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Bunuel
What is the ratio of a two-digit number and its sum of digits?

(1) The ratio is the minimum possible.
(2) The number has 9 as its unit digit.

Let the ratio of 2-digit number and its sum of digit be
(10x+y)/(x+y) = 1 + (9x/(x+y)) (x cannot be 0)

(1) ratio is minimum

9x/(x+y) will be minimum if x = 1 and y = 9 i.e. 9/10 which will become 1 + 9/10 = 19/10

if x = 2 and y = 9 then ratio will be 18/11 (not minimum) and so on in all other cases it will exceed 1

Hence Sufficient

(2) y = 9
then
1+ (9x/(x+y))
but we do not know the value of x ( any digit from 1-9)

Not Sufficient

Hence A
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Let the number be \(ab\) => \(10a + b\)

question: \((10a + b) / (a + b)\) => \((9a + a + b)/(a+b)\) => \((9a/(9a + b)) + 1 =\) ?

Statement 1:
the above equation \((9a/(9a + b)) + 1\) is minimum when a = 1 and b = 9
so ratio = \((9/10) + 1\) is minumum => a = 1, b = 9 => number is 19 => sufficient

Statement 2: units digit is 9, but no idea about tens digit => InSuff

Answer (A)
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