<p>How do you solve this algebraically? I know it's easier to replace numbers....</p>
<p>Mary has d dollars to spend and goes on a shopping spree. First she spends 2/5 of her money on shoes. Then she spends 3/4 of what's left on a few books. Finally she buys a raffle ticket that costs 1/3 of her remaining dollars. What fraction of d is left?</p>
<p>This is rather hard to type out since it’s hard to separate numerator from denominator. But I’ll try:</p>
<p>Start with d. Take away 2/5.</p>
<p>(3/5)d. Take away 3/4 of 3/5, which is 9/20.</p>
<p>(3/20)d. Take away 1/3 of 3/20, which is 3/60, which is 1/20.</p>
<p>In the end, you have (3/20)-(1/20)=2d/20 (or d/10 or 1/10) left.</p>
<p>EDIT: For some reason, I have a nagging feeling that I screwed up somewhere. You might want to wait for another post that confirms my answer.</p>
<p>3/5 * 1/4 * 2/3 = 6/60 = 1/10</p>
<p>yup xiggi pretty much nailed that one…</p>
<p>Xiggi’s process is essentially the condensed version of what I explained. It’s kinda like finding the percent of a percent of a percent of a starting number, if you (OP) know how to do those kinds of questions.</p>
<p>weird that is what i did but when i multiplied on my calc, it came to be like 3/10? mayb i entered it in wrong?</p>
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<p>Not that it matters, but “3/5 * 1/4 * 2/3” is NOT the condensed version of the solution proposed earlier, which focused on “taking away” from the remaining quantities and followed the problem step by step. </p>
<p>Fwiw, it’s also good to remember that the easiest way to solve such problem is to start with a number such as 100, and do the “take away” steps as described by 314159265 but with real numbers. </p>
<p>But the OP already knew that.</p>
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<p>That is why it is best to leave the calculators for VERY hard problems. ;)</p>
<p>Is that your advice? to deal with problems like this with substitution of numbers? When is the right time to substitute numbers and when to use algebra?</p>
<p>also another question for thought:</p>
<p>Six chairs are placed in a row to seat six people. How many different seating arrangements are possible if two of the people insist on sitting next to each other?(</3 permutations)</p>
<p>This is a sneaky kind of permutation problem. There may be a better way, but here’s the first thing that I thought of:</p>
<p>In general, to arrange n things, you have n! ways. But since the two people have to be next to each other, it would be better to think of them as a single unit, and so now we have 5 things to arrange. That would make the answer 5!. But wait – for each of those arrangements, you get a new arrangement if the two people (who we are treating as one unit) were to switch places with each other. So we have to double our answer…</p>
<p>Final answer: 2 x 5! = 240</p>
<p>Plugging in real numbers instead of using algebra is one of the most powerful SAT math strategies. Usually, the algebraic solution is the most efficient, but in some cases the algebraic solution can be challenging and/or prone to error, compared to the numeric method. It is hard to come up with a general rule for when to use which method, but I would say: if you can’t see how to use algebra to solve a question, or you are finding the algebra tough going, try plugging in (easy-to-use) numbers.</p>
<p>sounds good… that’s sorta what i do now… i usually try to solve it algebraiclly first just for the sake of understanding the concept. i’m sure i will use the plugging in concept alot more often on actual tests if i don’t know how to do a problem or to check my answers over.</p>
<p>and Pckeller, your idea definitely makes sense… i would like some verification though from some other posters though but it sounds good.</p>
<p>plugging in values just ruins the fun!! :(</p>
<p>Here’s another way, longer but more traditional:</p>
<p>Think of it as a series of decisions. First you pick a seat for one of the two who want to be together. But you have to consider two cases…</p>
<ol>
<li><p>He picks an end seat…2 choices. Then his friend has one choice. The remaining people have 4, 3, 2 and then 1 choices. Total = 2x1x4x3x2 = 48</p></li>
<li><p>He picks a non-end seat…4 choices. Then his friend has two choices (one on either side of him). The remaining people still have 4, 3, 2 and 1 choice.<br>
Total = 4x2x4x3x2x1 = 192</p></li>
</ol>
<p>Case 1 and case 2 together make 240. </p>
<p>I like the other solution better, but at least they give the same answer.</p>
<p>As a variance, what if the question was "Six chairs are placed in a row to seat six people. How many different seating arrangements are possible if two of the people REFUSE on sitting next to each other? :)</p>
<p>Fwiw, I do not know how often you will see a problem of arrangements with constraints on the SAT. Such problems of combinatorics are mostly GRE or GMAT fodder.</p>
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<p>I suppose a better way of comparing the 2 solutions is that mine focused on taking away while Xiggi’s focuses on taking parts of parts of a part. In reality, it’s the same process, but the 2 solutions approach it from opposite sides.</p>
<p>"As a variance, what if the question was "Six chairs are placed in a row to seat six people. How many different seating arrangements are possible if two of the people REFUSE on sitting next to each other? </p>
<p>Well, since there are 6!=720 arrangements total, and we just found that in 240 of them, they do sit together, then there are 720-240=480 ways to keep these hostile people apart!</p>
<p>But I agree that this level of combinatorics is not on the SAT. SAT counting problems can almost always be done by brute force (listing) or by applying the “how many choices do I have for each decision, then multiply” approach. So if you need yet another reason to stick with college board material, be aware that it is easy to state a combinatorics problem that looks easy but is actually quite sophisticated.</p>
<p>^That problem (with the 6 chairs and 6 people) seems like a MathCounts problem to me. I haven’t competed in AIME so I can’t really compare that problem to an AIME problem.</p>
<p>Speaking of AIME, can I take the test individually or do I need to do it through a school?</p>
<p>the six chairs problem…</p>
<p>two people wanna sit next to each other…so for every possible seating arrangement, there will be four other people who can sit in any arrangement of the four remaining chairs (the six minus the two that the two lovebirds are sitting in)…the number of possible ways to arrange four people in four seats would be just 4 x 3 x 2 x 1 (of four possibilities of the first person…there’s 3 seats possible for the second person…of that, there’s 2 possibilities for the third person…etc)…</p>
<p>so you’ll have to find the possible ways of getting the two horny lovebirds to sit next to each other within the six chairs, and then multiply that by 4! (or the same as 4 x 3 x 2 x 1 = 24)…</p>
<p>so lets see how to arrange it for the two horndogs…you have six chairs arranged like the following:</p>
<p>1 2 3 4 5 6</p>
<p>lets call the two lovers Kitty and Stud…lets hold Kitty into one of the seats and see where Stud can sit based on Kitty’s location…obv if Kitty sits in chair 1 or 6 (the two end ones)…Stud can sit in either 2 or 6 (the one further in from the end chair)…if Kitty sits anywhere in the middle seats (2-5)…Stud can sit either to the left or to the right (two possibilities)…two end seats with one possibility each…and 6 - 2 = 4 seats with two possibilities each = 2 x 1 + 4 x 2 = 10 ways to sit Kitty and Stud next to each other so they can make out and perform fellatio on each other…</p>
<p>with 10 possibilities for Kitty and Stud…each one of them has the 4! = 24 arrangement for the other people that’ll watch them as they eat each other out and make out…so total number of arrangements = 24 x 10 = 240 total arrangements…</p>
<p>sorry but what does fwiw mean? and yes this was from barrons 2400; i thought it may be useful to know it anyhow.</p>