<p>How do you reliably solve these? I'm talking about the ones where it's like:</p>
<p>The least and greatest numbers in a list of 7 real
numbers are 2 and 20, respectively. The median of the
list is 6, and the number 3 occurs most often in the list.
Which of the following could be the average
(arithmetic mean) of the numbers in the list?
I. 7
II. 8.5
III. 10
(A) I only
(B) I and II only
(C) I and III only
(D) II and III only
(E) I, II, and III</p>
<p>Now my question is, is there any way to solve these types of problems other than raw plugging numbers in and hoping you get lucky and hit the special number combinations that give the answers?</p>
<p>Let’s see: they give you enough information to get this far right away…</p>
<p>2, 3, 3, 6, _ , _, 20 </p>
<p>and now you are looking for potential averages. just consider the extreme cases.</p>
<p>The two missing numbers can be no lower than 6. But they can’t be exactly 6 or 3 would no longer be the mode. But they can be reeeeealy close to 6.</p>
<p>This would give you say: 2, 3, 6, 6.0001, 6.0002, 20 for an average of just over 6.14. All of their choices are above this value.</p>
<p>The two missing numbers can be no higher than 20, and again, they can’t be exactly 20 either. </p>
<p>So say 2, 3, 3, 6, 19.999, 19.99999, 20 which comes out to an average just under 10.57. All of their choices are below this value.</p>
<p>The point is that the answer is E but you don’t actually have to find the numbers that make each choice work – you just have to know that it is in fact possible.</p>
<p>From the question, the given seven nos. must be </p>
<p>2 3 3 6 x y 20 </p>
<p>Now, each mean in I , II and III such as 7, 8.5 and 10 imply that sum of seven nos. given can be 49, 59.5 and 70</p>
<p>All the three sums are possible with different values of x and y between 6 and 20. Hence, All the three values can be mean.</p>
<p>P.S. >>>> </p>
<p>like take 7 as mean then Sum of nos. given=7 *7=49 </p>
<p>so , x+y= 49- 34 = 15 … 15 is possible by assigning valid values to x and y… such as 7 and 8 ( valid values are between 6 and 20 according to question)</p>
<p>Similarly, all the values of mean give a valid sum of x and y .</p>
<p>Ok, first you have to make a list of the numbers. At first you have:</p>
<p>2 _ _ _ _ _ 20</p>
<p>Then you know 6 is the median, so:</p>
<p>2 _ _ 6 _ _ 20</p>
<p>If 3 is the most common number, it must appear twice:</p>
<p>2 3 3 6 _ _ 20</p>
<p>Now it’s a simple matter of finding upper and lower limits for the average. In this case, using two sixes, and two twenties, and finding the average of them will give lower and upper bounds respectively. You find the average must be between 6 4/7 and 10 4/7, so the answer is e.</p>
<p>I don’t suppose it specified the list contained integers? </p>
<p>Try
2, 3, 3, 6, 18, 19, 20
Average is above 10. So 10 works. (just replace 18 with 17 and the average is exactly 10).
Try
2, 3, 3, 6, 7, 8, 20.
Average is exactly 7. So 7 works. I’m sure you can manipulate the numbers to get an average of 8.5.</p>
<p>It’s funny and cool how many people post useful answers here. When I am working with an individual student,I try to show them how to self-study, using internet resources including this site. (I know that many CC’ers swear by self study. In general, I agree, but many students need help getting started. I take them quickly through my book and then I show them something called “the blue book” – you may have heard of it.) I like to take them to this site, pick some posted math question at random, and have them admire all of the useful solutions that people have posted. We then look at the time the question was posted and the times the answers came in. This particular post is an extreme case!</p>
<p>But one more thing: I teach my students to google the problem before they post. Many of the questions have already been discussed multiple times over the years. In fact, that instruction should probably be in its own thread, stickied at the top.</p>
<p>Here’s another one for practice. I just wrote it a few days ago so it shouldn’t be all over the internet yet. Should be about a level 3 (medium difficulty).</p>
<p>The median of a particular set of 10 integers is itself an integer. Which of the following could be true?</p>
<p>I. The integers of the set are all identical.
II. The integers of the set are consecutive.
III. The 5th largest and 6th largest integers of the set are odd.</p>
<p>(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II, and III</p>
<p>Nah, it’s D. I is obviously true. For 10 numbers, the median is the average of the 5th and 6th, so II doesn’t have an integer for a median, and III does…</p>
<p>In a list of ten values, the median is the average the fifth and sixth numbers. The median has to be an integer. What is a characteristic of the average of … two consecutive numbers. The problem stated consecutive numbers, and not consecutive ODD numbers.</p>
<p>Ryan, how do you decide which are the 5th largest and 6th largest integers in a series if they are all identical? </p>
<p>Again, the problem writer used the term “largest” and not something such as the 5th or 6th integer, or the 5th or 6th listed integer. A largest intimates there is a … smallest. Can I and III be true at the same time … which is what AND means?</p>
<p>I is true because all the integers can be same and median can be the same integer.
II is false because consecutive integers in 5th and 6th position doesn’t yield an integer median.
III is true because odd integers at both 5th and 6th position gives an integer median.</p>
(i don’t want to defend an incorrect method)</p>