Search found 117 matches
- by awilhelm
Fri Feb 04, 2011 1:35 pm- Forum: Problem Solving
- Topic: Here's a tough one, help!
- Replies: 6
- Views: 1550
Amazing, thank you! One final question: I don't understand how you are able to express each woman's rate as 1/(t + 4) and 1/(t + 9) and their combined rate as 1/t. How does dividing by 1 give us their rate when we don't know the distance?
Thanks in advance.
- by awilhelm
Fri Feb 04, 2011 1:19 pm- Forum: Problem Solving
- Topic: Here's a tough one, help!
- Replies: 6
- Views: 1550
Awesome, Mitch. Thank you! Could I ask you to walk me through how you would solve this problem without guessing and plugging in? I know that's kind of a pain, but I would be very grateful!
- by awilhelm
Fri Feb 04, 2011 12:09 pm- Forum: Problem Solving
- Topic: Here's a tough one, help!
- Replies: 6
- Views: 1550
Tough question, need help!
Two women start walking toward each other from two ends of a bridge at the same time. They both leave at dawn. When they meet, it is 12pm. When the first woman reaches the other end of the bridge, it is 4pm. When the second woman reaches the other end of the bridge, it is 9pm. They both walk at cons...
- by awilhelm
Fri Feb 04, 2011 12:00 pm- Forum: GMAT Math
- Topic: Tough question, need help!
- Replies: 3
- Views: 2010
Here's a tough one, help!
Two women start walking toward each other from two ends of a bridge at the same time. They both leave at dawn. When they meet, it is 12pm. When the first woman reaches the other end of the bridge, it is 4pm. When the second woman reaches the end of the bridge, it is 9pm. Both women walk at different...
- by awilhelm
Fri Feb 04, 2011 11:41 am- Forum: Problem Solving
- Topic: Here's a tough one, help!
- Replies: 6
- Views: 1550
simple probability
A car dealer is testing six of his cars to determine if they run properly. The probability that they are defective follows a binomial distribution. The probability of a car being defective is .1. What is the probability that exactly two cars will be defective?
Thanks!
- by awilhelm
Tue May 12, 2009 12:54 pm- Forum: Problem Solving
- Topic: simple probability
- Replies: 2
- Views: 952
700 Q47, V39
My two cents: I believe Kaplan/PR are good for two strategies: plugging in and backsolving. Avoid them for anything else. The official guide, the GMAT prep software, and BTG are the only resources you need. None of the other companies is able to write practice questions exactly as they are on the te...
- by awilhelm
Mon Feb 02, 2009 6:06 pm- Forum: I just Beat The GMAT!
- Topic: 700 Q47, V39
- Replies: 0
- Views: 2028
GMAT PREP tricky
Is sqrt((x-3)^2) = 3 - x 1) x is not equal to 3 2) -x(abs val of x) > 0 Ans is B, but why doesn't the first statement give us enough info? The way I see it, the question is really asking us whether x-3 = 3-x. What option for x is there (other than 3) that satisfies the first statement? And statement...
- by awilhelm
Sun Feb 01, 2009 8:26 pm- Forum: Data Sufficiency
- Topic: GMAT PREP tricky
- Replies: 3
- Views: 1558
Confusing combination/permutation GMAT PREP
At a dinner party, 5 people are to be seated around a circular table. Two seating arrangements are considered different only when the positions of the people are different relative to each other. What is the total number of different possible seating arrangements?
Why isn't it 5!
- by awilhelm
Sun Feb 01, 2009 8:22 pm- Forum: Problem Solving
- Topic: Confusing combination/permutation GMAT PREP
- Replies: 1
- Views: 1119
- by awilhelm
Sat Jan 31, 2009 8:03 pm- Forum: Problem Solving
- Topic: Areas and Ratios
- Replies: 12
- Views: 5844
- by awilhelm
Sat Jan 31, 2009 3:07 pm- Forum: Problem Solving
- Topic: Probability - selecting socks
- Replies: 11
- Views: 1922
Hi,
Could anyone explain how to arrive at the answer without using the combination formula? Maybe using the "1 - P(not choosing any black socks)" approach? Thanks
- by awilhelm
Sat Jan 31, 2009 1:53 pm- Forum: Problem Solving
- Topic: Probability - selecting socks
- Replies: 11
- Views: 1922
I have a different answer. What about this approach? Let's assume that t = 2 That makes the area of the triangle = (2*sqrt3)/2 = sqrt3 If the area of the square is also equal to sqrt 3, then s must be equal to sqrt(sqrt3) So we have t = 2, s = sqrt(sqrt3) t/s = 2/sqrt(sqrt3) In answer B, we have 16/...
- by awilhelm
Sat Jan 31, 2009 1:43 pm- Forum: Problem Solving
- Topic: Areas and Ratios
- Replies: 12
- Views: 5844
- by awilhelm
Fri Jan 30, 2009 1:21 pm- Forum: Data Sufficiency
- Topic: Coordinate Geometry
- Replies: 5
- Views: 1330
- by awilhelm
Fri Jan 30, 2009 12:32 pm- Forum: Problem Solving
- Topic: geometry confusion
- Replies: 5
- Views: 1516