Math 205 Test 2 Key. 1. Do NOT write your answers on these sheets. Nothing written on the test papers will be graded
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1 Math 20 Test 2 Key Instructions. Do NOT write your answers on these sheets. Nothing written on the test papers will be graded. 2. Please begin each section of questions on a new sheet of paper. 3. Please do not write answers side by side. 4. Please do not staple your test papers together.. Limited credit will be given for incomplete or incorrect justification. Questions. Arithmetic with repeating decimals (a) Example i. (3) Convert 9,, to decimal ii. (2) Add 9 + as fractions iii. (3) Add we change to a common denominator with the common denominator we add as decimals. Use the same addition algorithm used for non-repeating decimals iv. () Do your two answers match? Yes. Based on the calculations above we know that 20/ v. () How many repeating digits were in the sum? There are two repeating digits.
2 (b) Example 2 i. (2) Convert, 8 to decimal
3 ii. (2) Add + iii. (3) Add + as fractions. + first we change to a common denominator + + with the common denominator we add 8. as decimals. Use the same addition algorithm used for non-repeating decimals iv. () Do your two answers match? Yes. Based on the calculations above we know that 8/ v. () How many repeating digits were in the sum? There are six repeating decimals in the sum. 3
4 (c) Example 3 i. (2) Convert 4 333, to decimal ii. (2) Add as fractions first we change to a common denominator with the common denominator we add iii. (3) Add as decimals. Use the same addition algorithm used for non-repeating decimals. iv. () Do your two answers match? Yes. Based on the calculations above we know that 84/ v. () How many repeating digits were in the sum? There are six repeating decimals in the sum. 4
5 (d) Extrapolate i. (4) If you add a number with n repeating decimal digits to a number with m repeating digits, how many repeating digits will be in the result? Explain why. The result (sum) will be a number with lcm(n, m) repeating digits. If one repeats every n and the other every m then they both repeat every lcm(n, m) so that will be the new pattern. ii. () What problem solving method has been used in the Arithmetic with repeating decimals section? The guess and check method has been used.
6 2. Counting Factors (a) (9) Produce a factor tree for each of the integers 3, 4,, 6,, 8, 9, 0, 30. You should factor into primes (b) (9) Write the prime factorization for each of the integers 3, 4,, 6,, 8, 9, 0, (c) (3) For each integer write the number of distinct prime factors and the number of times each prime factor is repeated. n prime factors 3 3 (once) 4 2 (twice) (once) 6 2 (once), 3 (once) (once) 8 2 (thrice) 9 3 (twice) 0 2 (once), (once) 30 2 (once), 3 (once), (once) (d) (9) List all possible factors for each of the integers 3, 4,, 6,, 8, 9, 0, 30. 3, 3 4, 2, 4, 6, 2, 3, 6, 8, 2, 4, 8 9, 3, 9 0, 2,, 0 30, 2, 3,, 6, 0,, 30 6
7 (e) (2) Consider 42. The prime factorization of 42 is Note that the factor 2 can be written This shows which leaves of the factor tree were used. Rewrite your table of possible factors, writing each factor in this form , , 22, , , , , 8 2 0, 22, 42 2, , 33, , , , , , , (f) (2) Find a pattern relating the number of leaves and the number of possible factors. For numbers with two factors there are (n + )(m + ) factors when the first prime factor is repeated n times and the second prime factor is repeated m times. This is just the area of the table we created. For numbers with three factors there are (n + )(m + )(p + ) factors when each prime factor is repeated n, m and p times respectively. This is the volume of the cube built from the table. (g) () Test your pattern on three integers not in the list above , 2, 3, 4, 6, , 2, 4, 8, , 2, 3, 6, 9, (h) () What problem solving method has been used in the Counting Factors section? This is the making a table problem solving method.
8 3. Solving Fractions (a) Proportion: x. i. (4) Multiply both expressions (sides of the equation) by some expression to clear the denominators. Next solve for the value x. To clear both denominators we can multiply both sides of the equation (proportion) by the product of their denominators: x. x. x x x. x 0. x 2. ii. (4) Find a common denominator for the two fractions (sides of the equation). Next solve for x. Because the denominators do not share any factors, the common denominator is lcm(,x) x (the product of the denominators). iii. (4) Cross multiply then solve for x. x. x x x. x x 0 x. x 0. x 2. x. x (). x 0. x 2. iv. () Compare the common denominator to the expression you used to clear the denominator. The expression used to clear the denominator is the common denominator. This explains why the denominator goes away when using it to solve the proportion. 8
9 (b) (4) Solving rational equations: x 3 4 x 2. i. Multiply both expressions (sides of the equation) by some expression to clear the denominators. Next solve for the value x. x 3 4 x 2. x 3 (x 2)(x 3) 4 (x 2)(x 3). x 2 (x 2) 4(x 3). x 0 4x 2. x 2. ii. (4) Find a common denominator for the two fractions (sides of the equation). Next solve for x. Because the denominators do not share any factors, the common denominator is lcm(x 2,x 3) (x 2)(x 3) (the product of the denominators). iii. (4) Cross multiply then solve for x. 4 x 3 x 2. x 3 x 2 4 x 2 x 2 x 3 x 3. (x 2) 4(x 3) (x 2)(x 3) (x 2)(x 3). (x 2) 4(x 3). x 0 4x 2. x 2. 4 x 3 x 2. (x 2) 4(x 3). x 0 4x 2. x 2. iv. () Compare the common denominator to the expression you used to clear the denominator. The expression used to clear the denominator is the common denominator. This explains why the denominator goes away when using it to solve the proportion. (c) (2) Compare the methods of solution in the proportion and rational equation problems. The methods are the same. Learning to solve proportions teaches skills needed in algebra. 9
10 4. Explaining Arithmetic Operations (a) Multiplication i. (6) Show the following multiplications using the area model: each edge of a rectangle is marked with the length of one of the numbers. 3, and ii. (6) Show the following multiplications using groups. Draw repeated objects and draw boxes around groups to illustrate the problem and its solution. 3, 6 3, and iii. (4) Problem Solving. Ute is using granite tiles to cover her kitchen counters. When she reaches the end of a row, she ends up needing to cut one of the square tiles in half. At the end of another row she needs to cut this half tile in half so that it forms a square tile. If the original tile was 2 square, what is the area of the small square tile formed? Because the tile was cut in half along both sides, each side length is 6. Thus the area of the small, square tile is 36 square inches. iv. (4) Problem Solving. Guido is collecting Tux Cards. Each pack contains 0 Tux cards. At a swap meet Guido obtains 3 and /0 packs. How many cards did Guido obtain? Because each pack represents 0 cards we have ( 3+ ) cards. 0 v. () Which of the explanations of multiplication would help most for each of the problem solving questions above? For the tile problem, the area illustration of multiplication fits well. For the card problem, the repeated groups illustration works well. 0
11 (b) Division i. (6) Show the following division exercises using the subtractive interpretation. Draw a box around objects to indicate that they form a group. 3, 0 3, and ii. (6) Show the following division exercises using the subtractive interpretation using the following presentation. Draw a rectangle whose length is the first number. The height is irrelevant. Mark off lengths that have the length of the second number. 3, 0 3, and
12 iii. (6) Show the following division exercises using the partative interpretation. Draw a box around objects to indicate that they form a group. 3, 0 3, and iv. (4) Problem Solving. Wolfgang enjoys making sausages. Today he made weiss bratwurst. When he stuffed the sausage he ended up with a total length of 2 meters. He ties off the sausage every centimeters. How many sausages does he make? First we need to change units. 2 meters is 200 centimeters. We are dividing 200 centimeters into lengths of centimeters He makes 3 whole sausages and has a /3 of a sausage left over. v. (4) Problem Solving. As a promotion a store is giving away sets of Tux cards to the first people to arrive. They have 22 Tux cards. How many sets are given away? They need to divide 22 cards into groups of They can give a set of cards to 4 people. The fifth person will only receive 2 cards (the remainder). vi. (2) Which explanations of division best match these two problem solving questions? The sausage problem is best illustrated using the subtractive interpretation. The box diagram question above matches it well. The cards problem is best illustrated using the subtractive interpretation. 2
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