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A class had 26 students in a test one student made 12 mistakes the rest of the class made less mistakes than him prove that at least 3 people in the class made the same number of mistakes?
Kidfat
You have 25 other people. If 2 people made the same number of each mistake from 0 to 11 this would only account for 24 people, so at least one particular mistake must have been made by 3 people.
0,0,1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10,11,11. This accounts for only 24 people...
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A class comprises 26 students During a mathematics test one student made 12 mistakes and the rest of the class made less mistakes than he did. Prove that there were at least 3 students in the class wh?
I got that question too. It's very badly worded. When they are asking for 3 students in the class that made the same mistakes, they are not asking for at least 3 people that made 12 mistakes. They're asking for u to prove that there were at least 3 people that made the same mistakes as eachother.
In a class of 25 pupils, 19 have scientific calculators and 14 have graphic calculators. If xpupils have both and y pupils have neither, what are the largest and smallest possiblevalues of x and y?
The largest value of X is, obviously, 14. There are only 14 graphic calculators, so if every student has who has a graphic calculator also has a scientific calculator, then there are 14. The smallest possible value of X is more complex. 19+14=33, which is how many students we would need for there to be no overlap. Since there's only 25 students, there must be some overlap. 33-25=8, the difference between how many students that we need for no overlap and the number that we have. So let's do some quick math to verify. The number of students who have only scientific, plus the number of students who have only graphic, plus the number who have both should be 25. (19-8) + (14-8) + 8 = 25. So 8 is the smallest possible overlap. Any less than that would require more than 25 students. We already know, at this point, what the smallest possible value of Y is: zero. We already demonstrated that there aren't enough students to spread these calculators out with zero overlap, so we already know that it's possible for every student to have at least one calculator, leaving 0 who have none. The most students who will have neither is the same scenario as the most students who have both: 14 people have both. 19-14 =5. So there's 14 people with both, leaving 5 with only a scientific calculator. 25-5-14 = 6 people who have neither in this case. 6 is the highest possible value of Y. In total: X is between 8 and 14. Y is between 0 and 6.
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Two fifths...(assuming all the people are students)
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If you choose one person, then consider your options for a second, you have nine people for them to play against. Then, considering the second person, excluding the match already played with the first person, you have eight possible matches. Following this all the way, the number of unique matches are: 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45
Common mistakes people make when using relevant-difference or common thread reasoning?
All of the above
A class comprises 26 students During a mathematics test one student made 12 mistakes and the rest of the class made less mistakes than he did. Prove that there were at least 3 students in the class wh?
I got that question too. It's very badly worded. When they are asking for 3 students in the class that made the same mistakes, they are not asking for at least 3 people that made 12 mistakes. They're asking for u to prove that there were at least 3 people that made the same mistakes as eachother.
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