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Bounce1987
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Bounce1987
There is a set of numbers 2, 4, 6, 8,10 and x. What is the value of x?

(1) x is a 2-digit positive integer.
(2) The average of these integers is equal to the median of these integers.

Statement 1: \(x\) can be any two digit no 10,20,30.... etc. Hence Insufficient

Statement 2: Average of the set \(=\frac{(2+4+6+8+10+x)}{6}=\frac{(30+x)}{6}=5+\frac{x}{6}\)

Median, if \(6<x<8\), then the Median would be \(\frac{6+x}{2}\) but if \(x>8\), then the Median would be \(\frac{6+8}{2} = 7\)

Now the two values of median when equated against the average will give two different values of \(x\). Hence Insufficient

Combining 1 & 2, we know for sure that \(x>8\)
.
Therefore \(5+\frac{x}{6}=7\) or \(x =12\)

Option C
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Bounce1987
There is a set of numbers 2, 4, 6, 8,10 and x. What is the value of x?

(1) x is a 2-digit positive integer.
(2) The average of these integers is equal to the median of these integers.


Value base Qs, x=? in set ={2,4,6..x}

St-1 x is 2 digit positive integer- x can be any 2 digit integer insufficient
St-2 the average is equal to median then consecutive integers x can be 0 or 12,insufficient

Combine- x can be only 12 sufficient
Answer is C
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I went through with a little easier approach. Please let me know if I am on the right track though.
There is a set of numbers 2, 4, 6, 8,10 and x. What is the value of x?

(1) x is a 2-digit positive integer.
(2) The average of these integers is equal to the median of these integers.

1) According to the first statement, x=11,12,13 or any number.
Hence insufficient
2)According to this statement, Average of these integers(mean)=Median
The above Statement is possible when the set of integers are consecutive integers
x=0,12
Not sufficient
1+2=x=12
Therefore C is the answer
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Bounce1987
There is a set of numbers 2, 4, 6, 8,10 and x. What is the value of x?

(1) x is a 2-digit positive integer.
(2) The average of these integers is equal to the median of these integers.

Statement 1: \(x\) can be any two digit no 10,20,30.... etc. Hence Insufficient

Statement 2: Average of the set \(=\frac{(2+4+6+8+10+x)}{6}=\frac{(30+x)}{6}=5+\frac{x}{6}\)

Median, if \(6<x<8\), then the Median would be \(\frac{6+x}{2}\) but if \(x>8\), then the Median would be \(\frac{6+8}{2} = 7\)

Now the two values of median when equated against the average will give two different values of \(x\). Hence Insufficient

Combining 1 & 2, we know for sure that \(x>8\)
.
Therefore \(5+\frac{x}{6}=7\) or \(x =12\)

Option C


Actually if x is 8 then median is 7 and and average is 6.66.. hence x= 8 does not satisfy the condition , only x= 0, or 6 or 12 satisfy statement 2, since there are 3 values hence insufficient .

1 and 2 combined
we have only one two digit integer, 12 hence sufficient.
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Bounce1987
There is a set of numbers 2, 4, 6, 8,10 and x. What is the value of x?

(1) x is a 2-digit positive integer.
(2) The average of these integers is equal to the median of these integers.


A quick way to solve would be:
#1 in itself is insufficient.
#2 this statement tells us the set is evenly spaced, so x must either be zero or 12.

Now considering statement 1, we can conclude x=12.
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Are there always three cases to consider:

1) x > median
2) x = median
3) x < median

?
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ghnlrug
Are there always three cases to consider:

1) x > median
2) x = median
3) x < median

?

Yes, in general that's a good set of cases to think about, when the problem gives you a set of numbers including an unknown value and starts asking you questions about the median. Either the unknown value is the median itself, or some other (smaller or larger) value in the set is the median. Or, if the set has an even number of terms (including the unknown value!), then you also have to think about the possibility that the median is halfway between the unknown and the value right above or below it.
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ccooley
ghnlrug
Are there always three cases to consider:

1) x > median
2) x = median
3) x < median

?

Yes, in general that's a good set of cases to think about, when the problem gives you a set of numbers including an unknown value and starts asking you questions about the median. Either the unknown value is the median itself, or some other (smaller or larger) value in the set is the median. Or, if the set has an even number of terms (including the unknown value!), then you also have to think about the possibility that the median is halfway between the unknown and the value right above or below it.

Thank you!
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There is a set of numbers 2, 4, 6, 8,10 and x. What is the value of x?

(1) x is a 2-digit positive integer.
(2) The average of these integers is equal to the median of these integers.

(F1) Not sufficient
(F2) Mean = Median implies the terms are in AP.
This implies X = 0(Lower limit) OR X = 12(Upper Limit)
Combining 1 & 2
X is a two-digit positive integer, So X =12
C
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There is a set of numbers 2, 4, 6, 8,10 and x. What is the value of x?

(1) x is a 2-digit positive integer.

can be anything from 10 to 99. NS


(2) The average of these integers is equal to the median of these integers.

if x is 6, then median is 6, average is 6
if x is 0, then median is 5, average is 5

NS

combined:

A. 2 digit number
B. consecutive integers will result in the same median/mean.

Therefore x = 12

C
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