If the average of four distinct positive integers is 60, how many inte

given
a+b+c+d= 240

#1
median of three largest no 51
and two largest integer sum = 190
so
b+c+d /3 = 51= b+c+c = 153
and c+d= 190
b = 190-153 ; 37

a+b+37+190 = a+b +227 = 240
a+b = 13
so a,b & c are <50
sufficient

#2
The median of the four integers is 50.
median = 50 ; b+c/2
two no are less than 50 and two over 50
IMO D

If the average of four distinct positive integers is 60, how many integers of these four are less than 50?

Let the numbers be a, b, c and d in increasing order

(1) The median of the three largest integers is 51 and the sum of two largest integers is 190.
So, median of b, c, d is 51 and hence c is 51. Now c+d=190, so a+b=240-190=50.
Thus each of a and b have to be less than 50 as a and b are positive integers.
Sufficient

(2) The median of the four integers is 50.
Note that Median is of even number of elements
So median is the middle of b and c, that is \(\frac{b+c}{2}=50\)..
As both b and c are distinct, b<50 and c>50..
Thus two numbers are less than 50.
Sufficient

D

Key word: four distinct positive integers

From 1
When median of 3 largest numbers is 51, this means that the series is
x y 51 139

The sum of all 4 numbers will be 240, Now when we subtract 190, will give value as 50

so we get 2 numbers less than 50

From 2, lets look at our key word

since 50 is the median this means, center term, when n is even, median = (2nd term + 3rd term)/ 2

so if median has to be 50, which we can get in many ways such as, 15+85, 45+55

So in all those cases we will get two values, Sufficient

D

Statement 2 is not sufficient...
Here is an example where the median is 50

x1=50
x2=50
x3=50
x4= 90

here the average is 60, the median is 50, and there are no numbers less than 50...
please correct me if I am wrong.

1) The integers can be 1,49, 51, 139 . There are no other possible set. Sufficient.

2) Since the median of even integers is the average of the middle two integers and here, all the integers are distinct, so we can assume that the first two integers are less than 50 and last two integers are more than 50. The only possible set is the one shown under statement 1. Sufficient.

D is the answer.

In my opinion this example you've given wouldn't be an accurate example for this question, because it specifically says "four distinct positive integers."

Takeaways:

I could easily understand that A was sufficient. (please refer to Bunuel's official explanation, that's the best, as always.)

The trick was with the second option (at least I felt that, as I took a bit longer than usual to mark the correct answer. I was almost about to mark A as the correct answer).

Note the term: "Distinct positive integers" (This is always something I happen to miss out, initially, especially when moving on from one statement to the other, in a data sufficiency question.)
Each number must be different.

Let's talk about the second statement:
we know a+b+c+d = 240

Now, we also know, the mean of the 4 numbers = 50.
That can mean:
(b+c)/2 = 50
or, (b+c) = 100
again, note the word: "distinct integers". As per the question, we should look at any other scenarios other than b = c.
Which means, b must be 49, or any number less than 49.
and c must be 51 or any number more than 51.
again, we must also note that a+d = 140.
However, we don't need to do too much of calculations.
We can simply take a to be 40, b to be 49 (max possible number for b), 51(least possible number for c), and 100 (d).

possible combinations:
40, 49, 51, 100 (average = 60, median = 50)
41, 48, 52, 99 (average = 60, median = 50)
35, 40, 60, 105 (average = 60, median = 50)
1, 2, 98, 139 (average = 60, median = 50)
and so on... Note that in each case, we can see that there are only 2 numbers that can be less than 50.

Also, remember: while calculating median, the numbers must be arranged in ascending order.

So, we have 2 numbers less than 50. (which is sufficient)

So, 1 and 2 both are sufficient, independently.
Correct answer: D

Let me know if there are any discrepancies in my explanation. (This problem looked really interesting to me, so I thought of explaining it here)