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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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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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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 c [#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 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.

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

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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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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 c [#permalink]

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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 c [#permalink]

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New post 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
Answer=16-4=12
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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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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 c [#permalink]

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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 c [#permalink]

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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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New post 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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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 c [#permalink]

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New post 17 Nov 2016, 15:16
General formula for odd numbers is 2n + 1 and even 2n

Assume that 2 is the least in the even set then 2+7 = 9 has to be the first in the odd set.

So 2n + 1 = 9 gives n = 4 so the index n of the 10th value for the odd set is (4 + 10) - 1 = 13 AND THE magical -1 occurs because the formula has "zero based" indexing. Hence value n for the 10th is 13 AND NOT 14.

Therefore 2(13) + 1 = 27. This means that for the ODD set min = 9 and max = 27 so mean = 18

The mean of the EVEN set is the median which is equal to 6 so the difference is 12 and Correct answer D
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Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]

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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 c [#permalink]

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New post 11 Dec 2016, 11:48
S = 10 consecutive interegers
T = 5 c. integers

S>7t


If the least integer of T is 1, than the least integer of S is 8.

The largest number of T is thus: 1+2x4=9, 9+1=10/2 = 5, so mean is 5
And for S is 8+2x9= 26, 26+8, 34/2=17, so mean is 17

The difference is 17-5, which is 12. Hence answer D
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Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]

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New post 07 May 2017, 03:24
Abhishek009 wrote:
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:


guess thats the easiest approach. However, we could also start with 0 and recognise that we are dealing with an evenly spaced set, hence median = mean
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Re: List S consists of 10 consecutive odd integers, and list T c   [#permalink] 07 May 2017, 03:24
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