In a class, the teacher wrote a set of consecutive integers beginning

Hi Prajat,

The length of time that you take to answer this question will depend greatly on how you organize your work and how you 'see' the math involved.

To start, we need to figure out the range of numbers that we're dealing with. We know that it's a consecutive series of integers between 1 and X, inclusive . We also know that by removing ONE of the numbers, the average of the group becomes 27 4/13. Looking at the answers, we're clearly removing a 'relatively small' number from the group, so the impact of removing that number probably accounts for the 4/13 increase in the average. Logically, the average of the FULL group of numbers is almost certainly 27, so let's work with that...

If the average is 27, and since that value is an INTEGER, then the group of numbers will be be 1 to 53, inclusive (26 numbers 'below' 27, 27 itself, and 26 numbers 'above' 27).

The SUM of those 53 numbers can be 'bunched' in the following way:

1+53 = 54
2+52 = 54
3+51 = 54
Etc.

So we have 26 'pairs' of 54 and the extra 27 "in the middle." This sum can be written as...

26(54) + 27 =
52(27) + 1(27) =
53(27)

Next, we're told that the average becomes 27 4/13 when we remove a value. Removing that value will leave us with 52 terms. The sum of those numbers can be written as...

(27 4/13)(52) =
27(52) + (4/13)(52) =
27(52) + 16
52(27) + 16

We now have the 'old sum' and the 'new sum', so we can calculate the difference...

53(27) - [52(27) + 16] =
1(27) - 16 =
11

Final Answer:

GMAT assassins aren't born, they're made,
Rich

My way harder but more "logical"
We have S = the sum of all integer from 1 to N
x = the missing integer

So as sum = average * Nb of terms
S-x= 27(4/13) * (N-1) (A)
As S-x should be an integer N-1 should be a multiple of 13.
So we have some possible scenario
N-1= 13 => N=14
N-1= 26 => N=17
N-1= 39 => N=40
N-1= 52 => N=53
....

Back to our formula (A)
x = S - 27(4/13)*(N-1)
x = (N(N+1))/2 - 27(4/13)*(N-1)

As x>0
(N(N+1))/2 > 27(4/13)*(N-1)
approx 27(4/13) by 27
(N(N+1))/2 > 27*(N-1)
N(N+1) > 54*(N-1)

So we see than N=53 is working (indeed 53*54>54*52)

So if N=53:
x = (53(54))/2 - 27(4/13)*(52)
x = (53(54))/2 - (27+4/13)*(52)
x = (53(54))/2 - 27*52-(4/13)*52
x = 53*27 - 27*52-(4/13)*52
x = 53*27 - 27*52-4*4
x = 27*(53-52) - 16
x = 27 - 16
x=11

hiphip hourrah

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