<li>If the average (arithmetic mean) of t and t + 2 is x
and if the average of t and t − 2 is y, what is the
average of x and y ?
(A) 1
(B) t/2
(C) t
(D) t + 1/2
(E) 2t</li>
</ol>
<p>Ok, what i did was solve for x,which gives me t+1, then solve for y=t-1
then the average, is ( (t+1)+(t-1))/2 = t
i got C, but i think my way is lil dumb, i feel like there’s a shortcut to this, because i spent a good 50 sec± on this, and adding and subtracting isn’t too safe, any shorts / ideas?</p>
<p>well if you notice that the average of x any y is actually, (t + t+2 + t + t-2)/4 and if you look closely the plus 2 and the minus 2 cancels out therefore its just 4t divided 4 and you get t.</p>
<p>Say t=8 so 8 and 8+2(10) makes the mean for x=9. the average for 8 and 8-2(6) is 7, so y=7, so the average of x(9) and y(7)=8, which we can see is the value of t. Hope that wasn't too bad of an explanation.</p>
<p>2(x+y)=[t+(t+2)+t+(t-2)]
x+y=2t
AVG.(x+y)=(x+y)/2=t</p>
<p>hope that helps</p>
<p>Well, since you were asking whether there was a quicker, less computational way, you could get it by thinking about it like this:</p>
<p>The average of t and t+2 has to be the number halfway between them. Since you know the numbers will go t, t+1, t+2 in order, it's easy to see (without actually calculating the average) that t+1 will be the average.</p>
<p>Same deal for t and t-2. The numbers, in order, will go t-2, t-1, t . . . so again, you can just "see" that t-1 will be the average.</p>
<p>Do the same process yet again to find the average of t-1 and t+1: in order, the numbers will go t-1, t, t+1 and so the average would have to be t.</p>
<p>Notice, it's the same as the calculations you did, but just takes advantage of the fact that the question is asking for the average of numbers that are really only separated by one term. So, you shouldn't have to waste time <em>calculating</em> those averages. In terms of speed, this question differentiates those who know the formula for calculating an average from those who "get" that the average is the halfway point and can determine that point without calculation, even in an algebraic situation. </p>
<p>So, keep that in mind if you're trying to speed up. You wouldn't "calculate" the average of 1 and 3, or 10 and 30 or 100 and 300 if you can see that the point in the middle is 2, or 20, or 200. The same is true in an algebraic context. </p>
<p>Of course, there are lots of other perfectly good suggestions here, and there's nothing wrong with doing the calculations if you can do them quickly enough to meet your standards. Personally, I am comfortable thinking of averages as halfway points. Others do the long work every time, but can do it quickly enough that it doesn't slow them down. Just pick a method that works for you.</p>
<p>I know everyone has answered this one in a better way, but I just plugged in 4 for t. I got it done in like 10-15 seconds with writing the work down.</p>
<p>Yeah plugging in is pretty much the best way to do most of these problems. If you have to remember one thing for test day, remember how to plug in and use it effectively. We can all do it algebraically, but there is room for errors, and we don't want that.</p>
<p>RahoulVA: "Yeah plugging in is pretty much the best way to do most of these problems."</p>
<p>I wouldn't be inclined to make a statement about which way is best for everyone. There's nothing wrong with solving it by plugging in, just like there's nothing wrong with any of the other methods offered.</p>
<p>You may not get points for "clever" but you do (indirectly) for "speed." I'm not trying to defend some of the algebraic solutions here because they're "cool" but because many people are able to answer this question strictly algebraically in their head, or by working it out on paper with adequate speed. The conversion to actual values and back is a waste of time for those people. For others, plugging in will be much quicker.</p>
<p>I think it's great to see so many examples posted, since there are so many different ways to think through this problem, and hopefully the OP has found something that works for him/her that will less time than the first attempt, algebraic or not.</p>
<p>Maybe it's just the morning I had, so sorry if this all sounds unnecessary! Sometimes I'm just hyper-sensitive to suggestions that there is one best way to do things because people are just so different. I'll go have a coffee now and chill out! :)</p>