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DS - Range Mean Median

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DS - Range Mean Median

by karthikpandian19 » Thu Jun 28, 2012 1:30 pm
S = {x, y, 2, 2, 6}

If all the members of Set S above are integers between 0 and 9, inclusive, and x ≤ y, what is the value of y?

The range of the set is 5 greater than its least element.
The average (arithmetic mean) of Set S is 2.2 greater than the median.


*****couldnt able to proceed as do not know how to make the initial prompt*****
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by Anurag@Gurome » Thu Jun 28, 2012 2:33 pm
karthikpandian19 wrote:S = {x, y, 2, 2, 6}

If all the members of Set S above are integers between 0 and 9, inclusive, and x ≤ y, what is the value of y?

(1) The range of the set is 5 greater than its least element.
(2) The average (arithmetic mean) of Set S is 2.2 greater than the median.
Statement 1: (6 - 2) = 4 is not equal to (5 + 2) = 7.
Hence, either 2 is not the least element of S or 6 is not the greatest element of S.
We have following scenarios
  • 1. x ≤ y ≤ 2 < 6
    6 is the largest element and x is the least element.
    Hence, (6 - x) = (x + 5) --> x = 1/2 --> Not possible as x is not an integer

    2. x ≤ 2 < 6 ≤ y
    y is the largest element and x is the least element.
    Hence, (y - x) = (x + 5) --> x = (y - 5)/2
    As x must be integer, (y - 5) must be even.
    Hence, y must be odd.
    Possible values of y are 7 and 9 as we assumed y > 6.

    There are few more scenarios possible but as we have already found two possible values of y, we can say that this statement is not sufficient
Not sufficient

Statement 2: If both or either of x or y are less than or equal to 2, then median of S will be 2. Hence, average of S is (2 + 2.2) = 4.2 --> Sum of all elements = 5*(4.2) = 21
Hence, (x + y) = (21 - 6 - 2 - 2) = 11
Hence, possible pairs of x and y is (2 and 9) as other possible pairs will result in both x and y greater than 2.

And if both of x and y are greater than 2, then median of S will be x.
Hence, average of S = (x + 2.2)
Hence, (x + y) = 5*(x + 2.2) - (6 + 2 + 2) = (5x + 11 - 10) = (5x + 1)
Hence, y = (4x + 1)
Possible pairs of x and y are (1 and 5) and (2 and 9). But in both cases x is less than or equal to 2 which is contradictory to our assumption.

Hence, only possible value of y is 9.

Sufficient

The correct answer is B.
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by karthikpandian19 » Thu Jun 28, 2012 6:00 pm
statement 1 explanation is difficult to explain
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by eagleeye » Thu Jun 28, 2012 7:12 pm
karthikpandian19 wrote:statement 1 explanation is difficult to explain
This is how I did it. May be this can help.

S = {x, y, 2, 2, 6}

We are told that all members are between 0 and 9.So,
x and y both lie between 0 and 9.
Also we are told that x<=y.

We need to find the value of y.

With that in mind let's look at the statements.

(1) The range of the set is 5 greater than its least element.
Here, if x=1,y=1 we have the range condition satisfied with elements 1,1,2,2,6 and 6-1=5.
Also, if x=1, y=2 we still have the range condition satisfied. 1,2,2,2,6 and 6-1 = 5.
Since we don't know if y is 1 or 2, this statement is insufficient.

(2) The average (arithmetic mean) of Set S is 2.2 greater than the median.
Let the median integer be a. Then we are given that:
(x+y+2+2+6)/5 = a + 2.2 => x+y+10 = 5a+11 => x+y = 5a+1.

Let's focus on the statement above with the original statements (ones that I bold-ed before) in mind.
1. We see that x+y can't be greater than 18. (Since the maximum value of x, y is 9 each).
2. Since a is the median of 5 digits, it must be one of the digits, since two 2s are already present, a must be greater than or equal to 2.

So we have x+y = 5a+1 (Where a is the median)
Case 1: a=2, In this case: x+y = 11. So (x,y) combination can be (2,9) , (3,8), (4,7) or (5,6).

For x=2, y=9 we have the set as 2,2,2,6,9. where median is 2 indeed. So this works.
For x=3, y=8 we have the set as 2,2,3,6,8. But wait, our median was 2. so this is not possible.
In the same way x=4, x=5 will give median as 4, 5 which contradicts our assumption of a=2.
So far, we only have x=2, y=9 as the possible value.

Case 2: a=3. Here x+y = 5a+1 = 16. (Where a = 3 is the median).
Here, x+y = 16. The only combinations that work are (x,y) = (7,9) or (8,8)
But we reject both of them with the same reasoning as we did in case 1 because in both cases, x turns out to be the median, when we had assumed 3 as the median.

Case 3: a=4, Here x+y = 5*4+1 = 21. But wait, the maximum value of x+y is 18. Hence all values of a greater than 3 are rejected.

So we are left with only 1 valid case where x=2, y=9. Hence this statement (even though a bit convoluted) is sufficient.

Hence, B is the answer.

Let me know if this helps :)
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by Anurag@Gurome » Thu Jun 28, 2012 7:22 pm
eagleeye wrote:(1) The range of the set is 5 greater than its least element.
Here, if x=1,y=1 we have the range condition satisfied with elements 1,1,2,2,6 and 6-1=5.
Also, if x=1, y=2 we still have the range condition satisfied. 1,2,2,2,6 and 6-1 = 5.
You are misinterpreting the statement.

The statement says, range is 5 greater than the least element.
So, (greatest - least) = range = (least + 5)
--> greatest = 2*least + 5

You are interpreting the statement as range = 5
karthikpandian19 wrote:statement 1 explanation is difficult to explain
Consider the following cases,
  • x = 1, y = 7 --> Range = (7 - 1) = 6 = (1 + 5)
    x = 2, y = 9 --> Range = (9 - 2) = 7 = (2 + 5)
Hence, y may be 7 0r 9 --> Not sufficient

In my first post I just explained how I reached my conclusion that these two sets of values are possible for x and y.
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by eagleeye » Thu Jun 28, 2012 8:01 pm
Anurag@Gurome wrote:
eagleeye wrote:(1) The range of the set is 5 greater than its least element.
Here, if x=1,y=1 we have the range condition satisfied with elements 1,1,2,2,6 and 6-1=5.
Also, if x=1, y=2 we still have the range condition satisfied. 1,2,2,2,6 and 6-1 = 5.
You are misinterpreting the statement.

The statement says, range is 5 greater than the least element.
So, (greatest - least) = range = (least + 5)
--> greatest = 2*least + 5

You are interpreting the statement as range = 5
You know what, I did misinterpret. It took a while before it dawned on me.

Thanks Anurag :)
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by karthikpandian19 » Thu Jun 28, 2012 8:16 pm
OA is B
karthikpandian19 wrote:S = {x, y, 2, 2, 6}

If all the members of Set S above are integers between 0 and 9, inclusive, and x ≤ y, what is the value of y?

The range of the set is 5 greater than its least element.
The average (arithmetic mean) of Set S is 2.2 greater than the median.


*****couldnt able to proceed as do not know how to make the initial prompt*****
Regards,
Karthik
The source of the questions that i post from JUNE 2013 is from KNEWTON

---If you find my post useful, click "Thank" :) :)---
---Never stop until cracking GMAT---
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