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

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

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

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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.

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

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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 c [#permalink]

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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 c [#permalink]

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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

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

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01 Feb 2014, 08:59
Expert's post
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.

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

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28 May 2014, 02:47
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
D!
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Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]

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

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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.

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 c [#permalink]

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25 Jan 2016, 00:32
Expert's post
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.

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 c [#permalink]

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25 Jan 2016, 02:20
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Expert's post
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 c [#permalink]

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22 Apr 2016, 12:07
can someone please post a couple more problems like this one? involving counting techniques for odds\evens, evenly spaced sets etc? thank you!
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List S consists of 10 consecutive odd integers, and list T c [#permalink]

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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
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List S consists of 10 consecutive odd integers, and list T c   [#permalink] 22 Apr 2016, 12:27
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