<p>In a certain class, 4/7 of the students are boys, and the ratio of the students older than or equal to 10 years to the students less than 10 years old is 2:3. If 2/3 of the girls are less than 10 years old, what fraction of the boys are older than or equal to 10 years?</p>
<p>If someone could elucidate this for me, that would be helpful!</p>
<p>I recommend that you make up numbers. But here are some things to think about as you do:</p>
<p>Make the total number divisible by 7 so that it will be easier to break into boys and girls. But also make it a multiple of 5 so that you will be able to break the girls into a 2:3 ratio. So you may want to start with 70 kids and then take it from there…</p>
<p>Let x be the number of students in that class. Then 4/7<em>x are boys and 3/7</em>x are girls. Let 2y be the number of students over 10 years and 3y the rest (so their ratio is 2:3) Then 2y+3y=5y=x =>y=x/5 So, there are 2x/5 students aged 10 or more and 3x/5 students less than 10 years old.
If 2/3 of the girls are less than 10 years old, then there are 2/3<em>3/7</em>x girls aged less than 10 (which equals 2x/7) then, the rest of children under 10 must be boys. so there are 3x/5-2x/7 boys under 10. Sooo the rest of the boys must be over ten => 4x/7-(3x/5-2x/7) are boys over 10.
Therefore, there are 4x/35 boys older than 10. (ratio is 4/35)</p>
<p>pick a total classroom size that is divisible by both 7 and 5… 35 kids</p>
<p>you are told there are…</p>
<p>20 boys
15 girls
14 kids 10 or older
21 kids under 10
10 girls under 10</p>
<p>you can easily figure out that there are… </p>
<p>5 girls 10 or older
9 boys 10 or older
11 boys under 10</p>
<p>the question asks for the fraction of boys 10 or older…</p>
<p>9/20</p>
<p>I think anyone good enough at these questions to put the algebra together quickly without making a mistake is probably good enough to brute force the question in half that time.</p>
<p>Only click on the solution image after you have struggled with the problem. Follow the above suggestion by pckeller regarding the choice for the total number of students. 70 is a good choice. </p>
<p>I agree with PK and SQ replies. I solved this using a couple of methods, here is the third one. Similarly to the above, I started with picking the number 70. I did, however, placed my numbers in a square with nine boxes. My paper looks likes this:</p>
<h2>O — Y — T</h2>
<h2>18 – 22 – 40 B</h2>
<h2>10 – 20 – 30 G</h2>
<h2>28 – 42 – 70 T</h2>
<p>Nothing too original. My second method was again similar but a bit different. As I based my grid to the number 1 and used fractions. And here is my paper:</p>
<h2>O --------- Y ----- T</h2>
<h2>9/35 – 11/35 – 4/7 B</h2>
<h2>1/7 ---- 2/7 ---- 3/7 G</h2>
<h2>2/5 ---- 3/5 ---- 1 T</h2>
<p>While this is not helpful (considering the correct and more instructive methods by PK and SQ, here is the way I solved it in the first place. My paper simply reads (minus the letters)</p>
<p>And from there, one reads the problem statement, and enters the correct ratio for the percentage of older boys over … all BOYS, not over all students! Divide by 4/7. :)</p>
<p>I have to concur with xiggi on the value of practicing with these questions. These are called overlapping sets problems(terminology does not matter) and so far I don’t recall seeing them on the SAT test. They are definitely tested on the GMAT, also used to be written by ETS, the same test writers as SAT. </p>
<p>I do recommend doing challenging problems specifically for students targeting high scores, but this one might be irrelevant to the SAT.</p>
<p>Here are two that are closer to the SAT content:</p>
<p>Problem#1(Official SAT, don’t remember the year and month)
In a certain parking lot that contains 200 cars, 50 percent of the cars are red, 60 percent are four-doors cars, and 70 percent have alloy rims. What is the greatest number of cars in the parking lot that could be green two-door cars with alloy rims. </p>
<p>Problem#2(2002 AMC 10P, I like AMC problems)
Participation in the local soccer league this year is 10% higher than last year.
The number of males increased by 5% and the number of females increased
by 20%. What fraction of the soccer league is now female?
(A) 1/3 (B) 4/11 (C) 2/5 (D) 4/9 (E)1/2</p>
<p>I’m going to reiterate my advice to avoid algebra, particularly for the average student. It just isn’t necessary and it complicates things in a way that makes it far more likely to do what SQ did… overlook precisely what the question is asking for. Apologies if I’m wrong but it looks like you didn’t notice his error at first either.</p>
<p>The question can be easily rewritten with integers instead of fractions and ratios once the student recognizes that the class size must be a multiple of 5 and 7 (probably the hardest part of the question). From there, only simple arithmetic is needed to fill in the blanks. </p>
<p>SQ’s method is correct of course and may be second nature for strong students with good number sense, but it makes the question harder (much harder for average students) than it needs to be. This reminds me of those rectangle questions where the sides are increased/decreased by a given percentage and you are asked for the percentage change in area. These are always level 5 questions because so many students try to solve it with algebra and screw up the setup. Just plug in starting length and width numbers and this question type becomes a level 2 difficulty question at most.</p>
<p>Agreed on Dr. Chung. Although I think this one is a college board rip-off with the numbers changed around.</p>
<p>YZ, you are correct regarding SQ error, but I am quite certain that it was a matter of solving the problem “on the fly” versus having the question in front of him.</p>
<p>Regarding the algebra versus other methods, the reality is that many great approached consist of “merging” both methods – if that makes sense. In the end, the methods should deliver the same results, and the difference will be one of time and accuracy. As one of the earlier posts did show, there is a danger in trying to build long equations and getting lost in the reasoning part. </p>
<p>Fwiw, please note that there is no difference between following the problem logically, and using the boxes I did suggest. It is a matter of presentation. As far as the algebra, please note that I did NOT recommend others to follow what I did, as the changes to mess up the equations are really high. </p>
<p>In the end, I believe that the biggest problem for many of us is to SHOW how to solve problems we solve easily, and sometimes with mathematical intuition. That is why, for instance, I try to show a couple of methods and approaches, including some I know are more circular. </p>
<p>There are times also that I deliberately pick numbers that I expect others to pick as well. In this case, the number was 70. I probably would have picked 35 in most circumstances.</p>
<p>I remember the car problem from a while back. Boy did that stump a lot of students.</p>
<p>As for algebra vs. making up numbers, I think my leanings are pretty well known. The stark reality is that many students arrive in 11th grade with algebra skills that are far from fluency. As a tutor, I have to remember to answer the question a student asks: “How would I solve this?” – not “How do YOU solve this?”. And even the students who are in fact fluent in algebra will often pick up speed by reading, thinking and playing with numbers. </p>
<p>The problem in this thread is an amped up version of what would be on the SAT. Chung defenders will say that that’s what makes it valuable. I’m staying out of that debate (too much conflict of interest). But starting with 70 and following the numbers where they lead (which is essentially what Xiggi’s chart did) is faster and safer than anything algebraic. The blue book is packed with examples that serve as more evidence in favor of that non-algebraic approach. In fact, some of them seem almost a parody – as if they were written by guys like me to provide support for the methods I recommend.</p>
<p>Oops, made the very mistake I tell students to watch out for! thanks everyone for pointing out the error. Fixed it.</p>
<p>@YZamyatin
I agree with your comments. How I do the problems does not always necessarily mean that is how others should go about it. When it comes to solving problems my belief is that every one likes to approach it differently and what is important is to come up with a set of methods or approaches that one is comfortable with. Because in the exam under pressure one needs to rely on their own set of solid steps. It is immaterial whether that approach relies on using algebra, substituting simple numbers, or even eliminating answer choices by logic. </p>
<p>In reality the high scorers have to be able to approach the problems in several ways and they need to be flexible as well. The SAT math is too easy and it does not require a considerable amount of problem solving skills, say compared to the American Math Competitions. </p>
<p>The issue is with students who struggle with algebra, unfortunately the algebra education is an Achilles heel in this country. This often translates in to all sorts of inane methods to solve problems that are fairly straightforward using algebra but require considerable gymnastics with numbers. </p>
<p>That is why we see a plethora of books recommending inane SAT strategies, tips, and tricks. This is nothing but coddling the students. To get the high scores there is no escaping knowing the content inside out. For average students these techniques would probably work as far as getting an average score.</p>
<p>I don’t see what’s so complicated about the parking lot problem. I got it in about 40 seconds, and then second guessed myself for about 2 minutes because I thought my method was too simplistic. I won’t spoil the answer.</p>
<p>Solved the other problem very quickly also because my intuitive guess turned out to be a lucky one. I won’t spoil that one either…</p>
<p>Each method will require its own aha moments of recognition. Strong students may not even be cognizant of the cognitive leaps they are making in places where other students just hit a wall. My goal, for any given question type, is to minimize both the number of aha moments required and the level of abstractness required to achieve that recognition. My anti-algebra (anti-math in general actually) position stems from my observations that using algebraic methods to attack certain types of word problems (percentage change, fractions, and ratios without notional amounts) introduces the greatest number of required aha moments as well as the most abstractness (just generally speaking of course). </p>
<p>I was reminded of this thread by today’s SAT QOTD. It’s the rectangle area problem I referred to in my earlier post. Only 33% correct which is almost as low as it ever gets. The given explanation, laden with algebra, illustrates the worst possible method for attacking this type of problem. Anyone who just picks random notional amounts for length and width and just plods through the arithmetic would be hard pressed to get this question wrong…</p>
<p>“The length of a rectangle is increased by 20%, and the width of the rectangle is increased by 30%. By what percentage will the area of the rectangle be increased?”</p>
<p>and the given explanation…</p>
<p>"Let l and w be the original length and width of the rectangle, respectively. If we denote by L and W the new length and width of the rectangle, respectively, then we have:</p>
<p>L = l + ((20 over 100) times l) = 1.2 times l and
W = w + ((30 over 100) times w) = 1.3 times w</p>
<p>So, the new area will be:</p>
<p>A = L times W = (1.2 times l) times (1.3 times w) = 1.56 times (l times w)
1.56 times (l times w) = (l times w) + ((56 over 100) times (l times w))</p>
<p>where (l times w) is the original area of the rectangle.
So the area is increased by 56%."</p>
<p>Yeah, that is messier than necessary. There are certainly lots of methods and lots of mathematical principles you need to know if you are going to get a top math score. But if you had to come up with one, single most useful, flexible, and even often (but not always) most efficient idea, here’s what gets my vote:</p>
<p>“Can you make up numbers that fit this problem?”</p>
<p>I’m not saying that it is a panacea, but as a tutor, it is the question that I ask most often when I am trying to help a student who has hit a roadblock. </p>
<p>BTW, other favorites:</p>
<p>“Can you read that again slowly?”</p>
<p>and</p>
<p>“Look at what they asked for. And now look at what they gave you. Why do you think they gave you that?”</p>
<p>And of course:</p>
<p>“Let’s look at the answer you chose. Could that be right? Let’s see…”</p>