MBA Spotlight – The Biggest MBA Fair is June 13-14
close
It is currently 04 Jun 2023, 07:47
Register
GMAT Club Tests
Decision Tracker
My Rewards
New posts
New comers' posts

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Go to My Error Log Learn more
Close

Request Expert Reply

Confirm Cancel

Set S contains 26 distinct natural numbers. When the elements are

21 posts Go to First Unread Post
Print view
NEW TOPIC
POST REPLY
Downloads
My Bookmarks
Reviews
SORT BY:
Date
Kudos
 1   2   
User avatar
L
EgmatQuantExpert
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3746
Set S contains 26 distinct natural numbers. When the elements are [#permalink] New post  03 Apr 2019, 02:51
3
Kudos
53
Bookmarks
Expert Reply
Quiz Dashboard Attempt only this question Same topic and source Same topic and difficulty Same source and difficulty Random Chronological Reverse chronological Learn more
00:00
A
B
C
D
E

Difficulty:

95% (hard)

Question Stats:

54% (03:18) correct 46% (03:13) wrong based on 434 sessions

Hide Show timer Statistics

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

ShowHide Answer
Official Answer
Official Answer and Stats are available only to registered users. Register/Login.

_________________
Free Trial: Get 25 AI powered videos | 400 practice questions | 8 free webinars

Case studies: How a personalized study plan can save you 60+ hours

MBA admits: M7 admits | INSEAD admits | ISB admits [40+ case studies]


Learn why e-GMAT has 10X more 700+ verified scores than other GMATClub partners – Read our 2000+ reviews
Signature Read More
Quiz Dashboard Attempt only this question Same topic and source Same topic and difficulty Same source and difficulty Random Chronological Reverse chronological Learn more
Most Helpful Reply
User avatar
L
EgmatQuantExpert
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3746
Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink] New post  07 Apr 2019, 06:10
7
Kudos
4
Bookmarks
Expert Reply

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


_________________
Free Trial: Get 25 AI powered videos | 400 practice questions | 8 free webinars

Case studies: How a personalized study plan can save you 60+ hours

MBA admits: M7 admits | INSEAD admits | ISB admits [40+ case studies]


Learn why e-GMAT has 10X more 700+ verified scores than other GMATClub partners – Read our 2000+ reviews
Signature Read More
User avatar
P
eswarchethu135
Senior Manager
Senior Manager
Joined: 13 Jan 2018
Posts: 281
Location: India
Concentration: Operations, General Management
Schools: EDHEC Sept 2019 Intake (A) Erasmus '20 IE Jan20 Intake (A)
GMAT 1: 580 Q47 V23
GMAT 2: 640 Q49 V27
GPA: 4
WE:Consulting (Consulting)
Reviews Badge
Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink] New post  03 Apr 2019, 03:58
5
Kudos
1
Bookmarks
By trial and error method,

We can find that the 23 consecutive numbers start from 12 through 35.

12+13+14+15+16+......+35 = 552

Now \(\frac{Sum_{26}}{26} = 30\)

\(Sum_{26} = 780\)

Since the first 23 numbers add upto 552 the remaining 3 numbers should add upto 228. As the first 23 numbers are from 12 through 35, out of the remaining 3 numbers in order to make a number highest possible value we have to make the other two numbers to the lowest possible value.

The lowest possible values of the 2 numbers among 3 are 36 and 37.

228 - 36 - 37 = 155 is the highest possible element in S.

OPTION: E
General Discussion
avatar
S
Franzhel
Intern
Intern
Joined: 19 Jul 2018
Posts: 33
Location: France
Concentration: Economics, Nonprofit
Schools: ESSEC '22
GMAT 1: 680 Q47 V35
GMAT 2: 710 Q49 V38
WE:Analyst (Manufacturing)
Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink] New post  03 Apr 2019, 06:13
1
Kudos
2
Bookmarks
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
avatar
B
MeBossBaby
Intern
Intern
Joined: 05 Mar 2018
Posts: 33
Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink] New post  03 Apr 2019, 10:44
1
Kudos
2
Bookmarks
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.
User avatar
L
warrior1991
Director
Director
Joined: 03 Mar 2017
Posts: 593
Location: India
Concentration: Operations, Technology
Reviews Badge
Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink] New post  03 Apr 2019, 19:31
eswarchethu135 wrote:
By trial and error method,

We can find that the 23 consecutive numbers start from 12 through 35.

12+13+14+15+16+......+35 = 552

Now \(\frac{Sum_{26}}{26} = 30\)

\(Sum_{26} = 780\)

Since the first 23 numbers add upto 552 the remaining 3 numbers should add upto 228. As the first 23 numbers are from 12 through 35, out of the remaining 3 numbers in order to make a number highest possible value we have to make the other two numbers to the lowest possible value.

The lowest possible values of the 2 numbers among 3 are 36 and 37.

228 - 36 - 37 = 155 is the highest possible element in S.

OPTION: E



Bunuel chetan2u

Is it necessary to figure out that the first 23 numbers are from 12 through 35

If yes then how??
_________________
--------------------------------------------------------------------------------------------------------------------------
All the Gods, All the Heavens, and All the Hells lie within you.
Signature Read More
User avatar
L
Archit3110
GMAT Club Legend
GMAT Club Legend
Joined: 18 Aug 2017
Status:You learn more from failure than from success.
Posts: 7515
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE:Marketing (Energy and Utilities)
GMAT ToolKit User Reviews Badge
Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink] New post  04 Apr 2019, 01:24
2
Kudos
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


EgmatQuantExpert wrote:
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

User avatar
L
Archit3110
GMAT Club Legend
GMAT Club Legend
Joined: 18 Aug 2017
Status:You learn more from failure than from success.
Posts: 7515
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE:Marketing (Energy and Utilities)
GMAT ToolKit User Reviews Badge
Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink] New post  04 Apr 2019, 01:26
warrior1991
please see my solution you would understand why first term has to be 13..

warrior1991 wrote:
eswarchethu135 wrote:
By trial and error method,

We can find that the 23 consecutive numbers start from 12 through 35.

12+13+14+15+16+......+35 = 552

Now \(\frac{Sum_{26}}{26} = 30\)

\(Sum_{26} = 780\)

Since the first 23 numbers add upto 552 the remaining 3 numbers should add upto 228. As the first 23 numbers are from 12 through 35, out of the remaining 3 numbers in order to make a number highest possible value we have to make the other two numbers to the lowest possible value.

The lowest possible values of the 2 numbers among 3 are 36 and 37.

228 - 36 - 37 = 155 is the highest possible element in S.

OPTION: E



Bunuel chetan2u

Is it necessary to figure out that the first 23 numbers are from 12 through 35

If yes then how??
avatar
B
HAPPYatHARVARD
Manager
Manager
Joined: 07 Apr 2018
Posts: 81
Location: United States
Concentration: General Management, Marketing
GMAT 1: 600 Q45 V28
GPA: 3.8
Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink] New post  12 May 2019, 19:53
mean of n consecutive numbers is same as the avg of n consecutive numbers which is equal to (first term+last term)/2

let first term be a and last term be a+22n then avg = [a+(a+22n)]/2= a+11 = 24.
hence firs term is 11, 23rd term is 35.
so in order to maximize the last term we can make the 24 th term as 36, 25th term as 37 and 26th term as 155.
User avatar
L
arvind910619
Director
Director
Joined: 20 Dec 2015
Status:Learning
Posts: 889
Location: India
Concentration: Operations, Marketing
GMAT 1: 670 Q48 V36
GRE 1: Q157 V157
GPA: 3.4
WE:Engineering (Manufacturing)
Reviews Badge
Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink] New post  31 Jan 2020, 11:08
EgmatQuantExpert wrote:

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




Is it necessary to find the highest value. It is not written in the question stem. Highest of the three value is to be found out. Does it have to be max value?
I lost my way in this question after finding the sum 228. Then chose 78, as the answer which is clearly wrong.
User avatar
S
meetnikhil
Intern
Intern
Joined: 11 Jun 2011
Posts: 20
Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink] New post  29 Jun 2020, 01:06
EgmatQuantExpert wrote:

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




Hi e-GMAT,

Question says that first 23 integers are consecutive.
Does if not mean that 24th integer is not consecutive with 23 numbers.
I mean is it not the case that 23rd is 35 then 24th should be 36 and not 35.. because first 23 integers are consecutive. In a way that 24th is not in line...
avatar
B
Anu5162
Intern
Intern
Joined: 21 May 2020
Posts: 9
Set S contains 26 distinct natural numbers. When the elements are [#permalink] New post  29 Jun 2020, 08:12
If we consider 36 and 37 as 24th and 25th number respectively, will it not negate the fact that only first 23 numbers are consecutive. In order to maintain that, the 24th number (smallest yet non consecutive to the series) will be 37 and 25th number will be 38

Shouldnt the answer then be 153 and not 155 ??

Posted from my mobile device
avatar
B
mjjuneja
Intern
Intern
Joined: 27 Apr 2020
Posts: 13
Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink] New post  07 Jul 2020, 04:25
EgmatQuantExpert wrote:

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




EgmatQuantExpert but when it is explicitly told that first 23 numbers are consecutive, can't we infer that following 24th number won't be consecutive with first 23?
avatar
B
Suneha123
Intern
Intern
Joined: 29 Jan 2019
Status:BDM
Posts: 21
Set S contains 26 distinct natural numbers. When the elements are [#permalink] New post  12 Sep 2020, 00:15
1
Kudos
Anu5162 wrote:
If we consider 36 and 37 as 24th and 25th number respectively, will it not negate the fact that only first 23 numbers are consecutive. In order to maintain that, the 24th number (smallest yet non consecutive to the series) will be 37 and 25th number will be 38

Shouldnt the answer then be 153 and not 155 ??

Posted from my mobile device


Exactly I'm thinking the same. EgmatQuantExpert please confirm.
avatar
B
chikki420
Intern
Intern
Joined: 15 Dec 2018
Posts: 9
Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink] New post  01 Oct 2020, 13:48
1
Kudos
EgmatQuantExpert wrote:
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




The summation of CONSECUTIVE, FIRST 23 terms out 26 DISTINCT terms is given by the formula

S = n/2[ 2a + ( n - 1 )d]

Where n is number of terms which is 23

a is first term of sequence which is unknown

d is common difference which is 1 since the numbers are consecutive

We have the average 24 and the number of terms 23 therefore summation becomes 23 * 24

Therefore, 23 * 24 = 23/2 [ 2a + ( 23 - 1 ) ]

a = 13 which is the first term of the sequence

nth term is given by the formula

an = a + ( n - 1 )d

Therefore 23rd term becomes 35.

Posted from my mobile device
avatar
S
Jihyo
Manager
Manager
Joined: 01 Apr 2021
Posts: 57
Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink] New post  05 May 2021, 04:15
mjjuneja wrote:
EgmatQuantExpert wrote:

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




EgmatQuantExpert but when it is explicitly told that first 23 numbers are consecutive, can't we infer that following 24th number won't be consecutive with first 23?


Hey.. I am no expert nor do I know that your confusion still remains..Still just notice the question. It is said that 'when the numbers of the set are in ascending order' it means that although we are told that the first 23 are consecutive , the last elements(let's say a,b,c) will greater than 35(more clearly 35<a<b<c) .Now if we have to maximize c we have to minimize a and b.For that very reason a and b have to be 36 and 37 respectively..
User avatar
L
KarishmaB
GMAT Expert
Joined: 16 Oct 2010
Posts: 13791
Location: Pune, India
Top Member of the Month
Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink] New post  29 Aug 2021, 21:51
1
Kudos
Expert Reply
EgmatQuantExpert wrote:
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 :)
_________________
Karishma
Owner of Angles and Arguments

Check out my Blog Posts here: Blog

For Study Modules, click here: Study Modules
For Private Tutoring, contact us: Private Tutoring
Signature Read More
avatar
S
Karan911
Intern
Intern
Joined: 16 Aug 2020
Posts: 24
Location: India
Concentration: General Management, International Business
GMAT 1: 670 Q48 V34
GPA: 4
WE:Consulting (Computer Software)
Reviews Badge
Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink] New post  29 Aug 2021, 23:28
VeritasKarishma wrote:
EgmatQuantExpert wrote:
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?
User avatar
L
KarishmaB
GMAT Expert
Joined: 16 Oct 2010
Posts: 13791
Location: Pune, India
Top Member of the Month
Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink] New post  30 Aug 2021, 22:18
1
Kudos
Expert Reply
Karan911 wrote:
VeritasKarishma wrote:
EgmatQuantExpert wrote:
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?


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,


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.
_________________
Karishma
Owner of Angles and Arguments

Check out my Blog Posts here: Blog

For Study Modules, click here: Study Modules
For Private Tutoring, contact us: Private Tutoring
Signature Read More
avatar
S
Vetrick
Manager
Manager
Joined: 19 Jul 2021
Posts: 55
Location: India
Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink] New post  31 Aug 2021, 01:26
EgmatQuantExpert wrote:

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.



Isn't it a little tricky to assume 36 and 37 as the lowest possible for two among the remaining three numbers , for if we do so then does it not make the first 25 nos in the ascending order as consecutive instead of the first 23 ?

Yes,the question doesn't say that the first 25 should not be consecutive but it does say the first 23 are consecutive. If we go by that then we will assume 37, 38 instead of 36 & 37 right ?

What am I missing here ?

Posted from my mobile device
_________________
THE X whys thread - https://gmatclub.com/forum/the-x-whys-1-why-2-whys-5-whys-etc-thread-in-quants-371079.html
Signature Read More
GMAT Club Bot
gmatclubot
Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink]
31 Aug 2021, 01:26
NEW TOPIC
POST REPLY
Downloads
My Bookmarks
Reviews
 1   2   
Moderators:
Bunuel
Math Expert
88617 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3152 posts

cron

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions