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Senior Manager  Joined: 21 Oct 2013
Posts: 411
The average (arithmetic mean) of four distinct positive integers is 10  [#permalink]

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7 00:00

Difficulty:   65% (hard)

Question Stats: 56% (01:46) correct 44% (01:56) wrong based on 148 sessions

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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
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Joined: 02 Sep 2009
Posts: 58427
Re: The average (arithmetic mean) of four distinct positive integers is 10  [#permalink]

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goodyear2013 wrote:
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

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 $$d=14$$.

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.
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: The average (arithmetic mean) of four distinct positive integers is 10  [#permalink]

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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.

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Rich
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Re: The average (arithmetic mean) of four distinct positive integers is 10  [#permalink]

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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.
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Joined: 12 Aug 2015
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Schools: Boston U '20 (M)
GRE 1: Q169 V154 Re: The average (arithmetic mean) of four distinct positive integers is 10  [#permalink]

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1
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

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Re: The average (arithmetic mean) of four distinct positive integers is 10  [#permalink]

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goodyear2013 wrote:
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
M13-16

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
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Many of life's failures happen with people who do not realize how close they were to success when they gave up. Re: The average (arithmetic mean) of four distinct positive integers is 10   [#permalink] 09 Feb 2019, 05:31
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