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Re: Set A currently has five numbers, and its mean is 918,273,645,321. If [#permalink]
1
motion2020 wrote:
iampratikthorat wrote:
Honestly I cannot explain what I did but I just saw the last 3 digits were different. So I subtracted the two numbers and got 666. Then later just divided that number by 6 coz the addition of new number makes the number of entries 6. I got the answer 111. Now I added 111 to the number which was initially the mean of 5 numbers. And Arrived at option C and to my surprise it turns out to be Correct. So despite solving it please explain what I really should have done and what is the real method for solving such sums ?

Your answer and logic are correct.


Oh, Thanks. Generally I approach most Mean(averages) questions as Sums of Entities. Basically, sum of entities = Mean * number of entities. This formula is quite underrated in my opinion
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Re: Set A currently has five numbers, and its mean is 918,273,645,321. If [#permalink]
HI , in this method i subtracted both the mean and new entity by 918,273,645 . I got the mean as 321 and the entity as 987 . I then multiplied 321 * 5 and got the answer as 1605 and added 987 to it and then divided by 6 and got the answer as 432 which is the last digit of option C .

Is my method correct?
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Set A currently has five numbers, and its mean is 918,273,645,321. If [#permalink]
1
Yes that can be done as well.

Similarly,

As this is a large no, take the last two \(21\) as the mean of five nos, so total will be \(105\).

Now new no \(987\), but take last two \(87\) and add them to \(105\), so the sum will be \(192\).

When divided by \(6\), the answer will be \(32\). Last two digits of the correct answer.


aishumurali wrote:
HI , in this method i subtracted both the mean and new entity by 918,273,645 . I got the mean as 321 and the entity as 987 . I then multiplied 321 * 5 and got the answer as 1605 and added 987 to it and then divided by 6 and got the answer as 432 which is the last digit of option C .

Is my method correct?


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