x is an integer greater than 7. What is the median of the set of integ

X>7 and the set (let it be S) contains 1 to X inclusive.

I normally read both the statements and start with the easier one. In this case, it is statement 2.

Statement 2: If the range is 20. Then X must be 21. Since X is the last term. In this case, we can easily find the median. So Sufficient.

Statement 1: If we know the average of the set, we can find the sum and number of elements in the set. But this is a tedious process to jot down all the numbers and find the sum.

But we can use statement 2 as a clue to solve this easily without crushing our brain. Since two statements never contradicts each other. If we plugin X as 21. We get the sum as 231(using the sum of n numbers formula) and average as 11.

Bingo!!

We can find the median from this statement as well. So Sufficient.

Answer : D: Both statement Individually sufficient to solve.
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we can solve statement 1 in the following manner

since number are consecutive integers. therefore there sum from 1 to x can be given as (x(1+x))/2; where x is the last term

now average is =11
thus (x(1+x))/2) =11x
1+x=22
or x=21

Sorry to ask stupid question,

I answered E because the question is not explicitly telling that this set is consecutive. So, I assume it's not. But how do you know that it is??

We are told that the set consists of integers from 1 to x inclusive, so the set consists of consecutive integers. For example, if x=8, then set is {1, 2, 3, 4, 5, 6, 7, 8}.
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D.

1) Avg = 11
sum of consecutive integers from 1 to x (inclusive) = x(x+1)/2
avg = {x(x+1)/2*x} = 11
or x = 21
so median can be calculated. sufficient.

2) range = 20
x = 21
again media can be calculated. sufficient.
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For an evenly spaced out set, the mean == median.
here the set is space of consecutive integers.
Stmt1: Sufficient.
Stmt2: x-1= 20; x=21. Avg==mean =11. Sufficient.
Therefore D.
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Statement 1: AM(average ) of an arithmetic progression = (a+L) /2
a ------> first term of the series
L ------> last term of the series
This gives 11= (1+x)/2 ( since 1 is the first term and x is the last term of the series)
therefore x= 21
so sufficient to find the median

Statement 2: range is 20 therfore x=21
Sufficient

Answer: D

)Statement 1. If a set has equally distant elements i.e. if the set is in A.P., then the mean of the set is also the median of the set.
Here, we are given the mean of the set and we are told that the set is in A.P. So, the median would also be 11. Hence, Sufficient.
Statement 2. Range of a set = (Biggest element – smallest element)
We know that the smallest no is 1 and the range of the set is 20. So, the biggest number should be 21.
Now, We can easily find the median of the set consisting of integers from 1 to 21.
The median is 11. Hence, Sufficient.

Probability of picking 1 vowel from Set#1 = 2/5
Probability of not picking any vowel from Set#1 = 3/5
Probability of picking 1 vowel from Set#2 = 1/6
Probability of not picking any vowel from Set#2 = 5/6
Probability of picking atleast 1 vowel= P(2 vowels) + P(vowel from Set#1 & consonant from Set#2) + P(vowel from Set#2 & consonant from Set#1)
Probability of picking atleast 1 vowel = 2/5 x 1/6 + 2/5 x 5/6 + 3/5 x 1/6 = 15/30 = 1/2.

Can we always consider "the set consists of integers from 1 to x inclusive" as "a set of consecutive integers"? I really got stuck here. I thought the integers are not evenly spaced and can be repeated.

Can you suggest any material talk about this idea? Thank you.

Bunuel
should it be mentioned "the set consists of ALL integers from 1 to x inclusive" OR it's a standard definition of consecutive integers? can't it be such as "if x=8, then set is {1, 2, 3, 7, 8} .. so on ?

Yes, integers from 1 to x inclusive, include all integers from 1 to x inclusive.
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Given: x is an integer greater than 7.
Asked: What is the median of the set of integers from 1 to x inclusive?

(1) The average of the set of integers from 1 to x inclusive is 11.
Since the set consists of consecutive integers from 1 to x inclusive.
Average = Median = 11
SUFFICIENT

(2) The range of the set of integers from 1 to x inclusive is 20.
Range = Largest - Smallest = Largest - 1 = 20 ; Largest = 21
Median = (21+1)/2 = 11
SUFFICIENT

IMO D

Statement 1: Set = {1,2,3.......x-1,x}
if average is 11 for a set of consecutive integers starting from 1, then the max deviation from mean on the left side of mean = max deviation from mean on right side of mean; => 11-1 =10 is the max deviation on the left side which has to be balanced by 10 as on the right; so x-11=10 or x=21; so set is {1,2,3...21}; hence we can calculate median.

I is sufficient.

Statement 2: X=21; So clearly sufficient.

­Where is it mentioned that it is a set of consecutive numbers?

­Your doubt is addressed in this thread. Please review it carefully. Hope it helps.­
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