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Is the median of the 3 different integers equal to the average (arithm

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Is the median of the 3 different integers equal to the average (arithm  [#permalink]

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New post 27 Jun 2017, 15:51
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Is the median of the 3 different integers equal to the average (arithmetic mean) of them?

1) The median of the 3 integers is 19.

2) The range of the 3 integers is 19.
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Re: Is the median of the 3 different integers equal to the average (arithm  [#permalink]

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New post 27 Jun 2017, 18:22
1
aazt wrote:
Is the median of the 3 different integers equal to the average (arithmetic mean) of them?

1) The median of the 3 integers is 19.

2) The range of the 3 integers is 19.



let three integers be (a,b,c)
is a+c =2b

(1) b=19
a,c can be any integers
not suff

(2) c-a = 19--(a)
let b be any odd integer = 19
then if a+c =2b then it must satisfy a+c =38--(b)

solving (a) and (b)
c = 57/2 = not integer

similarly if b =even let =10
then if a+c =2b then it must satisfy a+c =20--(c)
then solving (a) and (c)
c =39/2 --not an integer

thus both cases are not giving any integer value (ie odd + even /2 =not integer)

thus suff

Ans B
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Re: Is the median of the 3 integers equal to the average (arithmetic mean)  [#permalink]

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New post 06 Jul 2017, 07:53
1
Is the median of the 3 integers equal to the average (arithmetic mean) of them?
Statement 1) The median of the 3 integers is 19

Yes, if the three numbers are 19,19,19.
No, if the three numbers are 2,19,21.

This statement is not sufficient.

2) The range of the 3 integers is 19

If the first number is integer x, the third number will be integer (x+19). Let the second number be the median and also the mean of the three numbers = m

So,
x+x+19+m = 3m
2x+19 = 2m
m = x+9.5
As we know m should be an integer, but above we see m is not. Hence the answer is No to the original question stem. This statement is sufficient.

Answer is B
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Is the median of the 3 different integers equal to the average (arithm  [#permalink]

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New post 06 Sep 2017, 12:04
rohit8865 wrote:
aazt wrote:
Is the median of the 3 different integers equal to the average (arithmetic mean) of them?

1) The median of the 3 integers is 19.

2) The range of the 3 integers is 19.



let three integers be (a,b,c)
is a+c =2b

(1) b=19
a,c can be any integers
not suff

(2) c-a = 19--(a)
let b be any odd integer = 19
then if a+c =2b then it must satisfy a+c =38--(b)

solving (a) and (b)
c = 57/2 = not integer

similarly if b =even let =10
then if a+c =2b then it must satisfy a+c =20--(c)
then solving (a) and (c)
c =39/2 --not an integer

thus both cases are not giving any integer value (ie odd + even /2 =not integer)

thus suff

Ans B


Can any one help me understand why have we taken even and odd numbers to test the second statement? abhimahna
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Re: Is the median of the 3 different integers equal to the average (arithm  [#permalink]

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New post 13 Sep 2017, 02:50
TheMastermind wrote:
Can any one help me understand why have we taken even and odd numbers to test the second statement? abhimahna


Hi TheMastermind ,

Personally, I would never prefer to solve in this way.

I think a very well explanation has bee given here.

Feel free to ask if you have any questions.
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Re: Is the median of the 3 different integers equal to the average (arithm  [#permalink]

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New post 30 Jan 2019, 05:48
1
aazt wrote:
Is the median of the 3 different integers equal to the average (arithmetic mean) of them?

1) The median of the 3 integers is 19.

2) The range of the 3 integers is 19.

\(a < b < c\,\,{\rm{ints}}\)

\(b\,\,\mathop = \limits^? \,\,{{a + b + c} \over 3}\)


\(\left( 1 \right)\,\,b = 19\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {18,19,20} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {18,19,21} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\)


\(\left( 2 \right)\,\,c - a = 19\,\,\,\, \Rightarrow \,\,\,\,\left( {a,b,c} \right) = \left( {a,b,a + 19} \right)\)

\(b\,\,\mathop = \limits^? \,\,{{2a + b + 19} \over 3}\,\,\,\, \Leftrightarrow \,\,\,\,2b\,\,\mathop = \limits^? \,\,2a + 19\,\,\,\, \Leftrightarrow \,\,\,\,b\,\,\mathop = \limits^? \,\,a + {{19} \over 2}\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\)

\(\left( * \right)\,\,\,a\,\,{\mathop{\rm int}} \,\,\,\, \Rightarrow \,\,\,\,\,a + {{19} \over 2}\,\, \ne {\mathop{\rm int}} \,\,\,\, \Rightarrow \,\,\,\,\,b \ne a + {{19} \over 2}\,\,\,\,\,\,\left( {b\,\,{\mathop{\rm int}} } \right)\)


The correct answer is therefore (B).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: Is the median of the 3 different integers equal to the average (arithm   [#permalink] 30 Jan 2019, 05:48
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