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The arithmetic mean of the 5 consecutive integers starting with 's' is

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The arithmetic mean of the 5 consecutive integers starting with 's' is  [#permalink]

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New post 13 Mar 2016, 12:42
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Re: The arithmetic mean of the 5 consecutive integers starting with 's' is  [#permalink]

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New post 13 Mar 2016, 13:39
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(5s+10)/5=s+2=a
(9s+54)/9=s+6=4+a
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Re: The arithmetic mean of the 5 consecutive integers starting with 's' is  [#permalink]

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Re: The arithmetic mean of the 5 consecutive integers starting with 's' is  [#permalink]

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Re: The arithmetic mean of the 5 consecutive integers starting with 's' is  [#permalink]

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New post 16 Mar 2019, 09:57
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Bunuel wrote:
The arithmetic mean of the 5 consecutive integers starting with 's' is 'a'. What is the arithmetic mean of 9 consecutive integers that start with s + 2?

A. 2 + s + a
B. 22 + a
C. 2s
D. 2a + 2
E. 4 + a


We see that a = s + 2 since (s + 2) is also the median of the 5 consecutive integers starting with s (recall that the mean and median of any number of consecutive integers are equal). For the 9 consecutive integers that start with (s + 2), the median (or mean) is s + 2 + 4 = s + 6. Since a = s + 2, then s + 6 = s + 2 + 4 = a + 4. So (a + 4) is the mean of the consecutive integers that start with (s + 2).

Answer: E
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Re: The arithmetic mean of the 5 consecutive integers starting with 's' is  [#permalink]

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New post 15 Aug 2019, 02:12
ScottTargetTestPrep wrote:
Bunuel wrote:
The arithmetic mean of the 5 consecutive integers starting with 's' is 'a'. What is the arithmetic mean of 9 consecutive integers that start with s + 2?

A. 2 + s + a
B. 22 + a
C. 2s
D. 2a + 2
E. 4 + a


We see that a = s + 2 since (s + 2) is also the median of the 5 consecutive integers starting with s (recall that the mean and median of any number of consecutive integers are equal). For the 9 consecutive integers that start with (s + 2), the median (or mean) is s + 2 + 4 = s + 6. Since a = s + 2, then s + 6 = s + 2 + 4 = a + 4. So (a + 4) is the mean of the consecutive integers that start with (s + 2).

Answer: E


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One question:

since this is an evenly spaced set, we can say that (first+last)/2=median=mean, hence: s+s+4=2a or s+2=a

but we cannot do that operation with real numbers, right?

so in other words if we use the bookend method we cannot get the total sum of the integers:

Lets assume we have set A: {40,41,42,43,44,45,46,47,48,49,50}

Then (40+50)/2=45 but the sum of the set is not 45*2, yet the sum of the set form this question is (s+s+4)*2, so when can we deduce the sum of the set with the bookend method?
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Re: The arithmetic mean of the 5 consecutive integers starting with 's' is  [#permalink]

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New post 15 Aug 2019, 19:46
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ghnlrug wrote:

ScottTargetTestPrep

One question:

since this is an evenly spaced set, we can say that (first+last)/2=median=mean, hence: s+s+4=2a or s+2=a

but we cannot do that operation with real numbers, right?

so in other words if we use the bookend method we cannot get the total sum of the integers:

Lets assume we have set A: {40,41,42,43,44,45,46,47,48,49,50}

Then (40+50)/2=45 but the sum of the set is not 45*2, yet the sum of the set form this question is (s+s+4)*2, so when can we deduce the sum of the set with the bookend method?


You can use the bookend method to find the sum of the elements for any set, but you need to get your numbers right.

The sum of the set from the question is not (s + s + 4)*2; let's just add the elements: s + s + 1 + s + 2 + s + 3 + s + 4 = 5s + 10. It's true that the average a is equal to (s + s + 4)/2; but multiplying this average (which also equals a) by 2 will only give you the sum of the elements that you averaged, namely s and s + 4. Since a is also the average of the set {s, s + 1, s + 2, s + 3, s + 4}; if you multiply a by the number of elements in this set (which is 5), you will get 5a and this is really the sum of the elements in this set. We just found that the sum of the elements is 5s + 10 above and we have 5s + 10 = 5(s + 2) = 5a.

For the other set you provided, i.e. {40, 41, ..., 50}, the average is indeed (40 + 50)/2 = 45; but if you multiply this average by 2, you will obtain the sum of the elements you averaged, namely 40 + 50 = 90. Since there are 11 elements in your set, if you multiply the average by 11, you will get the sum of the elements in the whole set. If you add all the elements 40 + 41 + ... + 50, you'll see that the sum you obtain will equal 11*45.

So, the answer to your question is, you can use the sum = average * quantity formula for any set; just make sure you are multiplying with the correct "quantity".
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Re: The arithmetic mean of the 5 consecutive integers starting with 's' is  [#permalink]

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New post 15 Aug 2019, 21:58
Bunuel wrote:
The arithmetic mean of the 5 consecutive integers starting with 's' is 'a'. What is the arithmetic mean of 9 consecutive integers that start with s + 2?

A. 2 + s + a
B. 22 + a
C. 2s
D. 2a + 2
E. 4 + a


Given: The arithmetic mean of the 5 consecutive integers starting with 's' is 'a'.

Asked: What is the arithmetic mean of 9 consecutive integers that start with s + 2?

The arithmetic mean of the 5 consecutive integers starting with 's' is 'a'.
(s + s +1 + s+2 + s +3 + s +4)/5 = a
5s + 10 = 5a
s + 2 = a

What is the arithmetic mean of 9 consecutive integers that start with s + 2?
a + a + 1 + a +2 + ..... a + 8 = 9a + 36 = 9x
x = a + 4

IMO E
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Re: The arithmetic mean of the 5 consecutive integers starting with 's' is   [#permalink] 15 Aug 2019, 21:58
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