theoneo

Man, that's the November 1995 test, not May 1996. I was searching my book cause I knew I'd seen that second problem somewhere. I just did this test a couple days ago, so here it goes:

17) What is the slope of a line that passes through the origin and the point (-2, -1)?
slope = (y2 - y1)/(x2 - x1), (x1, y1) = (0, 0), (x2, y2) = (-2, -1)
slope = (-1 - 0)/(-2 - 0) = (-1)/(-2) = 1/2 --> B

18) Julie has cats, fish, and frogs for pets. The number of frogs she has is 1 more than the number of cats, and the number of fish is 3 times the number of frogs. Of the following, which could be the total number of these pets?
frogs = cats + 1, fish = 3frogs
total = fish + cats + frogs
You want to find the total, so you can just start adding them together and simplifying through substitution
= (3 x frogs) + cats + (cats + 1) <-- substituted values of fish and frogs
= (3 x (cats + 1) ) + cats + (cats + 1) <-- substitutded value of frogs again
= 3 x cats + 3 + cats + cats + 1
= 3 x cats + 2 x cats + 4
= 5 x cats + 4
Since the number of cats must be an integer greater than 0, the total can be 5(1)+4 = 9, 5(2)+4 = 14, 5(3)+4 = 19, etc. Notice how the only answer choice that fits is E)19.

19) If x is an integer, which of the following could NOT equal x^3?
Don't be fooled by A, since you can take the odd root (3, 5, etc) of a negative number, but not an even one (2, 4, etc). In this case, you could just start from 0 and work your way up until the cube of the number is greater than 27. so 0^3 = 0, 1^3 = 1, 2^3 = 8, 3^3 = 27, 4^3 = 64. Logically, the cube root of 16 should be between 2 and 3, but no INTEGER value works, so D) 16 is the answer.

21) The incomplete table above categorizes the members of a ches club according to their age and status. During a tournament, each member of the club plays exactly one game with each of the other members. How many games of chess are played between amateurs 20 years or older and professional under 20 years old during the tournament?

First you need to finish the chart. 5 goes in the top right box and 3 goes in the bottom left box in order to satisfy the totals. So you need to figure out how many games were between the 3 young professionals and the 5 old amateurs. If you don't know this already, memorize it: the number of games between a group of x people and a group of y people in which each person competes with every single person in the other group once but never with someone in his/her own group (in other words, exactly what this problem is) is equal to x times y. In this case, that's 3 x 5 = 15 games. To check that, you could write A, B, C, D, and E in order to represent the five amateurs and draw three lines from each to represent one game with each professional, or vice versa, and count the number of lines. The answer is C.

22) A bag contains a number of pieces of candy of which 78 are red, 24 are brown, and the remainder are yellow. If the probability of selecting a yellow piece of candy from ths bag at random is 1/3, how many yellow pieces of candy are in the bag?
probability = chance of happening / all possibilities, which is the same as p(x) = x/total. You can use thiss as a formula. You know the probability p(Y) is equal to 1/3. 1/3 = Y/(R + B + Y). How can we simplify this? By getting rid of the R and B variables. Just substitute their actual values. Now 1/3 = Y/(78 + 24 + Y) = Y/(102 + Y). Now you can just solve for Y: 1/3 = Y/(102 + Y) --> (1/3)(102 + Y) = Y --> 34 + Y/3 = Y --> 34 = Y - Y/3 --> 34 = 2Y/3 --> 102 = 2Y --> Y = 51, which is B.

I hope this helped!