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I came across this quiz question in a forum today. I would like to ask for your opinion of this notorious mathematics question, and also, to share.

PSLE 2005 Question

This question came out in Singapore's PSLE examination 2005. Basically, PSLE is one of the most important exams of one's life in Singapore. It literally determines your future, to a huge degree. How did such a question slip past the examiners, we would never know.

As for my real question, is that, is there a valid answer to this question ultimately? I mean, there can be more then one answer, but can it be argued in such a way, that one answer can ultimately be proven to be true?

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Please feel free to edit my title. I am not sure what to put as a title. – Yellow Skies Nov 21 '12 at 16:15
Singaporean Dude: Since you are explicitly asking for "opinions" about the problem, ("I would like to ask for your opinion of this notorious mathematics question, and also, to share.") perhaps you can add the "community wiki" tag? I have not problems with it being posted here, but is asking for a poll of sorts, and might invite debate/argumentation. – amWhy Nov 21 '12 at 16:18
Well, I am looking for a logic based answer to this question. Something along the lines of using logical operators to prove that 1 argument fails and the other argument succeeds. I do not believe that there can be two answer for a question. It's like saying, there are two different values for an area. – Yellow Skies Nov 21 '12 at 16:20
Oh, and the correct answer would of course be that the assumptions (i.e., the numbers together with the sketch) are contradictory, so there really is no correct answer in the given multiple choice options. – Lukas Geyer Nov 21 '12 at 16:22
It means the problem is underdetermined. It's not entirely unlike the fact that $x^2 - 3 x + 2 = 0$ has two solutions... – Zhen Lin Nov 21 '12 at 16:32
up vote 2 down vote accepted

Triangle $B=QVX$ and $XUQ$ are congruent, triangle $A=SWX$ and $XUQ$ are similar and by comparing their areas, the scaling factor is 2. Hence $XU$ is $\frac23$ of $SR$ and $RU$ is $\frac13$ of $RQ$, thus the the area of rectangle $C$ is $\frac29$ of the big rectangle, i.e. $20\text{ cm}^2$. By the same argument, rectangle $TXVP$ also has area $20\text{ cm}^2$. But then for the total area we obtain $80$ instead of $90\text{ cm}^2$, i.e the given information is inconsistant. That means that suitable reasoning might arrive at any other value for the area of $C$ (ex falsum quodlibet).

We conclude that the problem is ill-posed or we made an unwarranted assumtion. However, the assumptions were essentially that $SXQ$ and $WXV$ are straight lines. If we drop these, there are a lot of degrees of freedom, enough to allow a whole range of different areas to be obtainable for $C$.

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It's "ex falso quodlibet", if you inflect "falsum" correctly... (The preposition "ex" calls for the ablative.) – Zhen Lin Nov 21 '12 at 19:57

Assume line segment $SXQ$ bisects the rectangle $PQRS$, and assume that all line segment intersections form right angles except those intersections involving segment $SXQ$. All areas are in square centimeters:

The total area is $90$. Let's label the rectangle $PVXT$ as having area $D$. The desired area $C$ is given by $C = 90 - 2A - 2B - D$. We know $A$ and $B$, so we need to find $D.$ Considering the triangle $PQS$, with area $90/2 = 45$, we can solve for $D$ with $D = 90/2 - A - B = 25$, thus $C = 25$.

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