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by Night reader » Wed Dec 01, 2010 9:32 pm
If the range of the set containing the numbers x, y, and z is 8, what is the value of the smallest number in the set?

(1) The average of the set containing the numbers x, y, z, and 8 is 12.5.
(2) The mean and the median of the set containing the numbers x, y, and z are equal.
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by gmatmachoman » Wed Dec 01, 2010 10:08 pm
Night reader wrote:If the range of the set containing the numbers x, y, and z is 8, what is the value of the smallest number in the set?

(1) The average of the set containing the numbers x, y, z, and 8 is 12.5.
(2) The mean and the median of the set containing the numbers x, y, and z are equal.
Bit tricky :

From stem : Range = 8

St 1:

X+Y+Z+8 = 50

X+Y+Z = 42

Not sufficient
St 2 : Mean = Median

Combining st 1 & st2 & the stem data that range is 8 we need to figure couple of combination where X+Y+Z= 42.

One combination is (10,14,18)

So smallest number is 10.


Pick C
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by Rahul@gurome » Wed Dec 01, 2010 10:19 pm
Solution:
Let us assume that x is the smallest value and z is the largest value.
So z - x = 8.
We need to determine the value of x.
Let us first consider statement (1) alone;
So (x+y+z+8)/4 = 12.5
Or (x+y+z+8) =50.
Or x+y+z = 42.
Now the above information is not sufficient to give the value of x.
So (1) alone is not sufficient.
Next consider statement (2) alone.
Now the median is y.
So (x+y+z)/3 = y.
Or x+z = 2y.
Even this is not enough to give the value of x.
So (2) alone is not sufficient.
Next combine both the statements together and check.
On combining we get that z-x = 8, x+y+z = 42 and x+z = 2y.
On solving we get x = 10.
So both statements together are sufficient to answer the question.
The correct answer is (C).
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by Testluv » Wed Dec 01, 2010 11:45 pm
gmatmachoman wrote:
Night reader wrote:If the range of the set containing the numbers x, y, and z is 8, what is the value of the smallest number in the set?

(1) The average of the set containing the numbers x, y, z, and 8 is 12.5.
(2) The mean and the median of the set containing the numbers x, y, and z are equal.
Bit tricky :

From stem : Range = 8

St 1:

X+Y+Z+8 = 50

X+Y+Z = 42

Not sufficient
St 2 : Mean = Median

Combining st 1 & st2 & the stem data that range is 8 we need to figure couple of combination where X+Y+Z= 42.

One combination is (10,14,18)

So smallest number is 10.


Pick C
Fantastic approach, Govi. When mean=median, all the numbers in the set are equally spaced. So, to find the only combination that worked, we could have just divided 42 by 3, giving us 14. 14 is the middle number. We then add 4 to it, and subtract 4 from it to find the two other numbers which will be equidstant from 14, and whose difference will be 8.
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by goyalsau » Thu Dec 02, 2010 3:59 am
Night reader wrote:If the range of the set containing the numbers x, y, and z is 8, what is the value of the smallest number in the set?

(1) The average of the set containing the numbers x, y, z, and 8 is 12.5.
(2) The mean and the median of the set containing the numbers x, y, and z are equal.
Guys, Please Correct what i am doing Wrong over here,

Set is x , y , z & 8
12.5 *4 = 50

Sum of x + y + z + 8 = 50

We know the sum of Four Numbers should be 50, and we know one Number is 8

Lets Consider 8 as the Smallest Number Then the Highest Number must be 16 { AS we know Range is 8 }

8 , ____, ____ 16 8 + 16 = 24
Means remaining Two Terms must add to 26 , It is possible In many Cases as 13,13 and 12,14, and 11, 15 But in all the cases Smallest Number will always be 8

Lets assume that the smallest number is less than 8 May be 7

7 , 8 , ______, 15 { Range is 8 so the largest number is 15 , Sum of 7 + 8 + 15 = 30 , Only one Place is Left that and 20 can not come at that place because range is 8 }

So the smallest Number Has to be 8

Please Correct me what i am doing wrong over here...

According to me it should be A,
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by Night reader » Thu Dec 02, 2010 5:18 am
simply:
the range of the set containing the numbers x, y, and z is 8

only three numbers in the set, you take four numbers ...
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by Rezinka » Tue Dec 14, 2010 9:35 pm
The question says :

If the range of the set containing the numbers x, y, and z is 8, what is the value of the smallest number in the set?

Whay are we asuming that these are the only numbers in the set? x,y,z could be 3 numbers from a set of 8 numbers.
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by HPengineer » Wed Dec 15, 2010 12:28 am
I was fooled as well thinking there are more numbers in the set then just x y an z...
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by thebigkats » Thu Dec 16, 2010 4:27 pm
HI:
I came to the solution using same method as another posting:
If the range of the set containing the numbers x, y, and z is 8, what is the value of the smallest number in the set?

(1) The average of the set containing the numbers x, y, z, and 8 is 12.5.
(2) The mean and the median of the set containing the numbers x, y, and z are equal.

Prob - range of set is 8 meaning that largest - smallest = 8. Assuming x,y,z in that order - z-x = 8

STATEMENT 1:
Average of x,y,z,8=12.5 meaning that (x+y+z+8)/4 = 12.5 ==> x+y+z=42
Using above inference from problem - x+y+(x+8)=42 ==> 2x+y = 34
Now x and y are within range of 8. So there can only be so many values satisfying the equations.
If we plug in x = 8 then z=16 and y = 18 - doesn't work (y must be smaller than 16 per our assumption. So we need to raise x)
If we plug in x = 9 then z = 9+8= 17 ==> y = 16
If we plug in x=10 then z= 10+8=18 and y = 14
If we plug in x= 11 then z= 11+8=19 and y = 12
If we plug in x=12 then y=10 which is incorrect because y must be > than x per our assumption

So x,y,z can be 9,16,17 _or_ 10,14,18 _or_ 11,12,19

STATEMENT 2:
Mean = median = 14. So set #2 is correct and smallest value is 10

Of course I liked the other person;s solution better where poster directly got to mean of 14 (42/3) and then equally spaced with range of 8 meaning 10 and 19 as other numbers :-) - although this rule only applies to sets with odd no of members
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by prachich1987 » Thu Dec 16, 2010 8:42 pm
Night reader wrote:If the range of the set containing the numbers x, y, and z is 8, what is the value of the smallest number in the set?

(1) The average of the set containing the numbers x, y, z, and 8 is 12.5.
(2) The mean and the median of the set containing the numbers x, y, and z are equal.
I would go for C
What is the OA?
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by Zerks87 » Thu Dec 30, 2010 10:35 am
We know that the range of set {x,y,z} = 8

We want the smallest value in the set {x,y,z}

(1): It says the avg. of set {x,y,z,8}= 12.5, therefore (x+y+z+8)/4 = 12.5
I think those who answered A confused the set in the stem with this one. Since we have no other information though it is impossible to know what the range is and therefore find the smallest value. INSUFF.

(2): says the median and mean of {x,y,z} are equal but this does not help us to find what the smallest value would be. since 9+1/2 = median/mean of 5 and 10+2/2 = 6. Both those possibilities have ranges of 8 and the median could equal the mean. INSUFF

(1) + (2): Since we know that the mean and median are equal and the avg. of {x,y,z,8} = 12.5, we can extrapolate that since x+y+z+8 = 50 that x+y+z = 42, since 42/3= 14 and we know the median and mean are equal that means the lowest number must be 10 and highest 18, which maintains the range of 8. SUFF
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by samrendra4u » Sat Jan 22, 2011 7:50 pm
Night reader wrote:If the range of the set containing the numbers x, y, and z is 8, what is the value of the smallest number in the set?

(1) The average of the set containing the numbers x, y, z, and 8 is 12.5.
(2) The mean and the median of the set containing the numbers x, y, and z are equal.
Let x<y<z
then z -x =8(a)

(1) alone
x+y+z =42(b)
insufficient
(2) alone
mean & median are equal implies the set is equispaced.
y= x+d; z= x +2d
putting above in(a) ,we get d=4
insufficient

combining both
putting value of d in (b),we get x=10


So ans is c.Both combined.
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by del@btg » Wed Apr 27, 2011 9:44 pm
Same thing struck to me....the question never limits the set to x,y & Z only.....
Experts opinion please....
Rezinka wrote:The question says :

If the range of the set containing the numbers x, y, and z is 8, what is the value of the smallest number in the set?

Whay are we asuming that these are the only numbers in the set? x,y,z could be 3 numbers from a set of 8 numbers.
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by ranjithreddy.k9 » Sat May 28, 2011 6:33 pm
My pick option C
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by saurabh2525_gupta » Sun Aug 21, 2011 6:56 am
St 1 : This gives us the information x+y+z+8/4 = 12.5, so x+y+z = 42.
Also the range of the set = 8.
Lets randomly take x as largest and z be smallest.
This means x-z = 8.
Three variable and two equations, cannot be solved. Also no other information is provided wherein we can plugin values and solve for the variables

So, Insufficient.

St 2 : Mean = Median, the numbers in the set are equally spaced.
So the middle number "y" is the average. The numbers x and z are at equal distance from it. Since the range is 8, the largest number x = y+4
and the smallest number z = y-4.

So, Insufficient

Taking st 1 and st 2 together, we get x+y+z = 42, x= y+4 and z =y-4. We can derive the three values.

So answer is C.

Alternate explanation: Since the set is equally spaced mean = (x+y+z)/3 = 14 => y = 14
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