## Is this a 6th grade geometry question?

The PSLE (Primary School Leaving Examination) occured in Signapore recently and it has some bloggers grumpy:

I swear, kids these days are learning stuff that are way more difficult than what we used to do back in primary school. Blame it on the big paper-chase that seems essential for surviving in this urban jungle we call Singapore, but really, by the time my kids come round to primary school, I don’t think I’ll be able to tutor them like my mother did to me.

Line segments AD, BD, and CD are congruent. What is angle ABC?

Here’s the other question that has commenters sounding off:

Jim bought some chocolates and gave half of it to Ken. Ken bought some sweets and gave half of it to Jim. Jim ate 12 sweets and Ken ate 18 chocolates. The ratio of Jim’s sweets to chocolates became 1:7 and the ratio of Ken’s sweets to chocolates became 1:4. How many sweets did Ken buy?

Simultaneous equations are required for a solution.

Also, one extra note from Wikipedia regarding the test:

The hard questions in the papers were to “filter” the average and below average students, as claimed by the Ministry of Education, only those who are truly the best can get good aggregate scores above 250.

### 11 Responses

1. You can do the chocolates/sweets problem with a “smart” guess-and-check. You know that each person starts off with the same number of sweets as the other, and the same number of chocolates as the other. And the number of chocolates each one starts with must be a multiple of 7. And when you subtract 18 from it, you must end up with a multiple of 4 — a big enough multiple of 4 that you can subtract 12 from a quarter of it (a quarter of the C-18 figure being the original number of sweets). So… a clever 12yo could solve this. It only took me a few guesses to hone in once I figured out how to set it up.

I don’t know the purpose of the exam, but is it a problem if there are a few problems on the exam that are “too hard” and that only a handful of students will solve? It seems to me that that give you the ability to identify the truly gifted math students, as well as to see who has achieved proficiency.

• It’s possible nearly all the students who got it right used guess and check. When you make a problem difficult enough with an “easy out” students will always take the easy out.

2. The kite looks not to have a unique solution. Set the three lengths at 2, just for concreteness. Fix BD, and let segment AC grow (and move south). The lengths can stay fixed, but the angle ABC grows….

• I have been wondering if some info has been lost from the transition from test to the blogs, but the comments all seem to be along the lines of the problem being impossible.

3. The other blog that has claimed a solution is indeed correct. That’s because the solution is based on the assumption that the outer figure is a square.

(Here’s why $\angle ABC = 150^{\circ}$. Suppose $\angle ABD = x$. Then, from congruency of appropriate triangles, we have $\angle ABC = 2x$. Now, note that $AC = AD = CD$ since the outer figure is a square. This allows us to compute $\angle CAD$ in two ways. It is not hard to show that we have $\angle CAD = 60^{\circ} = 2x - 90^{\circ}$, whence we have $\angle ABC = 2x = 150^{\circ}$.)

Of course, the way you stated the problem, it does not assume that the outer figure is a square, in which case, there is no unique solution as correctly pointed out by jd2718.

• For what its worth, my opinion is that I can imagine how this problem can be used to “filter” the above average (read: very good) students from the average ones or below. And, I even suspect that such “filters” are common devices employed on math test papers in most Asian countries. It should be kept in mind that this problem appeared on a ‘school leaving examination’, which again, I presume, is a common phenomenon in Asian countries. Could someone let me know if such is also the case here in the US?

• There (seem to be) two possible approaches:

1. Toss in some extremely difficult problems only the elite students will get.

2. Toss in enough high-medium difficulty problems that only the elite students will get every problem right.

On multiple choice tests the tendency (quite understandably) is towards #2. Typically (due to budget) the US tests are majority multiple choice.

The 8th grade geometry question I posted earlier bothers me in this respect because I’m guessing the majority of students short circuit the intended logic and use the fact the picture is drawn accurately.

• Hi,

The problem, as presented in the PSLE, has the following givens:
Let ABCD = square (A in top left hand corner, and name the other corners clockwise)
Let MN, which you currently label as AC, be parallel to AC
Let MN be perpendicular to PQ, which you currently label as BD
MN=PQ=NQ
What is <MPN?

Re < the original geometry problem posed
I just finished a series of all-day mathematics workshops with some teachers, and I asked when students learn equilateral triangle, sum of the angles of a triangle, isosceles triangle, etc., and I was told they learn these properties in gr.3/4. What doesn't occur in the U.S. is asking kids to use what they are taught, as occurs in the <MPN problem. I particularly like this problem, because elementary students (and teachers) can figure it out with elementary mathematics knowledge.
Thanks for including this problem.
patsy

4. I don’t think this is much of a test question, but I would love for my students to work it out by doing constructions (by hand or on a program like geogebra or geometer’s sketch pad).

Where do you get these question? I love the ones you’ve been posting!

• Thank you!

Every source has been different. This one I bumped into entirely by accident on the Internet, the 8th grade one I ran across because I was doing research on the TIMSS, and if you’re referring to some of my other posts they are either adapted from books or coming out of my own head.

• Why do you need construction to solve the problem?
Patsy