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by GmatKiss » Mon Aug 22, 2011 2:23 am
IMO:C

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by AbhiJ » Mon Sep 05, 2011 10:17 am
good question

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by saketk » Sat Sep 24, 2011 12:53 pm
Statement 2 tells us that mean=median. This implies that the numbers are equally spaced. We can't do much with this information alone. INSUFFICIENT

Statement 1 says -The average of the set containing the numbers x, y, z, and 8 is 12.5 i.e. x+y+z=42

also from the question we know that the range of the set containing the numbers x, y, and z is 8.

i.e. difference between largest and smallest number is 8. let x be the smallest number and Z be the largest number

X=a
y=a+d
z=a+2d ... where d is the common difference

we know that a+2d-a=8 or d=4

also we know that x+y+z=42
or 3a+3d=42
or a+d=14
or a=10

Therefore, the answer is C [both statement together are sufficient to answer]

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by leumas » Mon Oct 03, 2011 7:42 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.
Another way:

q: What is least value of 3 integer series with 3 numbers and range 8.

1. It says x+y+z = 42
We know range is 8 and Average is 14.
It can be 10+14+18 or 11+12+19, but we cannot arrive at one value for least number. (NS)

2. Now There's a rule:

For an evenly spaced set of integers, mean and median are equal if the total numbers are odd.

This implies - x+(x+4)+(x+8) = 42
Or 3x+12 = 42.

For each value of X, we can have the range 8 and three integers. Ex: 1,5,9 Or 2,6,10 etc., etc.,(NS)

Together:

3x+12 = 42 and this shows x=30/3 which is 10, the answer is 10 Sufficient!

Pick "C"

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by parul9 » Fri Oct 28, 2011 11:25 am
Given: Range (biggest number - smallest number) = 8

From 1 : (x+y+z+8)/4 = 12.5
=> x+y+z = 42.
We can't find the smallest no. from this.
Insufficient.

From 2 : mean and median is same.
This means that the numbers are evenly spaced.
Since the range is 8, the numbers can be written as - x-4, x, x+4.
But we still can't find the value of the smallest no.

Combining 1 and 2,
(x-4) + x + (x+4) = 42

One unknown variable in one eqn. We can get the answer.
So the answer should be C.

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by vaibhavgupta » Tue Nov 08, 2011 11:59 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.
IMO C

combining is the only way we could find out the value of the smallest :)
If OA is A, IMO B
If OA is B, IMO C
If OA is C, IMO D
If OA is D, IMO E
If OA is E, IMO A

FML!! :/

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by bpdulog » Wed Nov 09, 2011 12:04 pm
I chose C off the top of my head.
NO EXCUSES

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by olylo » Tue Nov 22, 2011 4:16 am
My answer is E

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by immaculatesahai » Wed Dec 14, 2011 2:40 am
quite straight forward ... ans is C.

From problem we know that largest- smallest= 8. Let x be the smallest number and z be the largest number. We need to find out value of x.

Statement 1: (x+y+z+8)/4 = 12.5
x+y+z= 42
Mean= 14

But, we can do anything further now.

Statement 2: Tells us that mean and median are equal. Hence we know that the set is an equally spaced set. i.e. z-y= y-x

Hence, z=y+4 and y=x+4 (because range is 8)

But we dont have any concrete value, by which we can find out the solution.

Combing 1 and 2.

We now know that value of y is 14. and z= 18 and x= 10.

C wins.

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by ronnie1985 » Sat Feb 04, 2012 5:54 am
Let the numbers be x, y and z in increasing order. x = min z = max
z-x = 8.
Q. x = ?
S1. x+y+z = 42 => 2z+y = 50
Not Sufficient
S2. y = (x+y+z)/3 => 2y = x+z
2z = 8+2y => z = 4+y
Comb. The numbers are in AP and d=4. the sum of numbers is also known hence can be solved for finding the first term of AP
Sufficient.
(C) is answer.
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by ronnie1985 » Fri Mar 30, 2012 10:04 am
(C) is ans
QED
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by Mayankapoor » Thu Apr 26, 2012 1:55 am
C is the answer as both statements are required to get the value.

What is OA

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by ronnie1985 » Thu Apr 26, 2012 10:21 am
Max = M and Min = m, let
R = M-m = 8 (Given)
S1. Avg = 12.5
Sum of 3 numbers is obtained. Cant say about minimum

S2. Mean = Median
Cant say which is max or min


Comb. Mean = Median = 12.5
Sum of 3 numbers = 37.5 and sum of extremes = 37.5-12.5 = 25

M+m = 25
M-m = 8 Solvable

Sufficient
(C)
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by spartacus1412 » Mon Apr 30, 2012 11:54 pm
from statment 1 x+y+z =42
since, range is 8
inorder to find the smallest possible number in the set. numbers should be of the a,a+8 and a+8, hence the range of 8 is maintained.

now x+y+z = 42.
therfoe, a+a+8+a+8 = 42
3a = 26 since 26 is not divisible by 3, the possible format of numbers is a, a+7,a+8

now, a+a+7+a+8= 42
3a = 27
a = 9
hence, numbers are 9, 16, 17.
smallest possible is 9.

so is'nt statement 1 alone sufficent?
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by shantanu86 » Tue May 01, 2012 6:23 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.
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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