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When 15 is appended to a list of integers, the mean is increased by 2.

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Math Expert
Joined: 02 Sep 2009
Posts: 58428
When 15 is appended to a list of integers, the mean is increased by 2.  [#permalink]

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19 Mar 2019, 01:36
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Difficulty:

65% (hard)

Question Stats:

53% (03:01) correct 47% (02:32) wrong based on 32 sessions

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When 15 is appended to a list of integers, the mean is increased by 2. When 1 is appended to the enlarged list, the mean of the enlarged list is decreased by 1. How many integers were in the original list?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

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Re: When 15 is appended to a list of integers, the mean is increased by 2.  [#permalink]

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19 Mar 2019, 05:45
2
Bunuel wrote:
When 15 is appended to a list of integers, the mean is increased by 2. When 1 is appended to the enlarged list, the mean of the enlarged list is decreased by 1. How many integers were in the original list?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

x = sum of integers
y= no of integers
x+15/y+1 = x/y + 2
and
x+15+1/y+2 = x/y+2-1
solve we get x=4
IMO A
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Joined: 11 Oct 2018
Posts: 21
Location: Germany
Re: When 15 is appended to a list of integers, the mean is increased by 2.  [#permalink]

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19 Mar 2019, 16:31
Bunuel wrote:
When 15 is appended to a list of integers, the mean is increased by 2. When 1 is appended to the enlarged list, the mean of the enlarged list is decreased by 1. How many integers were in the original list?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

$$x=$$ Sum of integers
$$y=$$ number of integers

$$I. \frac{x+15}{y+1}=\frac{x}{y}+2$$ or $$\frac{x+15}{y+1}-1=\frac{x}{y}+1$$

$$II. \frac{x+16}{y+2}=\frac{x}{y}+1$$

$$\frac{x+15}{y+1}-1=\frac{x+16}{y+2}$$

Cross-multiply

$$(y+2)(x+15)-(y+2)(y+1)=(y+1)(x+16)$$
$$x=y^2+4y-12$$

Plug in in $$I$$

$$\frac{y^2+4y-12+15}{y+1}=\frac{y^2+4y-12}{y}+2$$

Simplify

$$\frac{(y+1)(y+3)}{(y+1)}=y+6-\frac{12}{y}$$

$$y+3-(y+6)=-\frac{12}{y}$$
$$y=4$$

IMO A

That approach is really time consuming. I hope there are some faster approaches.
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Re: When 15 is appended to a list of integers, the mean is increased by 2.  [#permalink]

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21 Mar 2019, 17:53
Bunuel wrote:
When 15 is appended to a list of integers, the mean is increased by 2. When 1 is appended to the enlarged list, the mean of the enlarged list is decreased by 1. How many integers were in the original list?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

We can let n = the number of integers in the original list and a = the average of these integers. We can create the equations:

(na + 15)/(n + 1) = a + 2

and

(na + 15 + 1)/(n + 2) = a + 2 - 1

Solving the first equation, we have:

na + 15= (n + 1)(a + 2)

na + 15 = na + 2n + a + 2

13 = 2n + a

13 - 2n = a

Solving the second equation, we have:

(na + 16)/(n + 2) = a + 1

na + 16 = (n + 2)(a + 1)

na + 16 = na + n + 2a + 2

14 = n + 2a

Substituting a = 13 - 2n into 14 = n + 2a, we have:

14 = n + 2(13 - 2n)

14 = n + 26 - 4n

3n = 12

n = 4

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Joined: 01 Feb 2017
Posts: 242
Re: When 15 is appended to a list of integers, the mean is increased by 2.  [#permalink]

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27 Mar 2019, 11:49
Change in Average = value added above old average / n+1

1st Case: Original set (Avg A) to Enlarged set
2= (15-A) / (n+1)
A+2n = 13

2nd Case: Enlarged set (Avg A') to Super Enlarged set
-1= (1-A') / (n+2)
A'-n= 3

We are given that A'=A+2
So,
Equation 1: A+2n = 13
Equation 2: A'-n = A+2-n = 3 >>> A-n = 1
Solving, we get 3n = 12 and so n=4

Ans A
Re: When 15 is appended to a list of integers, the mean is increased by 2.   [#permalink] 27 Mar 2019, 11:49
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