<p>Page 835
Second edition college board blue book</p>
<p>17) For all positive integers j and k, let j(R)k be defined as a whole number remainder when j is divided by k. If 13(R)k = 2, what is the value of k?</p>
<p>18) The average (arithmetic mean) of the test scores of a class of (p) students is 70, and the average of the test scores of a class of (n) students is 92. When the scores of both classes are combined, the average score is 86. What is the value of p/n? (p divided by n)</p>
<p>Question 17:
The question is looking for you to solve for k.
It tells you that when J is divided by K, the resulting whole number remainder is 2.
Therefore through the 13(R)K=2 given, you know that 13 divided by a number (K) has a remainder of 2.
13/11 would have a remainder of 2, so the answer is 11.</p>
<p>Question 18:
This question requires a basic knowledge of arithmetic mean and total values.
Since you know the average of class P is 70, the total value would be 70p (average score multiplied by amount of scores)
Therefore, total value of class N is 92n.</p>
<p>set up an equation:
(70p+92n)/(n+p) =86
70p+92n=86(n+p)
70p+92n=86n+86p
solve for p/n:
3/8</p>
<p>Some additional solutions:</p>
<p>(17) The question can be rephrased as “The remainder is 2 when 13 is divided by what number?” Well, when 13 is divided by 11 the remainder is 2. Indeed, 13 = 11(1) + 2. So the answer is 11.</p>
<p>Remark: If you do not see that the answer is 11 right away, just start dividing 13 by various positive integers and keeping track of the remainders.</p>
<p>13 = 2(1) + 11. So when 13 is divided by 2 the remainder is 11.
13 = 3(1) + 10. So when 13 is divided by 3 the remainder is 10.
13= 4(1) + 9. So when 13 is divided by 4 the remainder is 9.
…
and so on</p>
<p>Visual solution: Draw 13 circles on your paper. We want 2 to be left over. So we need to take 11 of them. In other words, divide 13 by 11 to get a remainder of 2.</p>
<p>(18) The sum of the test scores of the class of p students is 70p.</p>
<p>The sum of the test scores of the class of n students is 92n.</p>
<p>Adding these we get the sum of the test scores of both classes combined: 70p + 92n.
We can also get this sum directly from the problem.</p>
<pre><code> 86(p + n) = 86p + 86n.
</code></pre>
<p>So we have that 70p + 92n = 86p + 86n.</p>
<p>We get p to one side of the equation by subtracting 70p from each side, and we get n to the other side by subtracting 86n from each side.</p>
<p>6n = 16p</p>
<p>We can get p/n to one side by performing cross division. We do this just like cross multiplication, but we divide instead. Dividing each side of the equation by 16n will do the trick (this way we get rid of n on the left, and 16 on the right).</p>
<pre><code> p/n = 6/16 = 3/8.
</code></pre>
<p>So we can grid in 3/8 or .375.</p>
<p>Note that I used the Intermediate Strategy of Changing Averages to Sums. See the following thread for further explanation of this strategy (post #14):</p>
<p><a href=“http://talk.collegeconfidential.com/sat-preparation/1475829-sat-math-strategies.html[/url]”>http://talk.collegeconfidential.com/sat-preparation/1475829-sat-math-strategies.html</a></p>
<p>Actually, there is a graphical solution that requires a simple understanding of rations. </p>
<p>Write down tge following</p>
<p>70 … 86…92</p>
<p>The first group accounts for a change of 16 and tge second 6. The answer is 6/16.</p>
<p>@xiggi – that’s true and quick. Some might find it cryptic. Can I expand on that a little?</p>
<p>When a bunch of numbers have a certain average, it’s because the ones that are higher contribute a “surplus” that balances out the “deficits” caused by the ones below the average.</p>
<p>For example: if you took 3 tests and got 88’s on all 3, for the next test to bring your average up to a 90, you need…well, you have accumulated 2 points of deficit from each of the 3 88’s for a total of 6 points. So a 96 will balance the books.</p>
<p>In OP’s example, the 70’s are all 16 points below the average and the 92’s are 6 points above. For the surplus to balance the deficit, you need 16 of the 6 point surpluses for every 6 of the 16 point deficits. In other words, the amounts vary inversely with the distance from the average.</p>
<p>I think that’s what Xiggi meant by: “70 … 86…92” :)</p>
<p>Thanks for making it less cryptic. I was in a rush. :)</p>