digitaljinx

No problem! good luck in your studies and let me know if you need anymore help with other math questions.

Hey red panda, here is the breakdown on how to solve this. since you have a 50% chance or 1/2 chance, this means its either going to rain or not rain (obvious i know :) ). Since you have 7 days as the period of time, you have 2 choices (rain or no rain) for each of the 7 days. 2x2x2x2x2x2x2. This gives you 128 possibilities max options. Now you need to find how many different arrangements of 4 days of rain out of 7 days you can have. This 4 out of 7 is the number of times its going to rain. You can have R, R, R, R, NR but you can also have NR, NR,NR, R,R,R,R. Since you dont want to have to go through all the different combinations of this, there is an easy formula to remember that involves combinations where order doesnt matter (which i showed above that order doesnt matter) the formula is 7 choose 4 or (Total objects in the list)!/((total objects - number of objects i want)! * number of objects i want!). pertinent to this question i have 7!/((7-4)!*4!) = 35. So, 35/128. Let me know if this makes sense or you want me to cover this in more detail.

E,b,c

The problem with this approach is that the balls are identical so 1,2,27 is the same as 2,1,27. Maybe?

nyssa - what about (1,3,26) (1,4,25)... and there are 14 just for this group. the next group starts at 2,2,26 and there are 13 options in this group. the pattern stops at (10,10,10) because the process will repeat.

are there answer choices with this problem? might be easier to solve that way but i got 78. There has to be an easier way to solve this than to list all the options, which is what i did?

Update on my GMAT studies: PR CAT #1 -570 Quant: 43 - 78% Verbal: 26 - 59% PR CAT #2 - 550 Verbal: 26 - 63% Quant: 39 - 73% PR CAT #3 - 600 Verbal: 31 - 68% Quant: 42 - 81% My Quant is getting a lot easier from a time perspective and I am missing really easy errors. My Verbal has improved because of SC1000 and CR bible. It is my current weakness holding me back from a good score. Any advice on how to improve verbal more?

The same logic you applied to statement 2 can be applied to statement 1 - all the numbers can go in the negative as well because 0+3 = 3 and now since 0 is in the set the only way to get 0 is -3+3. Now -3 is in the set and the same problem with option 2 is displayed in option 1. I'm going crazy over this question for no reason probably. I appreciate your help. Thanks

The answer is B and here's why: I break everything down to its prime factors 1)p is a multiple of 15 and k is a multiple of 6. so the smallest it can be is 90/210 which is not a multiple. now the question is whether they will converge at all? write everything as a prime of its factors and it is known now that k and p converge at multiples of 90. (3*3*5*x)/2*3*5*7) -> (3*x)/(2*7) and we can see that x can be 14 (or the 14th multiple of 90) which would cancel out the 7*2 if broken down into its prime factors. Also, I use x as a placement variable to show that KP is a mutiple of 90 but we dont know which. Please ask me if you have any questions on breaking this down. 2)k is a mutiple of 35 and k is a multiple of 6 and now I have to find where these converge. find the lowest common multiple. write k in terms of its prime factors. k1= 3*2 and K2=7*5 and since these values dont share any common factors, the lowest multiple will be when multiplying all these values together giving us 210. Now refer back to the question, 210*p/210 (use p in its variable form since we dont know what it is). It shows that no matter what p is, we will have a multiple of 210.

hmm I guess I will have to accept that A is true even though I'm not sure I agree completely. amitabh, if 0 is in A, then how did 0 get into the set since -3+3 = 0, and now because of this -3 is in the set. this arithmetic sequence would continue in the negative direction as well.

I guess my confusion lies with where does 3 come from? in part 1 3 comes from 0 ( 0+3 = 3, the number already in the set). and since 0 is not a multiple of 3 both would be incorrect answers.

princeton Review pg 424 # 3 The members of the newest recruiting class are taking a test and those who score in the bottom 16% will have to retest. if the scores are normally distributed and have a mean of 72, what is the score at or below which the recruits will have to retest? 1) there are 500 recruits 2) 10 recruits scored 82 or higher with option 1, all i know i can tell is how many people will have to redo the test, but not what that score is with option 2, I dont know what the other members got on their exam, or how many members there are, so I can't tell what the 16% cutoff score is Now I am down to 2 options C or E and I'm not sure what to do next. I believe 1 way to solve this question is to use standard deviation to determine the percentile spread (How many standard deviations is 82 from 72). Is there a faster way to solve a problem like this?

I am working in the princeton review book in math bin 3 and I am not sure how to solve problem 9 on page 419. The question states: if P is a set of integers and 3 is in P, is every positive multiple of 3 in P? 1)for any integer in P, the sum of 3 and that integer is also in P 2)For any integer in P, that integer minus 3 is also in P what I am confused about is option 2, for the integer 3 to be in the set, isn't it safe to assume that 6 is in the set also and that this arithmetic sequence continues? to get 3 I must have had 6, since 6-3=3 is also in the set. To get 6 i must have had 9 ect. Can anyone help me to understand this? Thanks

I am planning to take it oct. 25 as well. I am using the critical reasoning bible, OG, and princeton review along with practice tests that I have downloaded from this site. PR: 570 Good luck to all

Does anyone have a link to quantitative/Verbal shortcut documents? I am having trouble finishing practice tests and I am looking to improve my speed with tricks. Thanks