{"id":5139,"date":"2013-03-18T09:12:39","date_gmt":"2013-03-18T13:12:39","guid":{"rendered":"http:\/\/www.manhattanprep.com\/gre\/blog\/?p=5139"},"modified":"2019-08-30T16:45:52","modified_gmt":"2019-08-30T16:45:52","slug":"the-5-lb-book-regular-quant-theory-problem","status":"publish","type":"post","link":"https:\/\/www.manhattanprep.com\/gre\/blog\/the-5-lb-book-regular-quant-theory-problem\/","title":{"rendered":"The 5 lb. Book: Regular\u009d Quant Theory Problem"},"content":{"rendered":"

\"GREWe’ve got another problem for you from our new book, the 5 lb<\/span>\u00a0Book of GRE<\/span>\u00a0Practice Problems<\/span><\/a>. The book contains more than 1,100 pages of practice problems (and solutions), so you can drill on anything and everything that might be giving you trouble.<\/p>\n

This regular\u009d problem solving question asks us to pick one correct answer (other variations might ask us to select more than one answer or to type in our own answer). Give yourself approximately 2 minutes to finish (or make a guess).
\n<\/p>\n

If x<\/em> is odd, all of the following must be odd EXCEPT:<\/p>\n

(A) x<\/em>2<\/sup> + 4x<\/em> + 6<\/p>\n

(B) x<\/em>3<\/sup> + 5x<\/em> + 3<\/p>\n

(C) x<\/em>4<\/sup> + 6x<\/em> + 7<\/p>\n

(D) x<\/em>5<\/sup> + 7x<\/em> + 1<\/p>\n

(E) x<\/em>6<\/sup> + 8x<\/em> + 4\u009d<\/p>\n

\u00a9\u00a0ManhattanPrep, 2013<\/p>\n<\/blockquote>\n

\u00a0<\/p>\n

This is what we call a theory<\/em> question. No actual numerical solution exists \u201c we have no idea what x<\/em> is, nor can we ever calculate it. Rather, the question is asking about number theory: in this case, about the characteristic odd.\u009d<\/p>\n

There are two main approaches to this kind of problem: (1) Test Cases and (2) Use Theory. We’ll take a look at both in this article.<\/p>\n

Before we dive in, a general rule: using theory is often faster as long as you fully get\u009d the theory. If you don’t, then you’re likely to make a mistake, in which case the advantage (saving time) isn’t going to matter.<\/p>\n

So, as a general rule, use theory when you feel totally comfortable with the theory being tested by the problem. Otherwise, test cases.<\/p>\n

Test Cases<\/h3>\n

When we test cases, we’re really just picking actual numbers to test out in the problem. \u00a0(Note: On a must\u009d be true or could be true question type, that can mean testing numbers multiple times until we get down to one answer!)<\/p>\n

How do we test cases here? First, set up a little chart:<\/p>\n\n\n\n\n\n\n\n\n
\n

Answer<\/strong><\/p>\n<\/td>\n

\n

when x<\/em> = 1<\/strong><\/p>\n<\/td>\n

\n

odd?<\/strong><\/p>\n<\/td>\n<\/tr>\n

x<\/em>2<\/sup> + 4x<\/em> + 6<\/td>\n1 + 4 + 6 = 11<\/td>\n\n

yes<\/p>\n<\/td>\n<\/tr>\n

x<\/em>3<\/sup> + 5x<\/em> + 3<\/td>\n1 + 5 + 3 = 9<\/td>\n\n

yes<\/p>\n<\/td>\n<\/tr>\n

x<\/em>4<\/sup> + 6x<\/em> + 7<\/td>\n1 + 6 + 7 = 14<\/td>\n\n

NO<\/p>\n<\/td>\n<\/tr>\n

x<\/em>5<\/sup> + 7x<\/em> + 1<\/td>\n1 + 7 + 1 = 9<\/td>\n\n

yes<\/p>\n<\/td>\n<\/tr>\n

x<\/em>6<\/sup> + 8x<\/em> + 4<\/td>\n1 + 8 + 4 = 13<\/td>\n\n

yes<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\u00a0<\/p>\n

Awesome! Only one, answer C, wasn’t odd, so that must be the correct answer. I showed all of the work here, but when the clock is actually ticking, stop after answer C: if it’s not odd, then it has to be the correct answer.<\/p>\n

We’re done. Why bother with the theory (or any other) way? Let’s move on.<\/p>\n

Wait just a second. Notice something \u201c what if we had had to plug in, say, x <\/em>= 2? Or, even worse, x<\/em> = 3?<\/p>\n

(Why do I say that plugging in 3 would be even worse?\u009d Right, because of those ugly-looking exponents. I don’t know about you, but I don’t want to bother with 26<\/sup>, let alone 36<\/sup>!)<\/p>\n

So let’s take a moment to figure out how we could still do this one if we didn’t want to try real numbers because they were too ugly. First, each answer choice contains one term that represents raising x<\/em> to a power. The problem says that x<\/em> is odd. Are there any rules or patterns about what will happen when raising an odd number to a power?<\/p>\n

Yes! Squaring and cubing and so on are just ways of multiplying x<\/em> by itself. x<\/em>2<\/sup> = (x<\/em>)(x)<\/em>, x<\/em>3<\/sup> = (x<\/em>)(x)<\/em>(x)<\/em>, and so on. What happens when we multiply odd numbers together?<\/p>\n

odd \u00c3\u2014\u00a0odd = odd (always!)<\/p><\/blockquote>\n

In other words, the first term for every answer choice will always be odd.<\/p>\n

Let’s start to play this all out with answer A:<\/p>\n

(A) x<\/em>2<\/sup> + 4x<\/em> + 6\u009d<\/p><\/blockquote>\n

Okay, x<\/em>2<\/sup> is odd. What about 4x<\/em>? An even times an odd always equals an even! Finally, the number 6 itself is even, so we have:<\/p>\n

Odd + Even + Even<\/p><\/blockquote>\n

Take each piece sequentially. The first two are Odd + Even.\u009d This will always equal an odd number, so the first two terms combine to produce an odd number. Now plug that into the remaining piece:<\/p>\n

(Odd + Even) + Even =<\/p>\n

Odd + Even =<\/p>\n

Odd!<\/p><\/blockquote>\n

Okay, answer A is odd. The question wants us to eliminate<\/em> the ones that are odd, so cross off answer A.<\/p>\n

Test out the other answers in the same way:<\/p>\n\n\n\n\n\n\n\n\n
\n

Answer<\/strong><\/p>\n<\/td>\n

\n

Theory Step 1<\/strong><\/p>\n<\/td>\n

\n

Theory Step 2<\/strong><\/p>\n<\/td>\n

\n

odd?<\/strong><\/p>\n<\/td>\n<\/tr>\n

x<\/em>2<\/sup> + 4x<\/em> + 6<\/td>\n(Odd + Even) + Even<\/td>\n\n

Odd + Even<\/p>\n<\/td>\n

\n

Odd<\/p>\n<\/td>\n<\/tr>\n

x<\/em>3<\/sup> + 5x<\/em> + 3<\/td>\n(Odd + Odd) + Odd<\/td>\n\n

Even + Odd<\/p>\n<\/td>\n

\n

Odd<\/p>\n<\/td>\n<\/tr>\n

x<\/em>4<\/sup> + 6x<\/em> + 7<\/td>\n(Odd + Even) + Odd<\/td>\n\n

Odd + Odd<\/p>\n<\/td>\n

\n

Even!<\/p>\n<\/td>\n<\/tr>\n

x<\/em>5<\/sup> + 7x<\/em> + 1<\/td>\n(Odd + Odd) + Odd<\/td>\n\n

same as B<\/p>\n<\/td>\n

\n

Odd<\/p>\n<\/td>\n<\/tr>\n

x<\/em>6<\/sup> + 8x<\/em> + 4<\/td>\n(Odd + Even) + Even<\/td>\n\n

Odd + Even<\/p>\n<\/td>\n

\n

Odd<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\u00a0<\/p>\n

Again, I showed the work for all five answers, but stop after choice C (that is, as soon as you realize you’ve got a not-odd answer).<\/p>\n

The correct answer is C.<\/em><\/p>\n

Key Takeaways for Theory Problems:<\/strong><\/p>\n

(1) Recognize them: these problems ask about a characteristic of something but don’t actually provide enough information to calculate one definitive set of numbers. In this problem, the value for x<\/em> really could be anything, as long as it’s odd.<\/p>\n

(2) One approach: Test Cases. This is often more straightforward than Approach #2 but might take longer. Try allowed\u009d numbers (for example, in this case, we could only pick odd numbers for x<\/em>) and cross off answers. If more than one answer is left, try another, and maybe another \u201c until you have one answer left or you decide to guess.<\/p>\n

(3) Another approach: Use Theory. This is often faster than Approach #1 but more prone to mistakes \u201c so don’t use this approach unless you feel 100% confident about the underlying theory.<\/p>\n

\u00a0<\/p>\n

\u00a9\u00a0ManhattanPrep, 2013<\/p>\n

<\/div>\n","protected":false},"excerpt":{"rendered":"

We’ve got another problem for you from our new book, the 5 lb\u00a0Book of GRE\u00a0Practice Problems. The book contains more than 1,100 pages of practice problems (and solutions), so you can drill on anything and everything that might be giving you trouble. This regular\u009d problem solving question asks us to pick one correct answer (other […]<\/p>\n","protected":false},"author":6,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7,9],"tags":[34,271],"yst_prominent_words":[],"class_list":["post-5139","post","type-post","status-publish","format-standard","hentry","category-how-to-study","category-math-gre-strategies","tag-5lb-book","tag-quant"],"_links":{"self":[{"href":"https:\/\/www.manhattanprep.com\/gre\/wp-json\/wp\/v2\/posts\/5139","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.manhattanprep.com\/gre\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.manhattanprep.com\/gre\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.manhattanprep.com\/gre\/wp-json\/wp\/v2\/users\/6"}],"replies":[{"embeddable":true,"href":"https:\/\/www.manhattanprep.com\/gre\/wp-json\/wp\/v2\/comments?post=5139"}],"version-history":[{"count":1,"href":"https:\/\/www.manhattanprep.com\/gre\/wp-json\/wp\/v2\/posts\/5139\/revisions"}],"predecessor-version":[{"id":6956,"href":"https:\/\/www.manhattanprep.com\/gre\/wp-json\/wp\/v2\/posts\/5139\/revisions\/6956"}],"wp:attachment":[{"href":"https:\/\/www.manhattanprep.com\/gre\/wp-json\/wp\/v2\/media?parent=5139"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.manhattanprep.com\/gre\/wp-json\/wp\/v2\/categories?post=5139"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.manhattanprep.com\/gre\/wp-json\/wp\/v2\/tags?post=5139"},{"taxonomy":"yst_prominent_words","embeddable":true,"href":"https:\/\/www.manhattanprep.com\/gre\/wp-json\/wp\/v2\/yst_prominent_words?post=5139"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}