S96-08

Official Solution:


The first step is to figure out what region \(Q\) represents. Let's consider the boundary of region \(Q\) by ignoring the "less than" part: \(x^2 + y^2 = 100\). This equation represents a circle in the \(xy\)-plane, centered on the origin, with a radius of 10 (the square root of 100). Thus, region \(Q\) consists of all points on or inside this circle. We are asked whether point \((a, b)\) lies inside this region. We can rephrase the question by substituting \(a\) for \(x\) and \(b\) for \(y\): do the variables \(a\) and \(b\) always satisfy the inequality \(a^2 + b^2 \le 100\)?

Statement (1): INSUFFICIENT. Relatively quickly, we can find a point on the line \(a + b = 14\) that does not satisfy the inequality. Choose \(a = 0\). Then \(b = 14\), and the sum of the squares is 196, which is greater than 100. Thus, in this case, \((a, b)\) would not fall within region \(Q\).

However, can we find any point on or within the circle? If we make both \(a\) and \(b\) equal 7, then the sum of their squares is \(49 + 49 = 98\), which is less than 100. We could also choose \(a = 8\) and \(b = 6\), which gives us the sum of squares \(64 + 36 = 100\). Either case satisfies the inequality, and so \((a, b)\) in these cases would fall within region \(Q\).

Statement (2): INSUFFICIENT. The condition that \(a\) is greater than \(b\) is not very restrictive. We can find points that meet this condition both inside and outside the circle. For instance, \((1, 0)\) is within the circle, but \((101, 100)\) is not.

Statements (1) and (2) together: INSUFFICIENT. The case from statement 1 in which both \(a\) and \(b\) equaled 7 is no longer valid, but the case of \(a = 8\) and \(b = 6\) still works. Thus, we have a point satisfying both statements that lies on the circle. In fact, some suitable points lie within the circle, such as \((7.5, 6.5)\). However, we can still find suitable points that lie outside the circle - for instance, \((14, 0)\).


Answer: E