Tuesday, November 21, 2017

My students solved it better than me :)

This is an interesting problem that was given to about 30 Maths Teachers in one of the PD workshops that I attended recently.  I must confess that almost all of us, baring very few, struggled quite a lot and for quite a long to find its solution.. In fact, many of us could not even arrive at the desired solution :) However, I had a gut feeling that my students 'will' be able to reach the destination, and that too in a 'proper' way....

Yes, I was quite sure that my students would do this problem-solving much 'better than me' !  ..... Why?

Because I have been taking care and effort since past 2 years, to see to it that I do not pollute their minds or damage their mathematical thinking by feeding them 'my' methods or by enforcing upon them some 'standard' methods directly....

Makes you reflect?  :-)

So here is the problem:


So, I told this incident to my students while giving this problem... (that it was given to teachers and we all took lot of time, etc).. Naturally, the first thing they asked me was-

"Sir, could you solve it?"

"Yes, I could..."

I could see one of them saying to the other, "'Sir' will 'obviously' solve it....!"

"Wait ! It took me lot of time and effort.... and I was not so happy with my work.... However, I feel you will do better than me....and hence I am giving this to you.... "

Some of them smiled at me, while others had already dived into the problem.. :)

They sought clarification about some requirements mentioned in the problem... I clarified those with examples... and then there was silence for about 10 minutes...

I told them mid-way that they can even work in pairs or groups if they wish.... But none of them did.... It seems they wanted to give their own shot for some time..

Would suggest you to give a try before reading their solutions :)
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Two of them - Sahil and Rohit - told me that they are on the verge of completion... have eliminated many possibilities out of the total 120 and are working on the final six ones ....

I understood what they were saying / doing... Did you?  :-)

Soon, they declared that ---- "There is No solution."

I felt Waaaow.... but contained my excitement and rather probed them if they have a proof to support their claim...

Yes sir, we can prove it...

I asked them to pen down their solution systematically while others were still solving... 

In some time, Vaishnavi too declared her accomplishment and this was followed by the rest of the class in next 10-15 minutes....

Every student who was done, was directed to 'write down' their solution / approach in a comprehensible form...

Only when everyone claimed that they have solved it, we began with the Whole-Class Discussion...

Rohit pushed Sahil for sharing.... I had seen both of them working together in the end....

Sahil's work:



He says that there is only one way to make 14 and 13. For 14 - 2,3,4,5 should be together while for 13 - 1,3,4,5 should be together. Hence he concluded that 1 and 2 should be on the two extremes.

So this leaves 3,4,5 in the middle zone which gives rise to 6 possibilities, as listed by him.

He then ruled out each of these possibilities with reasoning. For example: 

Option 1 ----  1,3,4,5,2 does not yield 6
Option 2 ---- 1,3,5,4,2 does not yield 7

and so on.....

Since none of the options work in the favor of given conditions, so the problem does not have any solution.

I asked others if they want to ask anything to Sahil.... All agreed with his work...

Jeetu said that his approach was also very similar to this....

Tanvi came forward. This is her work:..




She too first found out that 1 and 2 will be on the two extremes.

Later, she wrote down all the possible ways of making every number from 6 to 15 (though she later said there was no need to write for 13 to 15 again :)  

She then cancelled out all the options which used both 1 and 2 to make a number.

Then she started with 6-sized bogie. She chose to test one of the options of 6 i.e. 5+1
This would mean placing 5 besides 1. This forces 7 to be made using 4,3. However the option (5,3) for making 8 fixes the position of 3 at the center and hence 5 next to 1.

Thus the train looks like this:  1,5,3,4,2

This structure cannot however make a small train of size-10. 

So this means the initial assumption of 6 to be made using 5 and 1 would be wrong, thus leaving only (4,2) for 6. 

Again, sizes 7 and 8  will fix the positions of 3 and 5 respectively, eventually and surprisingly leading to the same arrangement 1,5,3,4,2 which does not help us for giving 10.

Thus any of the options of 6 leads to an impossibility for the size 10 and hence it is not possible to make such a train....

Again, I looked at the class for their agreement/ disagreement. .. All agreed...

Vaishnavi came up with her approach:




Her initial approach is same as that of Tanvi.... Listing down and ruling out the options having 1 and 2 together. However then she chose to work from the other end (bigger numbers), unlike Tanvi who worked from 6.

Done with 15,14 and 13, she said 12 will also be possible with 3,4,5 anywhere in the center.....  1  ? ? ? 2

Analyzing the only remaining option for 11 i.e. 5,4,2, its clear that 3 will have to be next to 1. This rules out two more options (5,1) for 6 and (5,4,1) for 10.....    1 3 ? ? 2

Analyzing the only remaining option for 10 i.e. (5,3,2) its clear that its impossible to make 10 because 3 and 2 are already 2 spaces apart.

Thus its not possible to complete this order.

All agreed to this.

Kanchan said that her approach is also somewhat similar to Vaishnavi's (starting from the bigger numbers).. This is her work:



She first fixed up 1 and 2 in the two extremes like others. However, she did not list all the possibilities for 6 to 14 like her two peers.

After finishing off  the work of 12 to 15, she listed the two options for 11:
(5,4,2) and (1,2,3,5)

She ruled out the latter (since 1 and 2 are apart) and hence the former got fixed. This meant either 4 or 5 should be next to 2.  .... (note that Vaishnavi had directly concluded that 3 will be next to 1 using this case)

She then listed down the three options for making 10 and one-by-one ruled out those too with appropriate reasoning
(5,4,1) -- Not possible as 5 and 4 will be next to 2.
(1,2,3,4) -- Not possible as 1 and 2 are 3 spaces apart
(2,3,5) -- Not possible as (2,4,5) have to be together to make 11 as discussed above.

Thus it is not possible to make 10 and hence this problem has no solutions.

Remember, I told you in the beginning - about my confidence that my students would be probably solving this better than me... And I am so happy and even proud that they did so !!

You want to see my work?  This is it :))



How about you trying this problem with your students / children?

Could you arrive at the correct solution on your own? What/ How was your approach?
I will be happy if you can share your or your students' approach with me, if its different than any of the above....

What are your views about the thought processes of students above? 

Waiting to hear from you, 

Thanks and Regards

Rupesh Gesota

PS: These students are studying in class-7 and 8 in a navi-mumbai based marathi medium government school.. I am working with them on their Maths Enrichment since past couple of years after their school hours. To know more, check the link:
www.supportmentor.weebly.com

Thursday, November 2, 2017

Simple Puzzle (Tin of Biscuits) - multiple approaches

I was sure they will crack this puzzle quickly, but I was more curious to know their multiple approaches.

"A tin full of biscuits weighs 5 kg 200 gm. The same tin half full of biscuits weighs 3 kg. Calculate the mass of empty tin."

Almost all of them were done in about a minute.
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Give this problem a try before you read the solutions below.

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Tanvi's approach:

I took one more tin half full of biscuits. So two half-filled tins weigh 6 kg. But a fully filled tin weighs 5 kg 200 gms. So the difference in these two weights corresponds to the weight of tin i.e. 800 gms.

Kanchan's approach:

I halved the weight of fully filled tin, thus leading to 2600 gms. Now, the other half filled tin weighs 3000 gms. So the difference 400 gms corresponds to half the weight of tin. So the tin weighs 800 gms.

Vaishnavi's approach:

She solved this algebraically. Let the weights of tin and fully filled biscuits be T and B gms resp.

T + B = 5200 ...... (1)
T + 0.5 B = 3000  ....... (2)
T + 0.5 (5200 - T) = 3000   .......... (from 1)
T + 2600 - 0.5 T = 3000
0.5 T = 400
T = 800

To this Rohit responded,

Look at your 1st and 2nd equations. We can clearly see that 0.5 B = 2200.  
Thus B = 4400 and hence T = 800

How did YOU solve this problem? Was your approach different than any of the above?

Which approach did you like the most?

How about trying this with your students/ children? Would love to know their approach :)

Thanks and Regards

PS: Students belong to marathi medium government school (class-7 and 8) based at Navi-Mumbai. To know more about their Maths Enrichment program, check this link:
www.supportmentor.weebly.com


Monday, October 23, 2017

Simplifying Algebraic Fractions : Part-1


Could you find the mistake in her first step of simplification of LHS expression?

I am sure many algebra teachers would agree with me that this is one of the most common mistakes students do while simplifying algebraic expressions. 

So, why would be they doing so? What could be the cause(s) for this effect? Why is it that this nonsense does not seem nonsense to them? 

(I am intentionally keeping these questions open now, with the hope that some of us would probably pause and try to find the answer for these, though reflection, research or discussion with our peers.)

So, this was the problem:
Actually, I had seen Poonam committing this mistake i.e. 'cancelling' the term x^2  (i.e. x squared) from Numerator (Nr) and Denominator (Dr), however I chose to not stop her to point out this mistake at that moment and I chose to keep my mouth shut and just allow her to go ahead with this mistake....  (Why?)

Soon, she found herself stuck up and asked me to help.. I don't know what struck me I rather asked her this question:

"I see you have cancelled out x^2 in Nr and Dr... Why don't you then even cancel out 'x' from Nr and Dr ?"

She replied immediately - "No sir, we cannot do this... Their signs are different..."

"oh ok... Is that the reason?"

"Yes", she answered with a pinch of reluctance this time.

I thought for a while... and again threw the ball back to her court,

"Okay... so how about canceling the two terms (1 + x^2) from Nr and Dr?"

"How can we do that? They are far away... There is 'x' in between them..."

"Oh ok.... Can we rearrange the addends in an addition expression? I mean, can we write 
(1 + x + x^2) as (1 + x^2 + x)?"

Poonam got what I meant this time and re-wrote the Nr and Dr expressions this way:


As you can see that, now she had no problem in cancelling out the two (1 + x^2)  ...Interesting, isn't it?

I just thought of pointing out one more thing to her ---

"Could we have cancelled out even the '1's while cancelling out x^2 from Nr and Dr?"


She reacted to this simple idea in such a way, as if I had drawn her attention to some Wonder..    :-) :-)

"yes sir, we could have easily simplified the expression there itself, instead of this long business of first rearranging and then cancelling (1+x^2)...."

So having stuck up in her first way (canceling just x^2), she now simplified the problem - the second way i.e. by cancelling (1+x^2) instead of just x^2   ;-)


Now, if you carefully check her new solution (on the right side), then you will find many more (serious) mistakes..

Can you count and tell me how many? Possible reasons for these mistakes?

I could have pointed out these errors to her..... However, at this moment I chose to stay 'out of this zone of new errors'.....Why?  (perhaps, addressing this set of errors calls for another round of discussion with her and hence an another post as well ;)

Meanwhile, you can think of the method(s) one can adopt to help her find these mistakes....

She again found herself stuck up, like last time, after simplifying the problem to some extent.. She cried for help....

Now I thought to withdraw myself and rather involve my another student -- to take over this game.... I intentionally picked up Kanchan, told her to have a look at both the solutions of Poonam so far..  She went through her work and then gave a big smile to me,... I knew that she will be able to recognize the trap I had laid for Poonam.... After all, she too had been dragged into such traps before :-)

Kanchan forged ahead, thought for a while and wrote this on the board:
I don't know what made her think of these four numbers (3,10,15,9)... This is how she instructed Poonam then:

"Now simplify this example using both the ways... with cancelling and without cancelling.."

Poonam just solved this 13 / 24 .... Surprisingly she was not working on 3 and 9 in Nr and Dr here, like the way she was cancelling out 'same addends' 'x' or (1+x^2) in the original problem....

While Kanchan was unable to understand this contradicting behavior of Poonam in both these problems, a teacher would probably understand this duality..... May be Poonam thought that the two addends in Nr and Dr can be cancelled out only when they are same, else not..... So here, Kanchan was expecting Poonam to simplify 3 and 9..... But Poonam was not doing so (because both addends were different).... It was interesting to watch both of them fixed up in this lock...  :-)  :-)

Kanchan then looked at me for help... I decided to give in easily this time..... Hence just told her to explain while solving.....

This is how she proved that the method used by Poonam (cancelling out / simplifying addends in Nr and Dr) changes the value of given fraction....and hence is incorrect.


While it was a delight to listen to Kanchan as to why 13 / 24 and 11 / 18 are unequal, I could see that Poonam was mostly accepting what Kanchan was firmly and quickly asserting.....

According to Kanchan, 13/24 and 11/18 were the two most simplified forms (gcd of 13 and 24 is 1   and   gcd of 11 and 18 is 1) and they were clearly not equal.... However Poonam was not much ready yet to digest this level of understanding of fractions and hence this type of reasoning was not going much deep into her.....

If Kanchan had taken either of these 2 following routes, then it would have been easier for Poonam to understand:

(a) Different reasoning:  13/24  is 1/24 more than half while 11/18 is 1/9 more than half. Thus both are unequal, hence one cannot simplify the addends.
(b) Different numbers: Choose the numbers such that it becomes easier to compare the two fractions.

So I decided to intervene now with (b) rather than (a).... and hence suggested Kanchan to make some changes in her numbers.... Replace 10 by 9 in Nr...  (What made me think of this replacement?)  Why did I choose strategy (b) over (a) ?

This replacement made Kanchan's job easier:


So this is the way she explained to Poonam now:

If you simplify the fraction (3+9)/(15+9) without any modifications, then it is equal to 12/24 which is 1/2 i.e Half

While, if you simplify the two addends 3 and 9 in Nr and Dr as 1 and 3 respectively, then the fraction becomes equal to 10/18 = 5/9 which is more than half...

Hence you cannot simplify the addends in the Nr and Dr of a fraction.

This was pretty easier than the previous one... However I wanted to ensure if Poonam had understood this.... 

"So did you get this now?"

She replied - "Sir, how come Kanchan knows 5/9 is more than half?"

I looked at Kanchan.... She began -

"Look Poonam.... Why do you say 12/24 is half?"

"because 12 is half of 24..."

"So similarly if we have 9 in the Dr then whats required in the Nr to make it Half?'

She thought for a while and answered in a low-confident tone --- " 4.5 "

"yes...correct.... So if 4.5 / 9 is Half then, what about 5 / 9?"

There was a Big Smile on Poonam's face now.... and so on Kanchan's  :-) :-)

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Little did they realize that their game was not yet over....

"Wait Kanchan.... You have showed her that she cannot cancel or simplify the "addends" in the the Nr and Dr of a fraction... But what about the reason for why we CAN cancel or simplify the "factors" in the Nr and Dr of a fraction?"

She understood what I meant.... and hence again made another example to make this point:


I was very happy with the way she achieved this...  She explained how the same factors in the Nr and Dr.  (x^2) come together to equal to 1, and hence the simplified fraction formed (5/4) by the remaining factors multiplied by this 1, does not change the value of simplified fraction (1 x  5/4  =  5/4 )..... and thus we casually say that we are 'cancelling out' the same factors..... whereas actually or mathematically, we are multiplying and dividing by the same factor, thus multiplying by 1..........

I thought to reinforce this idea using only numbers now..... So I gave her this example:


I further asked her, what if the problem were (18 x 5) / ( 9 x 4)?

This is how she simplified it:

(18 / 9) x (5 / 4) = 2 x  (5 / 4) = 10 / 4

I could feel the joy of understanding radiated from her face.......

Do you or your students simplify the fractions with such deeper understanding or by simply cancelling out the factors in top and bottom using the 'times tables'? 

How about you triggering such a math discussion in your class, may be with few students, if not all?

Both of them had got so engrossed in this fault finding that they had forgotten that the actual problem (solving for x) however still remained unsolved...   :-)

So now, Can you solve the main problem and share your solution?

But more importantly, if you are a teacher or teacher educator or researcher, I would love to hear from you, your views about this post........your responses to many reflective questions raised in this post (in blue font)....

Regards

Rupesh Gesota

PS: Both the students hail from marathi medium government school, based at Navi-Mumbai. To know more about the maths enrichment program run for them, check the website www.supportmentor.weebly.com


Sunday, October 8, 2017

Relooking at stack of Tables (Triangle Numbers)

The  manner  in which tables were stacked up in our classroom that day , it suddenly caught my attention.. I counted them 4+5+6= 15 and Aha ! It's a triangular number..  I was somehow amused by the fact that how come 4+5+6=15? because the triangular expression for 15 is 1+2+3+4+5 



I usually include my students too in such investigations when they are around.. and hence this seemingly trivial question was posed to them as well... 

They started staring at this structure ... And soon, one of them - Vaishnavi responded: 

"Yes, it's easy," she said, " I can visualize this.."

I asked her to explain and this is how she had restructured the given structure in her mind... 



As you can see , she had formed the triangle arrangement of 1+2+3 in the top 3 layers.. and then shifted one and two blocks from the second last layer to its upper and lower layers respectively to form 4 and 5 respectively, thus getting the common expression for 15 = 1+2+3+4+5

Her spatial flexibility was worth appreciating.. but what absolutely delighted me was the visualization of my other student - Kanchan.. This is the way she had imagined:


She noticed that the rightmost column had 6 tables and it's previous two columns had 4 and 5 tables... So she rearranged 6 tables mentally as 1+2+3 before the stacks of 4 and 5 tables , so as to form the triangular equation of 15= 1+2+3+4+5

I asked her how did she know that the last column of 6 could be arranged as 1+2+3, then she immediately replied that 6 is a Triangular number...! 


So I asked her with curiousity that will we able to do this restructuring for every 3 consecutive numbers, say for 5+6+7 also?

She thought for a while and said: No, it's only possible when the last number is Triangle number...

"Why do you say so?"

"Because only then we will be able to rearrange the last column (number) in terms of triangular form before it's previous columns.."

"Okay... So can you tell me what would be the next possible case of consecutive numbers?"

It didn't take much time for her to work out that the expression would be 5,6,7,8,9,10.. 

"Can you explain how?"

"Sir, 10 is the triangle number and hence can be expressed as 1,2,3,4 before 5,6,7,8,9...."

"Hmm... Good thought.... Can we generalize this then?"

She looked at me for more clarity...

"Means... If the number in the last column is Nth triangular number, then what should be the sequence of consecutive numbers , how many numbers, and what triangle number will be eventually formed?"

I knew she would have understood my query... After about half a minute, she said this...

"If the last Nth Triangle number is K, then the sequence should start from N+1 and go on till K.. This will add up to (K-1)th Triangle number..."

And this was just awesome for me....

Do let me know your views about this exploration which was triggered by a casual observation of stack of tables :)

Thank you... 

Wednesday, August 30, 2017

Interesting Geometry problem (6 Rectangles puzzle) - Solved in various ways

I came across this interesting problem and thought to share this with my students...

Students started working on this and after about 5 minutes, one of them was ready with her solution...

I would suggest you to try solving this problem on your own first, and when you are ready with your solution, you can read further to see how these students have 'seen, approached and solved' it....

Method-1
Vaishnavi's 1st attempt

Her approach was to count all the sides of 6 rectangles (222 x 6) and remove the ones that are not to be counted. So according to her, the answer was 999. I think, you would have figured out the mistake she did. 

Yes, she forgot that the overlapping sides need to be removed twice and not just once..

But, I am glad she could quickly think and work out her error after overhearing the final answers of two of her peers, while they were explaining me :)

So this is how she finally presented her solution to the class. Note her work below. See her erased work in the yellowed part. Here, she explained that how the red segments are actually two segments when the rectangles are separated and hence need to be removed twice. 

Vaishnavi's 2nd attempt

If you are wondering what her 222 and 111 represents, then this is what she had done..

She has combined the two red breadths and two red lengths (see the 4 sides with vertical strokes) to form one rectangle (222)......what remains is one length (made up of two pieces) and one breadth which is same as half the perimeter i.e. 111

Method-2

Tanvi's explanation to me


Tanvi's explanation to the class

To explain her approach, let me just make another diagram:

Diagram - to understand Tanvi's approach

Her two 444s represents perimeters of four rectangles..  Now she understands that this sum 'includes' the segments that were not to be included --- the overlapping segments. 

From here, she used the same approach as that of Vaishnavi to get 222 and 111. Thus she finally gets 999 - 333 = 666

Method-3

Siddharth's method
He "traveled along the boundary" of the shape and counted the lengths and breadths of rectangles it included...  He calculated 6 lengths and 6 breadths, thus amounting to thrice the perimeter which leads to 666. 

Method-4

Sahil's method

To understand Sahil's method, let me re-draw the figure again...  

Diagram - to understand Sahil's method

I think the above diagram would be self explanatory...

Yes,.... this is how he beautifully saw and made three rectangles out of all the pieces on boundary to get three equivalent rectangles thus amounting to 666.

I thought at least one of the two remaining two students would solve the problem using my method.. However, Rohit's method was same as that of Sahil and Laxmi too had used the same approach as that of Siddharth....

So I asked them if they wanted to know how I solved this problem...

And there was a loud Yes !  :-)

Method-5




So the perimeter = 6 breadths + 6 lengths = 222 x 3 = 666 

They understood the "sliding" that I had done.... But I thought to interrogate them further....

"How can I be sure that when I slide upper block to the left such that pt. D coincides with C, then F too will exactly coincide with E?

Rohit jumped in...

"Sir its obvious..... that x = a...."

"There is nothing obvious in Maths, Rohit... Can you prove this statement ?"

He began...

k = y + a

k = x + y

k = a + b

So, y + a = x + y = a + b

So,  a = x   and  y = b

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Two students were absent when we did this problem.. So they solved it the next day... One of them, Jitu, used the same approach as that of Siddharth......

While the other student, Kanchan, gave me a surprise !!

Method-6

Kanchan just removed the three bricks on the top layer and bottom layer. So according to her. the given shape gets converted into this equivalent shape for finding the perimeter: 
Diagram to explain Kanchan's method

So, perimeter of the given shape = 222 x 3 = 666

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1) Did you understand her method?

2) How did you solve this problem?

3) How about trying this with your students? Plz do share your classroom experience...

4) Which approach did you like the most? Why?

5) Plz share your views and comments on this article......


Thanks and Regards

Rupesh Gesota