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Forum profile page for GRE Math on http://www.urch.com.
This report page is the aggregated overview from a single forum: GRE Math , located on the Message Board at http://www.urch.com.
This forum profile page summarizes the general forum statistics such as: Users Activity, Forum Activity, and Top Authors, which are reported in either a table or graph below for a given reporting time period.
Additional forum profile information for "GRE Math " on the Message Board at http://www.urch.com is also shown in the following ways:
1) Latest Active Threads
2) Hot Threads for Last Week
Warning: These statistics are generated using 'best efforts' and can experience delays and reporting errors at times. Please note that such statistics do not constitute a forum's popularity and/or exact posting volumes at any given reporting period.
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Posting activity on GRE Math :
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3 Months
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306
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GRE Math Posting activity graph:
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Top authors during last week:
user's latest post:
Counting problem
Published (2009-11-10 16:44:00)
go the same because katie and christine can sit in 2ways ... and we have 9 people .. so now k+c=1 person and the other can sit in 8! ways .. so 2!* 8! ... hope we`re right =)
user's latest post:
GRE Barrons - best preparation...
Published (2009-11-10 17:05:00)
Quote: Originally Posted by Maximumbirne Hey Bane, just took the test last month and not yet happy with my result (Q 730). Unfortunately, I found out about this forum far too late - may I ask for another copy? Thanks a lot in advance. Cheers! M ok...
user's latest post:
Help Please? PP Question
Published (2009-11-05 10:29:00)
let's n=35 & d=7 multiple of d=7 is 7, 14, 21, 28, 35 just 5 multiple & n/d=35/7=5 so answer is n/d
user's latest post:
number property question
Published (2009-11-10 19:44:00)
Please help solve this question: What is the remainder when 123^2-123+123^2 is divided by 11? Can anyone also point out a good source of questions for problems based on number theory. I can use some practice. Thanks!
user's latest post:
difficult probabitlity
Published (2009-11-10 06:15:00)
I don't think this is particularly difficult, unless I've misread the problem. we want the probability of drawing a card and having it be something other than a heart or king. 4 kings total, 13 hearts total, but that double counts the king of hearts. so we have 13+4-1=16 "bad" cards. there are 52 total. so, probability of not getting them is (52-16)/52.
user's latest post:
Percentile problems: new kinda Q...
Published (2009-11-10 23:44:00)
Quote: Originally Posted by mlkovach There are 10,000 numbers, ranging from 20 to 80. If 62 is 60th percentile, then Col A: What percentile is 74 Col B: 70th percentile answer: D We have no information on the distribution. Consider some set with 6,000 20's , 62, 3,998 63's, 74 (74=99%tile) compared to 6,000 20's , 62, 74, 3,998 80's... (74=61%tile) Are we not supposed to assume normal distribution for all percentile...
user's latest post:
probability
Published (2009-11-10 19:35:00)
Quote: Originally Posted by Curly213 we have 2 cards of every color ( b, r ,y, g) so we want to choose 2 out of 8 randomly ...8C2 = 28 ... condition at least one blue .. is the same as 28- no blue .. no blue is 6C2 = 15 .... so 28-15 = 13 ... so probability of gettin at least one blue is 13/28 whats the o.a. hope i`ll did it the right way! Anyone else please confirm CB
user's latest post:
Another PowerPrep QC Question...
Published (2009-11-09 04:13:00)
I'm having trouble understanding the answer to this QC PowerPrep question: y^2 = 3x^2 + x + 2 Column A: The value of y when x = 2 Column B: -√10 answer: The relationship cannot be determined From what I can tell, when x = 2, y will = 4. Regardless, Column B is negative, which would make Column A greater. Why, according to ETS, is this problem impossible to solve with the given information? Is -√10 an imaginary number,...
user's latest post:
percentage problem (powerprep)
Published (2009-11-02 11:47:00)
Quote: Originally Posted by walt526 Let V=100 X after 3 years Year 0: 100 Year 1: 90 Year 2: 81 Year 3: ~73 Y after 6 years Year 0: 200 Year 1: 180 Year 2: 162 Year 3: ~146 Year 4: ~131 Year 5: ~118 Year 6: ~106 That took less than 1 minute (they key is to not worry too much about precise values). Of course, another way to do it would be to just look at Y after 3 years and realize that at 3 years to go (~146) it is much larger than the...
user's latest post:
How to get confidence in...
Published (2009-11-09 01:50:00)
there are two free practice tests on the GRE website. i don't know about free sources that have good tests, but you can usually buy study guides that have a CD of practice tests. every time you miss a question and review it, that is one less mistake you will make on test day
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Latest active threads on GRE Math ::
Started 1 day, 5 hours ago (2009-11-11 18:09:00)
by walt526
First digit can be one of 4 possible numbers: {1,2,3,4}
Second digit can be one of 6 possible numbers: {1,2,3,4,5,6,7} less the first one.
Third digit can be one of 5 possible numbers: {1,2,3,4,5,6,7} less the first two numbers
4*6*5=120
Started 1 day, 6 hours ago (2009-11-11 17:33:00)
by walt526
Since they're independent events, P(four consecutive busy)=1/3^4=1/81
Obviously, 1/4>1/81, so answer is B.
Started 1 day, 6 hours ago (2009-11-11 16:47:00)
by computer-bot
OK so there are 52 cards and 4 to choose so we have 52C4 ways to choose them. Also there are 13 cards of each symbol, so we can choose 13C1 for each. This gives us 13^4.
I am confused whether it be (13^4)/52C4 OR (13^4)/(52*51*50*49)
Since the question says that 4 cards are chosen randomly, the second one is more appealing to me.
Please help.
CB
Started 1 month, 1 week ago (2009-10-01 15:32:00)
by bane2
I strongly recommend you barron's 17th ed or upper.
barron's 12th ed is to old and to easy..
Besides barron's it will be good "Quantitative Aptitude" – R.S. AGARWAL
Look into your private messages (top right corner).
"Quantitative Aptitude" is only for those who want score 800. Barron's 17th ed and Kaplan are enough for 700-750.
Started 1 day, 22 hours ago (2009-11-11 00:49:00)
by susangre
Started 3 days, 9 hours ago (2009-11-09 13:57:00)
by bane2
Quote:
Originally Posted by nin999
How many questions does one have to answer correctly to get a 700 on the math section? Is a goal of getting 20 correct too low?
it depends not from quantity of questions, but from quality of questions..
Started 2 days, 7 hours ago (2009-11-10 16:28:00)
by Curly213
we have 2 cards of every color ( b, r ,y, g)
so we want to choose 2 out of 8 randomly ...8C2 = 28
... condition at least one blue .. is the same as 28- no blue ..
no blue is 6C2 = 15 .... so 28-15 = 13 ... so probability of gettin at least one blue is 13/28
whats the o.a. hope i`ll did it the right way!
Started 1 day, 18 hours ago (2009-11-11 05:25:00)
by walt526
Actually, I think that the answer should be 2442, not 2440.
Think about this in terms of +/-1.
The product of an odd number of -1's is -1 and the product an an even number of -1's is even (e.g. (-1)^3=-1 and (-1)^2=1).
2443 is an odd number, but 2443 is positive. So all 2443 integers cannot be -1 (i.e. (-1)^2443=-1). But 1*(-1)^2442=1.
It will also work for 1*1*1*(-1)^2440=1, but...
Started 1 year ago (2008-10-20 20:54:00)
by cssamal
i think answer would be D, as nothing can be determined from the data given. As the number are between 20 to 80. 80 is 100th percentile,so 74 would lie between 60th and 100th percentile. But could not be determined.
Started 2 days, 17 hours ago (2009-11-10 06:15:00)
by mlkovach
I don't think this is particularly difficult, unless I've misread the problem.
we want the probability of drawing a card and having it be something other than a heart or king.
4 kings total, 13 hearts total, but that double counts the king of hearts. so we have 13+4-1=16 "bad" cards. there are 52 total. so, probability of not getting them is (52-16)/52.
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Hot threads for last week on GRE Math ::
Started 1 month, 1 week ago (2009-10-01 15:32:00)
by bane2
I strongly recommend you barron's 17th ed or upper.
barron's 12th ed is to old and to easy..
Besides barron's it will be good "Quantitative Aptitude" – R.S. AGARWAL
Look into your private messages (top right corner).
"Quantitative Aptitude" is only for those who want score 800. Barron's 17th ed and Kaplan are enough for 700-750.
Started 1 week ago (2009-11-05 00:32:00)
by ace_gre
Q2.
To be divisible by 3, sum of digits should be divisible by 3.
No. of ways of choosing 5 numbers from 6: 6C5 = 6C1= 6
Combinations are as follows:
1) 0,1,2,3,4; 0+1+2+3+4 =10
2) 1,2,3,4,5; 1+2+3+4+5 =15
3) 2,3,4,5,0; 2+3+4+5+0 =14
4) 3,4,5,0,1; 3+4+5+0+1 =13
5) 4,5,0,1,2; 4+5+0+1+2 =12
6) 5,0,1,2,3; 5+0+1+2+3 =11
Only number combinations of #2 and #5 are divisible by ...
Started 1 week, 1 day ago (2009-11-04 11:45:00)
by grefreak
1..relationship cant be determined
2..30
Started 1 day, 6 hours ago (2009-11-11 16:47:00)
by computer-bot
OK so there are 52 cards and 4 to choose so we have 52C4 ways to choose them. Also there are 13 cards of each symbol, so we can choose 13C1 for each. This gives us 13^4.
I am confused whether it be (13^4)/52C4 OR (13^4)/(52*51*50*49)
Since the question says that 4 cards are chosen randomly, the second one is more appealing to me.
Please help.
CB
Started 1 day, 22 hours ago (2009-11-11 00:49:00)
by susangre
Started 6 days, 2 hours ago (2009-11-06 21:31:00)
by Curly213
if i understand it right ... u wanna kno why (z^2-4) is an absolute value?
I think because u cant take the sqrt of a number that results to an negative number .... so its an absolute value because it is positive ... hope i helped u!
Started 6 days, 4 hours ago (2009-11-06 19:14:00)
by susangre
From which book this is????????
Started 1 year ago (2008-10-20 20:54:00)
by cssamal
i think answer would be D, as nothing can be determined from the data given. As the number are between 20 to 80. 80 is 100th percentile,so 74 would lie between 60th and 100th percentile. But could not be determined.
Started 4 days, 7 hours ago (2009-11-08 16:38:00)
by Curly213
Hey susan , now i think u `ve got the basic concepts... now its time for u to do as many as excercices u can and by the way stop the time for doin 1 ... i think if u can do one exercices till 1:30 min u will do fine on the real gre ... hope this helps
Started 3 days, 9 hours ago (2009-11-09 14:17:00)
by Curly213
this sign # stands for number ....so #6 means .. number 6 ..its just an abbreviation for number!
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