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

Question

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

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

KarishmaB · Accepted Answer

We can solve it in 30 secs using the basic concept of what mean is - the value that can replace all values in a list. 
When a new value 15 is added, the mean increases by 2. It means 15 has enough extra (above mean) to give 2 to each member of the list and itself.
When a new value 1 is added, the mean decrease by 1. It means 1 has enough deficit (below mean) to take away 1 from each member of the list and still be 1 less. 
Looking at the difference between 15 and 1, we can say that we will likely have 4 or 5 numbers in the original list.

Say if we have 4 numbers in the original list, 15 has 10 extra (2 for each of the 4 members and for itself) and hence original mean must be 5. When we add 15, the mean becomes 7 and now we have 5 numbers in the list. 
Now when 1 is added to the list, it is 6 less than the mean 7 and hence all 5 numbers give 1 to it to bring it up to 6 and the avg reduces to 6. It works. 
Hence, we had 4 numbers in the list. The numbers given are very manageable so that we don't need to do much calculation, the hallmark of a good GMAT question.

Answer (A)

rsrighosh

Archit3110 · Answer

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

mt7777 · Answer

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.

ScottTargetTestPrep · Answer

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

Answer: A

Shobhit7 · Answer

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

rsrighosh · Answer

Hi  KarishmaB

I also tried to approach a similar method earlier but took algebraic approach later. Doing so, i got the answer in -ve. I am unable to understand where I am going wrong. I am pretty sure I made mistake in the second equation

Let us consider original mean of the list as m.
Then we can consider the list as {m, m, m, m....... (n times)}

When 15 is appended to a list of integers, the mean is increased by 2. 
Hence list becomes
{m, m, m, ..... (n times), 15}
As this list has mean as m+2, hence going by your approach, I can say that total of  2n s (2 + 2+ 2 +.... (n times)) is carry forwarded from  m s to  15  
Hence,  15+2n=m+2

When 1 is appended to the enlarged list, the mean of the enlarged list is decreased by 1 
hence the list becomes
{m, m, m, ..... (n times), 15, 1} and its mean is m+2-1=m+1
Again following the same rule, I can say that  n  (1+1+1...(n times)) is carry forwarded from all  m s to 15 and  m  is carry forwarded from 1 to 15.
Hence we can otherwise say  15 + n - m = m+1

So if i solve for n using the equations  15+2n=m+2  and  15 + n - m = m+1 , i am getting the answer as  n = -4

KarishmaB · Answer

The equations will be 
15 - 2n = m + 2
(15 gives away 2 to each of the n elements so that it becomes m + 2 which is the new average)

and
1 + (n + 1) = m + 1
(All the n+1 elements (including 15) give 1 to 1 so that it comes up to the new average which is m+1)

When you solve these, you get m = 5, n = 4

Tejassharmaa · Answer

Can you please explain how mean came out to be 5?

KarishmaB · Answer

Assuming there were 4 people,
15 has 2 extra for each of the 4 people and for itself (that is why mean goes up by 2). So 15 has 10 more than mean. This tells us that mean must be 5 (that is how 15 has 10 more than mean)

PS31 · Answer

Posted from my mobile device

bumpbot · Answer

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

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

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