tmc_grad_09

What is the simplest and fastest operation to solve 7C4 or any combination for that matter? Edit: Found a pretty good site to explain the basics. Anybody wondering the same thing I was see this site Chapter 8 - Combinations and Permutations

Yeah I didn't like the problem at all. The first problem I found with the question is that it didn't state whether profits were distributed based on beg capital balance, ending balance, average balance, etc... I just took a stab at it assuming weighted balance, because they provided the months. I also made the assumption that sweat equity wasn't involved. After reading many different articles covering the GMAT I actually find this problem to be very harmful rather than the least bit helpful. From what I've gathered it isn't good to make any assumptions on the GMAT when answering a problem, which this example forces you to do. Instead, you use only the information provided.

My answer: 2/3 In this situation it's easiest to first find the probability of selecting all good apples, then subtracting that from one in order to find the probability of at least selecting one bad apple. P(all good)= (8/10)*(7/9)*(6/8)*(5/7)= (4/5)*(7/9)*(3/4)*(5/7)= (after you cancel tops and bottoms) (1)*(1/3)*(1)*(1)= 1/3 P(all good)= 1/3 P(at least one bad)= 1-P(all good) P(at least one bad)= 1-(1/3) P(at least one bad)= 2/3

My answer: 1/7 P(heads)= 6/7 P(tails)= 1 - P(heads) P(tails)= 1-(6/7) P(tails)= 1/7

My answer: 1/3 and ? (6/10)*(5/9)= 30/90= 1/3 Can't solve the second part.

My answer: d) 3400 5000*(3/12)= 1250 4500*(4/12)= 1500 3600*(5/12)= 1500 1250+1500+1500= 4250 1000/4250= 0.235.... 800/0.235= 3404 I couldn't do this without a calculator. Regardless, I have no faith that I approached it properly. I won't have the time or calculator to solve this during the actual test. Help please!

My answer: 1250 4500/3= 1500 1500*(5/6)= 1250

Anybody care to work these two out for me?

My answer: a) 10.5 x= 13.5-2(1.5) x= 10.5

My answer: c) 77 x-2(9)= 59 x-18= 59 x= 59+18 x= 77

My answer: a) 31,32,32,30,30,25,33,28 Again, I look at this to be another rule. I don't know the specific wording of the rule, but answer A contains the closest set of numbers in relation to one another.

My answer: d) 2 and 6 For this one I knew there was a rule involved, because there is no way they want you to work it all out. However, I don't know the rule so I grabbed my ti-83 and worked out the numbers to try and see the rule. The mean of the numbers is 4. Each pair is an equal distance from the mean. After running the numbers I got the answer to be 2 and 6. Those numbers immediately stood out to be the medians of the lower half and upper half of the number set. So I ran the same test using the numbers 0,2,4,6,8,10,12. I then added two more numbers 2 and 10 which are the median numbers of the lower half and upper half. This left me with the number set 0,2,2,4,6,8,10,10,12. For both number sets the standard deviation was 4. I assume the rule is that when you have a number set and then the same number set plus two more numbers, then the median number of each half will give you the closest new standard deviation to the original. Could somebody who knows chime in?

My guess: d) 75% I believe you are looking for the numbers in this range: (9.710.3) so that leaves us with this number set within the standard deviation: [9.9,9.9,9.9,10.0,10.2,10.2] which includes 6 of the 8 numbers so 6/8= 75%

For this one I got c) 74. 98-58= 40 40*(2/5)= 16 58+16= 74 I'm not sure if this is correct though. I just started studying and have many months to go so keep that in mind while reading any other answers I post.