{"id":6951,"date":"2014-01-28T13:04:59","date_gmt":"2014-01-28T18:04:59","guid":{"rendered":"http:\/\/www.manhattangmat.com\/blog\/?p=6951"},"modified":"2019-09-05T16:05:32","modified_gmt":"2019-09-05T16:05:32","slug":"want-a-51-on-quant-can-you-answer-this-problem","status":"publish","type":"post","link":"https:\/\/www.manhattanprep.com\/gmat\/blog\/want-a-51-on-quant-can-you-answer-this-problem\/","title":{"rendered":"Want a 51 on Quant? Can you answer this problem?"},"content":{"rendered":"

\"GMAT-problem\"Sequence problems aren\u2019t incredibly common on the test, but if you\u2019re doing well on the quant section, be prepared to see one. Now, you\u2019ve got a choice: do you want to guess quickly and save time for other, easier topics? Do you want to learn some \u201ctest savvy\u201d techniques that will help you with some sequence questions but possibly not all of them? Or do you want to learn how to do these every single time, no matter what?<\/p>\n

That isn\u2019t a trick question. Every good business person knows that there\u2019s a point of diminishing returns: if you don\u2019t actually need a 51, then you may study for a lower (but still good!) score and re-allocate your valuable time elsewhere.<\/p>\n

Try this GMATPrep\u00ae problem from the free test. After, we\u2019ll talk about how to do it in the \u201ctextbook\u201d way and<\/em> in the \u201cback of the envelope\u201d way.<\/p>\n

* \u201dFor every integer k<\/em> from 1 to 10, inclusive, the k<\/em>th term of a certain sequence is given by\u00a0\"Screen. If T<\/em> is the sum of the first 10 terms in the sequence, then T<\/em> is<\/p>\n

\u201c(A) greater than 2<\/p>\n

\u201c(B) between 1 and 2<\/p>\n

\u201c(C) between 1\/2 and 1<\/p>\n

\u201c(D) between 1\/4 and 1\/2<\/p>\n

\u201c(E) less than 1\/4\u201d<\/p>\n

First, let\u2019s talk about how to do this thing in the \u201ctextbook math\u201d way. If you don\u2019t want to do this the textbook math way, feel free to skip to the second method below.<\/p>\n

Textbook Method<\/em><\/p>\n

If you\u2019ve really<\/em> studied sequences, then you may recognize the sequence as a particular kind called a Geometric Progression. If not, you would start to find the terms and see whether you can spot a pattern.<\/p>\n

Plug in k<\/em> = 1, 2, 3. What\u2019s going on?<\/p>\n

\"Screen<\/p>\n

What\u2019s going on here? Each time, the term gets multiplied by -1\/2 in order to get to the next one. When you keep multiplying by the same number in order to get to the next term, then you have a geometric progression.<\/p>\n

This next part gets into some serious math. Unless you really just love<\/em> math, I wouldn\u2019t bother learning this part for the GMAT, because there\u2019s a very good chance you\u2019ll never need to use it. But, if you want to, go for it!<\/p>\n

When you have a geometric progression, you can calculate the sum in the following way:<\/p>\n

\"Screen<\/p>\n

Next, you\u2019re going to multiply every term in the sum by the common ratio. What\u2019s the common ratio? It\u2019s the constant number that you keep multiplying each term by to get the next one. In this case, you\u2019ve already figured this out: it\u2019s – 1\/2.<\/p>\n

If you multiply this through all of the terms on both sides of the equation, you\u2019ll get this:<\/p>\n

\"Screen<\/p>\n

Does anything look familiar? It\u2019s basically the same list of numbers as in the first sum equation, except it\u2019s missing the first number, 1\/2. All of the others are identical!<\/p>\n

Subtract this second equation from the first:<\/p>\n

\"Screen<\/p>\n

The right-hand side of the equation is always going to be just the first term of the original sum. The rest of the terms on the right-hand side of the two equations are identical, so when you subtract, they become zero and disappear.<\/p>\n

Solve for s<\/em>:<\/p>\n

\"Screen<\/p>\n

This value falls between 1\/4 and 1\/2, so the answer is (D).<\/p>\n

Back of the Envelope Method<\/em><\/p>\n

There is another way to tackle this one. At the same time, this problem is really tricky\u2014so this solution is still not an \u201ceasy\u201d solution. Your best choice might be just to guess and move on.<\/p>\n

Before you start reading the text, take a First Glance at the whole thing. It\u2019s a problem-solving problem. The answers are\u2026 weird. They\u2019re not exact. What does that mean?<\/p>\n

Read the problem, but keep that answer weirdness in mind. The first sentence has a crazy sequence. The question asks you to sum up the first 10 terms of this sequence. And the answers aren\u2019t exact\u2026 so apparently you don\u2019t need to find the exact sum.<\/p>\n

Take a closer look at the form of the answers. Notice anything about them?<\/p>\n

They don\u2019t overlap! They cover adjacent ranges. If you can figure out that, for example, the sum is about 3\/4, then you know the answer must be (C). In other words, you can actually estimate here\u2014you don\u2019t have to do an exact calculation.<\/p>\n

That completely changes the way you can approach this problem! Here\u2019s the sequence:<\/p>\n

\"Screen<\/p>\n

According to the problem, the 10 terms are from k<\/em> = 1 to k<\/em> = 10. Calculating all 10 of those and then adding them up is way too much work (another clue that there\u2019s got to be a better way to do this one). So what is that better way?<\/p>\n

Since you know you can estimate, try to find a pattern. Calculate the first two terms (we had to do this in the first solution, too).<\/p>\n

\"Screen<\/p>\n

What\u2019s going on? The first answer is positive and the second one is negative. Why? Ah, because the first part of the calculation is -1 raised to a power. That will just keep switching back and forth between 1 and -1, depending on whether the power is odd or even. It won\u2019t change the size of the final answer, but it will change the sign.<\/p>\n

Okay, and what about that second part? it went from 1\/2 to 1\/4. What will happen next time? Try just that part of the calculation. If k<\/em> = 3, then just that part will become \"Screen.<\/p>\n

Interesting! So the denominator will keep increasing by a factor of 2: 2, 4, 8, 16 and so on.<\/p>\n

Great, now you can write out the 10 numbers!<\/p>\n

\"Screen\u2026 ugh. The denominator\u2019s getting pretty big. That means the fraction itself is getting pretty small. Do I need to keep writing these out?<\/p>\n

What was the problem asking again?<\/p>\n

Right, find the sum of these 10 numbers. Let\u2019s see. The first number in the sequence is 1\/2 and the second is -1\/4, so the pair adds up to 1\/4.<\/p>\n

\"Screen<\/p>\n

Right now, the answer would be right between D and E. Does the sum go up or down from here?<\/p>\n

The third number will add 1\/8, so it goes up:<\/p>\n

\"Screen<\/p>\n

But the fourth will subtract 1\/16 (don\u2019t forget that every other term is negative!), pulling it back down again:<\/p>\n

\"Screen<\/p>\n

Hmm. In the third step, it went up but not enough to get all the way to 1\/2. Then, it went down again, but by an even smaller amount, so it didn\u2019t get all the way back down to 1\/4.<\/p>\n

The fifth step would go up by an even smaller amount (1\/32), and then it would go back down again by yet a smaller number (1\/64). What can you conclude?<\/p>\n

First, the sum is always growing a little bit, because each positive number is a bit bigger than the following negative number. The sum is never going to drop below 1\/4, so cross off answer (E).<\/p>\n

You keep adding smaller and smaller amounts, though, so if the first jump of 1\/8 wasn\u2019t enough to get you up to 1\/2, then none of the later, smaller jumps will get you there either, especially because you also keep subtracting small amounts. You\u2019re never going to cross over to 1\/2, so the sum has to be between 1\/4 and 1\/2.<\/p>\n

The correct answer is (D).<\/p>\n

As I mentioned above, you may decide that you don\u2019t want to do this problem at all. These aren\u2019t that common\u2014many people won\u2019t see one like this on the test. Also, you don\u2019t have to get everything right to get a top score. Just last week, I spoke with a student who outright guessed on 4 quant problems, and she still scored a 51 (the top score).<\/p>\n

Key Takeaways for Advanced Sequence Problems<\/strong><\/p>\n

(1) Do you even want to learn how to do these? Don\u2019t listen to your pride. Listen to your practical side. This might not be the best use of your time.<\/p>\n

(2) All of these math problems do have a textbook solution method\u2014but you\u2019d have to learn a lot of math that you might never use if you try to learn all of the textbook methods. That\u2019s not a problem if you\u2019re great at math and have a great memory for this stuff. If not\u2026<\/p>\n

(3) \u2026 then think about alternate methods that can work just as well. Certain clues will indicate when you can estimate on a problem, rather than solving for the \u201creal\u201d number. You may already be familiar with some of these, for instance when you see the word \u201capproximately\u201d in the problem or answer choices that are spread pretty far apart. Now, you\u2019ve got a new clue to add to your list: answers that offer a range of numbers and the different answer ranges don\u2019t overlap.<\/p>\n

* GMATPrep\u00ae questions courtesy of the Graduate Management Admissions Council. Usage of this question does not imply endorsement by GMAC.<\/p>\n","protected":false},"excerpt":{"rendered":"

Sequence problems aren\u2019t incredibly common on the test, but if you\u2019re doing well on the quant section, be prepared to see one. Now, you\u2019ve got a choice: do you want to guess quickly and save time for other, easier topics? Do you want to learn some \u201ctest savvy\u201d techniques that will help you with some […]<\/p>\n","protected":false},"author":6,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"yst_prominent_words":[],"class_list":["post-6951","post","type-post","status-publish","format-standard","hentry","category-how-to-study"],"_links":{"self":[{"href":"https:\/\/www.manhattanprep.com\/gmat\/wp-json\/wp\/v2\/posts\/6951","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.manhattanprep.com\/gmat\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.manhattanprep.com\/gmat\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.manhattanprep.com\/gmat\/wp-json\/wp\/v2\/users\/6"}],"replies":[{"embeddable":true,"href":"https:\/\/www.manhattanprep.com\/gmat\/wp-json\/wp\/v2\/comments?post=6951"}],"version-history":[{"count":1,"href":"https:\/\/www.manhattanprep.com\/gmat\/wp-json\/wp\/v2\/posts\/6951\/revisions"}],"predecessor-version":[{"id":17459,"href":"https:\/\/www.manhattanprep.com\/gmat\/wp-json\/wp\/v2\/posts\/6951\/revisions\/17459"}],"wp:attachment":[{"href":"https:\/\/www.manhattanprep.com\/gmat\/wp-json\/wp\/v2\/media?parent=6951"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.manhattanprep.com\/gmat\/wp-json\/wp\/v2\/categories?post=6951"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.manhattanprep.com\/gmat\/wp-json\/wp\/v2\/tags?post=6951"},{"taxonomy":"yst_prominent_words","embeddable":true,"href":"https:\/\/www.manhattanprep.com\/gmat\/wp-json\/wp\/v2\/yst_prominent_words?post=6951"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}