Posted on November 22, 2007 by shiunn - http://singaporewalk.wordpress.com/2007/11/22/psle-highest-score-in-17-years/
Natasha Nabila Muhammad Nasir from the Gifted Education Programme.
With an aggregate score of 294, Natasha – who is also the top Malay student – has the highest PSLE score in 17 years since the current PSLE system was implemented in 1991.
One of the top Chinese students is Leong Feng Ping from West Grove Primary, while the top Indian student is Vanessa Malishree Dharmaratnam from Raffles Girls’ Primary.
The top Eurasian student is Santa Maria Priscilla Nicole from CHIJ Katong Primary.
A total of 49,817 Primary 6 pupils sat for the PSLE this year – 798 pupils fewer than last year’s cohort.
Of these, 97.7 percent are eligible for secondary schools.
63.5 percent of these students made it to the Express course, 22.2 percent for Normal (Academic) course and 12 percent for the Normal (Technical) course.
A total of 1,152 pupils are assessed as not ready for secondary school in 2008 or more suited for vocational training.
I felt that it is truely amazing that someone can actually obtain 394 out of 400 for the exam, especially when there are languages papers and essays. Anyway, i must agree that Singapore is still an elitist society. Even articles are filling readers with statistics about how many students are going for which kind of courses be it special express or normal, this goes to show that the readers are also very interested to find out more about the situation about who can make it where in education of singapore. Well, be it good or bad it all depends on our own perspective.
Thursday, December 6, 2007
Saturday, November 10, 2007
Friday, November 9, 2007
My Final Post on 2007 PSLE Math Exam Paper
The questions reported in Straits Times are simplified!
Yes, if you have noticed, I have stopped posting for a while because I felt that justice was not done to our Primary 6 students.
Many claimed that they can solve the questions in 3 minutes, or did not find the question tough at all.
However, many readers were misled because the questions shown in the Straits Times are not asked in the exact manner as in the PSLE paper!!!
(of course, we cannot blame the Straits Times for not printing the real questions for they do not have the real one too!)
This is the newspaper report:
To help you see better, let's look at question 2 that was printed in the Straits Times:
6/14 of the chairs in a hall are in rows of 13. Half of the chairs are in rows of 7. There are 112 more chairs in rows of 7. The rest of the chairs are stacked up. Find the total number of chairs.
Now, take a look at the REAL question that was asked in 2007 PSLE Math paper:
There are some workers. They arranged chairs into short rows and long rows. There are 18 short rows and 7 long rows. Each long row has 16 more chairs than the short row. Half of the chairs are arranged into long rows while 3/7 of chairs are arranged in short rows. The rest of the remaining chairs are stacked in a corner.
a) What fraction of the chairs are stacked in the corner?
b) How many chairs are arranged in rows?
Do you see the difference?
Under exam environment, do you agree that the real question appears to be more difficult than the one shown in the newspapers?
Let's put all these pasts away and focus on what is ahead of us! (This will be my last post about 2007 PSLE paper)
Afterall, this is a changing world and we need to constantly change ourselves to meet the new demand. Continue to gear yourself up and strive for your best!
Yes, if you have noticed, I have stopped posting for a while because I felt that justice was not done to our Primary 6 students.
Many claimed that they can solve the questions in 3 minutes, or did not find the question tough at all.
However, many readers were misled because the questions shown in the Straits Times are not asked in the exact manner as in the PSLE paper!!!
(of course, we cannot blame the Straits Times for not printing the real questions for they do not have the real one too!)
This is the newspaper report:
To help you see better, let's look at question 2 that was printed in the Straits Times:
6/14 of the chairs in a hall are in rows of 13. Half of the chairs are in rows of 7. There are 112 more chairs in rows of 7. The rest of the chairs are stacked up. Find the total number of chairs.
Now, take a look at the REAL question that was asked in 2007 PSLE Math paper:
There are some workers. They arranged chairs into short rows and long rows. There are 18 short rows and 7 long rows. Each long row has 16 more chairs than the short row. Half of the chairs are arranged into long rows while 3/7 of chairs are arranged in short rows. The rest of the remaining chairs are stacked in a corner.
a) What fraction of the chairs are stacked in the corner?
b) How many chairs are arranged in rows?
Do you see the difference?
Under exam environment, do you agree that the real question appears to be more difficult than the one shown in the newspapers?
Let's put all these pasts away and focus on what is ahead of us! (This will be my last post about 2007 PSLE paper)
Afterall, this is a changing world and we need to constantly change ourselves to meet the new demand. Continue to gear yourself up and strive for your best!
Saturday, October 20, 2007
2007 PSLE Math - Letter From Today Newspaper Forum
Someone sent me this letter. This letter is from the mother of a TOP student.
Today newspaper forum - Letter from GUO WEIFU
My son, a top student, returned home shattered after difficult PSLE maths paper.
THIS year’s PSLE mathematics paper has caused many students undeserved anguish.
Their reactions, including that of the good and gifted ones, were one of despair and discouragement upon encountering problem sums that were set beyond their capabilities. Some even broke down right after the examination. My son is in a top class. Mentally strong and confident in mathematics, he returned shattered boy after the exam. I immediately checked with friends and teachers from other classes and schools, and the same story was heard. Although schools have been equipping pupils with the skills of applying mathematical concepts, many could not complete half or more of the problem sums.
Is there a huge gap between expectations and ability?
In the mind of a child, he has either failed or done badly. With his confidence ripped by the mathematics paper, how can a child recover in less than 24 hours to face the next paper?Tearing children apart mentally at a developmental stage snuffs out the joy of learning. The Singapore Examinations and Assessment Board (SEAB) had previously said “the exams were a means to differentiate students of diverse abilities by setting questions of varying difficulty levels”. It stated that an examination paper should be comparable in difficulty from one year to the next and “teachers know how to prepare their students for the examinations”.
SEAB also stated that different teams of experienced setters, moderators and examiners conduct several rounds of checking of each question and for the paper as a whole.
The fact different teams are involved is no consolation as it may be a systemic failure from pegging the paper at too high a standard. For some parents and students, education is seen as a source of mental torture.
All I ask is this: Please give the children of Singapore the confidence to face their future with a lion’s heart. Whaddaya think???
Today newspaper forum - Letter from GUO WEIFU
My son, a top student, returned home shattered after difficult PSLE maths paper.
THIS year’s PSLE mathematics paper has caused many students undeserved anguish.
Their reactions, including that of the good and gifted ones, were one of despair and discouragement upon encountering problem sums that were set beyond their capabilities. Some even broke down right after the examination. My son is in a top class. Mentally strong and confident in mathematics, he returned shattered boy after the exam. I immediately checked with friends and teachers from other classes and schools, and the same story was heard. Although schools have been equipping pupils with the skills of applying mathematical concepts, many could not complete half or more of the problem sums.
Is there a huge gap between expectations and ability?
In the mind of a child, he has either failed or done badly. With his confidence ripped by the mathematics paper, how can a child recover in less than 24 hours to face the next paper?Tearing children apart mentally at a developmental stage snuffs out the joy of learning. The Singapore Examinations and Assessment Board (SEAB) had previously said “the exams were a means to differentiate students of diverse abilities by setting questions of varying difficulty levels”. It stated that an examination paper should be comparable in difficulty from one year to the next and “teachers know how to prepare their students for the examinations”.
SEAB also stated that different teams of experienced setters, moderators and examiners conduct several rounds of checking of each question and for the paper as a whole.
The fact different teams are involved is no consolation as it may be a systemic failure from pegging the paper at too high a standard. For some parents and students, education is seen as a source of mental torture.
All I ask is this: Please give the children of Singapore the confidence to face their future with a lion’s heart. Whaddaya think???
Tuesday, October 16, 2007
PSLE Math - Reply By Singapore Examinations And Assessment Board
The Singapore Examinations and Assessment Board has made their reply on 2007 PSLE math paper.
What is your view? Do you agree with their reply?
Please send in your comments!
Regards,
Norman Tien
Sunday, October 14, 2007
PSLE maths questions stump pupils - Feedback Posted in The Straits Times Forum
Yes, this feedback by Mrs Piyali Roy was inline with the many feedback I received from my primary 6 students after their PSLE math paper.
Well, like I have mentioned prior to the PSLE examination, the trend observed from the many schools' math prelim paper showed that questions asked will no longer be testing their arithmetic skills only.
Let's look at one of the question asked (this is not the exact PSLE math question as I do not have the exact dimension and diagram)
2007 PSLE Math Question:
Tank A is filled with water to its brim.
Water is then poured from Tank A to Tank B till both shared the same height (see fig. 2).
What is this height?
What is this height?
This is an obvious "volume" question but many primary 6 students were 'frightened' by the way the question was being asked.
It looked so simple but yet, most students do not have the slightest clue where to begin!
Before we look at the solution, why not you all take out your pen and paper and start working on this.
(now, to be fair to those primary 6 students who sat for their PSLE, I want you to solve this question based on what a primary 6 child would know about the topic "Volume". In short, do not use what you have learned from your Secondary or Tertiary education to solve this question!)
Can you solve it? Was it easy to solve?
How long did you take to solve it?
What is your answer?
Actually, if one knows the concept well, we can solve this question EASILY by just applying the formula:
Height = Volume divide by Base Area
How long did you take to solve it?
What is your answer?
Actually, if one knows the concept well, we can solve this question EASILY by just applying the formula:
Height = Volume divide by Base Area
whereby Volume is the volume of water in Tank A, Base Area is the total base area of the 2 tanks!
Let's calculate:
Volume of water in Tank A = 50 x 40 x 60 = 120,000 cubic centimeters
Base Area of Tank A = 50 x 40 = 2000 square centimeters
Base Area of Tank B = 70 x 40 = 2800 square centimeters
Volume of water in Tank A = 50 x 40 x 60 = 120,000 cubic centimeters
Base Area of Tank A = 50 x 40 = 2000 square centimeters
Base Area of Tank B = 70 x 40 = 2800 square centimeters
Total Base Area of 2 tanks = 4800 square centimeters
Therefore, to find the height of water in both tanks,
height = 120,000 divide by 4,800 = 25cm
Answer: height = 25cm
height = 120,000 divide by 4,800 = 25cm
Answer: height = 25cm
So, did you get the same answer too?
I am proud of you if you did so, especially if you are a primary 6 child!
If you do have more of 2007 'tough' PSLE questions, I would appreciate if you could send them to norman@pslemath.com.
I will share with all my readers once I received your contribution.
Attention:
For parents whose child is in Primary 5, you would also like to visit www.pslemath.com to view the coming P5 Exam Preparation Prog I will be conducting this coming school holidays (PSLE marking days)
Sunday, September 30, 2007
Friday, September 28, 2007
Nanyang Primary School 2007 PSLE Math Prelim
Question:
At a school carnival, there were 520 more girls than boys. 1/8 of the girls and 20% of the boys left the carnival. In the end, there were 488 more girls than boys.
(a) Did more girls or boys leave the carnival? How many more?
(b) How many children were there at the carnival in the end?
Once again, we have another challenging question and in order to solve it, we are required to do a bit of logical thinking.
To solve part a, we need to perform 3 case studies first:
Case 1 – Same number of girls & boys left the carnival
Question:
At a school carnival, there were 520 more girls than boys. 1/8 of the girls and 20% of the boys left the carnival. In the end, there were 488 more girls than boys.
(a) Did more girls or boys leave the carnival? How many more?
(b) How many children were there at the carnival in the end?
Once again, we have another challenging question and in order to solve it, we are required to do a bit of logical thinking.
To solve part a, we need to perform 3 case studies first:
Case 1 – Same number of girls & boys left the carnival
If there are an equal number of girls and boys leaving the carnival, the difference will still be the same, 520.
Case 2 – More boys left the carnival
Case 3 – More girls left the carnival
If there are more girls than boys leaving the carnival, the difference will be smaller than 520.
(a) The question mentioned that after 1/8 of the girls and 20% of the boys left the carnival, there were 488 more girls than boys.
The difference given in this question is 488, which is smaller than the original difference of 520.
As such, can you identify which case study is closest to our question?
Yes, we should use case 3 (more girls left) because the new difference (488) is less than the original difference (520). Therefore, for above question, more girls left the carnival.
To calculate how many more girls than boys left the carnival, we need to analyze the 3 case studies again.
If the same number of boys and girls left the carnival, the difference would remain as 520.
Since there is a decrease in the final difference as compared to the original difference, therefore, using 520 - 488 = 32, we can justify that 32 more girls left the carnival.
This is illustrated in the diagram below:
(b) To solve part b and for illustration purposes, we assume 100% = 100u.
The strategy to solve part b is that you must be able to apply the technique of cutting the models separately.
To find 1/8 of the Girls model, you need to cut the 100u and 520 separately.
1/8 of 100u = 12.5u. If 12.5u left the carnival, 87.5u will remain.
1/8 of 520 = 65. If 65 girls left the carnival, 455 girls will remain.
Combining the 87.5u and 455 girls, the model of the Girls after 1/8 of them left will be as shown in the 'After' portion of the Girls model below.
To illustrate 20% of the Boys left the carnival, the 'After' portion of the Boys model shows 80u because 20u had left the carnival.
488 – 455 = 33
7.5u --> 33
Total units: 87.5u + 80u = 167.5u
167.5u --> 33/7.5 x 167.5/1 = 737
737 + 455 = 1192
Total units: 87.5u + 80u = 167.5u
167.5u --> 33/7.5 x 167.5/1 = 737
737 + 455 = 1192
From the model, we found that 7.5u = 488 - 455 = 33.
Therefore, 1u = 4.4
Since the total number of Boys and Girls = 87.5u + 80u +455 = 167.5u + 455, the total number of boys and girls at the carnival in the end is (167.5 x 4.4) + 455 = 737 + 455 = 1192.
Phew! That was tough, wasn't it?
Ok, I'll shall see you tomorrow, same time same channel :)
Norman Tien
http://www.pslemath.com
Wednesday, September 26, 2007
Christian Brothers School 2007 Prelim
Question:
There were 2 identical flights of steps. For the 1st flight of steps, James walked up some steps and ran 6 steps in 74s. For the 2nd flight of steps, James walked up some steps and ran 14 steps in 42s. How long will James take to walk up both flights of steps?
Do you find this question challenging? Well, I do.
This question actually appeared in year 2005 PSLE Math examination, if I recalled the year correctly.
In fact, ever since that year, many schools used this question either as work practices or as exam questions. Despite the fact that the students were taught the strategy to solve this question, I noticed many still do not understand the logics behind the solution and therefore, are still unable to solve it.
The following is my step-by-step solution.
First, let me show you the 2 identical flights of steps.
If James ran 6 steps and walked the rest of the steps, he will take 74 seconds.
If James ran 14 steps and walked the rest of the steps, he will take 42 seconds.
By comparing the two flights of steps below, I can logically say that the difference in time for James to walk up the 2 flights of steps is caused by walking up the 8 steps in the first flight instead of running up the 8 steps in the second flight.
If James took 32 seconds (74sec – 42sec) more to walk up 8 steps instead of to run up 8 steps, therefore, the extra time needed to walk up 1 step as compared to run up 1 step is 4 seconds longer. (Logic & Reasoning: Walking time is normally longer than running time)
To calculate the time taken to walk up the entire flight of steps, I will just need to convert the time to “run 6 steps” to walking:
If James needs 4 seconds more to walk up 1 step than to run up 1 step, therefore, he will need 24 seconds (6 steps x 4 seonds) more to walk up 6 steps than to run up 6 steps.
So, if James were to walk up the entire flight of steps, he will take 98 seconds (74 + 24).
If I calculate the time for James to walk up the second flight of steps, I should obtain the same time as the first flight of steps, that is 98 seconds.
Well, to prove it, let's work out the workings:
If James needs 4 seconds more to walk up 1 step than to run up 1 step, he will need 56 seconds (14 steps x 4 seconds) more to walk up 14 steps than to run up 14 steps.
Therefore, if James walks up the entire flight of steps, he will also take 98 seconds (42 + 56)!
Therefore to answer the question,
For James to walk up the two flights of steps, he will take 196 seconds (98 + 98).
Did my drawings help you visualize the question and solution better?
For you to master solving this question, my advice is that you should practice this question at least once after watching the solution.
Then, solve this question again 2 days later, but without watching the solution.
If you are able to do it, congratulations! …. because you have truly understood the logics of solving it!!
If you would like to be informed of my new post, please subscribe to my FREE mailing list at www.pslemath.com
Bye,
Norman Tien
http://www.pslemath.com/
Question:
There were 2 identical flights of steps. For the 1st flight of steps, James walked up some steps and ran 6 steps in 74s. For the 2nd flight of steps, James walked up some steps and ran 14 steps in 42s. How long will James take to walk up both flights of steps?
Do you find this question challenging? Well, I do.
This question actually appeared in year 2005 PSLE Math examination, if I recalled the year correctly.
In fact, ever since that year, many schools used this question either as work practices or as exam questions. Despite the fact that the students were taught the strategy to solve this question, I noticed many still do not understand the logics behind the solution and therefore, are still unable to solve it.
The following is my step-by-step solution.
First, let me show you the 2 identical flights of steps.
If James ran 6 steps and walked the rest of the steps, he will take 74 seconds.
If James ran 14 steps and walked the rest of the steps, he will take 42 seconds.
By comparing the two flights of steps below, I can logically say that the difference in time for James to walk up the 2 flights of steps is caused by walking up the 8 steps in the first flight instead of running up the 8 steps in the second flight.
If James took 32 seconds (74sec – 42sec) more to walk up 8 steps instead of to run up 8 steps, therefore, the extra time needed to walk up 1 step as compared to run up 1 step is 4 seconds longer. (Logic & Reasoning: Walking time is normally longer than running time)
To calculate the time taken to walk up the entire flight of steps, I will just need to convert the time to “run 6 steps” to walking:
If James needs 4 seconds more to walk up 1 step than to run up 1 step, therefore, he will need 24 seconds (6 steps x 4 seonds) more to walk up 6 steps than to run up 6 steps.
So, if James were to walk up the entire flight of steps, he will take 98 seconds (74 + 24).
If I calculate the time for James to walk up the second flight of steps, I should obtain the same time as the first flight of steps, that is 98 seconds.
Well, to prove it, let's work out the workings:
If James needs 4 seconds more to walk up 1 step than to run up 1 step, he will need 56 seconds (14 steps x 4 seconds) more to walk up 14 steps than to run up 14 steps.
Therefore, if James walks up the entire flight of steps, he will also take 98 seconds (42 + 56)!
Therefore to answer the question,
For James to walk up the two flights of steps, he will take 196 seconds (98 + 98).
Did my drawings help you visualize the question and solution better?
For you to master solving this question, my advice is that you should practice this question at least once after watching the solution.
Then, solve this question again 2 days later, but without watching the solution.
If you are able to do it, congratulations! …. because you have truly understood the logics of solving it!!
If you would like to be informed of my new post, please subscribe to my FREE mailing list at www.pslemath.com
Bye,
Norman Tien
http://www.pslemath.com/
Sunday, September 23, 2007
CHIJ Katong 2007 Math Prelim
Question:
In the figure below, ABC is a right-angled triangle. AB = 12 cm and
BC = 16 cm. Find the area of the shaded parts.
(Leave your answer in terms of π)
From the above question, everyone can easily tell that it is a Circle question.
Based on my analysis of all Circle questions that appeared in the 2007 math prelim exam papers, I have to retract my above sentence as these questions are no longer ‘free gift’ questions.
I bet for some parents, even after I have provided you with my answer, you might still not be able to understand how it was done!
To find the shaded area, all you need to do is this:
Can you understand my solution or are you now confused !!!!!?????
Well, for this question, the challenge faced by most primary 6 students is that they are not able to find the area of the shaded parts individually!
To solve this question, not only your foundation of Circles topic needs to be strong, you must be stronger in your Visual Processing Cognitive Skills (your ability to see, rotate and hold the diagram in your mind as a whole or in parts)!
Well, let me explain my solutions.
To help you to see better, first, I will split the diagram into parts and label these parts as shown below:
If I split the given diagram into 2 semi circles and add them up, do you agree with me that the total (in terms of parts) is equal to A, B, C, C, D, D, X & Y. In short, I will have one part A, one part B, two part C, two part D, one part X and one part Y.
Now, let us look at what parts this triangle contains:
To illustrate using A, B, C, D …., this is how it is done:
I hope I have added an additional tool for your child to tackle “tougher” questions on Circles.
Question:
In the figure below, ABC is a right-angled triangle. AB = 12 cm and
BC = 16 cm. Find the area of the shaded parts.
(Leave your answer in terms of π)
From the above question, everyone can easily tell that it is a Circle question.
I used to tell my students that circle questions for primary 6 level are actually ‘free gift’ questions as all students should be able to score for this type of question if they knew the two formulas well; the circumference of a circle = π x d and the area of a circle = π x r x r
Based on my analysis of all Circle questions that appeared in the 2007 math prelim exam papers, I have to retract my above sentence as these questions are no longer ‘free gift’ questions.
Please study the above question and see if you can solve it?
I bet for some parents, even after I have provided you with my answer, you might still not be able to understand how it was done!
Once you are done with you solution, check yours with mine.
To find the shaded area, all you need to do is this:
Can you understand my solution or are you now confused !!!!!?????
Well, for this question, the challenge faced by most primary 6 students is that they are not able to find the area of the shaded parts individually!
To solve this question, not only your foundation of Circles topic needs to be strong, you must be stronger in your Visual Processing Cognitive Skills (your ability to see, rotate and hold the diagram in your mind as a whole or in parts)!
Well, let me explain my solutions.
To help you to see better, first, I will split the diagram into parts and label these parts as shown below:
If I split the given diagram into 2 semi circles and add them up, do you agree with me that the total (in terms of parts) is equal to A, B, C, C, D, D, X & Y. In short, I will have one part A, one part B, two part C, two part D, one part X and one part Y.
Now, let us look at what parts this triangle contains:
(this triangle has one part C, one part D, one part X & one part Y)
Therefore, when I use the sum area of 2 semi circles to subtract the area of a triangle, I will end up with parts A, B, C & D and these four parts actually make up shaded parts of the diagram. cool right? ...hehe
To illustrate using A, B, C, D …., this is how it is done:
I hope I have added an additional tool for your child to tackle “tougher” questions on Circles.
I will continue to post more challenging questions till the final day before the PSLE. So, please remember to check out this site frequently.
Cheers
Norman Tien
Norman Tien
**If you have not subscribe to my FREE mailing list, please do so now by visiting http://www.pslemath.com/
Wednesday, September 19, 2007
2007 Math Prelim Papers
When I received the recent prelim exam papers, I was taken aback by some of the questions set. Not only were some of these ‘cheem’ questions extremely lengthy (some can have as many as 8 sentences in a question), the person reading them must be strong in their math concepts just to be able to piece the question together to ‘see’ the clues. Surprisingly, these “cheem” questions were not only from our elite schools but a growing numbers were originated from neighbourhood schools!
Judging from the growing trend of lengthy questions with many conditions attached, all students must not only know the Fundamental Concepts of each topic well, but must also be strong in their Cognitive Skills (the skills needed to process information mentally) in order to score well. These skills include Visual Processing (ability to rotate or flip a diagram in mind), Working Memory (ability to hold and process a group of information mentally) and Logic & Reasoning (ability to make see logics in questions).
I have selected some recent prelim questions for discussion as I personally felt that despite the fact that some students might be strong in problem solving, their learning styles and cognitive abilities might cause them to under perform for these questions.
Nanyang Primary School 2007 Math Prelim
Question:
A rectangle piece of paper is folded along AC at Corner B as shown in Figure 1 so that the line BC lies on the centre line.
To prove my point, referring to figure 2, you should spot a rhombus before folding DC. Therefore, in figure 3, angle ACD equals angle ADC (22.5 degree) as line AC equals line AD. Thus, angle CAD is 135 degree (180 – 22.5 – 22.5 = 135).
When I received the recent prelim exam papers, I was taken aback by some of the questions set. Not only were some of these ‘cheem’ questions extremely lengthy (some can have as many as 8 sentences in a question), the person reading them must be strong in their math concepts just to be able to piece the question together to ‘see’ the clues. Surprisingly, these “cheem” questions were not only from our elite schools but a growing numbers were originated from neighbourhood schools!
Judging from the growing trend of lengthy questions with many conditions attached, all students must not only know the Fundamental Concepts of each topic well, but must also be strong in their Cognitive Skills (the skills needed to process information mentally) in order to score well. These skills include Visual Processing (ability to rotate or flip a diagram in mind), Working Memory (ability to hold and process a group of information mentally) and Logic & Reasoning (ability to make see logics in questions).
I have selected some recent prelim questions for discussion as I personally felt that despite the fact that some students might be strong in problem solving, their learning styles and cognitive abilities might cause them to under perform for these questions.
Nanyang Primary School 2007 Math Prelim
Question:
A rectangle piece of paper is folded along AC at Corner B as shown in Figure 1 so that the line BC lies on the centre line.
Finally, it is folded along EF as shown in Figure 3 so that the line GD lies on the centre line.
Find angle b.
Can you imagine this question is actually an Angle question?
Frankly, I read this question 3 times but for every attempt, I was not able to complete the entire question because whenever I reached figure 3, I cannot proceed on as the diagram shown in the question is different from the one I had in my mental mind!
(What I was suffering from is commonly known as ‘mental block’. This will occur to most learners who are strong in their Logic & Reasoning skill. Whenever their perceived mental picture or logic differs from those shown in the question, the mind will become extremely ‘noisy’ and will be reluctant to move on until the differences are resolved.)
Finally, to help me move on, I took an A4 paper and started folding and only then, I was able to move on and complete the question as my folded paper revealed the same image as the one I had in my mind!
From an educator’s point of view, this question is a very good question as it tests a student’s true understanding of symmetry as well as the geometrical properties of triangles, four-sided figures and angles.
In order to solve this question, besides having a good grasp of the fundamentals of Symmetry, geometrical properties of Triangles, four sided figures and angles, a student must also be strong in their Visual Processing and Logic Reasoning skills.
1. To help you see and understand better, first fold an A4 paper into half as shown in figure 1
2. Next, fold a triangle ABC (as shown in figure 2). Noticed angle BAC and angle BCA are the same (line BC = line BA and triangle ABC is an isosceles triangle)
3. Then, fold such that line AC lies on the centre line of the A4 paper. You will notice that your folded paper is different from figure 3 because line AC and line AD should be of the same length! (this is the ‘cause’ of my mental block as instead of drawing line AC to be the same length as line AD, the picture showed line AC to be shorter than line AD)
To prove my point, referring to figure 2, you should spot a rhombus before folding DC. Therefore, in figure 3, angle ACD equals angle ADC (22.5 degree) as line AC equals line AD. Thus, angle CAD is 135 degree (180 – 22.5 – 22.5 = 135).4. Before you fold your paper to form figure 4, you will notice that line GD is parallel to line AC. Therefore, angle CAD equals angle ADG (135 degree) due to alternate angles.
5. Now, I believe you can confidently solve for angle b as it is the sum of angle ADG (135 degree) and angle ADC (22.5 degree).
Answer: angle b = 157.5 degree
_____________________________________________________________
Singapore Chinese Girls School 2007 Math Prelim
Question
When Mrs Lee was 40 years old, her son was twice her daughter’s age. Mrs Lee will be twice her son’s age when her daughter is 28 years old. How old will Mrs Lee be when her daughter is 20 years old?
Unlike the Nanyang question, I was ‘shocked’ when I first read this question, for what first comes to my mind is …….. ‘Is this a primary school question?’
I have a reason for feeling this way. My reaction was caused by my immediate choice of approach to solve this question. As there were too many variables or unknowns in this question, I used what most adults will use to solve this question, Algebra.
This was how I solve the question for my first attempt.
Let daughter’s age = x
When Mrs Lee is 40, daughter = x and son = 2x
‘y’ years later, when daughter becomes 28 years old
Mrs Lee will be 40 + y, daughter = x + y and son = 2x + y
From above, x + y = 28 --> equation 1
and since Mrs Lee will be twice son’s age, that is 2(2x+y)
therefore, 2(2x+y) = 40 + y --> equation 2
Using the above 2 equations and the substitution technique, I found daughter’s initial
age was 4 years old (x=4) and thus when she is 28 years old (16 years later),
Mrs Lee will be 40 + 16 = 56 years old
If you noticed my above solutions (does not matter if you don’t understand because you will be taught them when you are in secondary school), my default mode was to use Algebra to solve.
After solving, the next question that followed my mind was “How many primary 6 students, who were not taught algebraic expansion and simultaneous equations, can solve this question?”
Staring at the question for a while, I resumed the resource of a primary 6 student and re-approached the question using techniques like listing of tables, drafting of models and finally, after many minutes, I obtained the correct answer through Guess and Check (yeah)!
Not satisfied with the Guess and Check approach as the time taken was way too long, I began to explore the model drawing approach again. After overcoming the challenge of not being able to draw the models for this sentence, Mrs Lee was 40 years old, the son is twice the daughter’s age, I finally arrived the answer.
The following Step-by-Step model solution might not be easily accepted by some students, especially students who belong to the sequential learner group.
1. Construct the first set of models of ‘when Mrs Lee was 40 years old, son is twice the daughter’s age’ by JUST listing and not drawing Mrs Lee’s model first, drawing 1 unit for the daughter and 2 units for son. (for this step, sequential learners, students who need to do things in sequence, will feel uneasy when told not to draw Mrs Lee’s model first)
2. To draw the next step correctly, one needs to be strong in their language. ‘When the daughter is 28 years’, for this sentence, you will need to agree that you have to add some more years instead of subtracting a number of years from the current age. (you will really need to read the ‘English’ of the sentence, to pick up the present and past tense of the sentence for you to agree that it should be some years later from the initial age for her daughter to become 28 years old.)
My reasoning is highlighted in red:
‘When Mrs Lee was 40 years old, her son was twice her daughter’s age. Mrs Lee will be twice her son’s age when her daughter is 28 years old. How old will Mrs Lee be when her daughter is 20 years old?
Therefore, I shall construct a second set of models by adding an additional block to daughter and son to represent some years later. I shall name this additional block as ‘y’.
3. Following the condition of the sentence ‘When daughter is 28, Mrs Lee will be twice her son’s age’, I am now able to draw Mrs Lee’s model by copying twice the son’s model.
4. Now, you need to spot for patterns from the 2 sets of models. If you noticed the first and second set of the models, the model of the son and daughter in the first set is ‘y’ block shorter than the model of the son and daughter in the second set. Therefore, logically, the model of Mrs Lee in the first set must be ‘y’ block shorter than the model of Mrs Lee in the second set. Therefore, we can now construct the model of Mrs Lee in the 1st set.
5. When we begin to label the models, we will arrive the following equation:
4 units + y = 40 years
1 unit + y = 28 years
3 units = 40 – 28 = 12 years
Therefore, 1 unit = 4 years
The daughter was 4 years old when Mrs Lee was 40 years old.
For her daughter to become 20 years old, we need to add 16 more years to her
daughter. Therefore, Mrs Lee will be 40 + 16 = 56 years old when her daughter is
20 years old.
I hope my sharing has benefited you and your child. Before I end this posting, I would like
you print out an elite school’s 2007 math prelim paper for your child to work on. An answer
key is included for marking purposes. From the results, you can have a good gauge as to
how prepared your child is for the coming PSLE. I hope your child will benefit from this
2007 prelim paper.
To get a copy of a FREE PSLE 2007 Math Prelim Exam Paper, click here
Norman Tien
http://www.pslemath.com/
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