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List S consists of 10 consecutive odd integers, and list T consists of

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List S consists of 10 consecutive odd integers, and list T consists of  [#permalink]

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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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Originally posted by Bunuel on 30 Jan 2014, 01:57.
Last edited by Bunuel on 11 Mar 2019, 23:29, edited 1 time in total.
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Re: List S consists of 10 consecutive odd integers, and list T consists of  [#permalink]

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New post 30 Jan 2014, 01:57
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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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Re: List S consists of 10 consecutive odd integers, and list T consists of  [#permalink]

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New post 25 Jan 2016, 00:32
3
4
amanlalwani wrote:
Bunuel wrote:
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.



Hi Bunel,

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;"


Hi,
there are 10 consecutive odd numbers , means each number is 2 more than the previous number...
if the least number here is x, the next number will be x+2, third will be x+2*2...
and so on till 10th term= x+9*2..
also we can find this through arithmetic progression..
Nth term = first term + (N-1)d, d is the constant difference between two consecutive numbers..

2ND part..
"the mean of T will simply be the median or the third term: mean = (x - 7) + 2*2 = x - 3
in the second set, there are only five consecutive numbers so the median=mean=the central number, which is third number here..
the least integer in s is 7 less than T, so it will become x-7...
the third term here will be (x-7) + 2*2..same as nthterm above
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Re: List S consists of 10 consecutive odd integers, and list T consists of  [#permalink]

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New post 23 Jun 2014, 02:53
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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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Re: List S consists of 10 consecutive odd integers, and list T consists of  [#permalink]

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New post 30 Jan 2014, 02:54
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List S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. If the least integer in Sis 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

Sol: Let List T has the following members : 2,4,6,8 and 10
Then S has : 9,11,13,15,17,19,21,23,25,27

Now If we find the average of List T is 6 and average of List S is (19+17)/2 =18
So Ans is 12.

Suppose if we S also had 5 members and all the other condition remains same then Average of S would have been 13 and diferecne between the 2 would be 7 cause when the same number is added/subtracted from a given set then the average of the new set increases or decreases by the same number

So ans is D.

Average difficulty level of 650 is okay
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Re: List S consists of 10 consecutive odd integers, and list T consists of  [#permalink]

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New post 30 Jan 2014, 03:18
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We could do this by taking value for the lists
List T=-4,-2,0,2,4.Mean=0
List S=3,5,7,...21=>Mean=12;(21+3)/2 (S has started from 3 as -4+7=3)
Difference=12
Ans.D
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Re: List S consists of 10 consecutive odd integers, and list T consists of  [#permalink]

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New post 30 Jan 2014, 10:17
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Easy one.
Let us consider two set S and T.
1) T is the even consecutive set and S is odd consecutive set .
2) Least value of T +7=Least value of S

So if least value of T is 2 then least value of S is 9.

Its a series of even and odd consecutive integer.
So T 5th term= 2+4*2=10 ... Mean=(10+2)/2=6
Similarly S 10 th term = 9+2*9=27.... Mean=(27+9)/2=18

Difference is 18-6=12

answer is D
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Re: List S consists of 10 consecutive odd integers, and list T consists of  [#permalink]

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New post 28 May 2014, 02:47
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Since the least no. in S is 7 greater than the least no. in T, lets assume S starts at 7 so T will start at 0.
For S mean will be the average of 5th and 6th no.: {7, 9, 11, 13, 15, 17....} = (15+17)/2 = 16
For T mean will be the 3rd no. {0, 2, 4...} = 4
Answer=16-4=12
D!
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Re: List S consists of 10 consecutive odd integers, and list T consists of  [#permalink]

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New post 26 Aug 2015, 11:25
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Hi All,

This question can be solved by using Number Property Rules or by TESTing VALUES:

Here's how TESTing VALUES works:

List S: 10 consecutive ODD integers
List T: 5 consecutive EVEN integers
The least integer is S is 7 more than the least integer in T.

Let's say that….
T = {2, 4, 6, 8, 10}
Since the least integers in S is 7 MORE than the least integer in T…
S = {9, 11, 13, 15, 17, 19, 21, 23, 25, 27}

The average of T = 6
The average of S = 18

So, the average of S is 18 - 6 = 12 more than the average of T.

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Re: List S consists of 10 consecutive odd integers, and list T consists of  [#permalink]

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New post 24 Jan 2016, 23:49
Bunuel wrote:
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.



Hi Bunel,

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;"
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Re: List S consists of 10 consecutive odd integers, and list T consists of  [#permalink]

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New post 25 Jan 2016, 02:20
1
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

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


Since S is the list consisting of 10 consecutive odd integers we can put S={s, s + 2, s + 4, ...., s + 18}, where s is the least odd integer of S.
So the average of S is (10*s + 2+4+....+18)/10=(10*s + 90)/10= s+9.

Similarly we may put T={t, t+2, ..., t+8}, where t is the least even integer of T. So the average of T is (5*t + 2+ 4+ ....+8)/5 = t+4. s+9-(t+4)=s-t+5=7+5=12. So the answer is 12. ---> (D).
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Re: List S consists of 10 consecutive odd integers, and list T consists of  [#permalink]

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New post 22 Apr 2016, 12:27
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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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



Let the set of numbers be

S = { 9 , 11 , 13 , 15 , 17 , 19 , 21 , 23 , 25 , 27 }
T = { 2 , 4 , 6 , 8 , 10 }

Sum of the set S = 180
Mean of set T = 18

Sum of the set T = 30
Mean of set T = 6

So, The arithmetic mean of set S is 12 more than the mean of set T :-D :lol:
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Re: List S consists of 10 consecutive odd integers, and list T consists of  [#permalink]

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New post 29 Nov 2016, 16:43
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Bunuel wrote:

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


We can let x = the least integer in T. Thus, T contains the following integers: x, x + 2, x + 4, x + 6, and x + 8.

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.

Since each list is an evenly spaced set, the average of each list is the respective median. Since the median of the integers in T is x + 4, and the median of integers in S is [(x +15) + (x + 17)]/2 = (2x + 32)/2 = x + 16, the averages of the integers in T and S are x + 4 and x +16, respectively.

Therefore, the average of list S is (x + 16) - (x + 4) = 12 more than the average of list T.

Answer: D
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Re: List S consists of 10 consecutive odd integers, and list T consists of  [#permalink]

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New post 28 Sep 2017, 21:10
JeffTargetTestPrep VeritasPrepKarishma Bunuel Engr2012

Quote:
We can let x = the least integer in T. Thus, T contains the following integers: x, x + 2, x + 4, x + 6, and x + 8.



Is not a set of even consecutive no represented by 2x, 2x+2, 2x+4 .. ? Did we took 2 common to reach above step ?

I messed up taking first T as 2n, 2n+2 , 2n+4 ... and S as 2n+7, 2n+9... which was far more calculation intensive.

Please let me know flaw in approach ?
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Re: List S consists of 10 consecutive odd integers, and list T consists of  [#permalink]

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New post 11 Oct 2017, 05:29
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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

Problem Solving
Question: 70
Category: Arithmetic Statistics
Page: 70
Difficulty: 600


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Responding to a pm:

I would simply take an example since constraints are few.
"If the least integer in S is 7 more than the least integer in T"

S has odd integers so say it starts from 11.
11, 13, 15, 17, 19, 21 .... (10 numbers)
Average = 20 (middle of 19 and 21)

T will start from 11-7 = 4
4, 6, 8, 10, 12
Average = 8

So average of S is 12 greater than average of T.
Answer (D)
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Re: List S consists of 10 consecutive odd integers, and list T consists of  [#permalink]

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New post 12 Oct 2017, 23:33
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adkikani wrote:
JeffTargetTestPrep VeritasPrepKarishma Bunuel Engr2012

Quote:
We can let x = the least integer in T. Thus, T contains the following integers: x, x + 2, x + 4, x + 6, and x + 8.



Is not a set of even consecutive no represented by 2x, 2x+2, 2x+4 .. ? Did we took 2 common to reach above step ?

I messed up taking first T as 2n, 2n+2 , 2n+4 ... and S as 2n+7, 2n+9... which was far more calculation intensive.

Please let me know flaw in approach ?


There isn't a flaw in your approach. You can consider the first term of T as 2n and first term of S as 2n+7. Of course the more complicated your terms, more calculation intensive it will become. Since all you need is the difference between the averages, no matter how you take your integers, the answer will always be the same. So in such cases, I wouldn't take a variable at all.
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Re: List S consists of 10 consecutive odd integers, and list T consists of  [#permalink]

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New post 06 Jan 2019, 01:47
Let the set of numbers be

S = { 9 , 11 , 13 , 15 , 17 , 19 , 21 , 23 , 25 , 27 }
T = { 2 , 4 , 6 , 8 , 10 }

Sum of the set S = 180
Mean of set T = 18

Sum of the set T = 30
Mean of set T = 6
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List S consists of 10 consecutive odd integers, and list T consists of  [#permalink]

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New post 03 May 2019, 09:28
Hi,

here are my two cents for this question

Since we know that if we increase or decrease the data elements in a set by same value , Mean will increase or decrease by that value.

This means for this first five terms of S the mean is 7 more than that of T .

So Options (A and B are out )

If we Say mean of T is M then for first five terms of S mean is M+7.

Since we know that Set S is odd consecutive integers , In a set of five data elements third term will be mean . so third term is a+2d, here d =2 so third term is (a+4)
In a data set of 10 consecutive odd integers sixth term will be (a+5d) = (a+10)

Now if we see that a+10 is 6 more than a+4. Also on the same lines the next four data points will be 8,10,12,14 more than a+4. So this (6+8+10+12+14) will be equally divided among the all 10. So current mean is M+7 + \(\frac{(6+8+10+12+14)}{10}\)

this is M+7+5= M+12

So we have that Mean of S is 12 more than mean of T
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List S consists of 10 consecutive odd integers, and list T consists of   [#permalink] 03 May 2019, 09:28
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