course Mth 151 Question: `q001. Let A stand for the collection of all whole numbers which have at least one even digit (e.g., 237, 864, 6, 3972 are in the collection, while 397, 135, 1, 9937 are not). Let A ' stand for the collection of all whole numbers which are not in the collection A. Let B stand for the collection { 3, 8, 35, 89, 104, 357, 4321 }. What numbers do B and A have in common? What numbers do B and A' have in common?
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Given Solution: `aOf the numbers in B, 8, 89, 104, 4321 each have at least one even digit and so are common to both sets. 3 is odd, both of the digits in the number 35 are odd, as are all three digits in the number 357. Both of these numbers are therefore in A ' . &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I’m just not getting this one. Self-critique Rating: ********************************************* Question: `q002. I have in a room 8 people with dark hair brown, 2 people with bright red hair, and 9 people with light brown or blonde hair. Nobody has more than one hair color. Is it possible that there are exactly 17 people in the room? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No. Because there are 19 people in the room: 8 (dark brown) + 2 (red) + 9 (light brown or blonde) = 19
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Given Solution: `aIf we assume that dark brown, light brown or blonde, and bright red hair are mutually exclusive (i.e., someone can't be both one category and another, much less all three), then we have at least 8 + 2 + 9 = 19 people in the room, and it is not possible that we have exactly 17. Self-critique (if necessary): I solved the problem just without issue. But I was a bit confused at the end when it said there should be 17 but I was like there can’t be 17 if there are this many people in the room. My self-doubt gets me every time. Self-critique Rating: OK ********************************************* Question: `q003. I have in a room 6 people with dark hair and 10 people with blue eyes. There are only 14 people in the room. But 10 + 6 = 16, which is more than 14. How can this be? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 6 (dark hair), 10 (blue eyes). Of the 14 people, some will have blue eyes and dark hair. 6 of the 10 can have dark hair. 4 blue eyes left with a different hair color and 4 people that are unknown. 6(blue eyes and brown hair ) 4( blue eyes and unknown hair color) 4 (unknown hair and eye color).
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Given Solution: `aThe key here is that there is nothing mutully exclusive about these categories-a person can have blue eyes as well as dark hair. So if there are 2 people in the room who have dark hair and blue eyes, which is certainly possible, then when we add 10 + 6 = 16 those two people would be counted twice, once among the 6 blue-eyed people and once among the 10 dark-haired people. So the 16 we get would be 2 too high. To get the correct number we would have to subtract the 2 people who were counted twice to get 16 - 2 = 14 people. Self-critique (if necessary): I arrived at sort of the same answer. Self-critique Rating: 2 ********************************************* Question: `q004. In a set of 100 child's blocks 60 blocks are cubical and 40 blocks are cylindrical. 30 of the blocks are red and 20 of the red blocks are cubical. How many of the cylindrical blocks are red? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 60 (cubical) and 40 (cylindrical) 30 of the blocks (red) and 20 the red blocks (cubical) 30 (red) – 20 (red) = 10 (red cylindrical)
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Given Solution: `aOf the 30 red blocks 20 are cubical, so the rest must be cylindrical. This leaves 10 red cylindrical blocks. Self-critique (if necessary): I got distracted by the extra information of 60 cubical blocks and 40 cylindrical blocks at first. But I eventually got there. Self-critique Rating: 1 "