........(ii).
Putting the value of x from Eq (i). in Eq (ii). => 6 = x - y => 6 = 3y - y => 6 = 2y => y = 6/2 => y = 3
Putting the value of y in Eq (ii). => 6 = x - y => 6 = x - 3 => -x = -6 - 3 => -x = -9 => x = 9
So, Number = 10x + y => 10(9) + 3 => 90 + 3 => 93
Hence, the original number is 93.

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