Let N be a set of some positive integer numbers with an average of 20 and containing the number 80. The number may or may not be distinct. However if one number equal to 80 is removed, the average drops to 18. What is the largest number that can possibly be contained in that set?
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What have you tried? – TheHolyJoker Jan 19 '20 at 17:49
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I am confused whether this is possible or not? – AzizStark Jan 19 '20 at 17:52
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1Well, try to construct some examples that work. That might give you an idea as to how to approach the problem. – lulu Jan 19 '20 at 17:52
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Have you tried any examples? That might or might not "work"? – TheHolyJoker Jan 19 '20 at 17:53
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I tried with some random numbers but no luck – AzizStark Jan 19 '20 at 17:56
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So, try systematically. Suppose there are $n$ elements in the original set ${a_1, a_2, \cdots, a_{n-1}, 80}$. Work from there. – lulu Jan 19 '20 at 17:58
2 Answers
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Hint. Step 1 is to find how many numbers are in the collection. Suppose there are $n$ initially. Then their total is $20n$. After removing 80, the total must be $20n-80$. There are now $n-1$ numbers with an average of 18, so the new total also equals $18n-18$. Hence $20n-80=18n-18$, so $n=31$.
Thus after removing the 80, we are left with 30 numbers totalling 540. Can you finish?
almagest
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average of 620 by 31 is 20. Now, I am trying to find the largest possible number – AzizStark Jan 19 '20 at 18:20
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@AzizStark Yes! You could have 29 1s and one 511 giving a total of 540. – almagest Jan 19 '20 at 18:28
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20 N - 80 = 18( N - 1 )
2N = 62
N = 31
30 * 18 = 540
So, 31 * 20 = 620
Largest Possible Number is: 620 - 80 - 1 * (Remaining Numbers)
i.e X = ( 620 - 80 - 1 * 29 )
Answer is : 511 !
AzizStark
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