College Discussion - View Single Post - Blue Book Math Pg. 521 #12

xiggi · 03-07-2006, 02:26 PM

Hmmm, I don't know why this problem has to be made more complex than it really is. And, FWIW, I have to support Andre simple approach. P is known and so is P+N. This means that N is known. There is no need to go further than than 3 + N = 5 or 3 + 2 = 5.

The problem statement, "A list of numbers consists of "p" positive and "n" negative numbers. If a number is picked at random from this list, the probability that the number is positive is 3/5. What is the value of "n/p"?"

is not much different from rephrasing it as "If five numbers are either positive OR negative numbers, and you remove ALL of the three positives ones, how many negatives number are there?

or even simpler, "A bag contains five balls that are red or blue. If you remove ALL of the three red balls, how many blue balls are there?

There is a reason the SAT is known to test all materials from K to 12. In this case, it just happens to be a K problem.

03-07-2006, 02:26 PM	#8
xiggi Senior Member Join Date: Aug 2004 Posts: 7,015	Hmmm, I don't know why this problem has to be made more complex than it really is. And, FWIW, I have to support Andre simple approach. P is known and so is P+N. This means that N is known. There is no need to go further than than 3 + N = 5 or 3 + 2 = 5. The problem statement, "A list of numbers consists of "p" positive and "n" negative numbers. If a number is picked at random from this list, the probability that the number is positive is 3/5. What is the value of "n/p"?" is not much different from rephrasing it as "If five numbers are either positive OR negative numbers, and you remove ALL of the three positives ones, how many negatives number are there? or even simpler, "A bag contains five balls that are red or blue. If you remove ALL of the three red balls, how many blue balls are there? There is a reason the SAT is known to test all materials from K to 12. In this case, it just happens to be a K problem.