[Pri20150402PTSMAP] A Staircase with Higher Steps

Question

Introduction

     This pertains to the sum of consecutive numbers with constant skips.  I set this question

to illustrate the heuristic of looking for patterns [H04].  It is similar to this question, except that now the numbers jump or skip by  2  instead of just   1.  The more knowledgeable reader will doubtless recognise this to be an arithmetic progression.  The challenge now is how can a primary school pupil do it without having learnt about any more advanced mathematics or algebra, relying purely on pattern recognition.

Solution

     As in the previous solution, imagine the sum as a series of vertical bars.  The numbers all jump by  2  this time.  Because the jump amount  2  is constant, you see a nice staircase pattern (shown in violet).  Each step of the staircase is of height  2  units.  If we make a copy of it and turn it upside-down (shown in green), the two staircases join together nicely to form a rectangle.  Notice that  101+3 = 99+5 = 97+7 = ... etc and they are all equal to  104.  If we know the number of columns, we can work out our desired sum.  How many columns are there?

     The number of columns is the same as the number of terms in  our sum.  OK, but then how many terms are there?  How to calculate this?  Let us look at a few simple cases first [H10. Simplify the problem].

Let us try to observe the pattern.  Note that the size of each skip is always  2.  If there are  2  terms, it is just  3  and  5,  there is one skip of  2.  From  3  to  7,  there are  3  terms, there are two skips of  2  each.  From  3  to  9,  there are  4  terms,  the difference is  6  and there are  3  skips.  From  3  to  11,  there are  5  terms,  the difference is  8  and there are  4  skips.  If you go from  3  to  13,  the net jump is  10  and there are  5  skips  and  6  terms.  We can tabulate the data into a table [H02] below:-

        skip size = 2

Start	End	Total Skip	# skips	# terms
3	5	5 – 3 = 2	2 ¸ 2 = 1	2
3	7	7 – 3 = 4	4 ¸ 2 = 2	3
3	9	9 – 3 = 6	6 ¸ 2 = 3	4
3	11	11 – 3 = 8	8 ¸ 2 = 4	5
3	13	13 – 3 = 10	10 ¸ 2 = 5	6

Do you notice some things?  [H04]

The total skip is the difference between the starting and ending numbers.

The number of skips is the difference divided by the skip size.

The number of terms is always one more than the number of skips.

Since our last term is  103,  the total skip is  101 – 3 = 98.  The number of skips is  98 ¸ 2 = 49.   So there are  50 terms  i.e.  50  columns.

Hence the size of our rectangle is  50 × 104.  But we only want half of this rectangle (shown in violet).   Hence the sum is  ½ × 50 × 104 = 2 600.

Ans:   3 + 5 + 7 + ... + 99 + 101 = 2 600

Summary

     This article illustrates the heuristic [H04 Look for pattern(s)].  Our first pattern we notice is the staircase pattern.  After making a copy and turning that around, we notice that it forms a rectangle, with columns of size  104  each.  Now we look for a pattern that enables us to find the number of columns, which is the number of terms in our sum.  We note that the number of terms is always the same as the number of skips, which is the same as the difference between the start and the end all divided by the skip size.  This enables us to solve the challenge in a way similar to my previous example.

Reflections
     Do you think this method will work for different starting numbers and different ending numbers?  For different skip sizes?  Why not set up your own similar question and try it yourself and see whether it works?

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H09. Restate the problem in another way

H10. Simplify the problem

Suitable Levels

* Primary School Mathematics

* GCE ‘O’ Level “Elementary” Mathematics (Number patterns, with algebra)

* GCE ‘A’ Levels H2 Mathematics (sequences and series, with algebra)

* IB Mathematics (sequences and series, with algebra)

* anyone who loves patterns and relishes a challenge

[Pri20150308CSR] Charitable Savings Ratios?

Question

Introduction

     Here is another one of those Singapore Mathematics problems that are two-variable simultaneous equations in disguise.  The key to solving this question quickly is to exploit the fact that the amount of donations are the same in this case.

Solution

     First, read the question and translate the information into a diagram [H02. Use a diagram / model].  I use different shapes (e.g. circle and square) to envelop the different types of units.

     Since we have ‘-80’  for both Sharon and Ryan [H04. Look for pattern(s)], we may deduce that 1 ‘circle’ unit  (5 minus 4 ‘circle’ units) is equal to 3 ‘square’ units (10 minus 7 ‘square’ units).  That means 5 ‘circle’ units is 15 ‘square units’.  [ H10. Simplify the problem]

     By comparison again, we realise that 5 square units is 80 [H05. Work backwards] and hence 15 ‘square’ units is 240.

Ans: Sharon’s savings was $240 at first.

Check

Before    $240   $192      5 : 4

After      $160   $112     10 : 7

Solution makes sense.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H10. Simplify the problem

[Pri20150306APC] Apples-with-Pears Comparison

Question

Introduction

     This question involves money and looks rather challenging, because there are a few unknown quantities.  To solve it, we use the concept of unit costs, so that we can make an “apples-to-apples” ... er ... I mean “pears-to-apples” J comparison of the prices.

Solution

     Since absolute dollar amounts are given, we can quickly solve [H11. Solve part of the problem] for the total costs of pears and apples as follows [H02. Use a diagram / model]:-

     Now we know that the total cost (in dollars) of pears is 45 and for the apples it is 50 (5 more than 45).  Although we do not know the absolute numbers of pears and apples, we know their ratio.  So let us write these down as, say, ‘square’ units. 

     Dividing the total costs by the numbers gives the unit costs, which we know only in ratio terms.  So let us use, say, ‘circle’ units to denote these.  But we know that the unit cost (in $) of an apple is 0.50 less than that of a pear.  And that is equivalent to 5 ‘circle’ units.  From here [H05. Work backwards], we quickly work out the cost of a pear (15 ‘circle’ units) as $1.50.

     Ta da!

Heuristics Used

H02. Use a diagram / model

H05. Work backwards

H11. Solve part of the problem

[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!

[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?

[Pri20150303SJM] Race to the Bottom

Question

Thinking / Planning

Noting that Sarah spent all her money in both scenarios and that 22´(18´something) = 18´(22´something) , I am going to first guess that Sarah takes 18 days and 22 days to spend all her money in the 1^st scenario and the 2^nd scenario respectively. [ Heuristics: H07. Use guess and check & H08. Make suppositions ]  Then I try to adjust my educated guess.

Johari spent more money in the second scenario.  The difference in his spending among the two cases is 6, so we need

                10´(number of days₂) – 12´(number of days₁) = 6

Solution

If Sarah takes 18 days and 22 days respectively to spend all her money, then

the difference in Johari’s spending among the two cases is

                10´(22) – 12´(18) = 4

´³/₂:         10´(33) – 12´(27) = 6

Ans (a): Sarah has $22´27 = $594.
Ans (b): Johary has $(21+12´27) = $345.

Commentary

If the amounts of money involved were in the trillions, we would have thought that Sarah and Johari are certain countries, wouldn't we?

Anyway, a parent from a Facebook parent-support group posed this question asking for a simple solution without using ratios.  This question seems to be the equivalent of a system of simultaneous equations in four variables.  A few of us tried various methods to solve it, but all were quite complicated.  I knew this can be solved using algebra, but struggled for some time to give a simple solution.

I guess all solutions (even the one above) would have some notion of “ratio” hidden in it.  The reason is: every time you multiply or divide by something, there is actually a ratio involved.  But I hope this solution is “easy” enough.

[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