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Can you calculate the original average of some numbers by knowing how the average changes when you add numbers to it?

For example:

Let's say we have a class of some old students (we don't know their age or the average or the number of them). Then we are told that if we add the age of e.g. Dan, 16 to the ages of the students, the average age drops by 10 years, and then when we add the age of Michael, 12 to the new average, it drops by another 8 years. Can we just from this information calculate the average age of just the students?

Thanks for even reading this post.

Response to John Hughes:

I already tried doing this, but I got stuck at the point where you are supposed to do something with the things you've just written down. The things that I had written down that I thought I could get the answer from were:

$u-u'=S/N-(S+16)/(N+1)=10$

and

$u-u''=S/N-(S+28)/(N+2)=18$

I don't even know if this is what I was supposed to figure out.

How I figured it out:

After doing what John advised me to do I came up with this.

$S=10N^2+26N$

$2S=18N^2+64N$

In the first one, I divided both sides by N and in the second one, I divided both sides by 2 and by N. And this is what I got:

$S/N=10N+26$

$S/N=9N+32$

It seems pretty obvious that you should then put these equations together because they both equal $S/N$. Just like this:

$9N+32=10N+26$

and when you edit the equation

$6=N$

So now when you know what N equals you can plug it in another equation.

$S=10*6^2+26*6$

$S=516$

$u=S/N=516/6=86$

And that means the original average age is 86. Thanks for the help!

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Very large hint, without being a complete solution

Let $N$ denote the (unknown) number of students.

Let $S$ denote the (unknown) sum of their ages. Then the average age of the students is

$$ u = S/N. $$

If we add Dan to the class, how many students, $N'$, will there be? And what will the sum, $S'$ of their ages be, in terms of $S$? And what does that make the average age $u'$ of this enlarged class in terms of $N$ and $S$? Once you know that, and set $$ u' = u - 10, $$ and then replace $u$ and $u'$ by the formulas for them, you get one equation in the two unknowns $N$ and $S$.

If we now add Michael to the class, we get yet another number, $N''$ of students, and yet another age-sum, $S''$. How are these related to $N'$ and $S'$ (or to $N$ and $S$)? We also get yet another average age, $u''$, and know know that $$ u'' = u' - 8 $$

That gives us a second equation in the unknowns $N$ and $S$. Perhaps you can take it from here.

(Suggestion: Carry out the ideas I've described here and edit your question --- click on the word "edit" below the question to do so --- and show what you've gotten; perhaps we can then help your further if you still need it.)

Post-comment addition

You've got \begin{align} 10 &= \frac{S}{N}-\frac{S+16}{N+1}\\ 18 &=\frac{S}{N}-\frac{S+28}{N+2} \end{align} and that's great. It's typical, in situations like this, to clear the denominators, i.e., to multiply through by $N$ and $N+1$ or $N + 2$, resulting in \begin{align} 10N(N+1) &= S(N+1)-(S+16)N\\ 18N(N+2) &=S(N+2)-(S+28)N \end{align} When you expand out the right hand sides, a funny thing happens here: there are several $SN$ terms, and they all cancel. So you get \begin{align} 10N(N+1) &= S-16N\\ 18N(N+2) &= 2S-28N \end{align} From the first equation, we can solve for $S$; we can then plug this into the second equation to get an equation involving only $N$. That's good...we might be able to solve it. But it's a quadratic...that's potentially bad, because maybe there'll be two equally valid solutions. Or maybe they won't be equally valid. Why don't you go ahead and see where you end up when you follow that plan?

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  • $\begingroup$ Could you please help me more? I've edited the question with my response. $\endgroup$ Apr 24 '19 at 17:27
  • $\begingroup$ See post-comment addition to help push you along a bit further. $\endgroup$ Apr 24 '19 at 22:47
  • $\begingroup$ I'm sorry I'm going to school rn, but I already did it. $\endgroup$ Apr 26 '19 at 4:33

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