<p>I got these from Killer Math from Princeton Review</p>

<li><p>39 points are placed inside or on the surface of a perfect sphere. If 60% of fewer touch the surface, what is the maximum number of segments which, if connected from those points to form chords, could be the diameter of the sphere?

a. 7

b. 11

c.13

d. 23

e. 38

Answer: B. 11</p></li>

<li><p>If Q, R, S, T, U, V are six unique points along a circle such that six chords QU, RS, ST, TU, UV, and VQ are equal in length, which of the following is the circles diameter?

a. QS

b. UT

C. SV

D. (QR+QS)/2

E. (UV+RS+TQ)/3

Answer: C</p></li>

<li><p>How many three-digit integers exist where the sum of all the digits is 7, and even and odd digits alternate?

a. 1

b. 2

c. 4

d. 6

e. 8

Answer: d

Is there a quick way of solving this besides listing all the numbers out?</p></li>

<li><p>On a number line Q=1 and R = -2. How many points on the number line are twice as far from R as they are from Q?

a. 0

b. 1

c. 2

d. 3

e. Cant be determined

Answer: C</p></li>

<li><p>At Julie’s Jewelry Inc, there are necklaces, bracelets, and rings for sale. If $30.00 were to be added to the price of a necklace, the cost would be the same as that of the sum of 90% of the price of a bracelet and 75% of the price of a ring. If $20 were to be added to the price of a bracelet, the price would be the same as that of the sum of 3/5 of the price of a necklace and 3/4 of the price of a ring. If the price of a ring is the same as that of the sum of 40% of the price of a necklace and 10% of the price of a bracelet, then how much do two rings cost?

a. $25

b. $50

c. $100

d. $150

e. $200

Answer: E

Is there a faster way of solving thing then just doing a system of equations?</p></li>

<li><p>Two parallel lines have equations ay+bx=4a and y-8 = 2x. Which of the following is the equations of a third line that is equidistant from these lines?

a. y = 2x - 3b/a

b. y = -ax/b+6

c. y = -bx/a - 6b/a

d. y = -bx/2a + -6b/a

e. y = 2x - 6b/a

Answer: a</p></li>

</ol>

<p>Any help would be appreciated. Thanks :D</p>