<p>It's too bad the Blue Book doesn't explain answers. Can someone help with these?</p>
<p>1) Page 412 # 18 - Esther drove an average of 45 MPH to work. She took the same route home at an average 30 MPH. She traveled a total of one hour to and from work. What is the distance each way?</p>
<p>Can some provide the calculation for this? </p>
<p>2) Page 408 number 5 - A round disc with a spinner has six equally sized sections numbered 1 through 6. They spin the spinner twice. The first spin is a and the second spin is b. What is the probability that a over b is greater than 1?</p>
<p>The good thing to understand is that all distance/rate/time questions never get more complicated than distance/rate/time. The question says the distance is the same both ways. that means E1 d1=(r1)(t1)=(r2)(t2)=d2 where d1 =to work Then like most all distance rate and time questions, the questions states the total time is 1 hour. This is the second equation you need. E2 t1+t2=1. All you do from their is solve for either time in the second equation and plug it back into the first equation. Ill leave you to it. T2=3/5. T1=2/5. Then you just multiply by the average speed to find the distance each way.
The second problem dealing with probability is as simple as counting out the possibilities. Ok well i can tell this is going to be a sum series answer. The real question is just how many times A is greater than B. Take each possibility for B. 1-6. When its 1. A=6,5,4,3,2. from then on it decreases by one so for b=2. A=4 ect. Theres a sum formula created by Gauss that states when starting by one. The sum from the numbers 1 to n is ((n)(n+1))/2 so the sum is just 15 because n in this case is 5(since its the max number of a's). and i think the spinner can have 6 possibilites for the first spin and 6 for the second for a total of 6*6=36</p>
<p>That one formula, n*n+1 all divided by 2 is extremely useful! One that I won't find in many books! Do you have any other useful formulas that I can look at? Also, for the distance one, Do you just say that T1=1-T2? For rate, you said average rate. Obviously you don't do the average of 30 and 45. You know that the distance is the same for each, so that's 2x? Then 1 hour, so what is it, 2x=(30+45)/2? I don't know about that considering I just said you don't average 30 and 45. Also, what do you mean by E1 and E2? Esther?</p>
<p>If I came across a question like this NOW, I might considering plugging in plugging in all the distances. First I would guess which one is closest(like C for instance or the average that is closer to 30(since you travel longer going at 30 mph), and plug that in. How would it work though, Your Distance=(again, how do you get average rate?) * 1, and see if it all works out right?</p>
<p>Solutions to these two questions, as well as other, are here: Consolidated</a> List of Blue Book Math Solutions, 3rd ed.
Many of the solutions are better than ones offered in the College Board online course or the Test Masters book.</p>