Math Final Exam Solutions - Spring Jaimos F Skriletz 1

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1 Math Final Exam Solutions - Spring Jaimos F Skriletz 1 1. Consider the universal set U = Z 10 = {1,2,3,4,5,6,7,8,9,10} and the following subsets. A = {2,4,6,8} B = {2,3,5,7} C = {2,3,4,8,9,10} Find the following sets. (a) A B A B = {2,3,4,5,6,7,8} (b) A C C = {1,5,6,7} A C = {6} (c) (A B) C A = {1,3,5,7,9,10} A B = {1,2,3,5,7,9,10} (A B) C = {2,3,9,10} (d) B C B C = {(2,2),(2,3),(2,4),(2,8),(2,9),(2,10), (3,2),(3,3),(3,4),(3,8),(3,9),(3,10), (5,2),(5,3),(5,4),(5,8),(5,9),(5,10), (7,2),(7,3),(7,4),(7,8),(7,9),(7,10)} 2. Consider the following statements about set theory, where A and B are subsets of a universal set U. If the statement is true for all possible sets prove it, otherwise provide a counter example to disprove it. (a) A A = U This is a true statement. First A and A both live in the same universe and thus A A U. Second if x U then either x A or x A (x is either in A or it is not in A). Thus U A A. Thus U A A. The above two statements prove that A A = U. (b) A B A B False. Let A = {1}, B = {2} then A B = {1,2} while A B =. Since nothing is in the empty set it is not possible for A B A B. (c) n(a B) = n(a) + n(b) False. Let A = {1,2,3} and B = {2,3,4}. Then A B = {1,2,3,4}. So n(a B) = 4 while n(a)+n(b) = 3+3 = 6.

2 Math Final Exam Solutions - Spring Jaimos F Skriletz 2 3. Define the following, be sure to use truth tables were required. (a) Conjunction: Given two statements P and Q, the conjunction is the compound statement P Q (P and Q). The truth of the conjunction is defined by the following truth table. P Q P Q T T T T F F F T F F F F (b) Disjunction: Given two statements P and Q, the disjunction is the compound statement P Q (P or Q). The truth of the disjunction is defined by the following truth table. P Q P Q T T T T F T F T T F F F (c) Negation: Given the a statement Q, the negation is the statement Q, whose truth is the opposite of Q. P P T F F T (d) Conditional: Given two statements, the conditional statement is the compound statement P Q. The truth of the conditional is defined by the following truth table. P Q P Q T T T T F F F T T F F T (e) Converse: Given the conditional statement P Q, the converse statement is Q P. (f) Biconditional: Given two statements P and Q, the biconditional is the compound statement P Q (P Q) (Q P).

3 Math Final Exam Solutions - Spring Jaimos F Skriletz 3 4. Determine weather the following arguments are valid or invalid. (a) No mathematics test is fun. All fun things are worth your time. No mathematics test is worth your time. This argument can be described using the following Euler diagram. (b) As shown the premises do not not force the region of math exams to be outside the worth your time region. Since the truth of the conclusion is not forced this is an invalid argument. If I get finical aid, I will register for classes. I registered for my classes. I received my finical aid. Let P be the statement I received my finical aid and Q be the statement I registered for class. This argument is of the form P Q Q P This is an invalid argument and is known as the fallacy of the converse. The conditional only pushes the truth from the first statement to the second, this argument requires the reverse of this (i.e. the conditional statement).

4 Math Final Exam Solutions - Spring Jaimos F Skriletz 4 5. Consider the rectangle ABCD and the diagonal AC that cuts the rectangle into two triangles ACB and ACD as shown. (a) Prove that the triangles ACD and ACB are congruent. Since ABCD is a rectangle, the following AB CD BC AD AC AC Thus by the side-side-side condition, the two triangles are congruent. (b) If the measure of angle ACD is 35, find the measure of ACB, CAD and CAB. Since ABCD is a rectangle, ADC = 90 The three angles of a triangle add up to 180, so DAC = = 55 Since the two triangles are congruent from part (a), all the angles are known, as shown below.

5 Math Final Exam Solutions - Spring Jaimos F Skriletz 5 6. Area and Perimeter. (a) Find the area and perimeter of the following figure (a rectangle with a semicircle attached to each end). This figure is created from a 6x10 rectangle and a circle of radius 3. The area of the rectangle is A R = 6 10 = 60 The area of the circle is Thus the total area is A C = πr 2 = π(3) 2 = 9π A = A R + A C = π square feet The length around the circle is P C = 2πr = 2π(3) = 6π This included with the two straight sections that are 10 feet each, gives the total perimeter to be P = P C = π feet (b) Find the area of a generic trapezoid whose two lengths are a and b and height is h as shown. hint: divide the area into multiple regions you know. Divide this up into two regions. Region one is a rectangle with length a and width h. The second region is both triangles which make a rectangle of length 1 2 (b a) and width h. The area of the first rectangle is A 1 = a h The area of the second rectangle (both triangles combined) is A 2 = 1 (b a) h 2 Thus the total area is A = A 1 + A 2 = ah (b a)h = 1 h(a + b) 2 The final expression comes from some algebra.

6 Math Final Exam Solutions - Spring Jaimos F Skriletz 6 7. A dart board is constructed out of an 8 inch circle. The dart board is split into two regions as shown. The first region, E1, is the bulls eye which is an 1 inch circle. The second is the otter ring, E2, which is 1 inch wide and formed by a 5 and 6 inch circle. (a) Let S be the whole dart board. Find the area for the regions S, E1 and E2. The area of the full circle is S = π(8) 2 = 64π The area of the inner circle is The area of the ring is E 1 = π(1) 2 = π E 2 = π(6) 2 π(5 2 ) = (36 25)π = 11π (b) If you have a uniform chance to hit any point on the dart board, what is the probability to hit the bulls eye, E1? P(E 1 ) = E 1 S = π 64π = 1 64 (c) If you have a uniform chance to hit any point on the dart board, what is the probability to hit either of the shaded regions, E1 or E2? Since E 1 and E 2 are disjoint, we have P(E 1 E 2 ) = E 1 + E 2 S = π + 11π 64π = = 3 16