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805+ (Hard)|   Word Problems|      
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Bunuel
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Brian is analyzing the performance of certain stocks. He finds that 09 Nov 2021, 04:00
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95% (hard)

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30% (03:21) correct 70% (03:06) wrong based on 89 sessions

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Brian is analyzing the performance of certain stocks. He finds that for the first fifteen days of a 30 day period, the daily closing price per share of Phonexpharma Inc. conformed to the function \(f(x)=0.4x+2\) , where x represents the day in the period. He found that for the last 15 days in the same period, the stock followed the function \(f(x)=(-\frac{1}{7})x+9\), where x represents the day in the period. Approximately what is the median closing price per share of Phonexpharma Inc. for the entire 30 day period?

A. 4.7
B. 5.55
C. 6.71
D. 8
E. 11.1
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Brian is analyzing the performance of certain stocks. He finds that 09 Nov 2021, 21:54
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Brian is analyzing the performance of certain stocks. He finds that for the first fifteen days of a 30 day period, the daily closing price per share of Phonexpharma Inc. conformed to the function \(f(x)=0.4x+2\) , where x represents the day in the period. He found that for the last 15 days in the same period, the stock followed the function \(f(x)=(-\frac{1}{7})x+9\), where x represents the day in the period. Approximately what is the median closing price per share of Phonexpharma Inc. for the entire 30 day period?

A. 4.7
B. 5.55
C. 6.71
D. 8
E. 11.1
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The value of stock in first 15 days will be 2.4,2.8,3.2.......7.6,7.8 (common difference 0.4)
The value of stock in last 15 days will be 47/7, 46/7,45/7.......34/7,33/7 or say 6.71, 6.57,6.43,.....4.85,4.71(common diff 1/7 or approx 0.14)

We have to find the average of 15th and 16th term when all the above values are arranged in ascending or descending order

From the options, we can easily discard D and E. And C can also be discarded as they are the extreme terms of the 2nd AP.

Option B
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Brian is analyzing the performance of certain stocks. He finds that 05 Nov 2025, 00:27
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Brian is analyzing the performance of certain stocks. He finds that for the first fifteen days of a 30 day period, the daily closing price per share of Phonexpharma Inc. conformed to the function \(f(x)=0.4x+2\) , where x represents the day in the period. He found that for the last 15 days in the same period, the stock followed the function \(f(x)=(-\frac{1}{7})x+9\), where x represents the day in the period. Approximately what is the median closing price per share of Phonexpharma Inc. for the entire 30 day period?

A. 4.7
B. 5.55
C. 6.71
D. 8
E. 11.1
Show more
The first 15 days of the 30 day period confirms the function : \(f(x)=0.4x+2\) , where x represents the day in the period.

so, x assumes the values from 1,2,3,4.... Till 15.

Let’s assume this to be CASE 1:

And, the values of

f(x) = 2.4 , when x=1

f(x) = 2.8, when x=2

f(x) = 3.2 , when x=3

then 3.6, 4.0, 4.4, 5.2, 5.6 , 6.0, 6.4, 6.8, 7.2, 7.6, 8.0

In 2.xx range, number of values = 2

In 3.xx range, number of values = 2

In 4.xx range, number of values = 2

In 5.xx range, number of values = 2

In 6.xx range, number of values = 3

In 7.xx range, number of values = 2

In 8.xx range, number of values = 1


Case 2:

The last 15 days of the stock followed the pattern of : \(f(x)=(-\frac{1}{7})x+9\), where x represents the day in the period.

Numbers which are multiples of 7 after 15 are: 21 and 28.

with x = 21, f(x) = 6

with x = 28, f(x) = 5.

so the values of x=29 and x=30, should be below 5.0 mark, thus ONLY 2 VALUES falling in the 4.xx range.

The number of values in the band less than 6.0 but greater than 5.0 , that’s within the 5.xx band is (28-22)+1 = 7 values.

And we can also infer that the values for, x =16 till x =20 lies within the 6.xx band. There are 5 such values in 6.xx range.


In the 6.xx range, the number of values = 5

In 5.xx range, the number of values = 7

In the 4.xx range , the number of values = 2

Median = (15th value +16th value)/2

There are 8 values in the 2.xx and 5.xx range in case 1.

including the 4.xx range from case 2, we get 2 more values, totalling to 10 values.

5.xx range values from case 2, which lies in the median slot of 15th and 16th value.

Thus by approximation, we can arrive at the median value to be 5.xx

Option B

5.55
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Brian is analyzing the performance of certain stocks. He finds that 13 Jun 2026, 19:52
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Since there are 30 values in total, the median is the average of the 15th and 16th values when all the values are arranged in either increasing or decreasing order. For convenience, we shall arrange the values in increasing order.

The values corresponding to the first 15 days are generated by Set A, which is already in increasing order.

Day010203040506070809101112131415
Set A2.44.06.08.0

The values corresponding to the last 15 days are generated by Set B, which is in decreasing order.
Day161718192021222324252627282930
Set B6.05.0

We reverse the order of Set B to obtain Set B′ in increasing order so that both sets can be examined together while determining the positions of the 15th and 16th observations.
Day302928272625242322212019181716
Set B'5.06.0



Now, compare Set A and Set B′ to determine the possible positions of the 15th and 16th observations.
Day010203040506070809101112131415
Set A2.44.04.24.85.25.66.08.0

Day302928272625242322212019181716
Set B'5.06.0

Now, observe the ranges represented by the answer choices:
Option A: 4.7, which lies between 4.0 and 5.0.
Option B: 5.55, which lies between 5.0 and 6.0.
Option C: 6.71, which lies between 6.0 and 7.0.
Option D: 8, which is the maximum value in Set A.
Option E: 11.1, which exceeds the largest value in the data set.

Option E can be eliminated immediately since the median cannot exceed the largest observation.
Similarly, Option D can also be eliminated because, in an ordered list of 30 observations that are not all identical, the median cannot be equal to the maximum value.

Next, determine the possible location of the 15th and 16th observations.

At 5.0,
Set A contributes 7 values less than or equal to 5.0.
Set B′ contributes 3 values less than or equal to 5.0.
Therefore, there are exactly 10 observations less than or equal to 5.0.
Hence, the 15th and 16th observations must both be greater than 5.0.


At 6.0,
Set A contributes 9 values strictly less than 6.0.
Set B′ contributes 9 values strictly less than 6.0.
Therefore, there are exactly 18 observations strictly less than 6.0.
Hence, the 15th and 16th observations must both be less than 6.0.

Thus,
5.0 < 15th observation ≤ 16th observation < 6.0.

Consequently,
5.0 < Median ≤ 6.0.

Among the given answer choices, only Option B lies within this interval.
Therefore, the median closing price per share is 5.55
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