I just took a GMAT practice test. I was able to figure out how to work all the ones I missed but the following two. If you can do it, please show me. Thanks. 1) (1/5)^M * (1/4)^18 = 1/(2(10^35)) one fifth to the Mth power times one fourth to the eighteenth power equals one over two time ten to the thirty fifth power. Fine M Answer is 35.... 2) You have to select a combo of two chairs and two tables from a store. There are 5 chairs total and 150 combination of 2 chairs and 2 tables. How many tables are there? Answer is 6
The first one is just exponential function rules, which I don't remember at the moment. For the 2nd one, if you know there are 5 chairs, you know there are 10 combinations of 2 chairs -- if the chairs are labeled A,B,C,D,E, A can be paired with B-C-D-E (4), B can be paired with C-D-E (3 -- it's already paired with A), etc. 4+3+2+1 is 10. The number of total combinations is the number of chair combinations x the number of table combinations (for each table combination, there are 10 pairings of chairs). Since the total is 150, you know the number of table combinations must be 15. Working backwards from the above, 5+4+3+2+1 is 15, and based on that same heuristic, it means the number of tables must be 6.
can take away the fractions to simplify to... 5^M * 4^18 = 2 * 10^35 break down 4 to 2^2 and 10 to 2*5.. 5^M * 2^36 = 2 * 2^35 * 5^35 5^M * 2^36 = 2^36 * 5^M M has to equal 35
I got another question, see if anyone can help. It's a permutation/combination problem. A group of 5 students bought movie tickets in one row next to each other. If Bob and Lisa are in this group, what is the probability that both of them will sit next to only one other student from the group? The Answer is 10% First, I assume you need to find out the total number of of seating arrangements. It's 5!=120. Then, I find out how many options are available so that they sit next to only 1 other student. 5!/(3!2!)=10... Obviously either this number or the first are incorrect, since 10/120 is not 10%. So, I tried to figure out all the possible ways that could sit, let L be Lisa, B be Bob, and x be the other students. LBxxx BLxxx xBLxx xLBxx xxLBx xxBLx xxxBL xxxLB BxxxL LxxxB That is still 10. I cant quite figure out what I'm doing wrong. I think it's probably the way I'm reading the question, so hopefully some fresh eyes can help.
i read it to mean that they each only sit next to one other person, including bob or lisa, in which case the only way they can do that is if they are both at one end of the row. there are 5!=120 arrangements. then you have 2 situations with them at each end of the row (bob in the 1st seat, lisa 5th or lisa in the 1st seat, bob 5th). in both cases, the 3 people in the middle can be in 3!=6 arrangements for a total of 12 arrangements with B and L next to one person. 12/120=10%.
Well, you got the correct answer, which counts, but I'm not sure if that's correct. They are asking for the probability that they will only sit next to 1 other person. I would think who they are sitting next to is irrelevant. Am I wrong?
the 10 examples you listed have B or L sitting next to 2 people in many of them because B and L count as people so you can't be sitting next to B and x or L and x. the only way they can both only sit next to one person is by being on the ends of the row. i know i brought them up in the last post, but the other 3 people are basically irrelevant. there are only 20 ways B and L can be sitting (5P2=20) and there are only 2 ways to get B and L on the ends of the row: LxxxB BxxxL 2/20=10%.
