[Pri20150306COH] The Cards of Hearts?

Question

Introduction

     For this question, I shall illustrate my technique of Distinguished Ratio Units to model the situation.  Read the question carefully and translate the information into a diagram [ using heuristic H02 ] like below:-

     I used three different shapes (circle, triangle and square) to envelop the numerical counts of the different kinds ratio units. Although we do not know how many circle units’ worth of cards Kelly had at first, we quickly notice that  9 circle units are equivalent to 3 triangle units, so that one triangle unit is the same as 3 circle units.  So Kelly had 2 circle units’ worth of cards [ heuristic H05 ], as depicted below:-

 

This allows us to answer part (a) of the question already, namely that the required ratio is 10 : 2 i.e. 5 : 1.  With different types of units, it is difficult to compare things.  However, note that in the exchange of cards, the total number of cards remains constant.  Taking the LCM of 12, 4 and 11 which is 132, we can change all the ratio units to a common type of unit [ H09], say ‘heart’ unit, based on the total being 132 ‘heart’ units.  To do that, we can multiply the numerical counts in columns 1 & 2 by 11, multiply column 3 by 33 and multiply column 4 by 12.  This is what we would get:-

     To answer part (b) of the question, we actually do not need to bother about columns 2 and 3.  Just focus on columns 1 and 4.  From column 4 we observe that 84 minus 48 which is 36 ‘heart’ units gives 72, so one ‘heart’ unit corresponds to 2.  The number of cards won by Kelly can be found by comparing the 84 ‘heart’ units and 22 ‘heart’ units highlighted in yellow.


  That means 62 ‘heart’ units and that corresponds to 134.  And we are done!

Answer (a)   5 : 1

              (b)   134

Heuristics Used

H02. Use a diagram / model

H05. Work backwards

H09. Restate the problem in another way

Thinking Back

     In this question, we have used Distinguished Ratio Units to model the given situation.  We worked backwards to find that Kelly’s initial holdings were worth 2 circle units.  Then we converted everything to a common unit (‘heart’ unit) based on the constant total of 132 heart units.  Once again, I © hearts!

[Pri20150306PPP] A Perspective for Perplexing Parallelograms

Question

Comprehension

     Let us first be clear about what is given in the problem.  I have indicated the numerical values of the areas in the diagram below.

Planning

     First we construct  HK  parallel to  AB  passing through  E.  [ Heuristics H02 & H09 ]  This makes the problem easier to solve because there are congruent triangles and there are connecting ratios along the sides of the parallelogram as well as along the diagonal.  Namely  BF : FA = BE : ED = BK : KC.  We denote the  Area of DBKE by  x, and observe that it is the same as  Area of DBFE  by congruency (you can shift and rotate  DBKE  to get  DBFE).   Area of DAHE = Area of DAEF = 1 cm²  because of congruent triangles.  [ H04 ]

     Let us try to figure out  x [ H11 ].  Making further observations [ H04 ], note that  Area of DADB = Area of DCDB (big congruent triangles) and Area of DHDE = Area of DGDE  (congruent triangles).  See diagram below.

     Because of all these congruent triangles, the parallelogram AHEF and EGCK are forced to have the same areas (area measures), even though they may not be congruent:-

      Area EGCK = Area DCDB – Area DGDE – Area DBKE

                           = Area DADB – Area DHDE – Area DBFE 

                           = Area AHEF = 2 cm².

We conclude that  Area DBKE = Area DBFE = ¹³/₅ – 2 = ³/₅  [cm²].  [ H05 ]

Here is the important thing:-

If triangles/parallelograms have the same height, their area ratio equals their base ratio.

     [ This is different for similar triangles/parallelogram. Do not confuse these two situations. ]

     Area DBFE : Area DAEF = BF : FA = BE : ED = BK : KC  = 3 : 5.

Now we can quickly fill in the areas of the remaining pieces as follows:-

      Area ABCD = 2 ´ Area DABD = 2 ´ (³/₅ + 2 + ⁵/₃) = 8 ⁸/₁₅  [cm²]   Done!

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H11. Solve part of the problem

Thinking Back

     After solving a problem (especially a difficult one), it is always good to think back and recollect what we have learnt by solving the problem.  We have used the following important geometrical facts

#1   If two shapes are congruent (that means you can shift, rotate and/or reflect so        that they coincide), then their areas are equal. #2   If triangles/parallelograms have the same height, their areas’ ratio equals their        bases’ ratio. #3   Sometimes, two shapes can have the same area even if they are not congruent.

      Drawing a construction line splits the diagram into various parallelograms and triangles with the same height, for which we can compare the ratios.  Using congruent triangles, we realise that two of the smaller parallelograms have the same area even though they are not congruent.  With the right perspective, we can deal with these perplexing parallelograms.  Using comparison, we found the area of the small triangular piece  x, and from there we use fact #2 above to work out the rest.

[Pri20150306MOD] A Telescope Too Far?

Question

Introduction

     This is a nasty multiple-choice question to be set for a primary 6 (~ grade 6) pupil for a test.  It could also be a time trap as the pupil might spend a lot of time to no avail just to try to score that miserable mark.  With advanced knowledge, we can solve this using telescoping sums (a.k.a. the method of differences), where many terms cancel and the sum can be shortened, much like a telescope.

     However, this solution looks like over-kill.  Is there a solution that is more accesible to a primary 6 pupil?  Read on!

A Simpler Solution?

     We observe that a denominator of  5  is common.  In fact, since every term has an even factor, we have a common factor of 2 in the denominator also.  That means every term in the sum can be expressed as ¹/₁₀ of something [ Heuristic H09 ].  We can quickly work out the first few partial sums and express each of them as  ¹/₁₀ of something.  [ Heuristic H04 ]

     Tabulating the results [ H02 & H03 ], we observe that each partial sum is  ¹/₁₀ of something slightly less than ½.  So we try to express them as ½ minus something.  How?  We can subtract to find out.  [ H05 ]  For example,  if  ⁵/₁₂ = ¹/₂ – ???   then   ??? = ¹/₂ – ⁵/₁₂ = ¹/₁₂.

     Later on, we notice   [ H04 again ] another pattern: the factors in the denominators of the subtracted quantity matches the last two denominators of the last term.  If we considered the full sum, then what would those factors be?  (See the part highlighted in yellow).  These would be 20 and 21, as per the last term of the series.  Using this pattern, we work out the required sum as follows

That is the answer!

Heuristics Used

H02. Use a diagram / model

H03. Make a systematic list

H04. Look for pattern(s)

H05. Work backwards
H09. Restate the problem in another way

[Pri20150303CRT] Three Pencils in the Haystack?

Question

A man had fewer than 500 pencils.  If he packed them into 28 packets, there was a remainder of 3 pencils.  If he packed them into 20 packets, he had 7 pencils left over.  How many pencils had he?

Solution

     Let us “shift” the problem [Heuristic H09].  Imagine if there were 3 pencils less.  Then the remainders would be 0, 4 when divided by 28, 20 respectively.  Noting that 4 is also a common factor of 28 and 20, we divide through by 4 and get a simplified problem [Heuristic H10].  We now look for a number whose remainders are 0, 1 when divided by 7, 5 respectively. [Heuristic H11]

We draw tables represent the problem. [Heuristic H02]

     Obviously, the desired number must be a multiple of 7.  First, let us try 7 [Heuristic H07].  It does not work.   Trying the higher multiples 14, 21, ..., we find that 21 fits the bill.  Now just multiply everything back by 4 and we find that 84 leaves a remainder of 0, 4 when divided by 28, 20 respectively.  This is the number 3 less than what we want.  So we “shift” back, and get 87 as an answer.

     However, the is not the only possible answer.  Note that the LCM of 28 and 20 is 140.  Since 140 is divisible by both 28 and 20, we know that if we keep on adding 140, there will not be any change in the respective remainders.  We do just that, giving us two other possible answers below 500, namely 227 and 367.

Ans: 87, 227 or 367

List of Heuristics Used

H02. Use a diagram / model

H07. Use guess and check

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Commentary

     This is another question involving remainders.  It is quite a tough job juggling the requirements when we have different remainders and the divisors have a common factor.  We solved this problem by “shifting” (imagining that there were 3 pencils less) and reducing the problem (via division by 4).  Thus this problem can be solved in a relatively easy manner, without “cheating”by using advanced knowledge and without searching the whole haystack for the proverbial needle, so to speak.

     The person who set this question overlooked the fact that there could be multiple answers.  He/she listed the answer as 367 whereas a pupil who attempted this question got 227 which could be verified to be a correct answer.  Question setters should be cognisant of LCM and multiple answers, and specify a suitable range for the expected answer, if it is meant to be unique.  As “consumers” we should also be critical and creative thinkers not to blindly rely on the author’s “correct answer”.

[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.

[Pri20150306RCU] A Very Crowded Class

Question

Solution

     Let us write down the given information in a Ratio diagram.

     As you can see, we have ratios with different units, which seems difficult to solve.  However, notice that the number of boys stayed the same throughout.  We know that the LCM of 6 and 7 is 42.  So let us use another type of unit, say “heart” units, with the number of boys corresponding to 42 of these units.  This can be done by multiplying the first column by 7 and multiplying the second column by 6.  This is what we get

     With the “heart” units, now it is very obvious that one “heart” is equivalent to 2.  From here we easily deduce that the number of boys is 84.

Commentary

     This is a type of “problem” where one quantity (the number of boys) is kept constant while another (the number of girls) changes, giving rise to different ratios.  It is similar to the "Boys, Girls and Party" problem, and you can certainly solve this problem using the bridging method shown there.  However, here we exploit the fact that the number of boys stayed the same, and we use the LCM to create a common type of unit (“heart” unit).  Once this is done, we can easily compare the number of girls using this common unit, and then the problem unravels.  Don’t you © hearts?

     Anyway, talking about authenticity in mathematics problems ... the number of boys is already 84, if you work out the total i.e. including the girls, you get ... (Do This Yourself).  Won’t you find this class a little too crowded?

     The person who set this question should probably have moved the pupils to the auditorium, yes?