<p>15 If p and n are integers such that p>n>0 and p^2-n^2=12 which of the following can be the value of p-n?</p>
<p>I.1
II.2
III.4</p>
<p>Answer is only II but why!??! what's the way to solve this problem?</p>
<p>Also
20. If j,k and n are consecutive integers such that 0<j<k<n and the units (ones) digit of the product is what is the units digit of k?</p>
<p>I simply listed out consecutive integers...is that the fastest/best way?</p>
<ol>
<li>p^2 - n^2 = (p-n)(p+n) = 12
as you are already given the value of p-n, you can also find p+n
solve this set of 2 equations to find p and n. they should be integers.</li>
</ol>
<p>OHHHH i see, so you just assume that the given roman numeral you are testing is a possible answer then you solve for 2p and see if it is an integer? EXCELLENT!</p>
<p>14 There are 25 trays on a table in the cafeteria. Each tray contains a cup only, a plate only, or both a cup and a plate. If 15 of the trays contain cups and 21 of the trays contain plates, how many contain both a cup and a plate?</p>
<p>Whats the best way to solve problems like these?</p>
<p>I think the best way to solve problems like these is to use Venn diagram. Draw two ‘circles’, one representing the set of trays that have a cup and the other the set of trays that have a plate. Then, the intersection of the circles represents the trays that have both a cup and a plate. Call this number “x”. Then, the there are (15-x) trays with only a cup, (21-x) trays with only a plate, and x trays with both. But, as there are 25 trays, we must have (21-x) + x + (15-x) = 25 –> x=11.</p>
<p>ahhh okay. great.</p>
<p>"If j,k and n are consecutive integers such that 0<j<k<n and the units (ones) digit of the product is what is the units digit of k?</p>
<p>I simply listed out consecutive integers…is that the fastest/best way? "</p>
<p>That question is incomplete. You miss something here "the units (ones) digit of the product jn is 9. </p>
<p>Fwiw, the only way to have 9 in the units digit is for j<em>n to be 1</em>9 (see consecutive numbers that eliminate 3*3) </p>
<p>In this case, j is 9 and k is 10. Answer is thus zero.</p>
<p>ahhh sorry… but okay… that works too. find ways to create 9.</p>
<p>another Question</p>
<p>Tickets for a community play cost 2 for a child and 4 for an adult. If 200 tickets were sold for a total of 700 what was the ratio of the number of children’s tickets sold to the number of adults’ tickets sold?</p>
<p>answer is 1:3… wahts the best way to answer this?!</p>
<p>Step A. 2x + 4y = 700</p>
<p>Step B. x + y = 200 or
2x + 2y = 400</p>
<p>Step C. 2x + 4y = 700</p>
<h2>- 2x - 2y = -400</h2>
<p>2y = 300</p>
<p>Step D 2x = 400-300 = 100</p>
<p>y to x = 3 to 1</p>
<p>In reality, it takes a lot longer to type it out than setting up the equations. </p>
<p>Fwiw, there is a visual way to solve this problem, but it might confuse you. However, here it is: </p>
<p>Recognize that 200 tickets for 700 dollars means that the average ticket is 3.50. </p>
<p>Draw a quick line from to 2 to 4 with intervals at .5 </p>
<p>2**–2.5–3–**3.5–4. </p>
<p>From 3.5 to 4 there is ONE interval, and from 2 to 3.5 there are three intervals. This means that there are three times more tickets sold at 4 than at 2 dollars.</p>
<p>When you “see” something, the next step is to quickly check the math. In this case, 1 ticket at 2 plus 3 tickets at 4 equal 14 dollars, or exactly 4 tickets. </p>
<p>If that makes no sense, stick to the algebraic method. It’s fast and secure.</p>
<p>Ahh okay. yeah…my greatest fear is not recognizing that you can do it that way algebraically… lol</p>
<p>With these 2-equation, 2 unknown problems, it is often worth making up some numbers to get the feel of what is happening. You can end up finding the answer by trial and error in surprisingly little time.</p>
<p>In this case, just to get started, I tried making it 100 adult and 100 kids. 100 x 4 = 400, 100x2 = 200, total is $600. So I need more adult tickets.</p>
<p>The next numbers I tried were 150 adult, 50 kid…150x4 = 600, 50 x2 = 100, total is 700.<br>
So these are the right numbers and their ratio is 3:1</p>
<p>If you are afraid you won’t think of the algebraic method, this is an alternative approach that may work for you.</p>
<p>
</p>
<p>Agree completely. </p>
<p>Actually, I did exactly that when I first saw the problem. Although it was obvious that 100/100 would not be the answer, the numbers were easy to manipulate.</p>
<p>Regarding the numbers to plug in, after a while everyone develops a feel for the best ones to use. Many times, the problem (or the answers) offer great hints.</p>
<p>Unfortunately, there are problems that, albeit being similar, do not lend themselves to plugging-in numbers. </p>
<p>Here is an example based on an older SAT question: </p>
<p>96 students take Calculus. The Honor Class averaged 92 on the final exam, and the regular class averaged 83. If the combined average was 89, how many students were in the regular class?</p>
<p>Surprisingly, this can solved in about 15 seconds without a calculator.</p>
<p>^This one doesn’t need algebra either!</p>
<p>Now it is helpful to know a useful fact about averages: if a bunch of numbers have a certain average, then the “surpluses” have to balance the “deficits”. In this case, the honors kids beat the average by 3. The regular kids fell 6 points below the average. For the surpluses to balance the deficits, we’ll need 2 honors kids for every one regular kid. So the 96 kids must be split in a 2:1 ration – that’s 64 honors and 32 regular.</p>
<p>I know that algebra can be useful on the SAT. But I also know that these playful, mess-around-with-numbers methods also work well. And for smart kids who are not algebraically fluent, these methods are a life-saver.</p>
<p>^^–^^</p>
<p>Bingo, PCK! I know you know that, but I hope some students will also pick it up. Hence, the additional posting. </p>
<p>The surplus/deficits is similar (in fact it is the same) to the visual method I described before. I used intervals but I could have written it up with numbers:</p>
<p>Average cost of ticket: 3.5
Adults => 4.00 - 3.50 = .50
Child => 2 - 3.50 = -1.50
Ratio => 3/1</p>
<p>In this case, it is (as you wrote)
Average GPA: 89
Honor => 92 - 89 = 3
Regular => 83 - 89 = -6
Ratio => 2/1 </p>
<p>It just happens that for the first problem, the plugs will work because the numbers are easy. For the school average problem, plugging in numbers would be a waste of time. Same thing for setting equations. All will yield the correct answers, but some students will be penalized for spending too much time on an easy question --which, of course, is EXACTLY what ETS/SAT expects. </p>
<p>This is why practicing existing SAT questions is so important to find out why the test is a REASONING test.</p>
<p>EXCELLENT ADVICE!!! practice makes perfect.</p>