[Pri20150306TTT] The Tricky Triangle

Question

Introduction

     This is another one of those tricky primary school mathematics questions involving areas.  A perfunctory glance at the area seems to suggest there are four pieces.  Later you might realise that you can think of it as two quarter-circles with two little 45°-45°-90° isosceles triangles removed. 

Plan

     Our plan will be to first find the areas of the two quarter circles and then to subtract the areas of the isosceles triangles.  This is our usual divide-and-conquer strategy [ Heuristics H10 & H11 ].  Notice that the two quarter-circles can be rearranged [H09] into a semi-circle with radius 10 cm.  Simple enough.

What about the two exised triangles?  Notice that the longest sides  (the sloping sides) of the triangles (highlighted in green) are each equal to the radius  10 cm  of the quarter-circles, simply because they, by touching the arcs, are themselves also radii of the quarter-circles.

However, the problem seems to be that we do not know the base and the height of each triangle.  Examiners for Singapore Primary School mathematics like to set this sort of questions involving areas of isosceles right-angled triangles, in which you are given only the length of the hypotenuse (the longest side).  How to tackle this kind of situation?  By using our imagination!

Imagine that the two triangles are brought together.  This forms a larger right-angled isosceles triangle.  However, now you realise it is half of a 10 cm by 10 cm square.  You can also imagine turning the triangle around until one of the 10 cm sides is horizontal.  Treating this as the base, the height of the triangle is 10 cm.  Either way, you are able to solve it and get the same answer.

     All that is left now is to subtract this from your area of the semi-circle found earlier.

Solution

   Shaded Area [in cm²]

= Area of two quarter-circles – area of two triangles

= Area of semi-circle – area of combined triangle

= ½ ´ p ´ (10) ²  – ½ ´ 10 ´ 10

= 50p  – 50

Ans: Shaded area = 107.08 cm².

Heuristics Used

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Commentary

     We solved this problem by breaking it down into smaller problems.  Since areas are unchanged when you shift them, or turn them, or reflect them, we are able to arrange the two quarter-circle pieces into one semi-circle.  We can also combine the two triangles into a larger triangle for which we know the base and the height.  By breaking down the problem and transmuting these smaller problems into equivalent problems, our task becomes much simpler, allowing us to get the solution quickly.

Please refer to this similar problem.

[Pri20150220FSA] A Fishy Shaped Area

Question

Plan of Attack

     This problem looks difficult because the shaded area does not seem to look like any regular shape.  Is it a fish whose head is pointing in the top left direction and whose tail is in the bottom left direction?  Fortunately, this is not a Rorschach ink-blot test.

     As with all “area” problems in primary (elementary) school, we try to break down the unfamiliar shape into regular shapes (e.g. parts of circles, squares, triangles, rectangles).  It is basically a divide-and-conquer strategy (using heuristics H10 & H11).  If we look carefully, we realise that the required area consists of a semi-circle less a funny horn-shaped area, which I call ‘F’.  F is for funny, for want of a better description.  So I am going to find the area of the semicircle (which is half of a circle), then subtract the area of F.  We’ll worry about finding the area of F later.

Solution

     Area of semi-circle [in cm²]

= ½ ´ p ´ radius²

= ½ ´ p ´ (5) ² = ²⁵/₂ p   

     It is good to leave the calculation with  p  until the last step.

     OK, we are done with the first part.  [Heuristic H11]  Let us us tackle the next part, which is to find the area of  F.  Note that this is a 45°-45°-90° isosceles triangle minus a 45°-degree sector (which is one-eighth of a circle, because ^45°/_360° = ¹/₈).

     Area of F  [in cm²]

= Area of triangle – area of sector

= ½ ´ 10 ´ 10 – ¹/₈ ´ p ´ radius²

= ½ ´ 10 ´ 10 – ¹/₈ ´ p ´ (10) ²

= 50 – ²⁵/₂ p

     Let us combine our answers.  We need to subtract 50 – ²⁵/₂ p  from the area of the semi-circle.  If we subtracted 50 from ²⁵/₂ p, we would have over-subtracted.  So we need to add back ²⁵/₂ p.  Hence

    Required Area [in cm²]

= Area of semi-circle – area of F

= ²⁵/₂ p  – (50 – ²⁵/₂ p)

= ²⁵/₂ p  – 50 + ²⁵/₂ p

= 25p  – 50

Using the calculator’s value of p,  we obtain

     Required Area = 28.54 cm²  (to 2 decimal places)

Heuristics Used

H10. Simplify the problem

H11. Solve part of the problem

Commentary

     This difficult problem was solved by dividing the problem into smaller pieces and tackling each piece one at a time.  We break down a complicated shape into familiar shapes.  That is the secret.

Please refer to this similar problem.

[Pri20150302PBW] Pebbles on a Road Backwards

Question

A jar contained some pebbles. Nigel took out half of them plus 2 more for display. Jenny took out half of the remaining pebbles plus 1 more. Finally, Alice took out half of what remaining plus 5 more for her project. In the end, there were only 8 pebbles left. How many pebbles were in the jar at first?

Solution

We can trace what happens to the remaining pebbles at each stage, and represent the information given in a sequence as above.  Note that when “Nigel took out half of them plus 2 more for display”, we would have half and then two less the number of pebbles.  Similary for the Jenny’s and Alice’s stages.

                            

Now we just do the opposite of all the operations.  The inverse operation of ‘–5’ is ‘+5’, the inverse operation of ‘´ ½’ is ‘´ 2’ and so on.  Working our way backwards, we arrive at 112.

Ans: There were 112 pebbles at first. 

Commentary

     Remember: In mathematics, there is no such thing as "the only way" to solve problems.  There are many ways to skin the cat, as they say.  Other methods to solve this question include: using algebra, and “the branching method” (taught by some tutors/teachers) which tracks both the remaining pebbles as well as the ones taken away.

     If you do use algebra (of some sort), remember to use different letters e.g. v, w, ... etc to refer to different things.  Likewise, do not just write  ‘½’  to represent ½  a unit, otherwise it could be misinterpreted as literally the number ½.  I suggest you can surround the ‘½’ with different shapes (e.g. circle, square, triangle) .

Precision is very important in maths, as well as in life. You don't want an imprecise person to be your aircraft designer, accountant or doctor, do you?

Heuristics Used

H05. Work backwards

Suitable Levels

* Primary School Mathematics

* other syllabuses that involve whole numbers and fractions

* any learner who is up for a challenge

[Pri20150302MCP] Multiple Choice Questions with Penalty

Question

Solution

The ratio of the number of words spelt correctly to that of words spelt wrongly is 3 : 1.  I use triangle to surround the ratio units.  Multiplying by 10 and 4 respectively, we obtain 30 and 4 units respectively.  The score accrued is 30 – 4 = 26 units which corresponds to 130.  Dividing by 26, we get 1 triangle unit corresponding to 5.  And that is the answer required because it 1 triangle unit represents the number of words spelt wrongly.

Ans: 5

Check: 15´10 – 4´5 = 150 – 20 = 130.   Yes!  Woo hoo!!

Reflection

This question is similar to this fish and chicken cost problem, but there are no monetary units (e.g. dollars or cents) involved.  Also, because of the penalty for wrongly spelt words, we subtract instead of add.  We need to be mindful of the context of the question and to apply the appropriate arithmetic operation.

[Pri20150302FCW] The Cost of Fish and Chicken

Question

Solution

The ratio of the cost of fish to that of chicken is 1.5 : 1 or 3 : 2.  I choose this nice whole number ratio and write them in triangle units.  You can choose draw any shape you want to surround the ratio unit actually.  Multiplying by 3 and by 4 (the numerical dollar costs), we obtain 17 triangle units equal to 76.50.  From here we can divide by 17 to get the equivalent of 1 triangle unit, and then multiply by 3 to get the equivalent of 3 triangle units.

Ans: 1 kg of fish costs $12.50

Commentary

Once again my Ratio Units method solves this problem quickly and you can use this method to also present the answer in a matter that even the teacher who is marking this can understand J.  

P6HQWN001 Whole Numbers (Before vs After)

Introduction

     This is a question involving whole numbers, for primary 5 to 6 (» grades 5 or 6 for 11 to 12 year olds). Besides showing the full thinking process of solving this mathematics question with metacognition and heuristics, I use a Before-After Comparison Bar Model Diagram with big and small units (quite a common useful technique used in solving Singapore school word problems).  Once again, the planning and analysis stage (with the help of the bar model) is crucial, but once this is done, the calculations come out rather naturally.

Stage 1:  Understanding the Problem

Can you explain the problem in your own words?

     { Read the questions two or three times.  Make sure you can paraphrase it, not just parrot it.  Make sure you understand all the words and make sure you know what the problem is talking about }

What topic is this on?  What concepts are involved?

     Whole numbers, ratio, before-vs-after comparison.

What are you supposed to find?

     The total number of scouts and guides at the beginning of the camp (i.e. the “before” period).

Stage 2:  Planning the approach

What are some heuristic(s) you can use?

     · Bar-model diagrams

     · before vs after comparison

     · solve part of the problem (first)
     · restating the problem in another way

     · others as and when necessary

Let’s read the word problem sentence by sentence and try to translate the information into the bar diagram. “At the beginning of a camp, the number of boy scouts was five times the number of girl guides.”  How do you draw that?

     Well, I divide the working space into two.  On the top half, is the “before” stage.  On the bottom half is the “after” stage.  For each of the two stages, reserve two lines: one for boy scouts, the other for girl guides, and I label “boys” and “girls” accordingly.  For the “before” stage, draw five bars for the boys and one bar for the girls.

Figure 1. Interpreting the given sentences into the bar diagram.

Next sentence … “After  81  boy scouts left the camp, the number of girl guides was twice the number of boy scouts.”  How do you represent that on the diagram?

     Er … in the “after” stage, there are 81 boy scouts less, so I need to draw a shorter bar … but how short?  Anyway, only a number of boys went away but all the girls stayed put.  So I draw one bar for the girls in the “after” stage, exactly the same length as for the “before” stage.  But how to draw the bar for the boys in the “after” stage … hmmm …

Can you rephrase the problem?

     In the after stage, the number of girl guides is twice of the number of boys, so … er … the number of boys must be half of the number of girls.  So I draw a short bar for the boys that is half of the length of the girls’ bar.  Since the difference between the “before” and “after” numbers for the boys is 81, I draw a horizontal arrow spanner to indicate that.  I draw a vertical dotted line to help me compare these quantities more clearly.

Last sentence … “What was the total number of scouts and guides at the beginning of the camp? ”  How do you draw that?

     Total number … to represent that, I draw a vertical brace (or curly bracket) combining the boys and the girls in the “before” stage.  Then I put a question mark (‘?’) to remind me that this is actually what the question is asking for.

Now do you observe anything?

     I now have a short bar (a small unit) for the “after” boys.  The bar for the “after” girls is one big (original size) unit, which is two small units.  OK, let’s divide each big unit into two small units (using short dotted vertical dividers) and see what we get.

Figure 2. Further refinement of the bar diagram.

Now do you observe anything?

     Ah-hah!  (*epiphany* light-bulbs flashing in the brain)  The 81 actually corresponds to 9 small units!

Can you solve the problem now?

     Yes!  Yes!  Yes!

Stage 3:  Execution

Figure 3. Doing the calculations

Stage 4:  Evaluation

Is your answer correct?  Does it fit the conditions of the posed problem?

     Using 1 small unit « 9 people, and looking back at the diagram, we calculate as shown below (a rough pencil working will do, and this should not take very long).  We systematically check against all the given clauses in each sentence of the given problem.

Figure 4. Checking the solution

Final Presentation

     The final presentation is short and sweet.  The hard part was actually the thinking and the observation, which was explained in painful detail above.

Figure 5. Final Presentation

Stage 5:  Reflection

What did you learn by solving this problem?

     I practised more and got better at the problem solving process.

     I learned to use metacognition to monitor myself and ask myself questions as I am mulling over the maths problem.  I also learned to be careful and to check my own work systematically.

     I learned how to use the following heuristics:-

     · Bar-model diagrams

     · before vs after comparison

     · solve part of the problem (first)

     · restating the problem in another way (especially when I get stuck)

     · Using big units and small units

What else did you learn that you can use in future problems of this sort?

     I learned to cut big units into small units.  I learned that if I can find the link between the big unit and the small unit and if I can figure out the small unit, then I can figure out everything else in the problem!  In future if I encounter this sort of problem, I can try out this powerful tactic.

      [ Remark: Actually, the drawing of the bar diagram is already part of the execution i.e. in practice, stages 2 and 3 merge, or shall we say go back and forth in tiny loops of interpretation followed by part drawing, followed by having greater understanding and insights, making observations, which leads to further planning and refinement of the diagram and so on, until the solution becomes obvious.  In problems involving big units and small units, the solution becomes obvious once we can figure out what the small unit represents.  ]

Is there another way to solve the problem?

Can you make up a similar problem?

Please write your comments below.

P6HQ2000S46 Ratio, Proportion and Percentages

Introduction

     This is a challenging question on the topic of ratio, proportion and percentages.  I shall again illustrate the process of solving this mathematics question with metacognition and heuristics, which are applicable for all levels, and all topics.  The Bar Diagram Model is a very well known heuristic.  But there are many other good problem-solving heuristics as well.  There is a key to solving this kind of question and in order not to spoil the fun, I shall let my readers have a go at it first.  Please think a bit about this problem, and then read on.

Stage 1:  Understanding the Problem

Can you explain the problem in your own words?

     Siti paid \$ 24 for some towels at a discount of 20%.  Now, with the discount, she got three more towels than without the discount.

What concepts is this problem testing you on?

     Ratio, proportion, percentages, price, unit price …

Can organise the given information?

     We can put the information into a table, like this:-

Figure 1.  Heuristics: ‘Use a table’ and ‘compare before/after’

What are you asked to find? 

(a) the number (quantity) of towels Siti bought with the discount

(b) the price of each towel (the unit price) before the discount

Stage 2:  Planning the Method of Attack

How are things related?

     Total Price = Quantity (how many she bought) x Unit Price (how much each cost)

This relation is similar to

     Distance = Speed x Time

Have you solved a similar problem before?  How did you solve it the last time?

     Yes, a Distance-Speed-Time problem.  We used a Triangle Mnemonic.

Figure 2.  Triangle Mnemonics and Analogous Thinking

How is this problem different?

     Instead of ‘Distance’, we have ‘Total Price’

     Instead of ‘Speed’, we have ‘Unit Price’

     Instead of ‘Time’, we have ‘Quantity’

     Ah!  Maybe we can use a similar Triangle Mnemonic for Total Price, Unit Price and Quantity.

Remember: What are you trying to find?

     The number of towels - the Quantity.  OK … so … if we use a finger to cover ‘Q’ we see ‘T’ over ‘U’.  That means

     Quantity = Total Price ÷ Unit Price
     But this problem looks more difficult, because there are so many unknowns …

Don’t give up.  Try some heuristics.  (emotional management)

Can you restate the problem in another way?  (using a heuristic)

     The discount of 20% means she paid 100% – 20% = 80% = ⁴/₅ of the original price.  Without the discount, she either would have paid more for the same number of towels, or for the same amount of money, she would have gotten less. …

Can you simplify the problem?  (using a heuristic)

     Suppose the unit price was halved.  Then Siti would get double or 2 times the number of towels.  What if … the unit price was ¹/₃ of the original?  Then she would get thrice (3 times) the number of towels.  If the price is ¹/₄ of the original, then she would get 4 times …  It looks like the cheaper the things get, the more you can buy … there is a pattern … the numbers seem to go the opposite way … the fraction seems to be inverted (the reciprocal) … why? … why? … why?  Oh!  It’s because

     Quantity = Total Price ÷ Unit Price

The division ‘÷’ causes the fractions to ‘turn over’.  I see!  J  So, since the price is ⁴/₅ of the original, Siti would have bought  ⁵/₄  of the number of towels without discount!

Can you draw a bar diagram for this?  (using the famous Singapore heuristic)

     Yes!

Figure 3. Before-After Comparison Bar Model

Now, can we solve the problem?

     Yes!  Yes!  Yes!  It is now very easy.

Stage 3: Execution

(a)      1 part    =  3 towels  (after discount: 3 more towels)

          5 parts   =  3 ´ 5  = 15 towels

(b)       4 parts =  3 ´ 4 = 12 towels  (before discount)

          12 towels  ¬¾®  \$ 24

              1 towel  ¬¾®  \$  2

Answer:
(a)  She bought 15 towels.

(b)  Before the discount, each towel cost  \$2.

Stage 4:  Evaluation

     Let us check the answer:-

Figure 4.  Checking the answer

     Are we correct?

     Yes!  The numbers fit nicely.

Final Presentation

       The final presentation of the solution can be rather succinct.  The foregoing discussion seems long because all the thinking process are explained in full detail.  The checking of the answers can be done in pencil as rough work.

Figure 5. Final Presentation

Stage 5:  Reflection

What did you learn by solving this problem?

     Although all stages are important, the hardest part was stage 2, the planning stage.  This involves trying various heuristics to look at the problem from different angles.  [This is the real mathematics.  Patience, observation and creativity are needed.]   The key to solving this problem was to note the inverse ratio relationship between quantity and unit price (given a fixed total price).  Once we managed to find the key, we draw the appropriate diagram and the rest of the calculations are pretty straightforward.

In future if you encounter a similar problem, how would you solve it?

     If there is a similar problem in future, I can use the 5 stage problem-solving process:-

     1. Understanding

     2. Planning

     3. Execution

     4. Evaluation

     5. Reflection (learn and transfer for use in future)

Use metacognition (thinking about my own thinking, ask myself questions) to guide myself through the five stages.  Manage my own emotions.  If I get stuck, do not give up.  Look at the problem in different ways.

I can use the following heuristics:-

     · Bar model diagram (yes, but there are others …)

     · drawing a table

     · comparing ‘Before vs After’

     · using a Triangle Mnemonic

     · thinking of a similar problem

     · simplifying the problem

     · rephrasing the problem in another way (e.g. convert percentages to fractions)

Dear blogger: why don’t you just present the answer?

     Many book authors and teachers already do just that.  But do pupils’ actually improve much?  Mathematics still remains a mystery to many pupils.  They think that certain people are born geniuses and that there is always a way that these geniuses somehow have the knack of finding the right steps straightaway (and they themselves cannot).  All they do is try to memorise the algorithm and reproduce it, regardless of whether they understand it.   There is research to show that using worked examples alone does not improve pupils’ mathematics much, and that heuristics and metacognition are actually very important in mathematical problem solving.

     “Give a person a fish; you have fed him/her for today.

       Teach a person to fish; and you have fed him/her for a lifetime”

     If you take the time mimic these heuristic and metacognitive processes, your maths will definitely improve by leaps and bounds.  You still need to know the basic concepts, which most teachers teach, but you need to learn to put them together.   I hope you get empowered in this process.