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
