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Redo hints to be better and add some color

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ss3 committed Aug 9, 2012
1 parent a201966 commit 4d249b5bc65850997cae1dff36dccfeec4de63b5
Showing with 20 additions and 21 deletions.
  1. +20 −21 exercises/pile_patterns.html
@@ -1,5 +1,5 @@
<!DOCTYPE html>
-<html data-require="math graphie interactive">
+<html data-require="math graphie interactive word-problems">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>Pile patterns</title>
@@ -177,44 +177,43 @@
</div>
<div class="solution" data-type="multiple">
In step 0, we have <span class="sol"><var>START</var></span> blocks. At each step, we add <span class="sol"><var>XS</var></span> more blocks.<br><br>
- So, at step number <code>S</code>, we have <br><span class="sol"><var>XS</var></span> <code>\times S +</code> <span class="sol"><var>START</var></span> blocks.
+ So, at step number <code>S</code>, we have<br><span class="sol"><var>START</var></span> <code>+</code> <span class="sol"><var>XS</var></span> <code>\times S</code> blocks.
</div>
<div class="hints">
- <p>The number of blocks in the diagram make an arithmetic progression. This means that at each step, the same number of blocks are added. We want to find a way to figure out how many blocks there would be at any step.</p>
- <p>First, we need to find out how many blocks there are at the beginning. To do this, look how many blocks there are in step 0 by dragging the slider all the way to the left.</p>
+ <p>The number of blocks in the diagram make an arithmetic progression. This means that at each step, the same number of blocks are added. We want to find a way to know how many blocks there will be at any step.</p>
+ <p>First, we need to find out how many blocks we start out with. Count how many blocks there are in step 0 by dragging the slider all the way to the left.</p>
<div>
- <p>We see that there are <code><var>START</var></code> blocks in step 0. Fill this in the answer box.</p>
+ <p>We see that there <var>plural("is", START)</var> <var>pluralTex(START, "block")</var> in step 0. Fill this in the answer box.</p>
<div class="graphie" data-update="main">
graph.slider.moveTo(1, 0);
</div>
</div>
<p>Next, let's figure out how many blocks are added in each step. The easiest way to do that is to see how many blocks are added from step 0 to step 1.</p>
<div>
- <p>There are <code><var>START + XS</var></code> blocks in step 1. To find how many blocks are added in each steps we just subtract this number from the number of blocks in step 0. So there are <code><var>START + XS</var> - <var>START</var> = <var>XS</var></code> blocks added in each step. Fill this in the answer box as well.</p>
+ <p>There are <code><var>START + XS</var></code> blocks in step 1. To find how many blocks are added in each steps we just subtract this number from the number of blocks in step 0. So there are <code><var>START + XS</var> - <var>START</var> = <var>XS</var></code> <var>plural("block", XS)</var> added in each step. Fill this in the answer box.</p>
<div class="graphie" data-update="main">
graph.slider.moveTo(3, 0);
</div>
</div>
<p>Last, we have to figure out how many blocks there would be at any step. We can answer this by thinking "If we are at some step <code>S</code>, how many blocks would there be?"</p>
- <p>Because this is an arithmetic progression, we can represent the number of blocks using an expression that looks like: <code>a \times S + b</code> where <code>a</code> and <code>b</code> are numbers that we have to fill in.</p>
+ <p>Well, let's start looking at how many blocks there are, and we'll see if there's a pattern.</p>
+ <p>Step 0: <span class="const"><var>pluralTex(START, "block")</var></span></p>
<div>
- <p>If we are at step 0, we can figure out what <code>b</code> should be. We know that there are <code><var>START</var></code> blocks in step 0. So let's plug into the equation:</p>
- <p><code>a \times 0 + b = <var>START</var></code></p>
+ <p>We know step 1 has <code><var>START + XS</var></code> blocks, but we can also write it as the number of blocks in step 0, plus the number of blocks we add from step 0 to step 1:</p>
+ <p>Step 1: <code>(</code>Number of blocks in step 0<code>) + <var>XS</var> =</code> <span class="const"><code><var>START</var> + </code></span><span class="lin"><code><var>XS</var></code></span> blocks</p>
</div>
+ <p>We can do the same thing with steps 2 and 3, because we know we add the same amount of blocks in each step:</p>
+ <p>Step 2: <code>(</code>Number of blocks in step 1<code>) + <var>XS</var> =</code> <span class="const"><code><var>START</var> + </code></span><span class="lin"><code><var>XS</var> + <var>XS</var></code></span> blocks</p>
+ <p>Step 3: <code>(</code>Number of blocks in step 2<code>) + <var>XS</var> =</code> <span class="const"><code><var>START</var> + </code></span><span class="lin"><code><var>XS</var> + <var>XS</var> + <var>XS</var></code></span> blocks</p>
<div>
- <p>Since <code>0</code> times anything is 0, we can turn this into:</p>
- <p><code>0 + b = b = <var>START</var></code></p>
- </div>
- <p>Now that we know that <code>b = <var>START</var></code>, we can figure out what <code>a</code> is by using what we know about step 1.</p>
- <div>
- <p>Plugging in again, we get:</p>
- <p><code>a \times 1 + <var>START</var> = <var>START + XS</var></code></p>
- </div>
- <p>By solving for <code>a</code>, we find that <code>a = <var>XS</var></code>. Do you see any similarities between what we got for <code>a</code> and <code>b</code> and the numbers we found above?</p>
- <div>
- <p>That's right, <code>b</code> is just the number of blocks at the beginning, and <code>a</code> is just the number of blocks we add in each step.</p>
- <p>We now have the answer. In step <code>S</code> the number of blocks is <code><var>XS</var> \times S + <var>START</var></code></p>
+ <p>Here, we can see a pattern! The number of blocks in each step can be represented as the sum of <span class="const"><var>pluralTex(START, "block")</var></span> and some number of groups of <span class="lin"><var>pluralTex(XS, "block")</var></span>. How many groups of <var>pluralTex(XS, "block")</var> are there in each step?</p>
+ <div class="graphie">
+ $(".const").addClass("hint_blue");
+ $(".lin").addClass("hint_orange");
+ </div>
</div>
+ <p>That's right, the same number as the step they are on!</p>
+ <p>So, we can combine the groups of blocks together to look something like <code><var>START</var> + <var>XS</var> \times S</code>, where S represents which step we are on.</p>
</div>
</div>
</div>

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