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Set T consist of 19 elements. The average of set T is L. If

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Set T consist of 19 elements. The average of set T is L. If  [#permalink]

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New post 08 Jul 2011, 14:44
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Set T consist of 19 elements. The average of set T is L. If a new element is added to the set and the average grows by K, what is the value of the element?

А. L (1+K/5)
B. L*K/100 – 20L
C. 20L (1 + K/100)
D. 20 (1+K/100) – 19L
E. L*K/5 – 19 L
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Re: Average of a set  [#permalink]

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New post 08 Jul 2011, 18:50
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I found the wording of 'growth' in this to be a bit ambiguous. To me, growth is not strictly percentage change, which seems to be going on here.

Change in average is K, so we need average before and post.

pre: \(L\)

post: \(\frac{19L + x}{20}\) where x is the the addition to the set which we solve for

Change in average, as k% = \((\frac{post}{pre} - 1)*100\)
\(\frac{k}{100} = \frac{\frac{19L + x}{20}}{L} - 1\)

Rearrange for x

\(x = 20L(1 + \frac{k}{100}) - 19L\)
\(x = L(1+\frac{K}{5})\)

A
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Re: Average of a set  [#permalink]

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New post 08 Jul 2011, 18:53
T1 + T2 +......T19 = 19*L.

T1+ T2 + ....T20 = 20*(L+K)

Hence T20 = 20*(L+K) - T1 + T2 +......T19 = 20*(L+K) - 19*L.

= 20K + 1L.

I cannot find this in the answers. And I am not able to simplify the answers to this form.

Btw this is assuming that you add K to the L and not a K% growth.
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Re: Average of a set  [#permalink]

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New post 08 Jul 2011, 20:47
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Yes, it is quite ambiguous. I assumed it meant K% growth, rather an an absolute change of K
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Re: Average of a set  [#permalink]

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New post 09 Jul 2011, 00:07
MSM05 wrote:
Guys, can you help me understand how to solve this problem:
Set T consist of 19 elements. The average of set T is L. If a new element is added to the set and the average grows by K, what is the value of the element?

А. L (1+K/5)
B. L*K/100 – 20L
C. 20L (1 + K/100)
D. 20 (1+K/100) – 19L
E. L*K/5 – 19 L


What am I missing?
Assume the 19 elements are 5 each ... hence the average of set T is 95/19 = 5= L
if a new element is added , say 25 .. the new average is 95+25/20 = 120/20 = 6
the average grows by 6-5 = 1 = K.
so the value of the element = 25 .. isnt it?

A 5*( 1+1/5) = 5*6/5 = 6
which is new average and not the new element.
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Re: Average of a set  [#permalink]

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New post 09 Jul 2011, 01:29
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sudhir18n wrote:
MSM05 wrote:
Guys, can you help me understand how to solve this problem:
Set T consist of 19 elements. The average of set T is L. If a new element is added to the set and the average grows by K, what is the value of the element?

А. L (1+K/5)
B. L*K/100 – 20L
C. 20L (1 + K/100)
D. 20 (1+K/100) – 19L
E. L*K/5 – 19 L


What am I missing?
Assume the 19 elements are 5 each ... hence the average of set T is 95/19 = 5= L
if a new element is added , say 25 .. the new average is 95+25/20 = 120/20 = 6
the average grows by 6-5 = 1 = K.
so the value of the element = 25 .. isnt it?

A 5*( 1+1/5) = 5*6/5 = 6
which is new average and not the new element.


Yes, your interpretation of the problem is correct. Grows by k means that the new average is "L+K". The question actually wanted to say grows by "k%" as pike corrected it.
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Re: Average of a set  [#permalink]

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New post 09 Jul 2011, 01:39
@sudhir18n, what you have done is correct.
I think question should say "average grows by K%" for A to be the answer.

In that case
sudhir18n wrote:
What am I missing?
Assume the 19 elements are 5 each ... hence the average of set T is 95/19 = 5= L
if a new element is added , say 25 .. the new average is 95+25/20 = 120/20 = 6
the average grows by 6-5 = 1 = K.
so the value of the element = 25 .. isnt it?

A 5*( 1+1/5) = 5*6/5 = 6
which is new average and not the new element.


K= 1/5 * 100 =20

A 5*(1+20/5) = 25.

Original author can you please verify? Thank you.
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Re: Average of a set  [#permalink]

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New post 09 Jul 2011, 08:25
Hello guys,

This problem is from iPAD/iPhone gmutclub app. I've cheked it today and I haven't found any "%" marks there. Also the right answer seems to be "A", according to the app.

Yeah, with "%" the problem and answers make sense... Thank you!
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Re: Average of a set  [#permalink]

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New post 09 Jul 2011, 16:51
Hmm I think the question needs to be changed on the gmat club app to be clearer.

As far as I know (although I could be wrong), the word 'growth' does not strictly imply % growth, which seems to be assumed by this question.
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Re: Set T consist of 19 elements. The average of set T is L. If  [#permalink]

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New post 18 Dec 2016, 17:37
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We can solve it via 2 methods =>

Method 1-->
Using \(Mean = \frac{Sum}{#}\)

Sum(19)=19L
Sum(20)=20*L(1+k/100)

Hence added value = 20L+kL/5 -19L = L+kL/5

Hence A

Method 2-->

Added value id increasing the mean by --> L+kL/100-L=kL/100
Hence added value = L+20*kL/100
Hence A



Great Question

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Re: Set T consist of 19 elements. The average of set T is L. If  [#permalink]

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New post 22 Mar 2018, 15:50
MSM05 wrote:
Set T consist of 19 elements. The average of set T is L. If a new element is added to the set and the average grows by K, what is the value of the element?

А. L (1+K/5)
B. L*K/100 – 20L
C. 20L (1 + K/100)
D. 20 (1+K/100) – 19L
E. L*K/5 – 19 L


We use the formula: average = sum/number, which can be re-expressed as: average x number = sum. Thus, we see that the sum of set T is 19L.

If one element, say x, is added to set T, the new sum is (19L + x), and there are now 20 elements in the set. We can create the equation:

L(1 + K/100) = (19L + x)/20

20L(1 + K/100) = 19L + x

20L + 20LK/100 = 19L + x

L + 20LK/100 = x

L + LK/5 = x

L(1 + K/5) = x

Answer: A
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Re: Set T consist of 19 elements. The average of set T is L. If  [#permalink]

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New post 26 Dec 2018, 14:24
MSM05 wrote:
Set T consist of 19 elements. The average of set T is L. If a new element is added to the set and the average grows by K, what is the value of the element?

А. L (1+K/5)
B. L*K/100 – 20L
C. 20L (1 + K/100)
D. 20 (1+K/100) – 19L
E. L*K/5 – 19 L




I agree with gijoedude. The answer is supposed to be 20K+1L.
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Re: Set T consist of 19 elements. The average of set T is L. If   [#permalink] 26 Dec 2018, 14:24
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