# Exam question

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Down here in NZ students have just recently sat a horrible algebra paper.  It appears that the examination authority didn't tell the teachers that changes were going to occur in the algebra exam so the teachers taught to what previous exams had asked for.  And all hell has broken loose.

So the attached graphic is a question that occurred in the exam.

I can do most of the paper (with a little bit of revision if I dig out the books) but the question above does not make sense to me.

So what do fellow experts make of it?  If anyone wants the full paper I can supply links.
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Owner (Aidellio)
Most Valuable Expert 2015
Commented:
Unless i'm missing something, just substitute a,b,c,d for the numbers in the pyramid

go through the same addition and you'll see that changing the order makes a difference

You'll also end up with an equation.

e.g.

a,b,c,d
a+b, b+c, c+d
a+b+b+c,b+c+c+d
a+2b+c+b+2c+d

final sum=a+3b+3c+d
Principal Software Engineer
Commented:

Line 1:  Let 1 = a, 3 = b, 5 = c, 7 = d.

Line 2:
4 = 1 + 3 = a + b, let this be g
8 = 3 + 5 = b + c let this be h
12 = 5 + 7 = c + d let this be  i

Line 3:
12 = 4 + 8 = g + h, let this be x
20 = 8 + 12 = h + i, let this be y

Line 4:
32 = 12 + 20 = x + y = g + 2h + i = a + b + 2b + 2c + c + d = a + 3b + 3c +d

Will the same answer be achieved?  In two specific cases, yes:

1. the numbers are reversed left to right
2. a and b are identical, c and d are identical, a and b are interchanged, c and d are interchanged
Quid, Me Anxius Sum?  Illegitimi non carborundum.

Commented:
Agreed.

You either get the same answer or a different answer depending on what you shift.

Swap 1 and 3 and the answer is different.  Swap 1 and 7 and the answer is the same.

I also suspect that you are supposed to use x rather than a, b, c, d etc.   There is a third part to this question which I'll post later ...

So the triangle becomes for the initial case

x, x + 2, x + 4, x + 6
2x + 2, 2x + 6, 2x +10
4x + 8, 4x + 16
8x + 24

But if I'm a student and I've got two conflicting scenarios which one do I pick?  And the marker is going to have to look at all of these permutations as well.
Principal Software Engineer
Commented:
"8x + 24" is a specific solution for this particular configuration only.  "a + 3b + 3c + d" is a more general solution and leads students toward the binomial theorem.  Both are perfectly correct.  I would have to say that the question is overly open to interpretation.
Owner (Aidellio)
Most Valuable Expert 2015
Commented:
I don't follow your conflicting scenarios. It all depends on whether the numbers are related or arbitrary.

a,b,c,d is just giving the option of having 4 independent numbers.

Using x as you've described creates a relationship. If you call the last digit x (7) them you end up with 8x - 24

Following that pattern and using x for each of the 4 positions you get:
8x + 24, 8x + 8, 8x - 8, 8x +24

So it's about determining if in fact the numbers have a relationship and if so from which position
Quid, Me Anxius Sum?  Illegitimi non carborundum.

Commented:
Well, the answer to (i) is either Yes or No depending on what is swapped.

The answer to (ii) Explain your answer is from what I can see is going through the algebra for the original case and the new case and showing the difference.  If the total for LIne 2 does not change then Line 4 does not change.  But I'm not really happy with doing the algebra as the answer for (ii).
Quid, Me Anxius Sum?  Illegitimi non carborundum.

Commented:
And the last question.

(iii) If Jason writes 4 consecutive numbers in order, what do you know about the numbers if the number at the bottom of the triangle is divisible by 3?

Owner (Aidellio)
Most Valuable Expert 2015
Commented:
that the first number in the sequence has to be divisible by 3.

x, x+1, x+2, x+3
...
8x + 12 at the bottom of the pyramid

if (8x +12) / 3 is to be a whole number then the 8x / 3 component means x needs to be a multiple of 3 OR zero
Commented:
It is a nice exercise. If I remember well, we use to make such kind of exercises under age of 14.
Just for my curiosity, for what age and class/grade is that exam in NZ?
Quid, Me Anxius Sum?  Illegitimi non carborundum.

Commented:
Yes.  And the fourth number also has to be divisible by 3.  (Unless the fourth number works out as 0)

You can discard the 12 as it is already a multiple of 3.  So the only way 8x can be a multiple of 3 is if x is a multiple of 3.  Or 0.  (I missed that).

And that explanation for (iii) is far better than any explanation I can find for answers I get for part (ii).
Quid, Me Anxius Sum?  Illegitimi non carborundum.

Commented:
>>  If I remember well, we use to make such kind of exercises under age of 14.

Sigh.  I missed out on that stuff.

>>  Just for my curiosity, for what age and class/grade is that exam in NZ?

Year 11.  Which is roughly 15 to 17 year old students.
Commented:
As Rob Jurd has rightly pointed out the sum is a+3b+3c+d which by setting a,b,c and d to 1 shows that it is a Pascal Triangle and consequently those rules and conditions will apply. Like, for example, it is symmetric in b and c and one can also swap a and d.

I find the question slightly annoying in that part ii talks about "the first line" and "Line 1". I suppose it was to avoid mentioning "Line 1" twice in the same sentence, but using two terms for the same thing always causes confusion.
Commented:

Part 9 (i)   Investigate what happens when Jason changes the order of the numbers in Line 1.
Does he get the same answer in Line 4?

This is an absolutely horrible question because the answer depends on the arbitrary choice of the new order.
There are 24 possible orderings.
Swapping  (1 and 7) or (3 and 5)  or both give the same answer by symmetry.
In this particular case, swapping (1 and 3) does too.

So by chance some students get to the second part, find confirmation, and sail on.
The others find contradiction and perhaps stumble.
Life is not fair, but exams should really try harder.

How is it possible that NZ hasn't discovered the joys of Standardized Multiple Choice Exams?
Quid, Me Anxius Sum?  Illegitimi non carborundum.

Commented:
>>  How is it possible that NZ hasn't discovered the joys of Standardized Multiple Choice Exams?

Not sure if that comment is in jest or not.

But anyway one more assessment task.

Don't want the answer but does the text adequately describe the objects concerned?
Commented:
No it doesn't because it introduces an unexplained "r" in the formula. That's the sort of thing which causes great confusion in E-E threads and results in mountains of posts.

It also begs the question : given a three-quarter pizza which part of the Quattro-Staggione  would you like not to have?
Commented:
Another horrible question.  You don't have enough information.
You need to know the angular extent of the missing piece or pieces.

You could have 3/4ths of an r-sized pizza or 1/12th of a 3r-sized one.
Any R > sqrt(3/4)*r could work.
Commented:
Either is too simple or I do not understand it and creates confusion, but why is it not just simple as r=sqrt((4*A)/(3*PI))?
There are some assumptions as follows:
-      The pizza is made round, circular, so its total area is PI*r*r. Then of course in the given text is ¾ of its total area.
-      Unsaid we may consider is cut from center along 2 radius, ¼ of it and removed, because this is the normal/common way to cut a pizza.
Then when we refer to the radius of that piece of pizza we think only at that r and not at something else. Of course we could cut the pizza in an infinite ways/shapes, but for their age we may suppose the normal cut.
Then as we have given the area A of a ¾ of pizza, it should be simple just to write the radius r based on the area A: r=sqrt((4*A)/(3*PI))
What did I miss?
Owner (Aidellio)
Most Valuable Expert 2015
Commented:
I'm with you viki2000 re A: r=sqrt((4*A)/(3*PI))

Remember this is an exam question for high school students that have previously learnt the formula for area of a circle so the absence of describing the equation is part of the test.  To me, it's also getting them to see a relationship from what they know (A=pi*r^2) and use general algebraic manipulation to get the formula in terms of r.  The 3/4 is only there to help them think outside the box and avoid the wrote learnt approach.
Commented:
You are reading the examiner's mind.

The intended question was probably something like
3/4ths of a pizza has an area of 75*pizza sq in

But it doesn't say you have 3/4ths of a pizza.
It says you have a piece of pizza and the area is 3/4 times something else.
Commented:
I understand now what are you saying, but let's be practical:
- your "something else" above is PI*r² and every examiner expects from the children to automatically connect in their mind and recognize the area of a circle, because we speak about areas anyway. That is an untold assumption, but a normal one for that exam.
-with that in mind, the child should imagine a circle which represents the pizza and from here it should be easy to think at 3/4 of it.
-in my opinion is not really an alien request and I think the only problem, which worth to object, would be for the pupils who never seen or ate pizza and never had the chance to see how is it cut, but even that does not really matter for formula, because is a mental construction to help imagine 1/4 and 3/4, supposing they learned fractions and know 4/4 is a full circle/pizza.
-in fact is just a substitution of the word circle with the word pizza, just to bring it in a real world, to give a practical meaning to the maths.
I do not want to be on the examiners side, but I have seen worse texts than this one. For the age of 15-17 it should be easy to avoid confusions, I mean, was never asked that child by its mother at the table: "Dear do you want a piece of pizza or 2? Do you want a quarter or half? " ?
Commented:
I agree that the answer is  r=sqrt((4*A)/(3*PI))

>>   I think the only problem, which worth to object, would be for the pupils who never seen or ate pizza

I disagree.  Usually the problem is to interpret the language to come up with the proper equation.
In this case, you have the proper equation and you have to ignore the distracting language, imagery, and choice of variables.
Aliens who have never heard of pizza, and conventionally use Q for area and z for radius will do the best.
Commented:
>>  How is it possible that NZ hasn't discovered the joys of Standardized Multiple Choice Exams?
Not sure if that comment is in jest or not.

In a multiple choice test, the examiners have to solve the problem and commit to a correct answer.
And they have to come up with three wrong answers.  How hard can that be?
http://mathlair.allfunandgames.ca/saterrors.php
Commented:
"Aliens who have never heard of pizza, and conventionally use Q for area and z for radius will do the best. "

Aliens may solve easy the problem using Q and z, but is not sure about their children at the puberty age...when their imagination goes wild in another direction...
Commented:
What are the metric units for pizza?
I am okay with km and kg, but I don't think I could ever be happy with a 30 cm pizza.
Quid, Me Anxius Sum?  Illegitimi non carborundum.

Commented:
Yes, I suspect the answer is probably r=sqrt((4*A)/(3*PI)) but it is NOT an answer I am happy with the way the problem is described.

I could argue that r by itself is an answer.

If the problem said that a circular piece out of a round pizza had an area as stated above where A and r are the Area and radius of the round pizza I would be happy.

>>  What are the metric units for pizza?

Pepperoni, mushroom, bacon, olives, anchovies.  Full, hungry, leftovers.
Commented:
"I could argue that r by itself is an answer."
No.
The request in the text of the problem is to write a formula.
Commented:
I note that the question concerning the pizza is part ii. What was in part i? This might provide a clue to part ii. I also note the terrible use of the passive subjunctive "could be used to find". Do they all talk like that in New Zealand?
Quid, Me Anxius Sum?  Illegitimi non carborundum.

Commented:
Find two exams attached with all the goriness within.

As for the use of the passive subjunctive, it could be, but I've tried to forget about verb tenses since I've dropped Latin, French and German from my need to know subjects.

Oh, and button and cube testing is the future.
exam2.pdf
exam1.pdf
Commented:
How long time is allowed for each of the exam tests above?
Quid, Me Anxius Sum?  Illegitimi non carborundum.

Commented:
One hour.
Commented:
Hmmm. I'm none the wiser. Given that the third part of that question is solve x²-3x-10=0 (factors of 10 = 1,10 or  2,5 whose difference is three =2,5) it must be REALLY simple. Probably there is something which is missing, because in both papers there is a question about a rectangle having an area dependent on n, and we are asked to determine the length of one side given the other, also an expression involving n. Quite simply a factorization problem but we are then asked the really strange question "what do you know about the value of n for this rectangle?" What AM I meant to reply here? "n" could be a parameter, but I find the concept of parameters to functions (only formulas are mentioned) rather strange at this level.

In Exam2 page 10 there is a groove question - it's a simple question regarding parabolas, with a rather annoyingly misleading picture of some sort of game with an HTTP link underneath. The link seems only to be there to remind the examinee that he has seen this problem before, since the HTTP address is completely - like the picture itself - irrelevant to the question.

I have had problems like this when my mouse was at school. Although the "technical" content at a particular age has not over time changed, the style it is presented in has. In my time we had questions involving trains and pulleys and Victorian things like that. Trying to help her, I got comments like "but teacher said..." and I could have throttled the teacher's neck.

If we therefore apply this sort of logic to the pizza problem, then we are expected to write :

Let R be the radius of the pizza then πR² is it's area, but this is equal to 3/4πr² so equating we get R=sqrt(3r/4) or if you like sqrt(3r)/2.

I wouldn't be the least surprised if the kids had had exactly this problem at school and if they were paying attention wouldn't need to even think out a solution.

I did my maths in France and our maths teacher got us to do again and again old French and Belgian papers from way back when (with trains and pulleys and so on), because, she said, these questions turn up again and again in slightly different forms. In fact the two papers presented here follow that pattern exactly.
Quid, Me Anxius Sum?  Illegitimi non carborundum.

Commented:
Except that this paper DIDN'T follow the pattern for previous papers.

Which is one of the main complaints about it.  Students who did previous years papers and the pre-tests for this paper got something which had no resemblance to what they had learned and done.

>>  What do you know about the value of n for this rectangle?

I think you can determine what a minimum value for n must be.  The quadratic is something like (n-a)(n-b)=0.  n must be positive and greater than the smallest positive of a and b.  And possibly the ratio of the two sides to each other.  Anyway that's my interpretation.  Could be wrong.

The groove thing I don't understand.
Commented:
>>  What do you know about the value of n for this rectangle?

Hmm. I wasn't thinking trivially! The smallest possible value in order to have positive area with positive lengths.

>>The groove thing I don't understand.

Commented:
I don't think (iii) was answered correctly:
(iii) If Jason writes 4 consecutive numbers in order, what do you know about the numbers if the number at the bottom of the triangle is divisible by 3?

Since the last number is a+3b+3c+d, it shows that a+d is divisible by 3.  I'm presuming here (correctly?) that the initial four numbers can be any integers with no relation between them.

3b and 3c will both be divisible by 3 if b and c are integers.  Therefore, a+d must be divisible by 3.

Try this with the first four numbers: 1 2 3 5
Second line: 3 5 8
Third line: 8 13
Fourth line: 21, which is divisible by 3.
Commented:
@BigRat
"Let R be the radius of the pizza then πR² is it's area, but this is equal to 3/4πr² so equating we get R=sqrt(3r/4) or if you like sqrt(3r)/2."
You missed π
Commented:
I tried to go through the exercises to see what I remember from school.
There are not easy, but not complicated, except probably the last one, which you already solved and discussed here, so I did not bother.
What I find difficult is to solve all in1h.
I did not know how long time it took me, because I was all the time interrupted by a call, a priority from a colleague and other small things. I will try later the 2nd exam with the clock in front of me, only a pen and paper. And editing in pdf is a nightmare  when come to equations making the time double.
I hope I did well and I passed the exam :)
exam1_solutions.pdf
Commented:
At (iii) I would answer in this way:
1st line x,  x+1,  x+2, x+3 because are 4 consecutive numbers, then the sum is:
2nd line 2x+1, 2x+3, 2x+5
3rd line 4x+4, 4x+8
4th line 8x+12
If 8x+12 is divisible with 3 then 8x must be divisible with 3, because 12 is divisible with 3.
If 8x must be divisible with 3 then x must be divisible with 3 because 8=2*2*2 only 2 as prime number.
So the answer is: the numbers on the 1st line are all divisible with 3.
Quid, Me Anxius Sum?  Illegitimi non carborundum.

Commented:
I'd have to do some revision of algebra before I had a go at this.  Where's my Schaums?  Those were damm good books.  Learned more out of those than what some of my fri**ing teachers taught.  And looking back at some of my teachers, they were useless.

@CompProbSolv

Numbers must be consecutive
1235 is not consecutive.  It is missing a 4.

@viki2000

It depends on how you look at the problem.  In BigRat's interpretation the Pi drops out.
Commented:
At (iii) above I was wrong: only the 1st (x) and the 4th number (x+3) is divisible with 3.
And why "Pi drops out"?

And by the way, here is your old Schaums 2nd edition (now at 6th):
http://mechanical.uonbi.ac.ke/sites/default/files/cae/engineering/mechanical/problems%20calculus.pdf
Quid, Me Anxius Sum?  Illegitimi non carborundum.

Commented:
Her method.

The area of the piece can be either A=3*Pi*r*r/4 where r is the radius of original pizza. or A=R*R*Pi where R is the radius of the piece.  This assumes that this piece is circular.

Then R*R*Pi=3*Pi*r*r/4
The Pi then drop out.
R*R=r*r*3/4
Take the square root of both sides
R=r*SquareRoot(3)/2  She forgot to take the r out of the SquareRoot.  Fail.  Bad BigRat.
Commented:
Well I wouldn't say "Bad" rat, but I must admit to getting the brackets wrong - which wouldn't be the first time. And, it seems, somebody else thinks I missed pi. Now I never miss my pis, because I mind my pies and queues.

As the author rightly points out, it assumes that the piece is circular. There is no reason to assume this, except for the possible fact that, in class, one had done such a calculation before.

It seems that these days there is a complete lack of vigour. There is another question, regarding computing acceleration which has caused a thread as long as this by sloppy specification of the problem.
Commented:
@dbrunton
I got her method.
She assumed R and r, so 2 circles, 2 circular pizzas.
Gosh! You guys never ate and cut pizza up to this age?
Since when you cut pieces of pizza circular?
Are you using a crystal glass up side down instead of a knife?
That is a really funny imagination :)
Commented:
I tried for my curiosity exam2 with paper and pen and a clock.
It took me 50min to go through all and that after I knew exam1, which made the life easier, because actually are the same exercises a bit changed.
I am a bit concerned about time only 60min when the child sees 1st time the exercises. It takes time to read, imagine, process, then to write, eventually to correct mistakes.
I think 1h is not fair enough for all these exercises, unless you did many similar in advance.
Commented:
@dbrunton:
"Numbers must be consecutive
1235 is not consecutive.  It is missing a 4."

You are correct, my error.  I missed the "consecutive" partly based on the initial numbers selected.

So... Rob was correct with one slight difference.  You don't need to add "or 0".  The complete answer is that "8x must be divisible by 3" (we're assuming integers here).  0 satisfies that condition without needing to call it out as an exception.

Regarding the time limit:  I see it a bit differently.  I came to the conclusion in my school years that test times should always be too short to allow anyone to complete the test.  If I get a test done in an hour and you take half an hour and we both get the same answers correct, shouldn't you get some bonus for having done it quicker?  That's how the real world works.
Commented:
Did you try to take the full exam by yourself and to count your timing?
Try that and see how fast you do it.
Maybe after 1h you change your opinion.

(iii) If Jason writes 4 consecutive numbers in order, what do you know about the numbers if the number at the bottom of the triangle is divisible by 3?
The text of the problem says “about the numbers”, referring at the initial numbers: x, x+1, x+2, x+3
I think the correct full answer is: only the 1st (x) and the 4th (x+3) number is divisible with 3.
Commented:
I don't think I was clear.  I'm inferring from your comment that I'm not likely to finish it within the alotted time.  My view is that this is proper.  I don't think that anyone should be able to finish the test.  This makes time a factor in your score, which I think is appropriate.
Commented: