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Challenge-statistics
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]
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]
Another way: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.
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"
_________________
Samuel
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.
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.
-
vaibhavgupta
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IMO CNight 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.
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!! :/
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!! :/
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.
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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ronnie1985
- Legendary Member
- Posts: 626
- Location: Ahmedabad
- Thanked: 31 times
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.
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.
Follow your passion, Success as perceived by others shall follow you
-
ronnie1985
- Legendary Member
- Posts: 626
- Location: Ahmedabad
- Thanked: 31 times
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Mayankapoor
- Newbie | Next Rank: 10 Posts
- Posts: 2
- Location: India
-
ronnie1985
- Legendary Member
- Posts: 626
- Location: Ahmedabad
- Thanked: 31 times
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)
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)
Follow your passion, Success as perceived by others shall follow you
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?
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?
Its do or die this time!
Practise, practise and practise.
Practise, practise and practise.
This is a great question.. But contrary to popular opinion here, I think the answer is [A]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.
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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