(a) Describe in words the elements of A U B, and list them. The elements of A U B
will include anything in either A or B or both, so those are positive multiplies of ...
Midterm 1 Practice Exam Solutions
12. Describe the set of natural numbers less than 40 that are divisible by 4 using (a) Set-builder notation
{x|0 < x < 40, x = 4k, for some k ∈ Z}
(b) List notation
{4, 8, 12, 16, 20, 24, 28, 32, 26}
13. Determine how many subsets a set of 2 elements will have. How many subsets will a set of 3 elements have? How many subsets will a set of 4 elements have? How many subsets will a set of n elements have?
A set with 2 elements, say they are a and b, will have four subsets, namely ∅, {a}, {b}, and {a, b}. A set with 3 elements, say they are a, b, and c will have eight subsets, namely ∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, and {a, b, c}. A set with 4 elements, say they are a, b, c, and d, will have 16 subsets, which I won’t list this time :) A set with n elements will have 2n subsets. (think of each element have two choices, to be in the subset or not, and there are n of those elements)
14. Let U = the set of whole numbers, A = {4, 8, 12, 16, 20, . . .} and B = {3, 6, 9, 12, 15, 18, . . .}. (a) Describe in words the elements of A ∪ B, and list them.
The elements of A ∪ B will include anything in either A or B or both, so those are positive multiplies of either 3 or 4, or multiplies of both. A ∪ B = {3, 4, 6, 8, 9, 12, 15, 16, 18, 20, . . .} (b) Describe in words the elements of A ∩ B, and list them.
The elements of A ∩ B are elements which must live in both A and B, so they must be positive multiples of both 3 and 4, that is, they are positive multiples of 12. A ∩ B = {12, 24, 36, 48, . . .}
15. Use shading of Venn diagrams to determine whether or not A ∪ B = A ∪ B. Label your diagrams and explain how you came to your conclusions.
A includes sections 1 and 4, B includes sections 1 and 2. So, A ∪ B includes sections 1, 2, and 4, which is all sections except the very middle section, which is A ∩ B. A ∪ B includes sections 2, 3, and 4. So, A ∪ B includes only one section, just Section 1 (none of the circle sections). 16. Let U = the set of students at your university E = the students who are elementary education majors S = the students who have scholarships G = the students who are active in student government (a) Represent E ∩ S ∪ G with a Venn diagram and in everyday English (i.e. in words). An elementary education major who neither has a scholarship nor is active in student government. Venn diagram includes only one section, the one which is ONLY E.
(b) Represent E ∩ S ∩ G with a Venn diagram and in everyday English (i.e. in words). A student who is active in student government and has a scholarship, but is not an education major. Venn diagram includes only one section, where S and G intersect, but do not include any part of E.
(c) Represent the following set with a Venn diagram and in symbols: A UAF student in student government who pays his own tuition and majors in elementary education. G ∩ S ∩ E Venn diagram includes only one section, where E and G intersect, but do not include any part of S.
(d) Represent yourself in words, with a Venn diagram, and in symbols. This answer clearly depends on you. I, personally, am not a student at all so there is no real representation via symbols or the Venn digram for me. 17. If set A has 32 elements, set B has 18 elements, and the number of elements in A ∩ B is 12, how many elements are in A ∪ B? Since A has 32 elements, but A ∩ B has 12, there must be 20 elements which are in A but not in B. There must, then, be 6 elements in B which are not in A. To count the total in the union, I would count those that are in only one of the two sets (20+6) as well as those that are in both sets (12). Therefore, A ∪ B has 20+6+12 = 38 elements. 18. The first five square numbers are 1, 4, 9, 16, 25. What is the next square number? Write a rule for the nth square number. The pattern is that we are taking the digits 1,2,3, etc. and squaring them. So, the next square number will be 62 = 36. The nth square number will be Sn = n2 . 19. The first five triangular numbers are 1, 3, 6, 10, 15. What is the next triangular number? Write a rule for finding the nth triangular number. We can first recognize the pattern that the differences between the numbers are increasing by 1 at each step. Therefore, the next triangular number is 15 + 6 = 21. Then, we notice that this means that we are really adding the numbers 1 +2+3, etc. until n for the nth triangular number. In class, we discussed that this is equal to, Tn = n(n+1) . 2 20. If a function is such that (1,3), (2,5), (3,7), and (4,9) are ordered pairs, what is a possible rule for the function? We can see that what is produced is actually outputs that are increasing odd numbers. So, to write a specific rule, I will start with the fact that an odd number can be written as 2n + 1. Then, I will see if this exact rule works out, or if I have to change my n value at all. For n = 1, 2n + 1 = 2(1) + 1 = 3. So, this rule works! f (n) = 2n + 1.
21. A printing company charges $10.00 to print the first 100 copies of a document and $0.08 for each copy beyond 100. (a) Write the formula for a function that gives the total printing costs given the number of copies to be printed if more than 100 copies are to be printed. Let x = the number of copies total. I will need to factor in that the first 100 are a set cost, so I will multiply the $0.08 per copy times only (x − 100). So, our total cost is, C1 (x) = 10 + 0.08(x − 100), where x > 100. (b) How much would it cost to print 145 copies of a document? [3pts] C1 (145) = 10 + 0.08(145 − 100) = 10 + 3.60 = $13.60. (c) Another company charges $0.09 for each copy, no matter how many copies you want printed. Compare the two companies. Which would you choose? Why? What is the breaking point for choosing one versus the other? I will use a new function C2 (x) = 0.09x to represent the cost for this company for x copies, where x ≥ 0. Then, I will set C1 and C2 equal to one another to find the breaking point between the two. I know that clearly I will choose the second company if I have a small number of copies, and the first if I have a very large number of copies, but I am not sure where the exact cut-off would be. If I try exactly 100 copies, I saw that C1 (100) = 10, but C2 (100) = 9, so at this point the second company is still cheaper, so my cut-off is above 100. I think I will try some algebra to solve this a bit more exactly, because I think guess and check might take quite awhile here. 10 + 0.08(x − 100) = 0.09x 10 + 0.08x − 8 = 0.09x 2 + 0.08x = 0.09x 2 = 0.01x 200 = x This means that the breaking point is 200 copies. When I have 200 copies to make, the costs are exactly equal. If I make less than 200 copies, I will choose the second company. If I need to make more than 200 copies, I will choose the first company. 22. What is the base five number for 176? We are starting from the decimal number 176 (this should be clear because it includes numbers bigger than 4, which are not allowed in base 5). To convert to base 5, I need to find out how many 1’s, 5’s, 25’s, and 125’s are in the number 179. Starting at the largest unit, there is one unit of 125. There is then 51 left over. There are two units of 25 in this, and one unit of 1. So, the base five number equivalent to 176 is 1201.
23. Write the base five number for 59.
We are starting from the decimal number 59 (this should again be clear because it includes numbers bigger than 4, which are not allowed in base 5). To convert to base 5, I need to find out how many 1’s, 5’s, 25’s, and 125’s are in the number 59. Starting at the largest unit, there are two units of 25. There is then 9 left over. So, there is one unit of 5 in this, and four units of 1. So, the base five number equivalent to 59 is 214. 5910 = 2145
24. Write 150 in base 6 and in base 7.
We start by denoting that in base 6, we have a ones place, a sixes place, a 36’s place, and a 216’s place (if we need it). The largest one that goes into 150 is 36, and there are 4 of those (making 144 of the 150). There are 6 left over, so that is one of the 6’s. 15010 = 4106 In base 7, we have 1’s, 7’s, 49’s, and 343’s. The largest one that goes into 150 is 49’s, and there are 3 of those (making 147 of the 150). There are 3 left over, so that must go in the 1’s place (not enough for a 7). 15010 = 3037
