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

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

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16 Feb 2011, 08:42
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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
[Reveal] Spoiler: OA

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16 Feb 2011, 08:56
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Sum of evenly spaced progression:
$$\frac{n}{2}(2a+(n-1)d)$$

$$Average = \frac{Sum}{n}$$
or
$$Average =\frac{1}{2}(2a+(n-1)d)$$

$$\frac{1}{2}(2a+(10-1)*2) - \frac{1}{2}(2(a-7)+(5-1)*2)$$

$$\frac{1}{2}(2a+(10-1)*2) - \frac{1}{2}(2(a-7)+(5-1)*2)$$

$$a+9-(a-7+4)$$

$$a+9-a+7-4$$

$$12$$

Ans: "D"
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16 Feb 2011, 08:59
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Baten80 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

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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17 Feb 2011, 04:51
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If you didnt know the formula for evenly spaced sets, you can pick numbers and solve this question very easily.

Pick first number of set T=2. Since number of elements = 5, the mean will be the middle term ie the 3rd term, which will be 6.

From the question, you can infer that first number of the set T, will be 2+7= 9. Since this set consists of even number of terms,ie 10, the mean will be the average of the middle 2 terms, ie 5th and 6th term, which are 17 and 19 respectively and their average will be 18= mean of the set.

Therefore, the difference between the mean of two sets= 18-6= 12.

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17 Feb 2011, 12:01
Fluke - when u say 2a - what do u mean?
shouldnt it be A1 (the first number in the series)?

thanks.
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17 Feb 2011, 17:24
Easy by picking numbers, but I didn't know that formula. Thanks.

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17 Feb 2011, 23:46
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144144 wrote:
Fluke - when u say 2a - what do u mean?
shouldnt it be A1 (the first number in the series)?

thanks.

You are half right;

It should be $$2*a_1$$ i.e. twice of the first term. "a" is another way of depicting $$a_1$$

The formula for Sum of n numbers in an evenly spaced progression is denoted by:
$$\frac{n}{2}(2*a_1+(n-1)d)$$

If the progression is: 2,4,6. Sum=12

Using formula:
$$a_1=2$$. First term of the series
$$n=3$$ Number of elements in the progression(series)
$$d=a_2-a_1=4-2=2$$

$$Sum=\frac{n}{2}(2a_1+(n-1)d)$$
$$Sum=\frac{3}{2}(2*2+(3-1)2)$$
$$Sum=\frac{3}{2}*8$$
$$Sum=12$$
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18 Feb 2011, 05:55
ye, i mixed with N formula. thanks.
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18 May 2011, 22:43
(2a + 2a + 8) /2

(2a+7 + 2a + 25)/2

leaves 12 after subtracting.
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19 May 2011, 18:57
I got 12 by solving but i think Bunuel's method is brilliant ( as always )

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

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23 Jun 2014, 01:53
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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

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

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29 Jun 2014, 21:14
PareshGmat wrote:
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

PareshGmat's solution is the quickest way. Anyway, here is mine:
T=(t1+t2+t3+t4+t5)/5
=[t1+(t1+1*2)+(t1+2*2)+(t1+3*2)+(t1+4*2)]/5
=[5t1+2*(1+2+3+4)]/5
=[5t1+2*(4*(4+1)/2)]/5
=[5t1+20]/5=t1+4
S=(s1+...+s10)/10
=[s1+(s1+1*2)+...+(s1+9*2)]/10
=[10s1+2*(1+...+9)]/10
=[10s1+2*(9*(9+1)/2)]/10
=[10s1+90]/10=s1+9
We have s1=t1+7
S-T=s1+9-t1-4=t1+7+9-t1-4=12 =>D
P/S: Sum of n consecutive integers:
n(n+1)/2

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

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26 Aug 2015, 10:25
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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.

[Reveal] Spoiler:
D

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

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18 Jun 2016, 00:21
the idea is to remember the rule for consecutive numbers whether even or odd that it will have median=mean=average(1st term+last term)....then for S and T series have a difference of 7 of first digit of series when creating hypothetical digits for series....this will result in exact solution

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

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19 Jun 2016, 01:21
Mean of 10 consecutive odd integers is 9 more than the least.
Mean of 5 consecutive even integers is 4 more than the least.

9 - 4 + 7 = 12.

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

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20 Jun 2016, 01:02
when we read consecutive odd/even integers...that implies mean(average)=median=average(1st term+last term).....make series S and T with hypothetical numbers according to the condition given....on setting up the numbers.....find A.M according to the method described above leading to the right solution

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

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23 Jun 2017, 08:52
Baten80 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

let s=least integer of S
t=least integer of T
mean of S=s+range/2=s+18/2=s+9
mean of T=t+range/2=t+8/2=t+4
substituting s-7 for t,
(s+9)-(s-3)=12
D

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

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26 Jun 2017, 16:53
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Baten80 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.

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

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28 Jun 2017, 00:05
This is can also by using the concept of evenly spaced numbers.
Let Set S be s-10, s-8, s-6.....s+6-- 10 numbers.
Total =(10s-10)=> Avg=s-1

Set T be e-4,e-2,e,e+2,e+4--5 numbers
Total=5e => Avg =e

diff of avg is s-1-e

Other statement for least numbers gives us (s-10)=7+(e-4)
>s-e=13

Putting it back to s-e-1= 12

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Re: List S consists of 10 consecutive odd integers, and list T   [#permalink] 28 Jun 2017, 00:05
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