<p>For #1, simply realize that total number of students studying both languages MUST be 15 (you can realize this by simply reading and interpreting the question) and that the question is only asking for the amount of students who study ONLY Italian, not including people who study both. So, this means that the singulars are 15 total. If the amount of students studying German is 3 more than of Italian, g + i = 15 in which g = i + 3, in turn gives us 2i + 3 = 15 where ‘i’ must be 6. The answer is 6.</p>
<p>For #6, it’s quite easy.</p>
<p>(sqrt a)^n is the same thing as a^(0.5n)</p>
<p>Square both sides = 25. </p>
<p>Answer is 1/25.</p>
<p>I might be asking for too much but I would greatly appreciated if someone could help me with this, it has been bugging me for several days now! </p>
<p>X_ _ X</p>
<p>The figure above represents four offices that will be
assigned randomly to four employees, one employee
per office. If Karen and Tina are two of the four
employees, what is the probability that each will be
assigned an office indicated with an X ?
(A) 1/16
(B) 1/12
(C) 1/6
(D) 1/4
(E) 1/2</p>
<p>also<br>
19. For all positive integers w and y, where w > y, let
the operation @ be defined by w@y= 2^w+y / 2^w-y For
how many positive integers w is w@1 equal to 4 ?
(A) None
(B) One
(C) Two
(D) Four
(E) More than four</p>
<p>X_ _ X Question</p>
<p>Use combinations. </p>
<p>How many ways can you arrange 2 people in 4 desks? =C(4,2) or nCr 4,2</p>
<p>How many ways satisfy the condition that those exact two desks are filled? =1</p>
<p>Probability= 1/(C(4/2))
=(2!2!)/4!
=1/6</p>
<p>[c]</p>
<p>@@@@@ Question</p>
<p>Use exponent rules.
(2^(w+y))/(2^(w-y))
=2^((w+y)-(w-y))
=2^(2y)
=4^y</p>
<p>Note that w is eliminated from the equation. When w@1, y=1. Therefore any value of w will satisfy the equation 4^1=4.</p>
<p>[e]</p>
<p>------------------1- German/Italian:</p>
<p>Question implies n(i∩g)=n(i∪g)/2</p>
<p>n(i∪g)=n(i)+n(g)-n(i∩g)
30=n(i)+n(i)+3-15
n(i)=21</p>
<p>n(i)-n(i∩g)=6</p>
<p>[a]</p>
<p>------------------2- 90n + 23p = 4523
p=1,n=50
n+p=51</p>
<p>[51]</p>
<p>------------------3-June children</p>
<p>Note that (29)(2)<89<(29)(3)</p>
<p>Implication: Some days have 2 children, some 3 children to minimize values.
Thus the number of “15th” children must be 4 to be greater than every other day.</p>
<p>[4]</p>
<p>------------------4- Trick question</p>
<p>[e]</p>
<p>------------------5- 2/x > 3</p>
<p>Retain positive property and minimize “x” to maximize left side.</p>
<p>[c]</p>
<p>------------------6- a^(n/2)=5 , 1/(a^n)=?</p>
<p>When more variables than initial equations are given, you may substitute any value for one of the variables.</p>
<p>a=5</p>
<p>Thus
n=2</p>
<p>[1/25]</p>
<p>------------------7- 10a + 10b = 32 , (a+b)=?</p>
<p>Factor
10(a+b)=32
a+b=3.2</p>
<p>[3.2]</p>
<p>------------------8- Volleyball</p>
<p>Combinations.</p>
<p>Games = 2(nCr 4,2)
=2(4!)/(2!2!)
=12</p>
<p>[12]</p>
<p>------------------9- (x-y)^x = y^x = 1 , x=?</p>
<p>Note that y^x=1 is satisfied only when x=0 or y=1 (Exponent rules)
x≠0, thus y=1</p>
<p>(x-1)^x=1
x≠0, thus (x-1)=1 (Exponent rules)</p>
<p>**</p>
<p>------------------10- (1/k)=0.13… , (k-1)/k=?</p>
<p>(k-1)/k
=1-(1/k)
=1-0.13…</p>
<p>Simple subtraction, as any digits other than 1 and 3 are insignificant</p>
<p>=0.86…</p>
<p>[d]</p>
<p>------------------11</p>
<p>b=5a
5=ab</p>
<p>Thus a=sqrt(1)
a<0, thus a=-1</p>
<p>Sequence becomes</p>
<p>5, -1, -5, 5, -5^2, -5^3, 5^5, -5^8, -5^13, 5^21</p>
<p>[e]</p>
<p>------------------12- (a+b)^1/2 = (a-b)^1/2</p>
<p>If b≠0 the equation is never satisfied.
If b=0 the equation becomes a=a and is always satisfied.</p>
<p>[a]</p>
<p>------------------------------------Final Answers------------------------------------
a
51
4
e
c
1/25
3.2
12
b
d
e
a</p>
<p>Thank you so much for explaining those, they make sense now :). I appreciate it!</p>
<p>you can solve #8 extremely quickly using the handshake tip from dr chungs book. In a game there is only 2 teams playing so the number of games in the first round is 4<em>3=12, but u have to divide by 2! to account for duplicate games with the same two teams so 12/2=6. 6 is the amount of games in the first round, but each team plays all the other teams twice, so the final answer is 6</em>2=12.</p>
<ol>
<li>1/25 </li>
<li>3.2</li>
<li>12</li>
<li>2</li>
<li>5</li>
<li>E</li>
<li>A</li>
</ol>