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Let the Units digit of the Number be Y and the Tens digit be X. The number can be represented as 10X + Y
According to the Question Stem: X + Y = 10
When the number is reversed we get 10Y+X as the new number.
According to the question stem: (10x+Y) - (10 Y + X) = 54 , Simplifying we get 9X - 9Y = 54 or 9 (X-Y) = 54 or X-Y = 6.
Putting this in the original equation we get two equations ; X + Y = 10 (1) and X - Y = 6 (2)
Solving 1 & 2 we get X = 8. Y comes out to 2 , and the Original Number is 82. The new number is 28 (A)
Reversing digits gives a number 54 less than original => difference in digits =54/9 = 6 Only option A satsifies this.
Ans A it is.
Can you further explain this in detail? How did you get 54/9 ? Where did the 9 come from?
This 9 is logical derivation. Difference between a 2 digit number and number with reversed digit is always 9 times the difference between digits. Any number 10a+b when reversed would be 10b+a , the difference would be 9(b-a)
Re: The sum of the digits of a two digit number.. [#permalink]
13 Jul 2016, 09:20
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