Maths Help!

Test Preparation SAT Preparation
1

<ol>
<li>At a used book store, each paperback book is priced
at $1 and each hardcover book is priced at $5. Ken
purchased 25 used books at the store for a total price
of $65. How many paperback books did Ken purchase?</li>
</ol>

<p>I tried using x + 5x = 65 but x then is not a whole number!</p>

<p>For sequence A, the /7th term is lOn + 58.
For sequence B, the nth term is 2^n.
15. The nth terms of sequences A and B are defined
above for all positive integers n. What is the least
value of n for which the nth term of sequence B
is greater than the nth term of sequence A ?
I got this right by plugging in numbers for n. Can it be done by the arithmetic or geo seq formula?</p>

<p>STUDENTS’ LOCKER LOCATION BY GRADE
Image: [ImageShack®</a> - Online Photo and Video Hosting](<a href=“ImageShack - Best place for all of your image hosting and image sharing needs”>ImageShack - Best place for all of your image hosting and image sharing needs)</p>

<ol>
<li>At Eastern High School, each student is assigned one
locker. The locker may be either in hall A, B, or C.
The partially completed table above shows the number
of students from each grade assigned a locker in each
of the three halls. If 20 percent of the students in grade
9 were assigned a locker in hall B, what is the total
number of students who were assigned a locker
in hall B ?</li>
</ol>

<p>I tried using 155+25= 180 (as 80% of the grade 9 students. But I don’t get a whole number for 1%.</p>

<p>AVERAGE TIME
PER SCHOOL WEEK
SPENT ON HOMEWORK
[ImageShack®</a> - Online Photo and Video Hosting](<a href=“ImageShack - Best place for all of your image hosting and image sharing needs”>ImageShack - Best place for all of your image hosting and image sharing needs)</p>

<p>The table above shows the average (arithmetic mean)
time a student spent on homework per school week
during each of 4 years of high school. In total, how
many hours did the student spend on homework during
4 years of high school? (Assume that 1 school week
contains 5 days and 1 school year contains 180 days.)
(A) 322
(B) 1,008
(C) 1,080
(D) 5,040
(E) 5,400</p>

<p>I got 5400 after 30 x 1800 (total hrs per week x no of weeks)
But its C.</p>

<pre><code> Image: ImageShack® - Online Photo and Video Hosting
</code></pre>

<p>In the triangle above, AB = AC = 5 and BC = 6.
What is the area of the triangle?
(A) 4
(B) 6
(C) 9
(D) 12
(E) 16</p>

<p>Its isosceles triangle but I can’t figure out how the height is determined. (Apologies if the diagram is unclear)</p>

2

<ol>
<li>You need two equations to solve this one:</li>
</ol>

<p>Paperback + Hardcover = Total Books
P + H = 25</p>

<p>$1(P) + $5(H) = $65
P + 5H = 65</p>

<p>You can use substitution to solve for P:</p>

<p>P + H = 25
H = 25 - P</p>

<p>P + 5H = 65
P + 5(25 - P) = 65
P + 125 - 5P = 65
-4P = -60
P = 15</p>

3

<ol>
<li>First, find the total number of ninth graders:</li>
</ol>

<p>Hall A 9th Gr + Hall B 9th Gr + Hall C 9th Gr = Total 9th Gr
155 + 20% of the total + 25 = total
155 + 0.2t + 25 = t
180 + 0.2t = t
180 = 0.8t
225 = t</p>

<p>There are 225 ninth graders. Now find 20% of 225 to fill in the number of ninth graders in Hall B:</p>

<p>0.2 x 225 = 45
There are 45 ninth graders in Hall B.</p>

<p>Find the total number of students in Hall B:
45 + 100 + 70 + 75 = 290</p>

4

<p>I can’t read 15 and the images for the last two are not showing up. But I can still solve the last one with the triangle.</p>

<p>The base of the triangle is 6. If you draw a line down the center of the triangle, this becomes the height. You now have two right triangles and can use the Pythagorean Theorem to solve. </p>

<p>The base of one of the small triangles is 3. They hypotenuse is 5. Therefore, the remaining side (b) is 4. If you don’t know this common triangle side ratio, run it through the PT (a^2 + b^2 = c^2). </p>

<p>Now find the area of the big triangle using base = 6 and height = 4.
1/2 bh = area
1/2 (6)(4) = 12</p>

5

<p>Thanks very much.</p>

<p>The sequences question - seq A is 10n+58, seq B is 2^n.</p>

<p>The table on average of school week is as follows:</p>

<p>Grade: No of hours
9 -> 5
10 -> 6 3/4
11 -> 9 1/2
12 -> 8 3/4</p>

<p>Just a point: For the students lockers, your workings led to the same answer I got but I was stuck when I realized I did not get a whole number to represent 1% of the the students - I got 2.25 which got me wondering if I was on the wrong track.</p>

6

<p>

</p>

<p>Plug-ins works in this case. However, you may wonder why TCB comes up with such a weird number in the form of 58? Then, there is the easy n^2 sequence, which is 2,4,8,16,32,64, 128, 256, etc. From the 10n+58, you know that you can discard everything that is smaller than 68. So, the starting point is 128, or n^7. </p>

<p>ETS loves to throw a bone at the people who can notice patterns. In this case, it is POSSIBLE to see the relation between 70 + 58 and 2^7 (both equal 128). Based on that the answer is 8.</p>

7

<p>

</p>

<p>You were ALMOST there. Although you got to the correct 30 (no need to divide by 4, btw) but multiplied by the number of days as opposed to the number of week (36 week per year)</p>

<p>The answer should be 30 x 36 or 1080. As I wrote above, it is circular to calculate 30/4 * 36 * 4.</p>

<p>PS When picking an answer such as 5,400 you might have considered that it represents 1,350 hours of homework a year … quite a bit considering 180 days of school. :)</p>

8

<p>Thanks xiggi!</p>

9

<p>

Lately I became a big proponent of a no-brainer approach when an efficient solution (like xiggy’s) does not come to you easy, but the brute power of TI crunches the question in seconds.</p>

<p>In your TI-8* enter y1=10x+58, y2=2^x.
2ND TABLE tells:
y2>y1 for the first time at x=8.
(2ND TBLSET: Table Start = 1, Table Step = 1, Auto, Auto)</p>

10

<p>^ sadly, i dont use the TI calculator… thanks anyway!</p>

11

<p>@gcf101 Wow that’s really a great strategy! Thanks!</p>

12

<p>^zachzm Why not? It makes some problems much easier.</p>

13

<p>A TIxx helps solving a very small number of problems, but it is never essential. However, for people who are preparing for the higher SAT and KNOW how to use the calculator, it is not a bad idea. You can see that gcf’s solution is pretty useful. </p>

<p>The biggest danger of using the TI is that one could enter numbers wildly with the secret hope of stimbling onto the correct answer. In so many words, it is a great tool but only in the hands of people who know how to use it well. Of course, most of those people do NOT need a calculator to solve the 9th grade level math that is tested on the SAT.</p>

14

<p>There are people who have difficult time solving
2^x+2^x+2^x+2^x=2^7
Enter y1=2^x+2^x+2^x+2^x, y2=2^7,
2ND TABLE: for x=5 y1=y2=128.</p>

<p>Another one.
p is a common factor of n+3 and n+10. What can p be?
Takes 2 seconds to solve it if you know what to do. If not - whip out your trustworthy TI-8* and enter
y1=x+3, y2=x+10.
2ND TABLE: y1=7 and y2=14 have 7 as a common factor. p=7. Done.</p>