Review Sheet for MATH 302A at UA
Review Sheet for MATH 302A at UA
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Date Created: 02/06/15
Math 302A Review sheet for Exam II Multiplication 1 Construct a word problem which illustrates 67 gtlt 93 using the area context of mathematics but does not speci cally use the area of a rectangle Solution Solutions will vary but the typical solution will involve lining up objects by rows and columns say 67 rows and 93 columns The term rows and columns are ne when describing this context a more techinical term would be that they are both arrays Imelda Marco is trying to store her 20000 pairs of shoes In her bedroom with a cathedral ceiling she stacks some equal sized boxes along the back wall She can t 93 stacks of boxes where each stack has 67 boxes How many boxes did she t along the back wall One key observation is that the units in both arrays are the same Another key observation is that these contexts can be broken down into rows and columns Observe in the problem above Imelda has 93 columns and 67 rows of shoes 2 Construct a word problem which illustrates 6 gtlt 4 using the Cartesian product model Solution Solutions will vary but the typical solution should have two lists of options that can be combined in order to determine the total number of possible solutions An example could be After a long night of working on their 302A problem set a group of students decided to order a pizza They decide they want one veggie topping and one meat topping If the veggie options are onion pepper olive broccoli mushroom and spinach and the meat options are pepperoni sausage hamburger and chicken how many different pizzas could they choose from I know mushroom is not a veggie Let s say it is listed as a veggie option by the pizza place 3 Compare Multiplication as repeated addition versus multiplication as area Solution Solutions will vary perhaps one avenue of comparison is to reinterpret multipli cation in the context of area into the context of repeated addition Multiplication as repeated addition can be illustrated in the grouping context in other words I add the number of objects in a group or partition a number of times equal to how many groups or partitions I have Multiplication as area can be illustrated as lining up my objects into rows and columns Imagine each column as a group then each column will be a group which has a number of objects in it equal to the number of rows So this problem is now set up as mulitplication as repeated addition It is not enlightening to observe that both contexts get you the right answer what I am looking for is a comparision of the physical models You can use an illustrative example so long as you do not introduce numbers eg you can describe multiplication as repeated addition by talking about toys in chests dollars in bank accounts etc just don t write 11 toys in 6 chests or 5 dollars in 7 bank accounts 4 Kelly was asked to evaluate 83 gtlt 76 This is what Kelly did a I know I need to multiply the 10s places together so that s 80 gtlt 70 5600 b I know I need to multiply the 1s place together so that s 3 gtlt 6 18 c I know I need to multply the 10s place with the ls place so that s 70 80 gtlt 3 6 150 X 9 1350 d Now I add all of these numbers and get 5600 18 1350 6968 What accounts for the difference between Kelly s answer and the correct answer How would you describe Kelly s mistake to a fellow teacher Solution Kelly got a little too carried away with multiplication not only did she multiply the 10s place 83 by the 10s place and 1s place of 76 which is correct but she also multiplied it by the 1s place of 83 She did the same thing for 76 she multiplied 70 and 6 and added it to her result So her solution is too large and it is off by 80 gtlt 3 70 gtlt 6 240 420 660 Compare the area context of multiplication usually represented as a grided rectangle versus the Cartesian product model of multiplication usually represented as the set of all possible combinations between two sets Solution Solutions may vary A key observation is that you can illustrate both contexts in the following way A key difference is that the rows and columns in the area context have the same units and so the above picture depicts a speci c arrangement of objects while in the Cartesian product context the rows and columns have different units from one another and so the above picture represents a table whose entries are the possible combinations from the two available lists Another observation one could make is to reinterpret one context into the other For instance the Cartesian product context could be viewed as follows Given two lists which you want to combine in order to count the total number of possible outcomes Take one list and write it vertically in order to create rows Take the other list and write it horizontally in order to create columns The number of entries you have is the answer To help illustrate this I have provided an example below onion pepper olive mushroom broccoli spinach pepperoni sausage hamburger chicken Division 1 Scaffolding Algorithm 0 Apply the scaffolding algorithm to 264 29 Solution Solutions may vary if you use repeated subtraction you will end up subtract ing 29 a total of 9 times and be left over with a remainder of 3 0 Apply the scaffolding algorithm to 302 29 Solution Solutions may vary but it may be easy to start by subtracting 290 from 302 in order to get an answer of 10 remainder 12 0 Which of these two examples do you think is easier using the scaffolding algorithm Explain your reasoning Solution Since 29 gtlt 10 is larger than 264 there may not be an easy benchmark to start at in the rst problem whereas that is not the case in the second problem 2 Name one possible bene t which the scaffolding algorithm has over the standard division algorithm Solution You do not need to always have the exact number of times the divisor goes into the current place value 3 Compare division as repeated subtraction versus division as evenly grouping a given quantity into smaller quantities Solution Solutions will vary One main difference is the units of the numbers given In repeated subtraction the numbers have to be the same units eg it does not make sense to repeatedly subtract children from candy bars in other words the action is to repeatedly remove an amount equal to the divisor from the dividend until what is remaining is too small to subtract from In contrast subtraction as partitioning you do not chop o instead you distribute the dividend into a number of partitions so the units of the numbers given is different It is important to highlight that o The numbers in a division as repeated subtraction context are the same units and you are looking to nd out how many times you can take away the divisor from the dividend o The numbers in a division as partitioning context are different units and you are looking to see how the dividend can be evenly distributed into a certain number of partitions 4 List two concepts a child would need to understand in order to correctly use the standard division algorithm Describe why these concepts are important when using the standard algorithm Solution Solutions may vary Partial solutions that would still need to be expanded include 0 Place value in order to understand why a digit of the quotient goes in a particular spot in the algorithm o Multiplication in order to determine how much to subtract from the dividend at any given step 0 Subtraction to see how much of the dividend is left over after taking away some amount you determined from the previous step in the algorithm 5 Ruben wanted to divide 174 by 15 This was his workreasoning a 15 and 15 is 30 b repeated subtraction of 30 174 i 30 144 i 30 114 i 30 84 i 30 54 i 30 144 114 84 54 24 c I subtracted 30 ve times and 30 is 15 plus 15 so it s like I subtracted 15 ten times d So the answer is 10 Explain what Ruben did wrong Solution Justi cation may vary Ruben did not realize that he could take another 15 away from what he claimed was the remainder His method could be repeated using the scaffolding algoruthm where i might become apparent to him why you still need to take another 15 away from 24 6 Find the missing digits DD3 32D 115 431 Solution Solutions may vary 0 I need DD3 gtlt 32D to end in 1 So the only choice for the box in 32B is 7 since 3 gtlt 7 21 and 3 times any other digit between 0 and 9 does not end in 1 0 Evaluate 115 431 327 to nd the other number is 353 7 Fantasia evaluated 2170 divided by 30 as follows a 2170 30 217 3 b QRI 31217 7 07 Fantasia concludes that the answer is 72 remainder 1 Is this correct How could you use partitions to explain why her remainder is wrong Solution Let us illustrate this problem using units Suppose she is trying to solve the question If I have 2170 pennies and I want to give each of person I meet 30 pennies how many people do I give money to What she then did is convert her pennies into dimes and now she will give 3 dimes to each person she meets The number of people she will give money to in this scenario does not change but the number of coins she has left over would be different In the rst case she would have 10 pennies and the second case she would have 1 dime A more general solution would be to say 0 Before dividing the dividend and divisor by ten Fantasia wants to partition 2170 ob jects into 30 groups of objects 0 After dividing Fantasia wants to partition 217 groups of ten objects into 3 groups of groups often 8 Repeated Subtraction and Partitive Word Problems a Construct a word problem that illustrates 48 6 using repeated addition Solution Solution may vary but make sure 48 is clearly the dividend and 6 is clearly the quotient ie the units match up b Construct a word problem that illustrates 54 9 using partitive ie through partition ing Solution Solutions may vary but make sure 48 is clearly the dividend and 6 is clearly the divisor ie 48 are your objects and 6 is the number of bins you want to evenly distribute the 48 objects into c Examine the objects used in the above word problems Are the objects representing the dividend similar to the quotient or the divisor Why do you think this is Solution Already answered this the divisor will have different units than that of the dividend and quotient If the dividend and the divisor have the same units then the quotient will be a constant with no units attached to it
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