Bunuel wrote:

Official Solution:

If \(a=0.xyz\), where \(x\), \(y\) and \(z\) are digits from 0 to 9, inclusive, is \(a \gt \frac{2}{3}\)?

First of all \(\frac{2}{3}\) is a recurring decimal 0.666...

(1) \(x+y \gt 13\). The least value of x is \(5 (5+9=14 \gt 13)\), so in this case \(x=0.59z \lt 0.66(6)\) but \(x=7\) and \(y=9\) is also possible, and in this case \(x=0.79z \gt 0.66(6)\). Not sufficient.

(2) \(x+z \gt 14\). The least value of \(x\) is \(6 (6+9=15 \gt 14)\), but we don't know the value of \(y\). Not sufficient.

(1)+(2) The least value of \(x\) is 6 and in this case from (1) the least value of \(y\) is \(8 (6+8=14 \gt 13)\), hence the least value of \(a\) is \(0.68z \gt 0.66(6)\). Sufficient.

Answer: C

Hi Bunuel,

Here in S1 the value of x and y has been taken as 5,9 and 7,9.

and in S2 the value of x and z has been taken as 6,9.

Then how have we arrived at the conclusion that the lowest value of x is 6? I am not able to understand the logic, may you pl elaborate on it. Also why cant the value assumed in both the statements be reversed, after all there is nothing in the question stem that suggests that x is greater/smaller than y or z.

Regards