Exam Puzzle: Will it be Disadvantageous for Jane?

[K Sengupta]

Jane took several exams in math during the academic year with grades such as a,b,c,d,e,f and once she did quite badly. When repeating the poor exam, the teacher offered that instead of the two grades, say f and A, she can have the average of the two grades.

Will it be disadvantagenous for Jane if (a+b+c+d+e+f)/6 > (f+A)/2?

NOTE:

I found this problem in a puzzle periodical.

The one line answer given in the said periodical just mentions that it would be indeed disadvantageous for Jane.

I do not know the precise methodology that would lead to the said solution, and I am looking forward to any comment from members on the foregoing matter.

 

[Jimmy Snyder]

Surely there is some information missing here. Suppose

a = b = c = d = e = 20, f = A = 10

Then 18.33... = (a + b + c + d + e + f)/6
10 = (f + A)/2
so the conditions are met. However,

(a + b + c + d + e + f + A)/7 = 17.14

(a + b + c + d + e + (f + A)/2)/6 = 18.33...

So it is advantageous to average the two grades.

 

[davee123]

Jane took several exams in math during the academic year with grades such as a,b,c,d,e,f and once she did quite badly. When repeating the poor exam, the teacher offered that instead of the two grades, say f and A, she can have the average of the two grades.

Will it be disadvantagenous for Jane if (a+b+c+d+e+f)/6 > (f+A)/2?

Let's say that M = a + b + c + d + e.

It will be disadvantageous to Jane IF the NORMAL grade she would have gotten is GREATER THAN the AVERAGED grade that the teacher is offering:

(M+f+A)/7 > (M+(f+A)/2)/6
6*(M+f+A) > 7*(M+(f+A)/2)
6M + 6f + 6A > 7M +3.5f + 3.5A
2.5f + 2.5A > M
0.5f + 0.5A > M/5
(f+A)/2 > M/5
(f+A)/2 > (a+b+c+d+e)/5

So, that's interesting. I get a slightly different outcome than in the problem, since we were hoping to get:
(f+A)/2 > (a+b+c+d+e+f)/6
or
(f+A)/2 < (a+b+c+d+e+f)/6

That means (in theory, if I did my basic algebra correctly), that there may be two instances where (f+A)/2 < (a+b+c+d+e+f)/6 is true, but one is disadvantageous, and one is advantageous. Huh. I'll have to come back to that later and see if it's correct.

DaveE

 

[davee123]

It will be disadvantageous to Jane IF
(f+A)/2 > (a+b+c+d+e)/5

So, yes, the problem is ambiguous, or misstated.

Proof:

Assume Jane's grades are:
67,80,92,75,86, and 90, and she's making up the 90.

Case #1, Jane gets a 68 on the makeup exam. This fits the problem because:
(a+b+c+d+e+f)/6 > (f+A)/2
81.667 > 79

Jane has two options.

Normal Grade:
(a+b+c+d+e+f+A)/7 = 79.71429
Averaged Grade:
(a+b+c+d+e+(f+A)/2)/6 = 79.83333

So, it will be ADVANTAGEOUS in this case for Jane.

Case #2, Jane gets a 72 on the makeup exam. This fits the problem because:
(a+b+c+d+e+f)/6 > (f+A)/2
81.667 > 81

Jane has two options.

Normal Grade:
(a+b+c+d+e+f+A)/7 = 80.28571
Averaged Grade:
(a+b+c+d+e+(f+A)/2)/6 = 80.16667

So, it will be DISADVANTAGEOUS in this case for Jane.

Conclusion: Nothing. The problem is ambiguous.

My guess is that the problem is misstated, because:
It will be disadvantageous to Jane IF
(f+A)/2 > (a+b+c+d+e)/5

Note, it appears that the problem is answerable if you assume that the grade she's making up (grade f) is actually *lower* or *equal* than the average of the rest of her grades. (I haven't proven that, but it looked that way in playing with the numbers)

DaveE

 

[Jimmy Snyder]

Assume Jane's grades are:
67,80,92,75,86, and 90, and she's making up the 90.

In the OP you will find that it is the lowest grade that is to be made up.

Note, it appears that the problem is answerable if you assume that the grade she's making up (grade f) is actually *lower* or *equal* than the average of the rest of her grades.

There is a counter-example in post #2 in this thread.

 

[davee123]

In the OP you will find that it is the lowest grade that is to be made up.

That's what's implied, true, but it isn't explicitly stated. It says "once she did quite badly". It does NOT say how she did the rest of the time at all-- you're just left to assume that she did worse on this one than she had done on other tests.

Note, it appears that the problem is answerable if you assume that the grade she's making up (grade f) is actually *lower* or *equal* than the average of the rest of her grades.

There is a counter-example in post #2 in this thread.

It may be wrong, I'm not sure (as I said I haven't tried to prove it)-- but to be a proper counter-example, you'd have to show how it's untrue, which means you have to give *two* examples, where you get answers that it's advantageous and disadvantageous.

What I'm saying is that I think (again, haven't proven it, it may be wrong, please prove or disprove it if you'd like):

If her "bad" grade is lower than her normal average, then:
(a+b+c+d+e)/5 > (a+b+c+d+e+f)/6 > (f+A)/2

And, that would mean it's actually *advantageous* to Jane, because, as I proved above, it will only be *disadvantageous* if:
(f+A)/2 > (a+b+c+d+e)/5

It would be interesting to see how (a+b+c+d+e)/5 and (a+b+c+d+e+f)/6 relate.. Hm... let's see:

Let M = (a+b+c+d+e)
(a+b+c+d+e)/5 = (a+b+c+d+e+f)/6
M/5 = (M+f)/6
6M = 5(M+f)
6M = 5M + 5f
M = 5f
M/5 = f
(a+b+c+d+e)/5 = f

Indeed! If (as predicted) her original grade that she's making up is *less* than the average of her previous grades, then the problem is solveable, and it will be *ADVANTAGEOUS* for Jane, because:

(a+b+c+d+e)/5 > (a+b+c+d+e+f)/6 > (f+A)/2

However, if her original grade that she's making up is *greater* than the average of her previous grades, then the problem is NOT solveable because:

(a+b+c+d+e+f)/6 > (a+b+c+d+e)/5 > (f+A)/2

DaveE

 

[Jimmy Snyder]

It says "once she did quite badly".

It also says:
"When repeating the poor exam, ..."
I agree with you, the wording should have been less ambiguous, but I think the intent is clear.