Bunuel
If a is a positive integer less than 3 and b is a positive integer less than 11, what is the value of the decimal expression of a/b when rounded to the nearest tenths?
(1) When the decimal expression of a/b is rounded to the nearest hundredths the value of the hundredths digit is 1.
(2) When the decimal expression of a/b is rounded to the nearest ten-thousandths the value of the ten-thousandths digit is 9.
0<a<3, so a can be 1 or 2.
0<b<11, so b can be anything from 1 to 10.
(1) When the decimal expression of a/b is rounded to the nearest hundredths the value of the hundredths digit is 1.
So \(0.x05\leq{\frac{a}{b}}<{0.x15}\)
The answer has more to do with b than a, as a has only 2 values.
Now, when b=1, 2, 5 or 10, we will have no hundredths digit..Example \(\frac{1}{10}=0.2 \ \ and \ \ \frac{1}{2}=0.5\)
when b=4, \(\frac{a}{4}=\frac{1}{4}=0.25 \ \ or \ \ \frac{2}{4}=0.5\)..\(b\neq{4}\)
when b=3, \(\frac{a}{3}=\frac{1}{3}=0.333 \ \ or \ \ \frac{2}{3}=0.6666\)...\(b\neq{3}\)
when b=6, \(\frac{a}{6}=\frac{1}{6}=0.166667 \ \ or \ \ \frac{2}{6}=0.3333\)...\(b\neq{6}\)
when b=7, \(\frac{a}{7}=\frac{1}{7}=0.14285 \ \ or \ \ \frac{2}{7}=0.284\)...\(b\neq{7}\)
when b=8, \(\frac{a}{8}=\frac{1}{8}=0.125 \ \ or \ \ \frac{2}{8}=0.25\)...\(b\neq{8}\)
when b=9, \(\frac{a}{9}=\frac{1}{9}=0.111 \ \ or \ \ \frac{2}{9}=0.2222\)...\(b={9} \ \ a=1\)
\(\frac{a}{b}=\frac{1}{9}=0.1111\)
answer = 0.1
sufficient
(2) When the decimal expression of a/b is rounded to the nearest ten-thousandths the value of the ten-thousandths digit is 9.
So \(0.xyz85\leq{\frac{a}{b}}<{0.xyz95}\)
We have already done the calculations above, and only value that fits in is a=1 and b=7 => \(\frac{a}{7}=\frac{1}{7}=0.14285\).
\(\frac{a}{b}=\frac{1}{7}=0.14287\)
answer = 0.1
Sufficient
D
It would be better if the statement II is 'the value of the ten-thousandths digit is 1', so that value of b is 9 in each case.