Question

one of the digit of a two digit number is twice the other digit the sum of the original number and the number formed by reversing the digits is 99 find the number

Answer 1

let the digit at ones place be x
digit at tens place= 2x
original number= 20x+x
=21x
reversed digit number= 10x+2x
=12x
so
21x+12x=99
33x=99
x=3
Number= 21x = 21×3=63

Answer 2

Hi friend!

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The two digits = x and y
Original number = 10x + y
Reversed number = 10y + x

One of the digits is twice the other:
y = 2x

Sum of original and reversed number is 99:
10x + y + 10 y + x = 99
$11x + 11y = 99 \\ 11x + 11(2x) = 99 \\ 11x + 22x = 99 \\ 33x = 99 \\ x = \frac{99}{33} \\ x = 3$

Original number =
$10x + y = 10x + 2x = 12x = 12(3) = 12 \times 3 \\ = 36$

Note: If we take 'x = 2y' instead of 'y = 2x' then we'll get the answer as 63.
Both the answers 36 and 63 are correct for this question.

Hope you found my answer helpful. Keep Smiling!