The average (arithmetic mean) of four distinct positive integers is 10

Question

The average (arithmetic mean) of four distinct positive integers is 10. If the average of the smaller two of these four integers is 8, which of the following represents the maximum possible value of the largest integer?
A) 12
B) 14
C) 15
D) 16
E) 17

OE
 Given: 0<a<b<c<d. Notice that a,b,c,d are distinct positive integers.
The average of four distinct positive integers is 10: a+b+c+d = 4*10 = 40;

The average of the smaller two of these four integers is 8: a+b = 2*8 = 6. So, 16+c+d = 40, which leads to c+d = 24.
We want to maximize d, so we should minimise c. The minimum value of c is 10, for a=7 and b=9. So, 10+d = 24 and d=14.

Query
 Hi, can anyone explain why a=7 and b=9, please. 
a and b are distinct positive integers.

M13-16

Bunuel · Accepted Answer

Given:  0 \lt a \lt b \lt c \lt d . Notice that  a, b, c  and  d  are  distinct  positive  integers .
 
The average of four distinct positive integers is 10:  a+b+c+d=4*10=40 ;
 
The average of the smaller two of these four integers is 8:  a+b=2*8=16 . So,  16+c+d=40 , which leads to  c+d=24 . 
 
We want to maximize  d , so we should minimize  c . The minimum value of  c  is 10, for  a=7  and  b=9 . So,  10+d=24  and  d=14 .

Answer: B.

As for your question: we have that a + b = 16. Since a and b are distinct integers, then a = b = 8 is not possible, so the least value of b is 9 (7 + 9 = 16). Hence the least value of c is 10.

Hope it's clear.

EMPOWERgmatRichC · Answer

Hi All,

This question is an example of a 'limit' question. To maximize the value of one variable, we will likely have to minimize the values of all of the others (while making sure to follow all of the 'restrictions' described in the prompt).

We're told that the AVERAGE of four DISTINCT POSITIVE INTEGERS is 10...

Since the average is 10, the SUM = 4(10) = 40. So to start, we know that we have 4 DIFFERENT POSITIVE integers that add up to 40.

Next, we're told that the average of the SMALLER two of these four integers is 8...

This means that the sum of those 2 integers is 2(8) = 16. Since the total sum is 40, the sum of the LARGER two integers is 40 - 16 = 24. Thus, these two integers are BOTH bigger than the smaller two integers AND they sum to 24.

We're asked which of the following represents the MAXIMUM possible value of the LARGEST integer.

To start, we have to take the smallest two integers and make them as small as possible AND make them DIFFERENT. Since the integers CANNOT BOTH be 8, one has to be 7 and the other has to be 9. This gives us....

7 9 _ _

The remaining two integers are BOTH greater than 9 and sum to 24. To maximize the largest value, we have to minimize the other value. That happens when the third integer is 10...

7 9 10 _

7+9+10 = 26, so the remaining (largest) integer is 14.

Final Answer:  B

GMAT assassins aren't born, they're made,
Rich

sowragu · Answer

Let the distinct number be A,B,C, and D

Its given A > B > C > D
also A + B + C + D =40 and A + B = 16 means C + D = 24.

Since the question ask for the largest possible number we should choose the least value for both A and B. We cant choose 8 & 8 bcos numbers has been mentioned as distinct. Hence the least possible value for A an B is 9 & 7.
then we can have the least possible value for C as 10 if C is 10 the most possible value for D is 14
Hence the answer B.

stonecold · Answer

Outstanding Question.
Here is what i did in this Question ->

Let the 4 distinct integers be ->
w1
w2
w3
w4

Now,Using -->

Mean = \frac{Sum}{#}

Sum(4) = 10*4=40

Also Average of w1 and w2 is 8
Hence the sum -> w1+w2=8*2=16

Now w1+w2+w3+w4=40
16+w3+w4=40
Hence w4=24-w3
To maximise w4,we must minimise w3

If the Question didn't mention the word "  DISTINCT  "then we would get => w4=24-8=16

But as the integers involved are different -> the case we need to condor is 7,9,10
Hence,w4=24-10=14

Hence B

KanishkM · Answer

Keyword = four distinct positive integers

a +b +c + d= 40

a+b =16, means a can be 7 and b can be 9

d = 24 - c------------------from (40-16)
24 -12, 12, cant be possible 
24-14, 10, can be possible, series becomes 7,9,10,14

24-15,9 cant be possible, because of Keyword.

B

AnirudhaS · Answer

let a<b<c<d
a+b+c+d=40...........(i)
a+b=16.............(ii)
therefore, c+d=24..........(iii)

If d=12, then c=12, CANNOT be possible as they are distinct integers
If d=14, c=10, this is possible where the 4 numbers are 7,9,10,14
If d=15, c=9, this is NOT possible because the best case for a,b would be 7,9
We don't need to check d=16 and 17 as they will make the cases worse than when d=15

Answer:  B

ScottTargetTestPrep · Answer

Solution:

Since the sum of the four distinct integers is 10 x 4 = 40 and the sum of the two smallest integers is 8 x 2 = 16, the sum of the largest two integers is 40 - 16 = 24. Since we want the maximum possible value of the largest integer, we can let the second largest integer be as small as possible. The second integer can’t be 9; otherwise, the sum of the two smallest integers would be at most 7 + 8 = 15 (recall that all the integers are distinct). However, if the second smallest integer is 10 (and the two smallest integers are 7 and 9), the largest integer will then be 24 - 10 = 14.

Answer: B

RastogiSarthak99 · Answer

Good question because it tests averages and maximization of numbers

Basically once you realize it is 16 + c + d = 40

it is time to make cases for a b c d = 40

you can try 1 15 16 8 (But d cannot be less than previous number in sequence)

So you arrive at 7 9 + c + d = 40

to maximize d, you minimize c.

7 + 9 + 10 + d = 40

d = 14 max.

Answer B.

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The average (arithmetic mean) of four distinct positive integers is 10

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