<p>I have a maths problem.....</p>
<p>LetA={all 3-digit positive integers with the digit 1 in the ones place},and let B ={all 3_digit positive integers with the digit2 in the pens place}</p>
<p>how many elements are there in AUB that are not in A∩B????</p>
<p>A.162
B.171
C.180
D.182
E.192</p>
<p>There are
9x10=90 elements in A,
9x10=90 elements in B,
180 elements in AUB.
9 elements are in both A and B; they are included twice in AUB, so the answer is 180-18=162.</p>
<p>I have only a small elaboration and small correction. First, the reason there are 90 elements in A is because you can freely choose for the hundreds and tens digits. The hundreds digit can be 1-9 (not 0), and so there are 9 choices. The tens digit can be 0-9, so there are 10 choices. There are a total of 9*10=90 elements in A. The same reasoning applies for B.</p>
<p>Let us denote the number of elements in A by |A| and the number of elements in B by |B|. Then, |A| = 90, |B| = 90. What is the size of |A u B|? At first glance, it should be |A|+|B|. But actually the union of two sets does not count the duplicates. There are |A ∩ B| elements that are in both A and B. They are counted twice when we add |A|+|B|, so we need subtract |A ∩ B| from |A|+|B| to get |A u B|. This is known as the principle of inclusion-exclusion, or PIE:</p>
<p>|A u B| = |A| + |B| - |A ∩ B|</p>
<p>In particular, the size of A ∩ B is 9. Why? Because if a number is in both A and B, it must have 1 in the ones place and 2 in the tens place. You are only free to choose 9 choices (1-9) for the hundreds place. Thus, the size of A u B is 90 + 90 - 9 = 171. Every element in A ∩ B is in A u B, so the number of elements in A u B that are not in A ∩ B is |A u B| - |A ∩ B| = 171 - 9 = 162.</p>
<p>This is identical to the reasoning used by gcf101, except with a different definition (I think more conventional) of the set union.</p>
<p>LetA={all 3-digit positive integers with the digit 1 in the ones place},and let B ={all 3_digit positive integers with the digit2 in the tens place}</p>
<p>how many elements are there in AUB that are not in A∩B????</p>
<p>A.162
B.171
C.180
D.182
E.192 </p>
<p>I found it off of CC, actually, lol. But I've never seen anything like it and can't find anything like it in my books (Blue Book, Kaplan). Can anybody explain terminology and how to solve further? (: Thank you soooo much. x</p>
<p>Here's an introduction to the terminology, all part of what is known as set theory.</p>
<p>A set is any collection of "elements" or "members." In the abstract, these can be anything, such as the set of US states. In mathematics, they are often a collection of related mathematical objects. In this case, A and B are sets of real numbers given each by a different rule: A contains all 3-digit positive integers with a 1 in the units place, and B contains all 3-digit positive integers with a 2 in the tens place.</p>
<p>We can define two operations on sets: the union and the intersection.</p>
<p>The union of two sets is another set, denoted A u B. This set contains all the elements of A and all the elements of B, without duplicates. For more information, see <a href="http://en.wikipedia.org/wiki/Union_(set_theory%5B/url%5D)">http://en.wikipedia.org/wiki/Union_(set_theory)</a></p>
<p>The intersection of two sets is another set, denoted by A n B. This set contains all elements that A and B have in common, without duplicates. If A and B have no elements in common, they are called disjoint, and A n B = the null/empty set.</p>
<p>Given this terminology, what ideas do you have?</p>
<p>So:
A = 90 terms (with restrictions)
B = 90 terms
...I figured out the restrictions, where do I go from there?<br>
I mean, more specifically, how do I find out AUB and AnB? If that could be explained, that would be the missing link. (:
How often do these pop up on the SAT?</p>
<p>Probably rarely, but if so, the questions are usually easy once you've gotten a handle on them (read: easy points).</p>
<p>So yes, if we denote the size of a set A by |A|, we have |A| = 90, |B| = 90, like you said. Now, you can reason. Usually, you can find the size of A n B easiest. How many elements do A and B have in common? Count them =)</p>
<p>Anyone confirm my answer of 162?</p>
<p>Yes, it is 162.</p>
<p>What do you mean...count them - the ones in common? I really just need a step-by-step...this is foreign territory for me! :P</p>
<p>Well, you have found that there are 90 numbers in A and 90 numbers in B. For a number to be in A, it must be of the form _ _ 1. For a number to be in B, it must be of the form _ 2 _. For a number to be in both A and B, what form must it be in? How many of them are there?</p>
<p>Okay, let's make this simple:</p>
<p>First, consider this line of numbers:</p>
<p>1, 4, 5, 6, 6, 7, 8, 9, 10</p>
<p>Now, the question is: How many numbers are there that aren't repeated? This is easy. There are 9 numbers to begin with, and there's one number that's repeated. Since the number 6 is repeated once, you need to subtract 2 from the initial quantity, which is 9, so 9-2 = 7, so there are 7 numbers in there that aren't repeated. Make sure you see how we got this answer. This method will be useful later. Now, something different:</p>
<p>Let's pretend that set A is made up of these numbers (all 2-digit numbers with 1 in 1's place):</p>
<p>11,21,31,41,51,61,71,81,91</p>
<p>and set B is made up of these numbers (all 2-digit numbers with 2 in 10s place)</p>
<p>20,21,22,23,24,25,26,27,28,29</p>
<p>Now, the question is, how many of these numbers are there if you combine sets A and B together and take out the numbers with repeats?</p>
<p>First, let's see how many there are if you combine A and B. You get 19 numbers. Okay, so imagine all 19 numbers lined up side by side in a line. Now, some of those numbers are repeats. How many repeats are there? Well, the ones that are repeats are the ones that satisfy both A and B. In other words, they are the 2-digit numbers with a 2 in the 10s place and a 1 in the ones place. Well, there's only one number that foots the bill: 21. Do you see why 21 would be in both sets A and B, and therefore be a repeat?</p>
<p>So, let's take out the number 21. But since there are 2 21's, we need to take out 2 from the initial quantity of 19, leaving 17. So there are 17 numbers when you lump A and B together and take out the numbers that are repeated.</p>
<p>What if there were 3 21's in that set? Well, then we'd take out 3 from the initial quantity, to get rid of all the numbers that are repeated. </p>
<p>Make sure all of what I just said makes sense before we go on.</p>
<p>Now, let's move on to the actual problem you are struggling with.</p>
<p>You've correctly asserted that A has 90 numbers and that B has 90 numbers. Now, just imagine all 180 numbers lined up side by side, the numbers from A on the left and next to that the numbers from B on the right.</p>
<p>Okay. Now, what you do you want to find? </p>
<p>You want to find the quantity of numbers left when you take out all the numbers that have repeats. For example, if the initial set is 1,4,4,5,6 the answer to the problem would be 3, because you take out the 2 4's from the initial quantity, 5.</p>
<p>Strategy: Find how many numbers are repeated, and take all those numbers from the line, and see how many numbers are left.</p>
<p>Okay, resort to my previous post if you're confused. My example is pretty much exactly how we're going to tackle this problem. The numbers that are repeated are the ones that are in both bags (A and B). So, how do you find these numbers? Well, if A is made up of the 3 digit numbers with 1 in the ones place, and B is made up of the 3 digit numbers with 2 in the tens place, what kind of numbers would be in A and B, and therefore are repeats?</p>
<p>Thanks, Tedjn, for correcting! I don't know what I was thinking when I posted my reply (I was in a rush to leave), but I got lucky: two wrongs made the right answer! :o
If only I used the Venn diagram (one of my favorite tools)...</p>
<p>I read a question a little differently (it's inconsequential though):
"digit 1" and "digit 2", I think, are two random digits d1 and d2 (they can be equal to each other); without loosing generality we can assume d1=d2=0.</p>
<p>Let's interpret sets A and B as intersecting circles.
Common area of these two circles is comprised of 9 integers:
100
200
300
400
500
600
700
800
900.</p>
<p>Circle A without this common area has 90-9=81 elements:
110, 120, ..., 190
210, 220, ... 290
...
910, 920, ...990.</p>
<p>Circle B without the common area also has 90-9=81 elements:
101, 102, ..., 109
201, 202, ... 209
...
901, 902, ...909.</p>
<p>Two circles together excluding common area have 81+81=162 elements.</p>
<p>Edit.
I don't think a question in this form (AUB, A∩B) will ever "pop up" on the SAT.</p>