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In a class, the teacher wrote a set of consecutive integers beginning Updated on: 10 Sep 2015, 22:09
Originally posted by Prajat on 10 Sep 2015, 16:13.
Last edited by Prajat on 10 Sep 2015, 22:09, edited 1 time in total.
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C
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95% (hard)

Question Stats:

34% (02:36) correct 66% (02:45) wrong based on 83 sessions

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In a class, the teacher wrote a set of consecutive integers beginning with 1 on the blackboard. Little Johnny came and erased one number. The average of the remaining numbers was \(27 \frac{4}{13}\). What was the number that Little Johnny erased?

(a) 7
(b) 9
(c) 11
(d) 13
(e) 15
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C
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In a class, the teacher wrote a set of consecutive integers beginning 10 Sep 2015, 22:49
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Prajat
In a class, the teacher wrote a set of consecutive integers beginning with 1 on the blackboard. Little Johnny came and erased one number. The average of the remaining numbers was \(27 \frac{4}{13}\). What was the number that Little Johnny erased?

(a) 7
(b) 9
(c) 11
(d) 13
(e) 15
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This question is based on this concept: When a number is removed from a list of consecutive integers, the maximum change in the average cannot be more than 0.5.

I have discussed it in detail here: a-set-consists-of-n-consecutive-integers-in-which-the-smalle-156393.html#p1304540

The average of consecutive numbers would be the middle number. If you erase one number, the average will never change by more than 0.5.
So initially the average must have been either 27 or 27.5.
Say, the average was 27. So there must have been 53 numbers from 1 to 53. When you erase one number, the rest of the 52 numbers effectively increase by 4/13 to increase the average by 4/13.
(4/13)*52 = 16.
So the number erased must have been 16 less than 27 i.e. 11. Fits perfectly.
Answer (C)
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In a class, the teacher wrote a set of consecutive integers beginning 10 Oct 2015, 11:44
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Prajat
In a class, the teacher wrote a set of consecutive integers beginning with 1 on the blackboard. Little Johnny came and erased one number. The average of the remaining numbers was \(27 \frac{4}{13}\). What was the number that Little Johnny erased?

(a) 7
(b) 9
(c) 11
(d) 13
(e) 15
Show more

As we know if a number is removed from the first n natural number then there is a maximum difference of 1/2 in the average of given numbers and the numbers after removal

so giving a view to the answer choices, we know that the average will increase
therefore initial average must be 27
we have average (n+1)/2 = 27
=> n = 53
so sum of first 53 natural numbers = 53 * 27 = 1431

after erasing one number we are left with 52 numbers and their sum = 52 * \(27 \frac{4}{13}\) = 1420

we get deleted number = 1431 -1420 = 11 (Answer)
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In a class, the teacher wrote a set of consecutive integers beginning 10 Sep 2015, 22:00
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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:
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In a class, the teacher wrote a set of consecutive integers beginning 31 May 2016, 12:10
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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

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In a class, the teacher wrote a set of consecutive integers beginning 14 May 2025, 04:42
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