# Exam Puzzle: Will it be Disadvantageous for Jane?

• K Sengupta
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.

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...

K Sengupta said:
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

Last edited:
davee123 said:
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:

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.

(a+b+c+d+e+f+A)/7 = 79.71429
(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.

(a+b+c+d+e+f+A)/7 = 80.28571
(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

davee123 said:
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.

davee123 said:
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.

Last edited:
jimmysnyder said:
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.

jimmysnyder said:
davee123 said:
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):

(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

davee123 said:
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.

## 1. What is the purpose of the exam puzzle?

The purpose of the exam puzzle is to determine whether or not it will be disadvantageous for Jane to take the exam. It is used to assess the potential risks and benefits of Jane taking the exam.

## 2. How does the exam puzzle work?

The exam puzzle works by evaluating various factors such as Jane's current knowledge and skills, the difficulty of the exam, and the potential consequences of taking or not taking the exam. These factors are then analyzed and compared to determine the overall disadvantageousness for Jane.

## 3. Who benefits from the exam puzzle?

The exam puzzle benefits both Jane and those responsible for administering the exam. For Jane, it provides insight into the potential risks and rewards of taking the exam. For the administrators, it helps them make informed decisions about whether or not Jane should take the exam.

## 4. Can the exam puzzle accurately predict the outcome?

The exam puzzle is not meant to predict the exact outcome, but rather to provide a general idea of the potential disadvantageousness for Jane. It takes into account various factors, but there are always other variables that may impact the actual outcome.

## 5. Is the exam puzzle a reliable tool?

The exam puzzle is a reliable tool in assessing the potential disadvantageousness for Jane. However, it should not be the sole factor in making a decision about whether or not Jane should take the exam. Other factors such as personal motivation and external circumstances should also be considered.

• Calculus
Replies
3
Views
427
• Introductory Physics Homework Help
Replies
25
Views
501
• Programming and Computer Science
Replies
17
Views
1K
Replies
15
Views
2K
• General Discussion
Replies
11
Views
2K
• Electromagnetism
Replies
11
Views
3K
• Introductory Physics Homework Help
Replies
18
Views
1K
Replies
1
Views
2K
• MATLAB, Maple, Mathematica, LaTeX
Replies
45
Views
16K
• General Discussion
Replies
4
Views
6K