An example of developing one's own "tricks"

Test Preparation
1

<p>Question:
In a class of 80 seniors, there are 3 boys for every 5 girls. In the junior class there are 3 boys for every 2 girls. If the two classes combined have an equal number of boys and girls, how many students are in the junior class?</p>

<p>A 72
B 80
C 84
D 100
E 120</p>

<p>Common answer:
Among the 80 seniors there are 3 boys for every 5 girls, so 3/8 of the seniors, or 30, are boys and 5/8, or 50, are girls. Among the juniors, 3/5 are boys and 2/5 are girls. If x stands for the total number of juniors, then (3/5)x are boys and (2/5)x are girls. The total number of senior and junior boys is 30 + (3/5)x. The total number of senior and junior girls is 50 + (2/5)x. The question says that these quantities are equal, so 30 + (3/5)x = 50 + (2/5)x. Solving this gives 150 + 3x = 250 + 2x, so x =100. </p>

<p>This question is rated hard by TCB. Finding the answer using regular algebra is not that hard and most CC students should be able to work through the problem. However, there ought to be a simpler way to approach the problem. </p>

<p>First, the numbers are pretty convenient:
80 students distributed 5 to 3 = that ought to be 50G to 30B. Nice round numbers. This means that the solution needs to have similarly round numbers.
Now, let's look at the proposed solutions. 72 and 84 will never yield nice round numbers (as it has to be divisible by 5). 80 is divisible but would yield multiples of 16. Same thing for 120, which would yield multiples of 24. Tada ... 100 yields nice round numbers (20). Let's try that: 40G and 60B. Bingo: Now we have 90 B and G. </p>

<p>This type of approach only requires a bit of attention to the numbers and a bit of effort to discover special yet simple relationship. </p>

<p>You could even push it a bit further and use simple logic -as in a puzzle:</p>

<p>There are 50G and 30B in the senior class, or a deficit of 20. In the junior class, for each group of 5, there is one more B. This means that we would need a certain number of groups of 5 to eliminate the deficit. How many? Simply enough, that is 20. So the junior class has to be 5 x 20 or 100.</p>

<p>Working through the algebra is not wrong, but it is nice to identify quicker solutions by paying attention to the relationships between the given numbers. </p>

<p>Every time you practice, make sure to analyze not only your errors but also your correct answers. Always ask yourself if you did not overlook a simple clue, a simple relationship that could yield the answer WITHOUT jumping on your dear TI or working ferociously through a set of equations. </p>

<p>Pay special attention to the PROPOSED answers. Remember that four answers HAVE to be conclusively wrong. Problems written by ETS are VERY GOOD in giving two to three -if not four- answers that are easily identifiable. Remember that it is a reasoning test not a math test. :)</p>

2

<p>Excellent post.</p>

3

<p>Awesome!!!</p>

4

<p>Bravo! I remember doing this question a couple days ago as one of CB's "Question[s] of the Day." I opted for the equation method. But your methods and ways of thinking about the problem are very interesting and often easier.</p>

5

<p>The approaches proposed by Xiggi are good, but there are still alternative ways of thinking about this question:</p>

<p>Solution #1:</p>

<p>Technique: working backwards (or plugging in the answers)</p>

<p>First, figure out that in the senior class, there are 30 boys and 50 girls, as above. Then, start with choice C, 84, and see if that works. You will get 50.4 boys and 33.6 girls for the junior class. Obviously, the choice will not work, but the point is to see which direction to go next. If you add the junior and senior class numbers up, you get 80.4 boys and 83.6 girls. There are not quite enough boys to make up the "deficit" yet, so you opt for a larger class size. Try choice D next, and voila, there is your answer.</p>

<p>Total number of choices you have to work through: 2.</p>

<p>Solution #2:</p>

<p>Technique: "Joe Bloggs" and logic</p>

<p>Think about "Joe Bloggs" choices. This is a hard question, so choice B, 80, must be wrong. Then, choices A and C are not divisible by 5, so these must be eliminated as well. Finally, you get down to either choice D or choice E, and you only have to check one of them to find the right answer (e.g., if choice E does not yield the right numbers, you can safely assume that choice D is correct without actually checking it).</p>

<p>Total number of choices you have to work through: 1.</p>

6

<p>I'm not saying that you're wrong...but why can you automatically throw out choice B on solution number 2 Godot. Thanks!</p>

7

<p>because 80 is the number in the problem...the number stated in the problem is NEVER the answer</p>

8

<p>even on an easy problem?</p>

9

<p>:( I feel stupid.

[quote]
Among the 80 seniors there are 3 boys for every 5 girls, so 3/8 of the seniors, or 30, are boys and 5/8, or 50, are girls. Among the juniors, 3/5 are boys and 2/5 are girls.

[/quote]

I have no clue HOW you got those fractions...</p>

<p>
[quote]
80 students distributed 5 to 3 = that ought to be 50G to 30B. Nice round numbers. This means that the solution needs to have similarly round numbers.
Now, let's look at the proposed solutions. 72 and 84 will never yield nice round numbers (as it has to be divisible by 5). 80 is divisible but would yield multiples of 16. Same thing for 120, which would yield multiples of 24. Tada ... 100 yields nice round numbers (20). Let's try that: 40G and 60B. Bingo: Now we have 90 B and G.

[/quote]

I hate whenever someone is explaining answers and they just make it look so easy. I could never figure to use that logic for real unless I've practiced it many times (and I haven't because I didn't think about it until now). Will all SAT questions like this have such nice round numbers? What happens when they don't? What kind of logic do we use then?</p>

<p>I am still dumbfounded.</p>

10

<p>Nevertheless, my common approach to an SAT math question is to quickly assess if there is a fast "straight math" way of answering it (there are exceptions, where I can see a "non-math" shortcut right away).</p>

<p>For this question, I think, there is a plain math plane route.
You need to do a couple of quick things in your head without much thinking - just from experience.</p>

<p>Step 1.
80 seniors, 3(B):5(G) = 30 boys and 50 girls.</p>

<p>Step 2.
Juniors, 3(B):2(G) means there are 3x(B)+2x(G)=5x juniors,
x students being 1 part out of 5.</p>

<p>Step 3.
(B) = (G)
30+3x=50+2x
x=20
5x=100.
That's D.</p>

<p><<<<<<<<<<<<<<<<<<<
(the "parts method" is touched upon on
<a href="http://talk.collegeconfidential.com/showthread.php?p=978007#post978007%5B/url%5D"&gt;http://talk.collegeconfidential.com/showthread.php?p=978007#post978007&lt;/a>,
starting at "I forgot what I meant to prove here";
boethian's solution is at
<a href="http://talk.collegeconfidential.com/showthread.php?p=973549#post973549%5B/url%5D"&gt;http://talk.collegeconfidential.com/showthread.php?p=973549#post973549&lt;/a&gt;)&lt;/p>

<br>


<br>

<p>As Xiggi and Godot both stressed, there is always more than one way of finding the right answer.
You need to try to find as many alternative solutions as you can, and then decide which approach works better for you. (Seems, I am echoing Xiggi's mantra here).</p>

11

<p>
[quote]
80 seniors, 3(B):5(G) = 30 boys and 50 girls.

[/quote]

How do you get that? I mean, if the # of seniros wasn't 80 how would you know how many boys and girls?</p>

12

<p>The "PARTS way":
boys:girls = 3:5 = 3 equal parts : 5 equal parts (think of a cooking recipe).
Total = 3+5 = 8 (equal) parts = 80 students.
1 part = 80/8 = 10 students.
Boys = 3 parts = 3<em>10 = 30.
Girls = 5 parts = 5</em>10 = 50.</p>

<p>Or do it the "fraction way":
Boys = 3 out of 8 = 3/8 of 80 = 3/8 * 80 = 30
Girls = 5 out of 8 = 5/8 of 80 = 5/8 * 80 = 50.</p>

<p>In general, if
x:y = m:n,
and
x+y = a,
then
x = a<em>m/(m+n),
y = a</em>n/(m+n).</p>

<p>Plug different numbers in and see how it works.
Make up your own questions.
Give them to you friends.
Math CAN be fun!</p>

13

<p>oh oh okay. That formula you just put at the bottom was confusing. I will make an effort to remember the fraction way if it does the same thing. THankyou!</p>

14

<p>No SAT I math question should merit such extensive discussion.</p>

15

<p>Well excuse me for not being a freakin' math genius. Don't make yourself sound so snobby! Its people like you who give CC such a bad rep.</p>

16

<p>quitejaded, you should not shy away from the formulas - just play with them for awhile, and they will start making sense.</p>

<p>How 'bout using that confusing one here:</p>

<p>Point P divides a segment AB into two parts.
AP:PB = 3:7</p>

<p>What's PB - AP, if AB=1 ?</p>

<p>(drawing "parts" would help).</p>

17

<p>3:7=m:n?
3+7=10
3=10<em>m/(m+n)
7=10</em>n/(m+n)
... ?</p>

<p>Fraction way...
3/10 (AP) * 1 = 3/10
7/10 * 1 = 7/10</p>

<p>2/5 is the answer. But I couldn't figure the formula.</p>

18

<p>m=3
n=7
a=1.
x=AP
y=PB
++++++++++++++++++++++++
Draw a segment AB and break it into 10 equal parts, then mark the point P.
See now?</p>

19

<p>Oh, gcf, please. That was much more confusing than just using the fraction way. :( Why are you pushing the formula? You don't think the fraction way will be sufficient?</p>

20

<p>Sorry for the pressure, quitejaded!</p>

<p>It's not so much about a formula itself as about visualizing it, and understanding it better that way.</p>

<p>I believe that visualization is the most powerful tool in teaching/learning math.</p>

<p>As long as you know how to apply fractions for this kind of questions, you are fine. Formula and drawings are just extras (with no credits).
Get some rest! I am about to.</p>