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555-605 Level|   Statistics and Sets Problems|                           
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patronumbeagle
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List S consists of 10 consecutive odd integers, and list T consists of 15 Jan 2025, 04:57
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An evenly space set is the one that have the same difference between the components of a set. And as this difference ("r") is the same, the set have the property that (1) the mean = Median and (2) the mean of the set equals to the mean of the smallest and the largest elements of the set.

Eg:


  • The first 5 consecutive numbers: {1,2, 3, 4, 5}
    r = 2-1 = 3-2 = 4-3 = 1
    Mean = (1+2+3+4+5)/5 = 15/5 = 3Median correspond to the 3° position, 3 is the medianSo, Mean = Median
  • The even numbers between 60 and 70 {62, 64, 66, 68}
    r = 64-62 = 66-64 = 2
    Mean = (62+68)/2 = 65Median in a set where the total numbers is even (here, 4) is obtained by taking the mean of the (Last Position/2) and the (Last Position/2 + 1) . In this case the 2° (4°/2) and 3° ( 2° + 1): (64+66)/2 = 65So, Mean = Median

Hope is clear! :)




PrabhatKC
Bunuel
SOLUTION

List S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. 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?

(A) 2
(B) 7
(C) 8
(D) 12
(E) 22

For any evenly spaced set median = mean = the average of the first and the last terms.

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;

The mean of T will simply be the median or the third term: mean = (x - 7) + 2*2 = x - 3;

The difference will be (x + 9) - (x - 3) = 12.

Answer: D.
­Um, What do you mean by evenly spaced set?
Pls elaborate!
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BrettonJames
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List S consists of 10 consecutive odd integers, and list T consists of 24 Jan 2025, 10:36
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Hi, will this way of getting to the solution by plugging in real numbers serve me well up until the higher level questions or should I start to try these problems in less pragmatic way?

PareshGmat
Picked up numbers:

T = 2 , 4 , 6, 8 , 10 (Mean = 6)

S = 9, 11, 13, 15, 17, 19, 21, 23, 25, 27 (Mean \(= \frac{17+19}{2} = 18\))

Difference = 18-6 = 12

Answer = D
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List S consists of 10 consecutive odd integers, and list T consists of 25 Jan 2025, 05:39
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It depends - for some higher level questions a quick way to get an answer would be by plugging numbers, while for some questions an algebraic way or approximation would result in an answer quickly. Before approaching the problem you will have to figure out what method would work best for you for that particular qs. Look at these methods as tools in your toolbox, the more the better.
BrettonJames
Hi, will this way of getting to the solution by plugging in real numbers serve me well up until the higher level questions or should I start to try these problems in less pragmatic way?

PareshGmat
Picked up numbers:

T = 2 , 4 , 6, 8 , 10 (Mean = 6)

S = 9, 11, 13, 15, 17, 19, 21, 23, 25, 27 (Mean \(= \frac{17+19}{2} = 18\))

Difference = 18-6 = 12

Answer = D
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List S consists of 10 consecutive odd integers, and list T consists of 27 May 2025, 10:25
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Any odd integer can be represented as a series in the form 2n-1,2n+1, 2n+3, 2n+5.....
Mean will be (5th term+6th term) divided by 2
(2n+7+2n+9)/2=2n+8

Even consecutive integers can be represented as 2a, 2a+2, 2a+4
Mean is 2a+4 since the middle term is equal to mean as the set has odd consecutive integers

diff= 2(n-a)+4
2n-1=7+2a (given in question)
2n-2a=8
Diff=8+4=12
Bunuel
List S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. 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?

(A) 2
(B) 7
(C) 8
(D) 12
(E) 22
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recusandaeeveniet
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List S consists of 10 consecutive odd integers, and list T consists of 28 Oct 2025, 04:29
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How do you come up with the "+ 2*2" as part of the mean calculation for T ( mean = (x - 7) + 2*2 = x - 3)?
Bunuel
SOLUTION

List S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. 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?

(A) 2
(B) 7
(C) 8
(D) 12
(E) 22

For any evenly spaced set median = mean = the average of the first and the last terms.

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;

The mean of T will simply be the median or the third term: mean = (x - 7) + 2*2 = x - 3;

The difference will be (x + 9) - (x - 3) = 12.

Answer: D.
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Bunuel
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List S consists of 10 consecutive odd integers, and list T consists of 28 Oct 2025, 04:36
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recusandaeeveniet
How do you come up with the "+ 2*2" as part of the mean calculation for T ( mean = (x - 7) + 2*2 = x - 3)?

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Because T has 5 consecutive even integers, the mean is the middle (3rd) term.

First term of T is x - 7.

To reach the 3rd term, you take two steps of size 2 (even numbers go up by 2): (x - 7) + 2 + 2 = (x - 7) + 2*2 = x - 3.
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List S consists of 10 consecutive odd integers, and list T consists of 28 Oct 2025, 04:40
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Thank you very much for your help! Highly appreciate it!
Bunuel


Because T has 5 consecutive even integers, the mean is the middle (3rd) term.

First term of T is x - 7.

To reach the 3rd term, you take two steps of size 2 (even numbers go up by 2): (x - 7) + 2 + 2 = (x - 7) + 2*2 = x - 3.
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