Math for Early Childhood Educators 15-Math-151 Judy Butler Assignment 1 October 8, 2008
1.1 Pg. 7 Problem 4 & 9 4. Place the digits 4, 6, 7, 8, and 9 in the circles to make the sums across and vertically equal 19. In order to arrive at the correct answer for this problem I used the guess and check method. I started with the number 4 and placed it in the middle circle. I then found it was relatively easy to guess at the remaining possibilities to arrive at the correct numbers. I placed the 6 in the top position;the nine in the bottom position and along with the 4, these numbers equal 19. That left the 7 and the 8. Adding those numbers with the 4 will also make 19. There is no other way to complete this problem. Any other combination of numbers will exceed 19. 6 8
4
7
9
9. Using each of 1, 2,3,4,5, and 6 once and only once, fill in the circles so that the sums of the numbers on each of the three sides of the triangle are equal. I also used the guess and check method to arrive at the correct answer for this problem. I started by using the largest number available and placing it in the top circle. This would be the 6. I then began adding the additional numbers together to determine which numbers would work best. I discovered that by using numbers that would equal 6, I could add them with the original 6 and both of the sides of the triangle would equal 12. If I placed these remaining numbers in the correct order, I could place the remaining 3 in the middle and the bottom line would also equal 12. 6 2
1
4
3
5
There is another solution to this problem. The numbers can be changed so that the sum of the sides equal 9.
1 6 2
5 4
3
I discovered that if I flipped the numbers in the large triangle and placed them in the small inner triangle in the same order, I would get a correct answer.
Pg. 19 Problem 16 How many posts does it take to support a straight fence 200 feet long if a post is placed every 20 feet? I began the answering process for this problem by drawing a straight line that I would use for my 200 feet of fence. I knew that I would have to use a post at the beginning of the fence. I then knew that if I divided 20 into 200 I would need 10 additional posts. This would mean that I would need 11 posts to complete this process. 1______2_______3________4_________5________6________7_______8_______9______10_____ 11
Pg. 44 Problem 6 A rectangle table seats 6 people, I person on each end and 2 on each of the longer sides. Thus, two tables placed end to end seat 10 people. (a) How many people can be seated if n tables are placed in a line end to end? I know that my first table can seat 6 people if it stands alone. If I am placing another table next to it to form a row of tables then I will be able to place up to 5 individuals at the first table and 4 at subsequent tables. To seat an unknown number of people my formula would be A+4n. (b) How many tables, set end to end, are required to seat 24 people? Using this same process I found that I would need 6 tables. I knew that by multiplying the number of the seats at each table by the number 6 I would have the correct number of 24. This would seat the 24 people and 2 seats would need to be empty, or you could not use the 2 end seats. 5+4+4+4+4+5
or
4+4+4+4+4+4
Review Pg. 74 Problem 21 How many books must you choose from among a collection of 7 mathematics books, 18 books of short stories, 12 chemistry books and 11 physics books to be certain that you have at least 5 books of the same type? I began this problem by writing down the types of books in a row. I then knew that on the first pull of books I would need to take 6 to assure that I had at least 2 books of 1 subgroup. I then pulled 6 more and starting where I left off I placed these books. I then had 3 books in my first subgroup and needed to pull 5 more to assure that I had 5 books in this first subgroup. 18 short stories
11
1
12 chemistry books
11
1
1
7 math books
1 starting point for second pull 11
1
11 physics books
1
1
11
last pull
11
I found that the answer would be that I would need to pull 17 books in order to get one subgroup that had 5 books. After I thought about this problem a little more I decided that I could use the Pigeon Hole method and just pull 4 + 4 + 4 + 4 + 1 and get the 17 books that I would need.