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

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e-GMAT Representative
Joined: 04 Jan 2015
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Set S contains 26 distinct natural numbers. When the elements are  [#permalink]

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03 Apr 2019, 02:51
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Difficulty:

95% (hard)

Question Stats:

38% (02:58) correct 62% (02:46) wrong based on 68 sessions

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

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03 Apr 2019, 03:58
2
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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Joined: 19 Jul 2018
Posts: 7
Re: Set S contains 26 distinct natural numbers. When the elements are  [#permalink]

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03 Apr 2019, 06:13
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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Re: Set S contains 26 distinct natural numbers. When the elements are  [#permalink]

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03 Apr 2019, 10:44
1
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.

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Joined: 03 Mar 2017
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Re: Set S contains 26 distinct natural numbers. When the elements are  [#permalink]

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

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04 Apr 2019, 01:24
1
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

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

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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??
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Joined: 04 Jan 2015
Posts: 3158
Re: Set S contains 26 distinct natural numbers. When the elements are  [#permalink]

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07 Apr 2019, 06:10
1
1

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.

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

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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.
Re: Set S contains 26 distinct natural numbers. When the elements are   [#permalink] 12 May 2019, 19:53
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