<p>its in the blue book on page 835 number 18. i dont know how to even start this question. i know the givens but.....its not making sense from there. help?</p>
<p>Average mean of both classes is (x+y)/(P+N)=86 where x is the sum of the scores of the class with P students and Y is the sum of the scores of the class with N students. We know the sum of the scores for each class is 70P and 92N. 70P = X and 92N=Y. Substitution yields (70P+92N)/(P+N)=86 -> 70P+92N=86P+86N -> 16P=6N => p/n=3/8 or .375 and we are done.</p>
<p>Problems of moving averages and ratios usually have a super simple answer that is derived from pure reasoning. </p>
<p>92 - 86 = 6
---------- — ==>> 6/16 or 3/8
86 - 70 = 16</p>
<p>Whoo, I did not know that xiggi is still active!</p>
<p>You’re right xiggi but, usually is not always and it is still very important to know how to mathematically solve the problems rather than using pure reason.</p>
<p>Just do it on your calculator. NOTE: the average is given which means the sum of the scores of ‘p’ students nor the sum of the scores of ‘n’ students are changed. (ie: 70 + 70 = 140, 65 + 75 = 140, 50 + 90 = 140)</p>
<p>70 + 70 + 70 + 92 + 92 + 92 + 92 + 92 + 92 + 92 + 92, add them up, get 946.
Divide it by 11 and get 86.
Count the number of scores for each, p = 3 and n = 8. Lastly, get 3/8 or .375.</p>
<p>While I usually recommend the “mess around with numbers on your calculator” approach, this time you could easily spend too much time before finding the 3:8 combination.</p>
<p>But Xiggi’s approach works with so many of the SAT’s problems about averages – it’s worth making sure you understand the idea behind it: when a bunch of numbers have a certain average, the “surpuses” have to balance the “debts”.</p>