Bunuel
A list contains 11 consecutive integers. What is the greatest integer on the list?
(1) If x is the smallest integer on the list, then (x + 72)^(1/3) = 4
(2) If x is the smallest integer on the list, then 1/64 = x^(-2)
Kudos for a correct solution. GROCKIT OFFICIAL SOLUTION:If we can determine the smallest integer on the list or a specific integer on the list when the list is written in increasing order, we can determine the greatest integer on the list.
1) Sufficient: We’re given one variable and one equation for the smallest integer on the list. That means we could solve for smallest integer and add 10 to find the greatest integer. If you don’t see this, consider:
\((x+72)^{\frac{1}{3}} = 4\)
Cubing both sides, x + 72 = 64. Then x + 72 = 64 and x = -8. Adding 10 to -8, the greatest integer is 2. Eliminate choices B, C and E.
2) Insufficient: If \(\frac{1}{64}=x^{-2}\) then \(\frac{1}{64}=\frac{1}{{x^2}}\) and \(x^2=64\), so x could be -8 or 8.
There are two different possibilities for the smallest integer on the list, so there must be two different possibilities for the greatest integer on the list. Statement 2) is insufficient, leaving
the correct answer choice as A.