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here is the problem:

on a certain game show, contestants throw darts at a circular target. The target has a radius of 10 meters. Within the target is another circular region, which is painted red and called "the red zone". If a dart is thrown and hits the red zone, the contestant gets more points. The radius of the red zone is 5 meters. If every dart thrown randomly hits the target, what is the probability that a dart hits the "red zone"?

A) 0
B) 1/4
C) 1/2
D) 3/4
E) 1

by the way, would you please also explain how you got the answer?

thanks

By Physicskid123 (Physicskid123) on Sunday, June 20, 2004 - 09:58 pm: Edit

The answer is B. The formula for probability would be (favorable outcomes)/(possible outcomes). Now your favorable outcomes would be the whole area painted red. The area would be (pi)(r^2) or 25pi. The total area would be the possible, and is pi(10^2) or 100pi. so favorable/possible=(25pi)/(100pi) or 1/4.

By Amethyst (Amethyst) on Sunday, June 20, 2004 - 10:01 pm: Edit

That's what I thought when I first looked at the problem, however, the area of the "none red area" is the total circle "MINUS" the area in the center, which would be 75pi.

By Amethyst (Amethyst) on Sunday, June 20, 2004 - 10:07 pm: Edit

anyone have any input?

By Kaoak (Kaoak) on Sunday, June 20, 2004 - 10:09 pm: Edit

That's true, but his answer is still correct. The formula is (favorable outcomes)/(possible outcomes), not (favorable outcomes)/(unfavorable outcomes). The area of the whole thing, red area and non-red area, are possible outcomes, so you use 100pi.

By Amethyst (Amethyst) on Sunday, June 20, 2004 - 10:12 pm: Edit

oh, I never thought is THAT way. I was being dogmatic. Thanks for clearing the confusion.

By Amethyst (Amethyst) on Sunday, June 20, 2004 - 10:16 pm: Edit

here is another question:

the sum of 8 positive integers is 31. If no individual integer value can appear more than twice in the list of 8 integers, what is the greatest possible value one of these integers can have?

A) 16
B) 15
C) 10
D) 9
E) 7

By Aim78 (Aim78) on Sunday, June 20, 2004 - 10:26 pm: Edit

You want every number to be as small as possible, except for one which will be as big as possible. And each number can repeat twice.

1+1+2+2+3+3+4+15 = 31

Answer is B

By Madd87 (Madd87) on Sunday, June 20, 2004 - 10:29 pm: Edit

1,1,2,2,3,3,4,(31-4-3-3-2-2-1-1)
1,1,2,2,3,3,4,15

15

B

By Amethyst (Amethyst) on Sunday, June 20, 2004 - 10:32 pm: Edit

oh, again, I misread the question, I thought no integer can appear TWICE. I am very frustrated at myself, how can I stop making stupid mistakes???!!!

By Amethyst (Amethyst) on Sunday, June 20, 2004 - 10:35 pm: Edit

okay:

If the first two terms of a geometric sequence are 5 and 10, what is the fourth term of the sequence?

A) 15
B) 20
C) 30
D) 40
E) 45

thanks

By Energy1 (Energy1) on Sunday, June 20, 2004 - 10:42 pm: Edit

40

the multiple is 2 between each term.

Thus, 10 x 2^2 = 40

By Amethyst (Amethyst) on Sunday, June 20, 2004 - 10:51 pm: Edit

f(x)=x^2-5

if y represents the range of f(x), which of the following represents all possible values of y?

A) y is less than 5;
B) y is greater than or equal to -5;
C) y is less than -5;
D) y is greater than 5;
E) all real numbers

By Redbarn (Redbarn) on Sunday, June 20, 2004 - 11:03 pm: Edit

Should be B.

By Amethyst (Amethyst) on Sunday, June 20, 2004 - 11:06 pm: Edit

can you explain how you got the answer?
thanks

By Benzinspeicher (Benzinspeicher) on Sunday, June 20, 2004 - 11:11 pm: Edit

B. All squared number values are positive, unless the number is 0, in which case the squared value would be 0 (that would be the lowest possible value of the x^2 part, and x could increase to infinity). 0-5=-5, so that is the lowest value of f(x). And again, as x^2 approaches infinity, so does x^2-5. So the answer is B.

By Loop123 (Loop123) on Sunday, June 20, 2004 - 11:11 pm: Edit

I agree with Redbarn. The least value that x^2 could possibly have would be zero, in which case f(x) would equal -5. as |x| increases, the value of f(x) gets larger.

By Benzinspeicher (Benzinspeicher) on Sunday, June 20, 2004 - 11:12 pm: Edit

did i mention that the term "range" refers to all the possible values of f(x)?

By Amethyst (Amethyst) on Sunday, June 20, 2004 - 11:16 pm: Edit

thanks for all you guys' help, here is the LAST question:

Q: On her biology test, Cathy answered 5/6 of the questions correctly. If Cathy answered 18 of the first 27 questions correctly, then the total number of questions on the test must be at LEAST:

A) 32
B) 36
C) 45
D) 48
E) 54

THANKS A LOT!

By Loop123 (Loop123) on Sunday, June 20, 2004 - 11:22 pm: Edit

I got E: 54
Correct me if I'm wrong, but I did the following:

Let x = the number of problems in addition to the 27 already answered.

(18+x)/(27+x) = 5/6

Cross multiply:

135 +5x = 108 + 6x

Solve:
x=27

Add those 27 problems to the original 27 to get 54.

By Amethyst (Amethyst) on Sunday, June 20, 2004 - 11:26 pm: Edit

THANKS !!!!!!

By Xgamerx13 (Xgamerx13) on Sunday, June 20, 2004 - 11:31 pm: Edit

wow im stupid i did it the guess and check method kind of, add 7 to both sides so that the right answers become 25, then just kept adding 6 to both sides, and arrived at 45/54 which is 5/6

By Dukedreamer (Dukedreamer) on Sunday, June 20, 2004 - 11:36 pm: Edit

Wow, either I am really smart, or you need to work on math...I think i am too good to answer this question (just kidding)

E- More than one way to do it..easiest would be the simple trial and error method.

She got 18/27 right or 9/27 wrong. which of the above lets you miss 9, and still maintain 83.333%
5/6 is about .833

32-27=5... 5+18=23...23/32=.72 (not quite)
Do this for every answer going from smallest to largest (the minute you get .833 BAM! bubble it in and move on)

I suck at explaining problems-sorry if it didnt help much...I tried!

By Shaman88 (Shaman88) on Monday, June 21, 2004 - 09:21 pm: Edit

Is this problem from the free NEW practice SAT from Princeton Review?