<p>If two different, nonparallel lines both intersect a circle, what is the minimum total number of points on the circle that intersect at least one line?</p>
<p>A. 1
B. 2
C. 3
D. 4
E. 5</p>
<p>John has a penny, a nickel, a dime and a half-dollar (50 cents). How many different ways can John select different coins to make different, non-zero amounts of money? </p>
<p>A. 4
B. 8
C. 15
D. 16
E. 24</p>
<p>For the first question, some quick trial-and-error shows that the answer is 2, B. This occurs when you have two lines that intersect at a point on the circle itself. One of the lines will pass through another point on the circle, and the other line can be tangent to the circle at the point of intersection. Another way you can get 2 is if you have two tangent lines. Each intersects the circle at one and only one point. Thus, 2, B.</p>
<p>For the second question, either each coin is selected or not selected. Because we are looking for how many different ways John can select these coins, we do 2<em>2</em>2*2 (one ‘2’ for each coin). This equals 16. Then we subtract 1 case because we must have a non-zero amount of money. Next, we check different combinations of the coins to see if any add up to the same amount. A quick trial-and-error again makes it clear that none of the possible amounts of money are repeated. The answer is 16-1=15, C.</p>
<p>For the first one, I’d say B) 2. Both lines could be tangent to the circle. It cannot be A) 1 because that would imply the two lines are the same.</p>
<p>For the second one, note that the amount of money is uniquely determined by the coin combination (e.g. 11 cents can <em>only</em> be made with a dime and a penny). In higher-level mathematics this is called an injection.</p>
<p>There are 2^4 = 16 ways to select any combination of four coins. However this includes the case where we select no coins, so we have to subtract 1. The answer is C) 15.</p>