Barron's math problem on rate

Test Preparation SAT Preparation
1

<p>Hey, I'm doing some exercises from Barron's SAT prep book and I don't really understand this one problem (Exercises on Word Problems, p. 509, #1):</p>

<p>In the afternoon, Judy read 100 pages at the rate of 60 pages per hour; in the evening, when she was tired, she read another 100 pages at the rate of 40 pages per hour. In pages per hour, what was her average rate of reading for the day?
(A) 45
(B) 48
(C) 50
(D) 52
(E) 55</p>

<p>The correct answer is B. The way Barron's says to do it is this:
Average rate of reading = (total # of pages)/(total time spent reading)</p>

<p>I also did the problem using the average # of pages and average time spent reading instead of the totals than Barron's said to use. It worked out to get the correct answer as well. But someone told me that if there were different numbers of pages (e.g., 100 pages and 200 pages), the rates would be weighted differently so only Barron's way (with the totals) would work.</p>

<p>I still don't understand how Barron's way would account for the different weights if you had two different numbers of pages. Can anyone explain this to me in a way that is conceptually intuitive (I don't need the exact numbers)? I would really appreciate it. I'm not good at math, so all of this is very confusing to me.</p>

2

<p>An easy way, not necessarily the most fair, if you will, is to multiply Speed 1 by Speed 2 by 2 and divide the product by Speed 1 + Speed 2.</p>

<p>In this case: Speed 1 = 40 Speed 2 = 60 </p>

<p>40x60x2=480 40+60=100</p>

<p>480/100 = 48</p>

3

<p>@Snappy: oooh okay. I’ve seen this formula floating around here but I didn’t realize it could apply.</p>

<p>Can you use this when you have the same number of pages (100 and 100) AND different number of pages (100 and 200)?</p>

4

<p>You could do:</p>

<p>100/60 = 1.6667</p>

<p>100/40 = 2.5</p>

<p>(100 + 100)/(1.6667 +2.5) = 47.999999</p>

5

<p>

</p>

<p>Let’s review the above steps … that would be the same as </p>

<p>2(100) / (100/60 +100/40) = ?
or
2(100) / [(100<em>60) + (100</em>40) / (60<em>40)] = ?
or
2(100) * (60</em>40) / [(100<em>60) + (100</em>40)] = ?
or
2(100)* (60<em>40) / 100</em>(60+40) = ?
**and finally **
2<em>60</em>40 / 60+40 = 48 :)</p>

<p>Oh, what’s happening here? The last line is the application of the formula. Could it be that formula simply eliminates the irrelevant and redundant information (number of pages) and yields the correct answer in one step? </p>

<p>The reality is that it is NOT needed to manipulate the “100 pages” in this problem.</p>

6

<p>@james112233: Thanks for your response. That’s what my Barron’s book said to do too.</p>

<p>@xiggi: Thank you also. This formula seems to be very helpful. You can use it whether you read 100 pgs both times or if you read 100 pgs then 200 pgs, right?</p>

7

<p>Dang i didn’t know this. very helpful!
I usually have like 20 mins left over so it’s not a problem to just do it in a long way haha
but it is still interesting</p>