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Could someone assist with the explanation for this question? Didn't follow how to answer with the corrected problem statement 2

A furniture manufacturer produced at least 3 cabinets each day for the past 30 working days. What was the median number of cabinets produced daily by the manufacturer for the past 30 working days?

Given that at least 3 cabinets were produced daily, the minimum total production is 3 cabinets/day * 30 days = 90 cabinets. To find the median of an even number of terms (30 days in this case), we average the two middle terms once the terms are arranged in order.

(1) For the past 30 working days, the manufacturer produced at most 6 cabinets each day.

If 6 cabinets were produced each day, the median would be 6. However, if 3 cabinets were produced for 29 days and 6 cabinets on one day, the median would be 3.

Not sufficient.

(2) For the past 30 working days, the manufacturer produced a total of 104 cabinets.

We already know that the minimum production over this period, with at least 3 cabinets produced each day, is 90 cabinets. This implies that there are 14 additional cabinets to account for in the total production.

Initially, our set of daily productions looks like this:

    {3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3}

To find the median, we need to determine if adding the extra 14 cabinets can change the middle values (15th and 16th terms) in this set. Even if we distribute these additional cabinets in the most balanced way possible, by adding 1 cabinet to 14 of the days, the set becomes:

    {3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4}

In this configuration, the 15th and 16th terms, which determine the median, remain unchanged at 3. Therefore, regardless of how the additional 14 cabinets are distributed, the median number of cabinets produced daily remains 3.

Sufficient.

Answer: B.
­


Hi Bunuel, this makes sense, my only confusion is what would the scenario have to look like for us NOT to be able to identify the median? During an exam under timed conditions it would be time consuming to draw this out as you did, and multiple scenarios of it at that to ensure the median would not change. Is there a trick to be able to identify when the median will be the same in every which way?


TIA
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Could someone assist with the explanation for this question? Didn't follow how to answer with the corrected problem statement 2

A furniture manufacturer produced at least 3 cabinets each day for the past 30 working days. What was the median number of cabinets produced daily by the manufacturer for the past 30 working days?

Given that at least 3 cabinets were produced daily, the minimum total production is 3 cabinets/day * 30 days = 90 cabinets. To find the median of an even number of terms (30 days in this case), we average the two middle terms once the terms are arranged in order.

(1) For the past 30 working days, the manufacturer produced at most 6 cabinets each day.

If 6 cabinets were produced each day, the median would be 6. However, if 3 cabinets were produced for 29 days and 6 cabinets on one day, the median would be 3.

Not sufficient.

(2) For the past 30 working days, the manufacturer produced a total of 104 cabinets.

We already know that the minimum production over this period, with at least 3 cabinets produced each day, is 90 cabinets. This implies that there are 14 additional cabinets to account for in the total production.

Initially, our set of daily productions looks like this:

    {3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3}

To find the median, we need to determine if adding the extra 14 cabinets can change the middle values (15th and 16th terms) in this set. Even if we distribute these additional cabinets in the most balanced way possible, by adding 1 cabinet to 14 of the days, the set becomes:

    {3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4}

In this configuration, the 15th and 16th terms, which determine the median, remain unchanged at 3. Therefore, regardless of how the additional 14 cabinets are distributed, the median number of cabinets produced daily remains 3.

Sufficient.

Answer: B.
­


Hi Bunuel, this makes sense, my only confusion is what would the scenario have to look like for us NOT to be able to identify the median? During an exam under timed conditions it would be time consuming to draw this out as you did, and multiple scenarios of it at that to ensure the median would not change. Is there a trick to be able to identify when the median will be the same in every which way?


TIA
­
The method explained by Bunuel is probably the most efficient way to do this. You don't have to write everything down but understand the fact that only the places after the median are our concern. Here, for instance, the median will depend on the elements in the 15th and 16th place. After filling all places with 3, we still have 14 elements left. Now the best way to put those 14 without touching our median values would be to fill all 14 in the last place. But we don't need to do that. We need the worst case, in which we distribute 1-1 element each, over places starting from the end. The worst we can do is fill all 14 places with 1 element each making places 17-30 have 4 elements. Since this was the worst we could do, we see that the median remains untouched.

Now coming to your original question, when will the median change? For this particular scenario, if the total would have been 105 (instead of 104), we would have been left with 15 elements (instead of 14). And the worst we could do with 15 is, again, to fill all spaces from the back with 1-1 element each. In this case, we see that the 16th place would become 4. And thus we would get 1 case where the median would change.

Hope that helped!­
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Hi Bunuel,

I have trouble understanding, how is it that we assume that manufacturer can start producing 4 pieces on furniture on 1st day or 17th day only, what if he started producing 4 pieces on 2nd day then median will (4+3)/2, if 3rd day median will be (4+4)/2.

Additionally the question also does'nt specify that manufacturer has gradualy increased/ decreased his production, he can also produce 3 on eventh day and 4 on oddth day
Bunuel
akshayferdi
Could someone assist with the explanation for this question? Didn't follow how to answer with the corrected problem statement 2

A furniture manufacturer produced at least 3 cabinets each day for the past 30 working days. What was the median number of cabinets produced daily by the manufacturer for the past 30 working days?

Given that at least 3 cabinets were produced daily, the minimum total production is 3 cabinets/day * 30 days = 90 cabinets. To find the median of an even number of terms (30 days in this case), we average the two middle terms once the terms are arranged in order.

(1) For the past 30 working days, the manufacturer produced at most 6 cabinets each day.

If 6 cabinets were produced each day, the median would be 6. However, if 3 cabinets were produced for 29 days and 6 cabinets on one day, the median would be 3.

Not sufficient.

(2) For the past 30 working days, the manufacturer produced a total of 104 cabinets.

We already know that the minimum production over this period, with at least 3 cabinets produced each day, is 90 cabinets. This implies that there are 14 additional cabinets to account for in the total production.

Initially, our set of daily productions looks like this:


{3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3}

To find the median, we need to determine if adding the extra 14 cabinets can change the middle values (15th and 16th terms) in this set. Even if we distribute these additional cabinets in the most balanced way possible, by adding 1 cabinet to 14 of the days, the set becomes:


{3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4}

In this configuration, the 15th and 16th terms, which determine the median, remain unchanged at 3. Therefore, regardless of how the additional 14 cabinets are distributed, the median number of cabinets produced daily remains 3.

Sufficient.

Answer: B.
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Hi Bunuel,

I have troible understanding, how is it that we assume that manufacturer can start producing 4 pieces on furniture on 1st day or 17th day only, what if he started producing 4 pieces on 2nd day then median will (4+3)/2, if 3rd day median will be (4+4)/2.

Additionally the question also doesnt specify that manufacturer has gradualy increased/ decreased his production, he can also produce 3 on eventh day and 4 on oddth day
Bunuel
akshayferdi
Could someone assist with the explanation for this question? Didn't follow how to answer with the corrected problem statement 2

A furniture manufacturer produced at least 3 cabinets each day for the past 30 working days. What was the median number of cabinets produced daily by the manufacturer for the past 30 working days?

Given that at least 3 cabinets were produced daily, the minimum total production is 3 cabinets/day * 30 days = 90 cabinets. To find the median of an even number of terms (30 days in this case), we average the two middle terms once the terms are arranged in order.

(1) For the past 30 working days, the manufacturer produced at most 6 cabinets each day.

If 6 cabinets were produced each day, the median would be 6. However, if 3 cabinets were produced for 29 days and 6 cabinets on one day, the median would be 3.

Not sufficient.

(2) For the past 30 working days, the manufacturer produced a total of 104 cabinets.

We already know that the minimum production over this period, with at least 3 cabinets produced each day, is 90 cabinets. This implies that there are 14 additional cabinets to account for in the total production.

Initially, our set of daily productions looks like this:


{3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3}

To find the median, we need to determine if adding the extra 14 cabinets can change the middle values (15th and 16th terms) in this set. Even if we distribute these additional cabinets in the most balanced way possible, by adding 1 cabinet to 14 of the days, the set becomes:


{3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4}

In this configuration, the 15th and 16th terms, which determine the median, remain unchanged at 3. Therefore, regardless of how the additional 14 cabinets are distributed, the median number of cabinets produced daily remains 3.

Sufficient.

Answer: B.

The confusion comes from mixing the actual sequence of production with the ordered list used for median.

It doesn’t matter on which days the manufacturer produced more than 3 cabinets — for median we only care about the values after sorting them in increasing order.

Since the minimum production per day is 3, the smallest possible total for 30 days is 30 * 3 = 90.

Given total production is 104, there are 104 - 90 = 14 extra cabinets.

This means - after sorting the daily productions - only the largest 14 values can be greater than 3.

The smallest 30 - 14 = 16 values must be exactly 3.

That’s why the 15th and 16th values (which determine the median) are both 3 - regardless of whether the manufacturer produced 4 cabinets on the 2nd day, 5th day, or any other day.

Median depends only on the ordered values, not on how production was distributed across the days.
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A furniture manufacturer produced at least 3 cabinets each day for the past 30 working days. What was the median number of cabinets produced daily by the manufacturer for the past 30 working days?

(1) For the past 30 working days, the manufacturer produced at most 6 cabinets each day.

(2) For the past 30 working days, the manufacturer produced a total of 104 cabinets.


Attachment:
44.png

Given: A furniture manufacturer produced at least 3 cabinets each day for the past 30 working days.

This means at least 3*30 = 90 cabinets were produced.

Asked: What was the median number of cabinets produced daily by the manufacturer for the past 30 working days?

The median for 30 numbers is the average of the 15th and 16th numbers when all numbers are arranged in increasing order.

(1) For the past 30 working days, the manufacturer produced at most 6 cabinets each day.

So number of cabinets produced each day ranged from 3 to 6. We don't know what the average of 15th and 16th numbers are when they are arranged in increasing order.
Not sufficient alone.

(2) For the past 30 working days, the manufacturer produced a total of 104 cabinets.

We know that at least 3 were produced everyday so 90 were produced in that way. This means 14 extra were produced on some of the days. Now, it doesn't matter how we distribute these 14 among the 30 days, the 15th and 16th number will always remain 3 only.
Say all 14 were produced on 1 of the days. Then when we arrange daily productions in increasing order, we will get: 3, 3, 3, ... 3, 3, 17 (15th and 16th numbers will be 3 only)
Say on 14 days, 1 extra was produced. Then when we arrange daily productions in increasing order, we will get: 3, 3, 3, ... 3, 3, 4, 4, ... 4 (15th and 16th numbers will be 3 only because only last 14 numbers will be 4 each))

Answer (B)

Here is another interesting Math type DS question discussion: https://youtu.be/Bhjj6ik0LO8
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