please help with confounding BB math problem

Test Preparation SAT Preparation
1

<p>In an effort to help S prep for the PSAT/SAT, I have managed to reteach myself the algebra I first learned 30 years ago and that S will learn this coming school year. However, there is one problem that, for the life of me, I cannot get. I would greatly appreciate any help. It is BB Test 5, Section 8 question 9 on page 669. It states as follows:</p>

<p>A regulation for riding a certain amusement park ride requires that a child be between 30 inches and 50 inches tall. Which of the following inequalities can be used to determine whether or not a child's height h satisfies the regulation for this ride?
a. |h-10|<50
b. |h-20|<40
c. |h-30|<20
d. |h-40|<10
e. |h-45|<5</p>

<p>In my mind they pretty much all work. Obviously, I am missing something.</p>

<p>While I am at it, is there any way that is easier to answer Test 1, section 8, number 16 (p.424) than the full page explanation offered by the CB?</p>

<p>Thank you smart people so very much!!!!!</p>

2

<p>Okay, I’ll try to explain this conceptually.</p>

<p>When you subtract two things, you are finding the difference between the two things, or the distance between the two things . . . like on a number line. For example, if you do 40 − 30, you get 10. Now if you do 30 − 40, you get −10, but the “distance” between the two numbers is still 10, since you don’t normally say something like “Those two things are −10 meters apart.” So to refine this formula for “distance,” you can use the absolute value signs: |30 − 40| = 10. Absolute value signs make any quantity positive. </p>

<p>So |h − 10| would represent the difference between h and 10. For example, if h is 8, then the difference would be 2. If h is 12, the difference would be 2. Both 8 and 12 are 2 integral units away from 10. So notice that this quantity, |h − 10|, encompasses not only a certain quantity to the LEFT of 10 on the number line (for example, 2 units to the left of 10 is 8), but also that quantity to the RIGHT of it (for example, 2 units to the right of 10 is 12). This is what the absolute value does.</p>

<p>Now, since we want to constrict “h” (height) to being between 30 and 50, we want to find the midpoint of the range (30−50), which is 40. This number is fitting because 30 is 10 units away and 50 is also 10 units away. We don’t want a number more than 10 units away. For example, if a number is 11 units away from 40, it is either 29 or 51, which is outside of the range. So we want a number less than 10 units away from 40. Hence, you get the inequality |h − 40| < 10.</p>

<p>A lot of questions on the SAT are conceptual. Knowing algebra, knowing methods, knowing techniques, etc., is vital to scoring well, but since the SAT is a Reasoning test, it calls for a lot of abstract thinking and good judgment. (Just wanted to note that.)</p>

3

<p>Since there all inequalities, for d it’s quite obvious to see it’s right if you know one of the rules. Since it says less than 10. On the right side write down -10< | h-40< 10. now just add 40 the right and add 40 the left. You end up with 30< | h | < 50. and this is exactly what the question was saying.</p>

4

<p>Oooooh! The lightbulb just went off. I was plugging in numbers between 30 and 50 and each worked out to fit the equation. But your way DOES make conceptual sense. DUH!!!</p>

<p>Thank you so much!</p>

5

<h1>16</h1>

<p>Using the formula, number of bees on day 10 = (10 x 10)/2 - 20 x 10 + k =
k - 150 - I</p>

<p>Let the day on which # of bees was equal to this be T. Then # of bees on Tth day =
(T^2)/2 - 20T + k - II</p>

<p>The expressions labeled I and II are equal. So,
(T^2)/2 - 20T + k = k - 150</p>

<p>Moving every term to left we have,
(T^2)/2 - 20T + 150 =0
Multiplying the whole equation by 2 we have,
T^2 - 40T + 300 = 0
This can be factorized as (T-10)(T-30) = 0
So, either T - 10 =0 or T - 30 = 0 i.e. T = 10 or T = 30</p>

<p>So, the day is 30.</p>

6

<p>ppathak4 - Thank you so much! That is the way I did it, too. But for some reason mine took up much more space! LOL. You made it look much more simple than it seemed when I did it.</p>

<p>Thanks folks - we’re beginning to get the “feel” for the test that everyone talks about after you’ve done enough problems and practice.</p>

<p>I really appreciate the prompt replies and support (and no one making me feel dumb!)</p>