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Set S contains 26 distinct natural numbers. When the elements are 03 Apr 2019, 02:51
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Set S contains 26 distinct natural numbers. When the elements are sorted in ascending order, the first 23 numbers are consecutive, and their average is 24. What can be the value of the highest element in S such that the average of all the elements present in S is 30?

A. 52
B. 78
C. 104
D. 153
E. 155


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E
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Set S contains 26 distinct natural numbers. When the elements are 03 Apr 2019, 06:13
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First since its consecutive numbers and their average is 24 we can get two things. They add up to 24*23=552 and the numbers go from 13 to 35.

Then we focus on the average that we want to get, 30 with 26 distinct numbers. So those numbers will have to add to 30*26 = 780. So the 3 numbers will have to sum 780-552= 228.

Since they ask for the highest number, we choose 36 37 and another one. So 228-36-37=155 and this is our 3rd numbers.

Therefore option E
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Set S contains 26 distinct natural numbers. When the elements are 03 Apr 2019, 10:44
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Let n1 to n23 be the 23 consecutive natural numbers, arranged in ascending order. Let n24, n25 and n26 be the remaining three numbers, in ascending order.

We know that for consecutive numbers, mean = median.
Thus, n12 = 24 and n23 = 35
Average = Sum/23 (for the 23 numbers)

thus Sum = 24 * 23 = 552

Now, for the average of n1 to n26 to be 30,
Sum = 26*30 = 780

Thus, 552+n24+n25+n26 = 780

Now, to find the maximize n26, n24 and n25 must be 36 and 37 respectively (since all numbers are distinct).

Thus, n26 = 155.

Answer E.
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Set S contains 26 distinct natural numbers. When the elements are 04 Apr 2019, 01:24
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sum of first 23 consective no
let first no be x
so we have
x+x+1+x+2....+x+22/23=24
or say ; 22*23/2 = 253
23x+253 =24*23
23x=299
x= 13
first term is 13 and 23rd term = 13+22 ; 35
and last term ; 13+26 = 39
so sum of first 23 terms = 23*24 ;
avg of all terms ; a,b,c are 24,25,26 th term
552+a+b+c=26*30
552+a+b+c= 780
a+b+c =228
now so as to get max value of C we need to minimiize value of a & b , since all are distinct no and in ascending order so b,c at best can be 36,37
36+37 +c= 228
C= 228-73 ; 155
IMO E


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Set S contains 26 distinct natural numbers. When the elements are sorted in ascending order, the first 23 numbers are consecutive, and their average is 24. What can be the value of the highest element in S such that the average of all the elements present in S is 30?

    A. 52
    B. 78
    C. 104
    D. 153
    E. 155

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Set S contains 26 distinct natural numbers. When the elements are 07 Apr 2019, 06:10
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Solution


Given:
In this question, we are given that
    • Set S contains 26 distinct natural numbers.
    • When the elements are sorted in ascending order, the first 23 numbers are consecutive, and their average is 24.
    • The average of all the elements present in S is 30.

To find:
    • The value of the highest possible element in S.

Approach and Working:
As the average of the least 23 numbers is 24, their sum = 23 * 24 = 552
Also, the sum of all the 26 elements = 26 * 30 = 780
    • Therefore, the sum of the last 3 elements = 780 – 552 = 228

Now, when the elements are arranged in ascending order, the first 23 elements are consecutive integers and their average is 24.
    • Hence, there should be 11 consecutive integers before 24, and 11 consecutive integers after 24.
    • Thus, the last of the 23 numbers = 24 + 11 = 35
    • Therefore, each of the remaining 3 numbers must be greater than 35.

Now, to maximise the value of the highest element, we should minimise the value of the other two elements.
    • Minimum possible value of the remaining two elements = 36 and 37
    • Therefore, the maximum possible value of the highest element = 228 – (36 + 37) = 228 – 73 = 155

Hence, the correct answer is option E.

Answer: E

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Set S contains 26 distinct natural numbers. When the elements are 29 Aug 2021, 21:51
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EgmatQuantExpert
Set S contains 26 distinct natural numbers. When the elements are sorted in ascending order, the first 23 numbers are consecutive, and their average is 24. What can be the value of the highest element in S such that the average of all the elements present in S is 30?

    A. 52
    B. 78
    C. 104
    D. 153
    E. 155

Show more

Avg = 24
Total 23 consecutive numbers. So 24 must be the 12th number (middle).
First 11 numbers must be 13 to 23. Next 11 numbers must be 25 to 35.

Since these are first 23 numbers, next 3 numbers must be greater than 35 so let's say the next two numbers are 36 and 37. Now we need the value of the greatest/last number such that avg is 30.

First 23 numbers are 6*23 = 138 less than 30.
36 and 37 are 6+7 = 13 more than 30.
So last number must be 138 - 13 = 125 more than 30 i.e. it must be 155.

Karan911 - This is how the method of deviations discussed here:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... eviations/
works its charm :)
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Set S contains 26 distinct natural numbers. When the elements are 29 Aug 2021, 23:28
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VeritasKarishma
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Set S contains 26 distinct natural numbers. When the elements are sorted in ascending order, the first 23 numbers are consecutive, and their average is 24. What can be the value of the highest element in S such that the average of all the elements present in S is 30?

    A. 52
    B. 78
    C. 104
    D. 153
    E. 155


Avg = 24
Total 23 consecutive numbers. So 24 must be the 12th number (middle).
First 11 numbers must be 13 to 23. Next 11 numbers must be 25 to 35.

Since these are first 23 numbers, next 3 numbers must be greater than 35 so let's say the next two numbers are 36 and 37. Now we need the value of the greatest/last number such that avg is 30.

First 23 numbers are 6*23 = 138 less than 30.
36 and 37 are 6+7 = 13 more than 30.
So last number must be 138 - 13 = 125 more than 30 i.e. it must be 155.

Karan911 - This is how the method of deviations discussed here:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... eviations/
works its charm :)
Show more

Hi VeritasKarishma,
I realised I made a mistake, so let the three numbers be a, b, c when added result in a deviation of +3 from the average, the extra amount they bring in gets divided over all of the numbers and each number gets an extra 3,

So letting the numbers be a,b,c, their deviations would from avg would be (a-24 + b -24 + c-24)/ 26 = 6, [ 6 i deviation from 24 to 30]

hence a + b + c = 156 + 72 = 228, now to minimize 2 out of these 3 , i Let them be 36 and 37, hence so the last number is 228 - (36 + 37) = 155, is that correct?

Also one question, if the numbers drop the average, we should be taking (a-24, b-24, c-24)/ 26 = -6 right?

I checked this on a smaller set using arbitrary integers and got the correct result :

for eg if i have 10,20,30, avg is 20, now if I need avg to be dropped to say 15,


so a-20/ 4 = -5, so a = 0,

is this correct?
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Set S contains 26 distinct natural numbers. When the elements are 30 Aug 2021, 22:18
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Karan911
VeritasKarishma
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Set S contains 26 distinct natural numbers. When the elements are sorted in ascending order, the first 23 numbers are consecutive, and their average is 24. What can be the value of the highest element in S such that the average of all the elements present in S is 30?

    A. 52
    B. 78
    C. 104
    D. 153
    E. 155


Avg = 24
Total 23 consecutive numbers. So 24 must be the 12th number (middle).
First 11 numbers must be 13 to 23. Next 11 numbers must be 25 to 35.

Since these are first 23 numbers, next 3 numbers must be greater than 35 so let's say the next two numbers are 36 and 37. Now we need the value of the greatest/last number such that avg is 30.

First 23 numbers are 6*23 = 138 less than 30.
36 and 37 are 6+7 = 13 more than 30.
So last number must be 138 - 13 = 125 more than 30 i.e. it must be 155.

Karan911 - This is how the method of deviations discussed here:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... eviations/
works its charm :)

Hi VeritasKarishma,
I realised I made a mistake, so let the three numbers be a, b, c when added result in a deviation of +3 from the average, the extra amount they bring in gets divided over all of the numbers and each number gets an extra 3,

So letting the numbers be a,b,c, their deviations would from avg would be (a-24 + b -24 + c-24)/ 26 = 6, [ 6 i deviation from 24 to 30]

hence a + b + c = 156 + 72 = 228, now to minimize 2 out of these 3 , i Let them be 36 and 37, hence so the last number is 228 - (36 + 37) = 155, is that correct?

Also one question, if the numbers drop the average, we should be taking (a-24, b-24, c-24)/ 26 = -6 right?

I checked this on a smaller set using arbitrary integers and got the correct result :

for eg if i have 10,20,30, avg is 20, now if I need avg to be dropped to say 15,


so a-20/ 4 = -5, so a = 0,

is this correct?
Show more

Karan911 -
Work only with deviations, not the actual numbers. It will be far easier.

I have 23 numbers with average 24.
I need to add 3 numbers (all greater than 35) to make the average 30.

Now think this way: If I were to add the three numbers each 24 only, the avg would stay the same i.e. 24.
But if I want the avg to go to 30, it means the 3 numbers bring 6 extra for everybody including themselves. So if I add three 30s, I still need another 23*6 = 138 extra to makes up the 6 extra for 23 numbers.
Since the smallest 2 numbers can be 36 and 37, I have already utilised 6+7 = 13 of the 138.
So the third number must be 138 - 13 = 125 more than 30 which gives 155.

Start your though process from the highlighted step. It brings a lot of clarity of the situation. I still do.

Quote:

"if the numbers drop the average..."
for eg if i have 10,20,30, avg is 20, now if I need avg to be dropped to say 15,
so a-20/ 4 = -5, so a = 0,
Show more

Think: I have 3 numbers with avg 20. If I add a number at 20, the avg doesn't change. But I need to add the number such that the avg goes down to 15. So the number must reduce 5 from everyone including itself. Hence, if I add the number 15, I still need to reduce 3*5 for the other 3 numbers. So the number I must add becomes 0.
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Set S contains 26 distinct natural numbers. When the elements are 25 Feb 2025, 10:53
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I do not think this is the right answer. It is clearly mentioned that 23 numbers are consecutive, in that case 36 cannot be the next number and to keep the sum to minimum, the other two numbers must be 37 and 38, which will give the largest value of S as 153.
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Set S contains 26 distinct natural numbers. When the elements are 26 Feb 2025, 15:45
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anirchat
I do not think this is the right answer. It is clearly mentioned that 23 numbers are consecutive, in that case 36 cannot be the next number and to keep the sum to minimum, the other two numbers must be 37 and 38, which will give the largest value of S as 153.
Show more

The sequence is, S = {13, 14, 15, 16,......., 23, 24, 25,........, 34, 35, \(a_{24}\), \(a_{25}\), \(a_{26}\)}

\(a_1\) = 13 and \(a_{23}\) = 35

Therefore, \(a_{24}\) = 36, \(a_{25}\) = 37, and \(a_{26}\) = 155
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Set S contains 26 distinct natural numbers. When the elements are 27 Feb 2025, 00:45
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Set S contains 26 distinct natural numbers. When the elements are sorted in ascending order, the first 23 numbers are consecutive, and their average is 24.

What can be the value of the highest element in S such that the average of all the elements present in S is 30?

Let the value of the highest element in S be x.

First 23 numbers in S = {13,14,15,..,24,25,.....,35}

To maximize x, we have to minimize other numbers.

Total of all elements in S = 24*23+36+37+x = 30*26
625+x=780
x = 780 - 625 = 155

IMO E
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Set S contains 26 distinct natural numbers. When the elements are 08 Aug 2026, 22:39
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Hi, here the question is 'What can be the value of the highest element in S?' ie what value can be possible for highest no out of the 26, NOT 'What can be the highest possible value of highest element in s?'

So by that logic, all of the answer choices should work- 153, 155, 104, 78

Since we have already established above 23rd term is 35, hence rest terms must be bigger than that and sum of the next 3 terms should be 228 as-
Sum of first 23 terms- 23*24- 552 & sum of sum of all 26 digits- 780
So, sum of last 3 terms (>35)- 780-552=228

& this can work with all the answer choices I've mentioned above as-
1. 78 - 24th term- 74, 25th term- 76, 26th term- 78 (all >35 & sum- 228, making overall average 30)
2. 104- 24th term- 61, 25th term- 63, 26th term- 104 (all >35 & sum- 228, making overall average 30)
3. 153- 24th term- 37, 25th term- 38, 26th term- 153 (all >35 & sum- 228, making overall average 30)
4. 155- 24th term- 36, 25th term- 37, 26th term - 155 (all >35 & sum- 228, making overall average 30)

I may be wrong in understanding the question, if so, please help me know my mistake
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Set S contains 26 distinct natural numbers. When the elements are 09 Aug 2026, 00:12
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As written, “what can be the value” is ambiguous, because other options such as 78, 104, and 153 can also be the highest element. The question should ask for the maximum possible value.

Archiving the question.

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