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

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At a hospital, babies are born every day for a certain number of days.  [#permalink]

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09 Mar 2017, 22:39
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52% (02:33) correct 48% (02:29) wrong based on 126 sessions

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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 median 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.

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At a hospital, babies are born every day for a certain number of days.  [#permalink]

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Updated on: 15 Mar 2017, 09:02
Hi,
thanks for posting!
Legend:
#b = number of born (number of people that were born on a given day);
S={#b} = set of all the #b measured (S contains 'n' numbers of born);
med(S)= median of the set S.

Problem:
every day: #b born, for n days; #b>6 on 20% of n days; is med(S={#b})<4?

(1) 75% of 80% of n days, #b<4;
(80% =percent of the days with less than 6 born=(1-20%))

(2) 50% of (# of days where #b>4, the #b was greater than 6); This means that 20% of the n days #b>6 and thanks to statement 2, 20% of the n days #b>4.

Now, to solve this I assumed n=20days (since we have 3/4 and x/5 fractions).

Statement (1):
3/4*(4/5)*(n) --> #b<4, hence for 12 days the #b is less than 4, which means that the median of the set S (set of the #b) is less than 4. Sufficient.

Statement (2):
if half of the time that #b>4 #b is actually greater that 6, then 20% of the n days #b>6 and 20% of n days #b>4;
20% of 20 = 4 days --> 4 days #b>6 and 4 days #b>4. Hence, on the other days, #b will be less then 4, which means that the median is less than 4. Sufficient.

I hope I haven't made any mistakes in my reasoning; this was the only way I could think about the solution.

Originally posted by carlopanieri on 13 Mar 2017, 04:01.
Last edited by carlopanieri on 15 Mar 2017, 09:02, edited 2 times in total.
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Re: At a hospital, babies are born every day for a certain number of days.  [#permalink]

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14 Mar 2017, 03:30
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Re: At a hospital, babies are born every day for a certain number of days.  [#permalink]

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14 Mar 2017, 05:45
1
ziyuen wrote:
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 median 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.

OFFICIAL SOLUTION

By solving con 1) & con 2) together,

Median number of babies born (about 50%)

shown above, con 1) = con 2). In other words,

For con 1), on 80% of the total days, less than 6 babies were born each day, and on 75% of that days, less than 4 babies were born each day, thus it becomes 80%(75%)=60%, and therefore on 60% of the total days, less than 4 babies were born each day.

For con 2), on 50% of the days that 4 or more babies were born each day, 6 or more babies were born each day, thus it becomes 4 or more ~ less than 6 = 6 or more babies, and therefore you get 20% each as shown above. Then, the median number of the babies born is always about 50% and according to the figure above, it is always less than 4, hence yes, it is sufficient. Therefore, the answer is D.
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Re: At a hospital, babies are born every day for a certain number of days.  [#permalink]

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14 Mar 2017, 05:49
abhisheksinha717 wrote:

Dear MathRevolution, Could you help to resolve?
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At a hospital, babies are born every day for a certain number of days.  [#permalink]

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28 Apr 2017, 01:31
ziyuen wrote:
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 median 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.

ziyuen
statement-1: % of days when #babies born was less than 4 = (80%*75%) = 60% of total time. So median < 4. Sufficient
statement-2: % of days when 4 or more were born = 2*20% = 40% of total time. So 60% of the time less than 4 were born. Hence median < 4. Sufficient
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Re: At a hospital, babies are born every day for a certain number of days.  [#permalink]

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28 Apr 2017, 06:32
1
1
ziyuen wrote:
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 median 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.

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.

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Re: At a hospital, babies are born every day for a certain number of days.  [#permalink]

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28 Apr 2017, 08:31
VeritasPrepKarishma wrote:
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.

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.
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Re: At a hospital, babies are born every day for a certain number of days.  [#permalink]

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29 Apr 2017, 02:03
2
mbsingh wrote:
VeritasPrepKarishma wrote:
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.

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.
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Re: At a hospital, babies are born every day for a certain number of days.  [#permalink]

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29 Apr 2017, 09:49
VeritasPrepKarishma wrote:
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.
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Re: At a hospital, babies are born every day for a certain number of days.  [#permalink]

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29 Apr 2017, 20:38
3
mbsingh wrote:
VeritasPrepKarishma wrote:
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)
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Re: At a hospital, babies are born every day for a certain number of days.  [#permalink]

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30 Apr 2017, 20:28
Thanks VeritasPrepKarishma . That's really helpful
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