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Four of the six numbers $$1867,\quad 1993,\quad 2019,\quad 2025,\quad 2109,\quad \text{and}\quad 2121 $$ have a mean of 2008. What is the mean of the other two numbers?

I would like to get help for this problem because i want to find a way to answer this problem without using guess and check. thank you :)

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Please make the body of your message self-contained, not relying on the subject for content. –  Arturo Magidin May 11 '11 at 20:00
6  
This is a great example of how math works (in an elementary example). A beginner will think (I'm guessing): compute -all- means of subsets of size four out of six, find one that is 2008, and then compute the means of what is left over. And of course this is too hard, too much computation. And the beginner would then just give up, but the more experienced would try -not- to do all that work. –  Mitch May 11 '11 at 21:51
    
Also, please don't use (homework) as the sole tag for your questions... –  J. M. May 12 '11 at 0:56

2 Answers 2

up vote 13 down vote accepted

Suppose $a,b,c,d$ are the four numbers whose mean is 2008. That means that $$\frac{a+b+c+d}{4} = 2008.$$

If $e$ and $f$ are the other two numbers, you want to find $$\frac{e+f}{2}.$$

Well, if you know $e+f$, you can figure out their mean (just divide by $2$). And you happen to know how much the sum of all six numbers is (you can just compute it). Can you use the information you have to figure out what $e+f$ must be?


Added. Given the comment by the OP below, it seems the hint above is insufficient.

You don't need to know the four numbers. Nobody is asking you what the four numbers are, or what the other two numbers are. All you need to know is what $e+f$ is. Because once you know what the sum of "the other two numbers" is, then you can just divide that sum by $2$ and get the mean.

Now, you know what $a+b+c+d+e+f$ (the sum of all six numbers) is: it's $$1867 + 1993 + 2019 + 2025 + 2109 + 2121 =12134.$$ (I'm not saying that $a$ is 1867; I'm just saying that adding all six numbers is 12134).

You also know what the sum of the four numbers whose mean is 2008 is: because you know that $(a+b+c+d)/4 = 2008$, so that means that the sum of the four numbers is $4\times 2008 = 8032$.

So: the sum of all six numbers is $12134$. The sum of the four numbers with mean 2008 is $8032$.

How much is the sum of "the other two numbers"?

And if the sum of the other two numbers it that much, how much is their mean?

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but what if i do not know the four numbers whose mean is 2008 because in the problem, you are not given the exact four numbers. You have to find them. I love the equations but how would use it for this problem? im alittle confused here –  Briana791 May 17 '11 at 15:54
    
@Briana791: The whole point is that you don't need to know the four numbers. You just need to know how much they add up to. I've added a fuller explanation. –  Arturo Magidin May 17 '11 at 16:04
    
ohhh ok im understanding you...thank you....just one more thing...what would be the sum of the other two numbers? –  Briana791 May 17 '11 at 16:09
    
nevermind i got the answer you just subtract the sum of the six numbers and the sum of the four numbers –  Briana791 May 17 '11 at 16:27
    
@Briana791: Please dont forget to accept an answer. –  user9413 May 17 '11 at 20:22

Hint: If $X$ is the mean of $N$ numbers then their sum is...?

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ok so this is what i did...I made x = 2008 and N = 4. 2008 x 4 = 8032 so the sum of the four numbers is 8032....what do i do from here to get the four numbers? –  Briana791 May 17 '11 at 15:57
    
@Briana791: Read the question: you don't need to get the four numbers. Nobody is asking you to find the four numbers. You don't care what the four numbers are. What you want to know is what the mean of "the other two numbers" is. –  Arturo Magidin May 17 '11 at 16:04

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