<p>hi.
its from the new BB. on pg 417.(test1. section 7, question number 12 , 14, and 16)</p>
<ol>
<li><p>Five different poitns A.B.C.D. AND E. lie on a line in that order. the lenght of AD is 4.5 and the length of BE is 3.5. If the length of CD is 2, what is one possible value for the lenght of BC ?</p></li>
<li><p>Each term of a certain sequence is greater than the term before it. the difference between any two consecutive terms in the sequence is always the same number. if the third and sixth terms of the sequence are 17 and 77, respectively , what is the eighth term?</p></li>
<li><p>a four-digit interger, WXYZ, in which W,X,Y, AND Z each represent a different digit, is formed according to the following rules.</p></li>
<li><p>X=W+Y+Z</p></li>
<li><p>W= Y+1</p></li>
<li><p>Z=W-5
what is the four-digit integer ?</p></li>
</ol>
<ol>
<li><p>I know i got this answer right ,but i cannot explain this one properly sorry.</p></li>
<li><p>For this problem, you should know the basic arithmetic sequence formulas.</p></li>
</ol>
<p>The formula you use for this problem is: The term = the first term + (the number of the term - 1)(common difference between each consecutive term)</p>
<p>In this problem you would set up a system of two equations as follows:</p>
<p>17= first term + (3-1)d , where d=common difference
77= first term + (6-1)d</p>
<p>Simplify a bit
17= first term + 2d
77= first term + 5d</p>
<p>By now you should see that this is a simple system of equations problem.
When you solve this system you should get d=20 and the first term=-23</p>
<p>The problem however asks for the eighth term, so plug this into the original equation, and you should get 117 which is the answer.</p>
<ol>
<li>I did not do this problem “legitly” and I simply plugged in a number for one of the variables say for X. Then you solve for the other variables until you get a logical number for each one. For example W cannot equal 50 since it is only 1 digit. Using this method you should get the answer.</li>
</ol>
<p>between B & E there is 7/2 , CD is 2 so there is a remainder of 3/2 for BC and DE combined, </p>
<p>thus any number 0<x<3/2 should work </p>
<p>I dont know why they give you the length of AD however, perhaps there is another way to solve it? either way, this way works perfectly fine AD is unnecessary information</p>
<p>oh & to help with #16 because pakrats method was, well it works but idk </p>
<p>the equations here </p>
<p>X= W+ Y+ Z
W=Y+1
Z=W-5</p>
<p>think about it this way, each number is an individual digit and none are negative,
so thinking through it
first- the greatest X can be is 9 and for the three other numbers to add up to 9 while one is 5 greater than another, one digit must be zero. that is Z
second- now that we know Z is zero & X is 9, we see that W and W-1 must add up to 9 so W must be 5, and Y which is W-1 must be 4 </p>