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

Question

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

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

Bunuel · Accepted Answer

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

(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.

PareshGmat · Answer

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

WoundedTiger · 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

(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

AKG1593 · Answer

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

vikrantgulia · Answer

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

NoHalfMeasures · Answer

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!

EMPOWERgmatRichC · Answer

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.

Final Answer:  D

GMAT assassins aren't born, they're made, 
Rich

amanlalwani · Answer

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 ;"

chetan2u · Answer

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

MathRevolution · Answer

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

(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).

Abhishek009 · Answer

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

JeffTargetTestPrep · Answer

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  /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

adkikani · Answer

JeffTargetTestPrep   VeritasPrepKarishma   Bunuel   Engr2012

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 ?

KarishmaB · Answer

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)

KarishmaB · Answer

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.

StringArgs · Answer

Hey  Bunuel , Is there a thread/directory for Arithmetic statistics questions. I'd really appreciate one.

Bunuel · Answer

20. Descriptive Statistics 
     Theory   
 Statistics Made Easy - All in One Topic! 
 GMAT Statistics 101 
 Math: Standard Deviation 
 How to Quickly Solve Standard Deviation Questions on the GMAT 
     Questions   
 Standard Deviation Problems 
 DS Questions 
 PS Questions

For more check:
 ALL YOU NEED FOR QUANT ! ! ! 
 Ultimate GMAT Quantitative Megathread

Malar95 · Answer

I was struggling with this Q (just started my Quant practice) and my short cut to this was 
7 (the difference in the least integer) + 5 (10 odd integers - 5 even integers should give a difference of one each) = 12
Correct me if my approach to this is wrong.
It was a bit of a guess game but I was trying to find a method that works for me.

PrabhatKC · Answer

Um, What do you mean by evenly spaced set?
Pls elaborate!

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

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GMAT Problem Solving (PS) Questions

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

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