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

Question

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.

rohit8865 · Answer

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

quantumliner · Answer

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

profileusername · Answer

Can any one help me understand why have we taken even and odd numbers to test the second statement?  abhimahna

abhimahna · Answer

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.

fskilnik · Answer

a < b < c\,\,{m{ints}}

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

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

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

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( * ight)} \,\,\,\left\langle {{m{NO}}} ightangle \,\,

\left( * ight)\,\,\,a\,\,{\mathop{m int}} \,\,\,\, \Rightarrow \,\,\,\,\,a + {{19} \over 2}\,\, 
e {\mathop{m int}} \,\,\,\, \Rightarrow \,\,\,\,\,b 
e a + {{19} \over 2}\,\,\,\,\,\,\left( {b\,\,{\mathop{m int}} } ight)

The correct answer is therefore (B).

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

Regards, 
Fabio.

minustark · Answer

The median and average are equal when the integers are at equal distance from each other. Here, from statement 1 we get to know that the median is 19. So to make the average same, the other two integers together has to be 38. But no such information is given . Insufficient.

From stmnt 2, we can take 3 integers ( 1, 12, 20) .Here mean is 11 and median is 12. Again lets take (2, 11, 21). Here also mean and median different. We can remember the rule of equidistant where if the difference between 1st and 2nd integer is k, then the difference between 2nd and 3rd integer will also be k e.g : 1st integer = a, 2nd integer = a +k and 3rd integer = a+ 2k, where k has to be an integer. Since here 2k =19, that makes sure that k is not an integer and for the reason these 3 integers cannot be equidistant from each other. So median and average will be different. Sufficient.

B is the answer.

bumpbot · Answer

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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

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