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Home » Mathematics » High School Basics (Ages 12-16)
 
Grade point average
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Arrange your tan
Riemann Hypothesis
Riemann Hypothesis

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#1
Grade point average
my problem
>
>

Chris and Pat have taken 40 courses together.
There are 100 points maximum available in each course per student.

For each course the grading system is:

A 90 -> 100
B 80 -> 89
C 70 -> 79

A 4.0
B 3.0
C 2.0

After the completion of these 40 courses (but no other courses than these 40 have
been taken by either student), Chris has a grade point average of 3.375, and a grade
point average of 2.500 is held by Pat.

Is it possible for Pat to have a greater total number of points across all 40 courses
as compared to the total points for Chris?

PostPosted: Thu Oct 29, 2009 6:19 am  Back to top 
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minime1235813
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#2
No they cant. Because: A = Average of points (given), P = All points added up (were finding this) N = Number of items (40)
A = P/N
For Chris, 3.375 = P/40; P = 135 so thats 135 points added up
For Pat, 2.5 = P/40; P = 100
135 > 100, so No.

PostPosted: Thu Nov 05, 2009 12:37 pm  Back to top 
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Arrange your tan
Riemann Hypothesis
Riemann Hypothesis

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#3
minime1235813 wrote:
No they cant. Because: A = Average of points (given), P = All points added up (were finding this) N = Number of items (40)
A = P/N
For Chris, 3.375 = P/40; P = 135 so thats 135 points added up
For Pat, 2.5 = P/40; P = 100
135 > 100, so No.


Arrange your tan wrote:

Chris and Pat have taken 40 courses together.
There are 100 points maximum available in each course per student.


No, you don't have the right understanding. With 40 courses apiece
and a maximum of 100 points per course, the students have their
totals out of 40\cdot100 = 4000 points apiece maximum.


* * * Open Hint/Discussion:

This situation can vary, but for relative simplicity, suppose that
Chris gets either an 80 or a 90 in the 40 courses, and that Pat
gets either an 89 or a 79 in the 40 courses.

Let x and y be appropriate positive integers for the next inequalities.

Then, for Chris's points to be less than Pat's points, you might have:

(x)(90) + (40 - x)(80) < (y)(89) + (40 - y)(79) . . . (*)

Look at the values for the letter grades. Chris gets the upper two
(4 and 3), and Pat gets the lower two (3 and 2), *if* you settle for
those grade limitations they are given as discussed as above.

(x)(4) + (40 - x)(3) > (y)(3) + (40 - y)(2)

This becomes:

y < x + 40 . . . (Though this is some upper limit, I did not make use of this in
. . . . . . . . . . . . . . my solution in a direct way.)

This becomes:

y \ge x + 5.

The values of x and y can vary.

End of the Open Hint/Discussion * * *

For my choice of an answer:
(The answer can vary.)
Click to reveal hidden content

Let x = 15
Let y = 20

Using (*) from above, x = 15 gets substituted on the left-hand side and
y = 20 gets substituted on the right-hand side:

[(15)(90) + (25)(80)] points versus [(20)(89) + (20)(79)] points

3350 points versus 3360 points

So Pat has more total points than Chris, but he/she has a lower grade
point average than Chris.


Grade point average work for Chris:

(15)4 + (25)3 = 60 + 75 = 135

135/40 = 3.375

-------------------------------------------------------------

Grade point average work for Pat:

(20)3 + (20)2 = 60 + 40 = 100

100/40 = 2.500


PostPosted: Fri Nov 06, 2009 8:33 pm  Back to top 
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