GMAT question 44-46

44.Alicia lives in a town whose streets are on a grid system, with all streets running east-west or north-south without breaks. Her school located on a corner, lies three blocks south and three blocks east of his home, also located on a corner. If Alicia is equally likely to choose any possible path from home to school, and if she only walks south or east, what is the probability that she will walk south for the first two blocks?

Solution:
We know that without any restriction she has to go 3 blocks south and 3 blocks east to get to work  Hence, we need to find the number of arrangements of the anagram SSSEEE  Total arrangements : 6!/3!*3! = 20   we know that she has to take 2 blocks south 1st. So, 1st 2 are fixed SS. The remaining 4 steps we will have to find the number of arrangements of SEEE  = 4!/3!*1! = 4  Hence Probability = No of fav/total = 4/20 = 1/5



45.The students at Natural High School sell coupon books to raise money after-school programs. At the end of the coupon sale, the school selects six students to win prizes as follows: From the homeroom with the highest total coupon-book sales, the students with the first-, second- and third-highest sales receive $50, $30, $20, respectively; from the homeroom with the second-highest total coupon-book sales, the three highest-selling students receive $10 each. If Natural High School has ten different homerooms with eight student each, in how many different ways could the six prizes be awarded? (Assume that there are no ties, either among students or among homerooms.) Write your answer as a product of primes raised too various power ( do not actually compute the number).

Solution:
Break down the problem into 3 decisions you have to make:  1) Select 2 homerooms,1 with Highest sales and 1 with 2nd highest sales.(order matters here)  2)Select 3 Students from the highest sales Homeroom, again here order matters as depending on the position the value of the prize differs  3)select 3 students from the 2nd highest sales homeroom, here the order doesnt matter as all 3 will receive the same prize.
 Decision 1) No of ways: 10p2 = 10!/8! = 10*9 = 2*5*3*3  Decision 2) No of ways: 8p3 = 8!/5! = 8*7*6 =  2*2*2*7*3*2  Decision 3) No of ways: 8C3 = 8!/3!*5! = 8*7 = 2*2*2*7  multiply all: we get (2^8 * 3^3 * 5 * 7^2)


46.A group of men and women gathered to compete in a marathon. Before the competition, each competitor was weighed and the average weight of the female competitors was found to be 120 lbs. What percentage of the competitors were women?  (1) The average weight of the men was 150lb.  (2) The average weight of the entire group was twice as close to the average weight of the men as it was to the average weight of the women.

Solution:
(1) INSUFFICIENT: This statement merely provides us with the average of the other subgroup - the men. We don't know what weight to give to either subgroup; therefore we don't know the ratio of the women to men.   (2) SUFFICIENT: If the average weight of the entire group was twice as close to the average weight of the men as it was to the average weight of the women, there must be twice as many men as women. With a 2:1 ratio of men to women of, 33 1/3% (i.e. 1/3) of the competitors must have been women. Consider the following rule and its proof.

RULE: The ratio that determines how to weight the averages of two or more subgroups in a weighted average ALSO REFLECTS the ratio of the distances from the weighted average to each subgroup's average.
Let's use this question to understand what this rule means. If we start from the solution, we will see why this rule holds true. The average weight of the men here is 150 lbs, and the average weight of the women is 120 lbs. There are twice as many men as women in the group (from the solution) so to calculate the weighted average, we would use the formula [1(120) + 2(150)] / 3. If we do the math, the overall weighted average comes to 140.  Now let's look at the distance from the weighted average to the average of each subgroup.  Distance from the weighted avg. to the avg. weight of the men is 150 - 140 = 10.  Distance from the weighted avg. to the avg. weight of the women is 140 - 120 = 20.   Notice that the weighted average is twice as close to the men's average as it is to the women's average, and notice that this reflects the fact that there were twice as many men as women. In general, the ratio of these distances will always reflect the relative ratio of the subgroups.
Answer is B