At a hospital, babies are born every day for a certain number of days.

For 20% of the days, 6 or more babies were born. So for 80% of the days, 1/2/3/4/5 babies were born.
Say there were 100 days. 6 or more babies were born for the last 20 days. Then in the first 80 days, less than 6 babies were born each day.
Is median less than 4?
Is the average of 50th and 51st term less than 4?

1) On 75% of the days that less than 6 babies were born, the number of babies born each day was less than 4.
75% of those 80 days when less than 6 babies were born, i.e. on 60 days, less than 4 babies were born. So the 50th and 51st terms will be less than 4. The median will be less than 4.
Sufficient.

2) On 50% of the days that 4 or more babies were born, the number of babies born each day was 6 or more.
On 20 days, 6 or more babies were born. This constitutes 50% of the days on which 4 or more babies were born. So on 20 days, 4 or 5 babies were born.
Hence on 60 days, less than 4 babies were born. So the 50th and 51st terms will be less than 4. The median will be less than 4.
Sufficient.

Answer (D)

Karishma,

If the question was asking about average and not median would the answer be C ? I read the question wrong and solved it for average and not median and now wondering if i did the math right or not.

Please specify exactly how you interpreted the question and solved it. Note that the average depends on the exact value of each element in the set. Just saying 50% are 4 or less would not be sufficient. You would need to know how many are 4, how many are 3, how many are 2 and how many are 1.

I read the question as -

At a hospital, babies are born every day for a certain number of days. If 6 or more babies were born for 20% of the total number of days, is the average number of babies born less than 4?

1) On 75% of the days that less than 6 babies were born, the number of babies born each day was less than 4.

2) On 50% of the days that 4 or more babies were born, the number of babies born each day was 6 or more.

Say there are 100 days. 6 or more babies were born on 20 days. Note that the number of babies born on these 20 days could be any number greater than 6 such as 20 or 50 or 120 etc. The minimum number of babies on these 20 days would be 120. There is no limit to the maximum number.

1) On 75% of the days that less than 6 babies were born, the number of babies born each day was less than 4.
On 80 days, less than 6 babies were born. Of these, 75% is 60 days. On 60 days, less than 4 babies were born. So on 60 days, you have minimum 60 babies born and maximum 180 babies born.
On the leftover 20 days, 4 or 5 babies were born so 80 or 100 babies.
The minimum average is (120 + 60 + 80)/ 100 = 2.6
The maximum average could be anything.
Not sufficient.

2) On 50% of the days that 4 or more babies were born, the number of babies born each day was 6 or more.
The 20 days when 6 or more babies were born make up 50% of the days when 4 or more babies were born. So for 20 days, 4 or 5 babies were born i.e. 80 or 100 babies
For 60 days, 1/2/3 babies were born. So on 60 days, you have minimum 60 babies born and maximum 180 babies born.
The minimum average is (120 + 60 + 80)/ 100 = 2.6
The maximum average could be anything.
Not sufficient.

Note that both statements give you the same information. So if they are not sufficient independently, they are not sufficient together.
Answer of this modified question would be (E)

median is at 50% days
1/5 total or 20% days had 6+

1) On 75% of the days that less than 6 babies were born, the number of babies born each day was less than 4. sufic

6+ = 20% days; so 80% days left * 75% had < 4 babies = 4/5*3/4 = 60% days;
since median is at 50% days, then its in the range of < 4 babies.

2) On 50% of the days that 4 or more babies were born, the number of babies born each day was 6 or more. sufic

6+ babies = 20% total days = 50% days that 4+ babies were born;
so, 4+ babies represents 20%*2=40% total days;
thus, remaining 60% days there were < 4 babies.

Ans (D)

Let the total number of days = 100
In ascending order, let the 100 daily birth rates be represented as follows:
1<-----------------------------------------------------80--81----->100
Daily birth rate of 6 or more = birth rates for 20 of the 100 days = blue portion above
Daily birth rate of less than 6 = birth rates for the remaining 80 days = red portion above

Statement 1:
Daily birth rate of less than 4 = 75% of the red portion above = green portion below:
1<----------------------------------------60--61-----80--81----->100
Since the median birth rate is contained within the green portion -- which represents a birth rate of less than 4 -- the answer to the question stem is YES.
SUFFICIENT.

Statement 2:
In the number line for Statement 1, notice the following:
green portion = daily birth rate of less than 4
red portion + blue portion = daily birth rate of 4 or more
blue portion = daily birth rate of 6 or more
The blue portion constitutes 50% of the daily birth rates of 4 or more, as required by Statement 2.
Implication:
Statement 2 implies the same number line as Statement 1.
Thus, the answer to the question stem is YES.
SUFFICIENT.

I'm unable to understand what St. 2 means- can anybody help to decode it?