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If the arithmetic mean of three positive integers is 8, what are the

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If the arithmetic mean of three positive integers is 8, what are the  [#permalink]

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New post Updated on: 16 Dec 2018, 00:02
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A
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If the arithmetic mean of three distinct positive integers is 8, what are the values of the integers?

(1) The largest integer is twice the smallest integer.
(2) One of the integer is 9.


The official answer was A. But I think both statements are sufficient. I need help on this question.

Originally posted by Khawasbnam1 on 14 Dec 2018, 03:39.
Last edited by Gladiator59 on 16 Dec 2018, 00:02, edited 2 times in total.
Renamed the topic and edited the question. Edit 2: Added distinct
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Re: If the arithmetic mean of three positive integers is 8, what are the  [#permalink]

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New post 14 Dec 2018, 03:51
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Is the answer is A?

Because it is not given that the three positive integer are distinct.
for statement :-1
Possibility:

5…9…10

6…6…12 so not sufficient.

St-2 not sufficient.

Combine:

only one set:

5...9....10.. Sufficient.
I think answer should be C.
What is the source?
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Re: If the arithmetic mean of three positive integers is 8, what are the  [#permalink]

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New post 15 Dec 2018, 23:38
1
Khawasbnam1 wrote:
If the arithmetic mean of three positive integers is 8, what are the values of the integers?

(1) The largest integer is twice the smallest integer.
(2) One of the integer is 9.


The official answer was A. But I think both statements are sufficient. I need help on this question.


Given,
a+b+c=24

RTF: a=? b=? c=?

Option-1: Let us consider a=largest and c= smallest, then a=2c
So, a+b+c=2c+b+c=24
=> 3c+b=24

Now from here, as we don't know the value of b, we can't determine the value of b either. So not sufficient.

Option-2: One of the integer is 9.
So it can be either a, b or c. So not sufficient.

Option-1 + Option-2:
a=2c
Now, if a=9, then c will not be an integer => a cannot be 9
If c=9, then a=18 => a+c=27 (>24) => c cannot be 9 either
Hence, b can only take the value of 9

So, 2c+9+c=24 => 3c=15 => c =5
a=10, b=9, c=5....Sufficient.

Hence, IMO OA is wrong and ans should be C.
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Re: If the arithmetic mean of three positive integers is 8, what are the  [#permalink]

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New post 16 Dec 2018, 00:03
Updated the question to match the OA.

Best,
Gladi

Khawasbnam1 wrote:
If the arithmetic mean of three positive integers is 8, what are the values of the integers?

(1) The largest integer is twice the smallest integer.
(2) One of the integer is 9.


The official answer was A. But I think both statements are sufficient. I need help on this question.

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Gladi



“Do. Or do not. There is no try.” - Yoda (The Empire Strikes Back)
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If the arithmetic mean of three positive integers is 8, what are the  [#permalink]

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New post 28 Dec 2018, 07:04
The answer is A.
As the numbers are distinct and the average of all three is 8.

Let three numbers be a,b,c where c is largest and a is smallest.


Stmt 1: c=2a
Therefore, a+b+c=24
3a+b=24

Lets check with a=1, then c becomes 2 and b becomes 21.
Not correct as 'c' has to be largest

Try with a=2, 3,4 with all the cases b becomes largest.

Only a=5 is the case when c=10 and b=9.

Thus, Stmt 1 is sufficient.
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Re: If the arithmetic mean of three positive integers is 8, what are the  [#permalink]

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New post 07 Feb 2019, 10:45
St 1: 3a+b = 24

check for a = 5, 6

for a= 5, b =9 , c =10

for a = 6, b = 8, c = 12

In both cases a is smallest and c is largest.

We need St. 2 to eliminate option 2.

Ans should be C.

Am I missing something here ?
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Re: If the arithmetic mean of three positive integers is 8, what are the  [#permalink]

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New post 07 Feb 2019, 12:42
1
Chanda15 wrote:
St 1: 3a+b = 24

check for a = 5, 6

for a= 5, b =9 , c =10

for a = 6, b = 8, c = 12

In both cases a is smallest and c is largest.

We need St. 2 to eliminate option 2.

Ans should be C.

Am I missing something here ?



6+8+12 = 26 <> 24
You would need 6, 6, and 12. In that case the numbers are not distinct, so only 5, 9, and 10 is possible from S1.
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Re: If the arithmetic mean of three positive integers is 8, what are the   [#permalink] 07 Feb 2019, 12:42
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