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The daily closing price of one barrel of oil for first 15 days of a

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KarunMendiratta
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The daily closing price of one barrel of oil for first 15 days of a [#permalink] New post   23 Sep 2021, 08:06
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The daily closing price of one barrel of oil for first 15 days of a month is given by \(f(x) = 0.4x + 2\). Whereas, for the last 15 days in the same period, the the closing price is given by \(f(x) = \frac{−x}{7} + 9\), where \(x\) represents the day in the period for both the functions.

Quantity A
Quantity B
Median closing price of one barrel of oil for the entire 30 day period
7.35


A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
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B
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Re: The daily closing price of one barrel of oil for first 15 days of a [#permalink] New post   25 Sep 2021, 11:02
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Explanation:

For the first 15 days let us calculate the minimum and maximum closing price

Min. price is when \(x = 1\)
\(f(x) = 0.4(1) + 2 = 2.4\)

Max. price is when \(x = 15\)
\(f(x) = 0.4(15) + 2 = 8\)

Now, let us calculate the minimum and maximum closing price for last 15 days

Min. price is when \(x = 30\)
\(f(x) = \frac{−30}{7} + 9 = 4.71\)

Max. price is when \(x = 16\)
\(f(x) = \frac{−16}{7} + 9 = 6.71\)

Now the median (M) for the 30 day period would be the average of 15th and 16th term arranged in ascending order
i.e. (2.4) .......... (4.71) (M) (6.71) .......... (8)
\(M = \frac{4.71 + 6.71}{2} = 5.71\)

Col. A: 5.71
Col. B: 7.35

Hence, option B
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Re: The daily closing price of one barrel of oil for first 15 days of a [#permalink] New post   24 Sep 2021, 08:55
To find the median we have to arrange the values in ascending order first. How would we even achieve this? I know plugging n the numbers won't work. What is the strategy behind this question?
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Re: The daily closing price of one barrel of oil for first 15 days of a [#permalink] New post   24 Sep 2021, 11:09
koala wrote:
To find the median we have to arrange the values in ascending order first. How would we even achieve this? I know plugging n the numbers won't work. What is the strategy behind this question?

to ans your question , let see -
30 days were given, to find median of 30 days is avg of 15th and 16th day.
we are provided with equations for 1st half of month and other 15 days for remaining half.
for first equation, put x=15 you will get 6 and put 15 in second equation you will 6.85
taking average 6 and 6.85 is less than 7.85 .
hence option b is your ans.
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Re: The daily closing price of one barrel of oil for first 15 days of a [#permalink] New post   24 Sep 2021, 20:20
void wrote:
koala wrote:
To find the median we have to arrange the values in ascending order first. How would we even achieve this? I know plugging n the numbers won't work. What is the strategy behind this question?

to ans your question , let see -
30 days were given, to find median of 30 days is avg of 15th and 16th day.
we are provided with equations for 1st half of month and other 15 days for remaining half.
for first equation, put x=15 you will get 6 and put 15 in second equation you will 6.85
taking average 6 and 6.85 is less than 7.85 .
hence option b is your ans.


When x = 15, you will get f(x) = 8
Why do we need to put x = 15 again in the second equation?
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The daily closing price of one barrel of oil for first 15 days of a [#permalink] New post   24 Sep 2021, 21:36
KarunMendiratta wrote:
void wrote:
koala wrote:
To find the median we have to arrange the values in ascending order first. How would we even achieve this? I know plugging n the numbers won't work. What is the strategy behind this question?

to ans your question , let see -
30 days were given, to find median of 30 days is avg of 15th and 16th day.
we are provided with equations for 1st half of month and other 15 days for remaining half.
for first equation, put x=15 you will get 6 and put 15 in second equation you will 6.85
taking average 6 and 6.85 is less than 7.85 .
hence option b is your ans.


When x = 15, you will get f(x) = 8
Why do we need to put x = 15 again in the second equation?

i am not sure, it will explain or justifying explaination....but let go with it..
i just assume that 15 days as one quantity and another 15 days which, i start number from 1 to 15. I did it because question says two half and start number so i start number from 1 to 15 and same way for another number. if you put x=30 in seond equation will still valuse about 4.75 and average of 4.75+6 is is less than 7.85..
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Re: The daily closing price of one barrel of oil for first 15 days of a [#permalink] New post   25 Sep 2021, 09:04
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Shouldn't we substitute 15 in equation 1(as the question mentions there are 2 15 days period) and then 1 in equation 2 and then find the mean.
Now eq.1 gives 8 and eq.2 gives \(\frac{62}{7}\) which gives us and average of \(\frac{59}{7}\) . So option A should be correct.
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Re: The daily closing price of one barrel of oil for first 15 days of a [#permalink] New post   26 Sep 2021, 00:21
yagneshward
Can you please explain here why the OA has to be Option (A)?
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Re: The daily closing price of one barrel of oil for first 15 days of a [#permalink] New post   26 Sep 2021, 00:58
The median for the 30 days will be the mean of the 15th and 16th days.

For 15th day, f(x)= 0.4x+2, where x = 15. ====> On substituting x=15 in the function, we get 8.

For 16th day, f(x) = (-x / 7 ) + 9, where x=16.==> On substituting x= 16 in the function, we get 6.714285714285714

The median will be ( 8 + 6.714285714285714 ) / 2 = 7.357142857142858

7.357142857142858 is greater than 7.35.

So, therefore Option A should be the right answer.
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Re: The daily closing price of one barrel of oil for first 15 days of a [#permalink] New post   26 Sep 2021, 01:25
yagneshward wrote:
The median for the 30 days will be the mean of the 15th and 16th days.

For 15th day, f(x)= 0.4x+2, where x = 15. ====> On substituting x=15 in the function, we get 8.

For 16th day, f(x) = (-x / 7 ) + 9, where x=16.==> On substituting x= 16 in the function, we get 6.714285714285714

The median will be ( 8 + 6.714285714285714 ) / 2 = 7.357142857142858

7.357142857142858 is greater than 7.35.

So, therefore Option A should be the right answer.


On substituting x= 16 in the function, we get 6.714285714285714 - Correct!
But is it the least value for the second equation?
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Re: The daily closing price of one barrel of oil for first 15 days of a [#permalink] New post   26 Sep 2021, 01:37
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Understood my mistake. Thanks for the explanation.
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Re: The daily closing price of one barrel of oil for first 15 days of a [#permalink] New post   26 Sep 2021, 01:38
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yagneshward wrote:
Understood my mistake. Thanks for the explanation.


Anytime :thumbsup:
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The daily closing price of one barrel of oil for first 15 days of a [#permalink] New post   12 Dec 2024, 07:54
KarunMendiratta wrote:
Explanation:

For the first 15 days let us calculate the minimum and maximum closing price

Min. price is when \(x = 1\)
\(f(x) = 0.4(1) + 2 = 2.4\)

Max. price is when \(x = 15\)
\(f(x) = 0.4(15) + 2 = 8\)

Now, let us calculate the minimum and maximum closing price for last 15 days

Min. price is when \(x = 30\)
\(f(x) = \frac{−30}{7} + 9 = 4.71\)

Max. price is when \(x = 16\)
\(f(x) = \frac{−16}{7} + 9 = 6.71\)

Now the median (M) for the 30 day period would be the average of 15th and 16th term arranged in ascending order
i.e. (2.4) .......... (4.71) (M) (6.71) .......... (8)
\(M = \frac{4.71 + 6.71}{2} = 5.71\)

Col. A: 5.71
Col. B: 7.35

Hence, option B


How do you know that 4.71 and 6.71 are the middle numbers?
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Re: The daily closing price of one barrel of oil for first 15 days of a [#permalink] New post   12 Dec 2024, 08:30
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Now the median (M) for the 30 day period would be the average of 15th and 16th term arranged in ascending order

i.e. (2.4) .......... (4.71) (M) (6.71) .......... (8)
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Re: The daily closing price of one barrel of oil for first 15 days of a [#permalink]
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