![]() ![]() ![]() So average of S is 12 greater than average of T. S has odd integers so say it starts from 11. "If the least integer in S is 7 more than the least integer in T" ![]() I would simply take an example since constraints are few. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Please vote for the questions themselves by pressing Kudos button Ĥ. Please vote for the best solutions by pressing Kudos button ģ. Please provide your solutions to the questions Ģ. We'll be glad if you participate in development of this project:ġ. GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions ProjectĮach week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution. Therefore, the average of list S is (x + 16) - (x + 4) = 12 more than the average of list T. Since the median of the integers in T is x + 4, and the median of integers in S is /2 = (2x + 32)/2 = x + 16, the averages of the integers in T and S are x + 4 and x +16, respectively. Since each list is an evenly spaced set, the average of each list is the respective median. Since the least integer in S is 7 more than the least integer in T, x + 7 = the least integer in S, and so S has the following integers: x + 7, x + 9, x + 11, x + 13, x + 15, x + 17, x + 19, x + 21, x + 23, and x + 25. Thus, T contains the following integers: x, x + 2, x + 4, x + 6, and x + 8. The least integer in s is 7 less than T, so it will become x-7. In the second set, there are only five consecutive numbers so the median=mean=the central number, which is third number here. "the mean of T will simply be the median or the third term: mean = (x - 7) + 2*2 = x - 3 Nth term = first term + (N-1)d, d is the constant difference between two consecutive numbers. If the least number here is x, the next number will be x+2, third will be x+2*2.Īlso we can find this through arithmetic progression. There are 10 consecutive odd numbers, means each number is 2 more than the previous number. I could not understand how x+9*2 is the final term and similarly "the mean of T will simply be the median or the third term: mean = (x - 7) + 2*2 = x - 3 " I listed the odd numbers into ten (10) groups. The difference will be (x + 9) - (x - 3) = 12. The mean of T will simply be the median or the third term: mean = (x - 7) + 2*2 = x - 3 So the mean of S will be the average of the first and the last terms: mean = (x + x + 9*2)/2 = x+9, where x is the first term If the least integer in S is 7 more than the least integer in T, how much greater is the average (arithmetic mean) of the integers in S than the average of the integers in T?įor any evenly spaced set median = mean = the average of the first and the last terms. List S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. ![]()
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