math problem

Test Preparation SAT Preparation
1

<p>The least and greatest numbers in a list of 7 real numbers are 2 and 20 and respectively the median is 6 and the number 3 occurs most often in the list which of the following could be the average of the members in the list?
I 7
II 8.5
III 10
A I only</p>

<p>B I and II only
C II and III only
E I and II and III
With explanation please also if I practices hard can I get a high score by Nov which is next week because my math score is really low it is 430 and how do I study beside Practicing</p>

2

<p>Lining your numbers up from least to greatest, the first is 2, the 4th (median position) is 6, there are at least 2 3’s, so they take spots #2 and #3. You’re now out of spots for any additional 3’s, so you know that spots #5 and #6 are two different (non-identical) numbers, and spot #7 is 20.</p>

<p>The total of the numbers you have is 34, so you need 15 more to get to a mean of 7, 25.5 more to get to a mean of 8.5, and 36 more to get to a mean of 10. You can use any two real numbers greater than 6 and smaller than 20 to come up with those totals. Real numbers include decimals - you aren’t limited to whole numbers.</p>

<p>You get 15 with 7 + 8, so I works.
You get 36 with 17 and 19, so III works.
You get 25.5 with 12.5 and 13, so II works.</p>

<p>I vote E.</p>

3

<p>Yup, E. Here’s why.</p>

<p>The average formula (very useful for SAT) is as follows:
(total sum)/(number)=(average)
For instance, if three students score a 70, 75, and 80 on a test, what is their average?
(70+75+80)/3=average
75=average</p>

<p>That’s pretty intuitive, but knowing the whole formula helps a lot on the SATs.
Now on to the actual problem.</p>

<p>The set is of seven real numbers. 2 is the first, 20 is the last. So we can start to fill in some blanks:</p>

<p>2 _ _ _ _ _ 20</p>

<p>The median is 6</p>

<p>2 _ _ 6 _ _ 20</p>

<p>The number 3 occurs twice. This means it HAS to be in between the 2 and the 6, because 6 is the median.</p>

<p>2 3 3 6 _ _ 20</p>

<p>And that’s the information we have. Now time to set up the average formula. I’m going to call the two numbers we don’t know x and y. First let’s try out option I. Remember, sum divided by the number gives you the average. Therefore:</p>

<p>(2+3+3+6+20+x+y)(sum of the numbers in the set)/7(amount of numbers in the set) = 7
(34+x+y)/7=7
34+x+y=49
x+y=15</p>

<p>Because of the way our numbers are set up, x and y have to fit in between 6 and 20. So we need to think of some numbers that fit between 6 and 20 and also add up to 15. 7 and 8 fit the bill, so I works, because your set can be 2, 3, 3, 6, 7, 8, 20.</p>

<p>Next we’ll try II, an average of 8.5. To save time, we already know that the sum of our known numbers is 34. </p>

<p>(34+x+y)/7=8.5
34+x+y=59.5
x+y=25.5</p>

<p>If college board wanted to be really mean with this problem, they would have told us that the set was composed only of integers. Thankfully, the problem gave us “all real numbers”, so decimals count. 25.5/2 gives us 12.75. The numbers 12.5 and 13 fit in our set, so II works. </p>

<p>Finally time for III, average of 10.</p>

<p>(34+x+y)/7=10
34+x+y=70
x+y=36</p>

<p>36/2=18. The numbers 17 and 19 fit in our set, leaving 20 as the highest number. Therefore III works also. The answer is E, all of them work.</p>

<p>Sorry for the longwinded reply. There are easy ways to get around this, but none of them are better than understanding averages!</p>

4

<p>This is a comparatively difficult question, jasmine95. If your math score is 430, I would suggest that you focus on the less difficult problems. Generally, students who are scoring 600 or below can improve their scores by making sure that they are answering correctly as many of the easy to medium difficulty questions as possible, while just omitting the harder questions.</p>

5

<p>We have</p>

<p>2 w x 6 y z 20 (in increasing order)</p>

<p>3 is a unique mode, so replace the two leftmost blanks with 3’s</p>

<p>2 3 3 6 y z 20</p>

<p>The average of the seven numbers is (2+3+3+6+y+z+20)/7 = (34+y+z)/7. We note that y and z are strictly between 6 and 20, and not equal to each other. Therefore</p>

<p>12 < y+z < 40 → 46 < 34+y+z < 74</p>

<p>Dividing both sides by 7, 46/7 < (34+y+z)/7 < 74/7. Therefore the average must be strictly between 46/7 and 74/7. All three choices (I, II, III) satisfy this constraint, so the answer is E.</p>

6

<p>P.S. I agree with QuantMech, you may want to work on some of the mid-range questions first. This question is comparatively hard for SAT math – it’s not too difficult, but could take a little long if you’re not familiar with algebra or topics such as mean/median.</p>

7

<p>But I heard that hard questions have higher marks on them</p>

8

<p>Hard ones are NOT worth more. So listen to Quantmech and rspence! Focus on the easy and medium questions. That way, you can do the single best thing to improve: slow down.</p>

9

<p>Yes, all the math questions are worth the same. You’d rather nail the easy/medium questions, then go to the hard ones, rather than spend half the time doing the last five problems without looking at easier ones.</p>

10

<p>yeah, hard probelsm aren’t worth more.</p>