Need some help with these math questions, thanks!

Test Preparation SAT Preparation
1

<li>How many numbers from 60 to 180 inclusive are divisible by 3?</li>
</ol>

<p>A. 20
B. 39
C. 40
D. 41
E. 60</p>

<li>Let k be the least of 4 consecutive positive integers whose sum is s. What is the sum of 8 consecutive integers, the least of which is k?</li>
</ol>

<p>A, s+16
B. s+22
C.2s
D. 2s+16
E. 2s+22</p>

<li>If x equally priced items cost a total of c cents, what is the cost, in dollars, of y of the terms?</li>
</ol>

<p>A. xc/y
B. yc/x
C. yc/100x
D. 100yc/x
E. 100xc/y</p>

<li>If f(x)= x^2+1,then f(f(x)) =</li>
</ol>

<p>A. x^4+1
B. x^4+2
C. x^4+x^2+1
D. x^4+2x^2+1
E. x^4+2x^2+2</p>

2

<p>1) a= 60 d=3 Tn= 180</p>

<p>60 + (n-1)3 = 180
n= 41
answer 41</p>

<p>2) List k, k+1, k+2,....
so for 4 numbers, 4k + 6 = s
then for 8 numbers, s + 4k + 4+5+6+7= 2s +16
answer D</p>

<p>3) One item c/x.
y items : yc/x in dollars thn yc/100x
answer C</p>

<p>4) (x^2 + 1 )^2 + 1 = X^4 + 1 + 2X^2 + 1 = X^4 + 2X^2 + 2
Answer E</p>

<p>I hope I did not make careless mistakes here. I am famous for careless mistakes in Math :)</p>

3

<p>Hi, I am not an 800 math SAT taker (yet anyways), and I'm not the best at explaining math, so everyone else should feel free to elaborate on this or correct me if either I am wrong or there are better ways of solving, but here goes:</p>

<p>1.D
3 x 20 = 60 and 3 x 60 = 180, therefore every number between 20 and 60 (inclusive) times 3 is every every multiple of 3 between 60 and 180. So, 60 - 20 = 40. But there is still one more simple, but important step: to find the number of terms between two given numbers, you must subtract the smaller number from the larger number and add 1. So, after adding 1, the answer is 41.</p>

<p>2.D
There may be a better way, but I solved it like this:
Four consecutive integers = N + (N + 1) + (N + 2) + (N + 3)
Simplified: 4N + 6 = S
The question asks for the sum of 8 consecutive numbers and tells us that the least integer for the first four consecutive integers is the same, therefore we just add four more:
(4N + 6) + (N + 4) + (N + 5) + (N + 6) + (N + 7) = 8N + 28, and this is our answer.
D matches this because
2s + 16 =
2(4N + 6) + 16 =
8N + 12 + 16 =
8N + 28</p>

<p>3.C
The easiest way to do this if you have trouble with these types is just to plug in your own numbers. Let's say x=5, since there are 5, and they all cost the same, so c=25 and we will say that y=2. 2x5= 10 cents, so our answer will be $0.10. Plugging in these values into the answer choices, only C matches.</p>

4

<p>4.D</p>

<p>f(x) = x^2 + 1
f(x^2 + 1) = (x^2 + 1)^2 + 1 = (x^2 + 1)(x^2 + 1) + 1 = x^4 + 2x^2 + 1</p>

5

<p>Low-tech approach - plugging in. </p>

<p>2.
Let k=1.
s = 1 + 2 + 3 + 4 = 10
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 10 + 5 + 6 + 7 + 8 = 36.
Only D produces 36:
2s + 16 = 20 + 16 = 36.</p>

<p>4.
Let x = 2.
f(x) = 2^2 + 1 = 5
f(f(x)) = f(5) = 5^2 + 1 = 26.
Since 26 is even, we need to test B and E only (A, C, and D expressions
produce odd numbers for any x).
E (and E only) works:
2^4 + (2)2^2 + 2 = 26.</p>