<p>Problem: <a href="http://i.imgur.com/dcPML.png%5B/url%5D">http://i.imgur.com/dcPML.png</a></p>
<p>I honestly have no idea how to approach this question. Any help?</p>
<p>The key to the problem is to reason that x+y and (x+y) /2 must be negative.
Why? Half of a positive number is less than the original positive number. Half of a negative number is greater than the original negative number.</p>
<p>Since (x+y)/2 should be halfway between x+y and 0, we can deduce that D is the zero mark. </p>
<p>Let’s suppose the distance between each tick mark is 1. Then (x+y)/2 is at -2, x+y is at -4 and x is at -6.
So y = +2 in order to satisfy x + y = -4 , where x = -6.
E is at position +2.
Answer: E</p>
<p>I like the geometric approach above!</p>
<p>An alternate way (algebraic) approach to arrive at the answer (of E) is to solve for x, y as follows.</p>
<p>Since the separation between the x and x + y is 2, we know that x + y = x + 2. So y =2. That’s actually all you need for this particular question.</p>
<p>You can also solve for x by observing that the separation between x + y and (x + y)/2 is 2. So: (x + y)/2 = (x + y) + 2, and solving you get x = - 6.</p>
<p>Quickest way to solve this one is to simply remember that the average (x+y/2) is exactly in the middle of x and y. Since x is 4 spots behind the average move 4 spots ahead and you arrive at E.</p>
<p>@fogcity, what if the distance between each tick mark was not 1?</p>
<p>Regarding “units”.</p>
<p>The question (from SirWanksalot) is “what if the distance between tick marks is not 1?” Good question! Clearly there are some units, implied or otherwise, associated with this question. The x and y are not abstract. They’re placed on a graph, and the graph has implied units. We can call the units “tickmarks” or “spots”, and all the numbers are then “tickmarks” or “spots” measured from the “0” position. So x = -6 is 6 tickmarks to the left of the “0”, and y = 2 is 2 tickmarks to the right.</p>
<p>@knowthestuff
Thanks, that was a really nice explanation! </p>
<p>@fogcity
That actually IS the approach I used. I got y=2, but then I couldn’t figure out how to get an answer from that. I guess I should have solved for x.</p>
<p>@sat100
Yeah, that seems like the fastest way to get it.</p>
<p>Thanks guys.</p>
<p>Here’s another problem:
<a href=“http://i.imgur.com/hZXAJ.jpg[/url]”>http://i.imgur.com/hZXAJ.jpg</a></p>
![http://i.imgur.com/hZXAJ.jpg[/url]](http://i.imgur.com/hZXAJ.jpg%5B/url%5D){kind=link}
<p>The answer I got was 64.</p>
<p>I got 68. Maybe a miscalculation since I did this all in my head. What’s the correct answer?</p>
<p>@SirWanksAlot
I figured it out. The answer is 125.</p>
<p>Basically the entire figure is a cube right? You can consider that it has X number of blocks on a side. The volume of the cube is X^3. And we are painting all of the faces. </p>
<p>Therefore, you can consider that you are essentially removing the entire outer layer. This is where it got tricky. When you remove the entire outer layer of the cube, you are essentially shortening each of the sides of the cube by 2. (Draw a figure and you’ll see).</p>
<p>So basically, Unpainted Volume = (X - 2)^3
Unpainted volume = 27(given)
27 = (x-2)^3
x = 5</p>
<p>5 is the length of the sides of the bigger cube. The question is asking us to find the total number of smaller blocks, so 5^3 = 125. 125 is the answer.</p>
<p>Good job. Thanks for that question.</p>