Set S contains 26 distinct natural numbers. When the elements are : Problem Solving (PS)

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EgmatQuantExpert

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Set S contains 26 distinct natural numbers. When the elements are [#permalink]

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.

B.

C.

D.

E.

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EgmatQuantExpert

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Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink]

07 Apr 2019, 06:10

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Solution

Given:
In this question, we are given that

consecutive

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.

Now, to maximise the value of the highest element, we should minimise the value of the other two elements.

Hence, the correct answer is option E.

Answer: E

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eswarchethu135

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Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink]

03 Apr 2019, 03:58

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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

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Franzhel

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Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink]

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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MeBossBaby

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Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink]

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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Archit3110

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Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink]

04 Apr 2019, 01:24

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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.

B.

C.

D.

E.

B

Suneha123

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Joined: 29 Jan 2019

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Set S contains 26 distinct natural numbers. When the elements are [#permalink]

12 Sep 2020, 00:15

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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.

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chikki420

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Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink]

01 Oct 2020, 13:48

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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.

B.

C.

D.

E.

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

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KarishmaB

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Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink]

29 Aug 2021, 21:51

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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.

B.

C.

D.

E.

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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KarishmaB

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Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink]

30 Aug 2021, 22:18

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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.

B.

C.

D.

E.

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.

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warrior1991

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Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink]

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??

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Archit3110

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Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink]

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??

B

HAPPYatHARVARD

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Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink]

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.

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arvind910619

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Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink]

31 Jan 2020, 11:08

EgmatQuantExpert wrote:

Solution

Given:
In this question, we are given that

consecutive

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.

Now, to maximise the value of the highest element, we should minimise the value of the other two elements.

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.

S

meetnikhil

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Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink]

29 Jun 2020, 01:06

EgmatQuantExpert wrote:

Solution

Given:
In this question, we are given that

consecutive

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.

Now, to maximise the value of the highest element, we should minimise the value of the other two elements.

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...

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Anu5162

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Set S contains 26 distinct natural numbers. When the elements are [#permalink]

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

B

mjjuneja

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Joined: 27 Apr 2020

Posts: 12

Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink]

07 Jul 2020, 04:25

EgmatQuantExpert wrote:

Solution

Given:
In this question, we are given that

consecutive

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.

Now, to maximise the value of the highest element, we should minimise the value of the other two elements.

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?

B

Jihyo

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Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink]

05 May 2021, 04:15

mjjuneja wrote:

EgmatQuantExpert wrote:

Solution

Given:
In this question, we are given that

consecutive

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.

Now, to maximise the value of the highest element, we should minimise the value of the other two elements.

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..

S

Karan911

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Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink]

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.

B.

C.

D.

E.

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?

S

Vetrick

Manager

Joined: 19 Jul 2021

Posts: 54

Location: India

Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink]

31 Aug 2021, 01:26

EgmatQuantExpert wrote:

Solution

Given:
In this question, we are given that

consecutive

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.

Now, to maximise the value of the highest element, we should minimise the value of the other two elements.

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

gmatclubot

Re: Set S contains 26 distinct natural numbers. When the elements are [#permalink]

31 Aug 2021, 01:26

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

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