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by Stuart@KaplanGMAT » Tue May 01, 2012 9:54 am
Hi!

You seem to have misinterpreted the question as "what is the smallest possible number in the set?", when in fact it's simply "what is the smallest number in the set?"

With (1) alone, there's no way to determine the actual value of x, y or z, so (1) is insufficient.

Stuart
shantanu86 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.
This is a great question.. But contrary to popular opinion here, I think the answer is [A]
Lets analyze..

(1) (x+y+z+8) = 4*12.5
=> average of x,y and z is 14

So one of the solution set which satisfies (1) is
(18,14,10)
Now to minimize the smallest number I decrease minimum and balance other two for mean to be 14

(17,16,9).. integral solution with 9 as smallest

(16.66,16.66, 8.66) .. non-integral solution with 8.66 as smallest

Hence (1) alone is sufficient and obviously (2) alone is not sufficient.
Therefore the correct answer is [A].

Hope it helps!!
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by miteshpant » Wed Jul 18, 2012 11:28 pm
Hi,

No where its said that the numbers are integers. So i think it must be E
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by optimist » Fri Jul 20, 2012 10:34 am
miteshpant wrote:Hi,

No where its said that the numbers are integers. So i think it must be E
Here, when you combine stmt 1 and stmt 2, you get:
x+y+z=42
assuming 'y' as the median, we get
x+z=42-14=28 ---1
and we know, x-z = 8 ---2

solving 1 and 2 we get x=18 and z=10

So, I don't think there is any question of the numbers being non integers..

Pls let me know if I have missed anything here

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by ashg84 » Tue Oct 16, 2012 1:16 am
Hi,

I agree with you.. The question is not asking for smallest possible but the smallest number in the set. Two things are entirely different. In my views the answer should be E as we can not find the smallest number with these two statement. Please suggest if my interpretation of the question is wrong.

What is the OA.

Thanks

Ashish Gupta \
Stuart Kovinsky wrote:Hi!

You seem to have misinterpreted the question as "what is the smallest possible number in the set?", when in fact it's simply "what is the smallest number in the set?"

With (1) alone, there's no way to determine the actual value of x, y or z, so (1) is insufficient.

Stuart
shantanu86 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.
This is a great question.. But contrary to popular opinion here, I think the answer is [A]
Lets analyze..

(1) (x+y+z+8) = 4*12.5
=> average of x,y and z is 14

So one of the solution set which satisfies (1) is
(18,14,10)
Now to minimize the smallest number I decrease minimum and balance other two for mean to be 14

(17,16,9).. integral solution with 9 as smallest

(16.66,16.66, 8.66) .. non-integral solution with 8.66 as smallest

Hence (1) alone is sufficient and obviously (2) alone is not sufficient.
Therefore the correct answer is [A].

Hope it helps!!
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by ashg84 » Tue Oct 16, 2012 1:23 am
My mistake, answer should be C.
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by rajeshsinghgmat » Wed Feb 20, 2013 1:45 am
C in answer.
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by Java_85 » Sat Sep 14, 2013 7:26 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.
A really nice question, IMO C because-->

(1) x+y+z+8=50 --> x+y+z=42
(2) 3y=42 --> y is 14 --> if 14 is the middle number, if the range want to be 8 then x y z should be 10 14 18 --> smallest number= 10
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by Vikramjeet Sharma » Sun Oct 12, 2014 5:26 am
Doubt:

Lot of people have mentioned that if Mean = Median for a given set then the set is evenly spaced. Here are two possibilities which satisfy mean = median condition:

First Set: 1,2,3
Second Set: 5,5,5

Now, does second set qualify for an evenly spaced set?
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by Brent@GMATPrepNow » Sun Oct 12, 2014 6:05 am
Vikramjeet Sharma wrote:Doubt:

Lot of people have mentioned that if Mean = Median for a given set then the set is evenly spaced. Here are two possibilities which satisfy mean = median condition:

First Set: 1,2,3
Second Set: 5,5,5

Now, does second set qualify for an evenly spaced set?
Yes, they both qualify as being equally/evenly spaced.
In the first set, each value is 1 greater than the term before it.
In the second set, each value is 0 greater than the term before it.

Also notice that, in the second set, the median = mean = 5

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by GMATGuruNY » Sun Oct 12, 2014 6:21 am
It should noted that the mean will be equal to the median for any set that is SYMMETRICAL about the median.

Examples:
10, 10, 10, 10, 10, 10, 10
7, 8, 9, 10, 11, 12, 13
4, 6, 8, 10, 12, 14, 16
5, 6, 9, 10, 11, 14, 15
1, 2, 5, 10, 15, 18, 19

All of the sets above are symmetrical about the median of 10.
In every case, mean = median = 10.
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by jaspreetsra » Mon Nov 03, 2014 3:01 pm
took more than 3 mins.:(
My answer is C.
Explanation:
x, y, and z
Let z be the largest and x be the smallest number.
z-x = 8
1)(x+y+z+8)/4 =12.5
So x+y+z = 12.5*4 - 8 =42 Not Sufficient
2)mean = median Not sufficient
From 1 and 2
mean = 42/3 =14
median 14
so x,y,z ---> x,14,z
x+14+z =42

z+x = 28 (1)
Also z-x =8 (range)(2)
Solve for z, and then for x.
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by nikhilgmat31 » Sun Jul 05, 2015 11:03 pm
Why we didn't considered negative numbers
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by Brent@GMATPrepNow » Mon Jul 06, 2015 5:44 am
nikhilgmat31 wrote:Why we didn't considered negative numbers
It is given that the RANGE = 8
Statement 1 tells us that the AVERAGE = 12.5

If the average is 12.5 and the range is 8, none of the numbers can be negative.

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by mahesham » Thu Jul 23, 2015 1:58 am
thebigkats wrote: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
Thanks for the explanation!!
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by Max@Math Revolution » Fri Jul 31, 2015 10:55 am
In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem.

Remember equal number of variables and equations ensures a solution.

In original condition, we have 3 variables (x,y,z) and 1 equation (range=8), then we need 2 more equations to solve the problem. So, the answer is C.
Why C? If you know our logic, you don't need to solve the problem but let me prove for you.

In (1) and (2), x+y+z+8=12.5*4=50. If median and mean are the same, x=y-4, z=y+4 (x<y<z). x+y+z=42 so mean=y=14.
Minimum value is 14-4=10. So, 10,14,18.

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