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Challenging Word Problems Help...


tammyw
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This is from Book 3. I saw the bar model and I still don't understand it. I used to be really great in math. I'm either getting really, really stupid, or math was too easy when I was a kid :tongue_smilie:

 

Abel saved 43 more nickels than dimes. After he spent 17 dimes, he had twice as many nickels than dimes. How many coins does he have left.

 

Like I said, I saw the bar model they drew, and I don't understand why it is the way it is, or how you mentally put it together. It's embarrassing to feel so ridiculously stupid when using a grade 3 math book.

 

ETA: I think we need extra bar model guidance. I don't think we really went over it enough. Are there any really great online resources to pound this home? Or is there a way to do these problems without the stinking bar models? I'm kind of hating math today.

Edited by tammyw
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Looked in our book and their bar diagram has one slight difference from the one I drew myself to answer the question before looking - I drew my dimes as the first amt (43 less than nickles), and then "removed" (marked out) the 17 less in my dimes bar to equal the 1/2 way point I drew on the nickel line. Does that help?

 

Sending confidence your way :grouphug:

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Don't let it throw you. If I could teach SM anyone can teach it. Well, sometimes it would be more correct to say that I facilitated rather than taught, but my son got through all 6 years. I am on my second time through with my younger.

 

Do a google search "singapore math bar method" and you will get many hits. Do the same for You tube. Keep doing them again and again, copying what it says in the books until it clicks. Do you have the HIG? There might be more information in there.

 

I cannot stress enough how amazed I am that I have taught all 6 years of SM. I did it, he did it, you can do it.

 

ETA: try this:

http://www.thedailyriff.com/WordProblems.pdf

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Oh, and I wanted to add that my son is not a fan of the bar method. Yes, there were times that he needed to use it. It is good that he understood how it works. But, he preferred to work with the numbers. In my HIG she often gave two solutions, one with the bar and one without. Almost all the time, my son used the numbers. He said it was faster and worked better for him.

 

If he got stuck or go an answer wrong I was the one who pulled out the bar method. I could illustrate the solution in a different way than what he had tried. Sometimes he would rework the problem using numbers only, after seeing the bar worked out.

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Abel saved 43 more nickels than dimes. After he spent 17 dimes, he had twice as many nickels than dimes. How many coins does he have left.

 

 

Well. you know that after he spent 17 dimes, he had twice as many nickels as dimes, so bars that looked like those below.

 

[NNNNN][NNNNN]

[DDDDD]

 

Therefore, before he spent 17 dimes, you could represent the coins as follows...

 

[NNNNN][NNNNN]

[DDDDD][17]

 

Further, at that point, the nickels exceeded the dimes by 43, or...

[NNNNN][NNNNN]

[DDDDD][17][43]

 

Therefore, [NNNNN] must equal 17+43=60. Then returning to the beginning, he ends up with 60+60=120 nickels and 60 dimes.

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Well. you know that after he spent 17 dimes, he had twice as many nickels as dimes, so bars that looked like those below.

 

[NNNNN][NNNNN]

[DDDDD]

 

Therefore, before he spent 17 dimes, you could represent the coins as follows...

 

[NNNNN][NNNNN]

[DDDDD][17]

 

Further, at that point, the nickels exceeded the dimes by 43, or...

[NNNNN][NNNNN]

[DDDDD][17][43]

 

Therefore, [NNNNN] must equal 17+43=60. Then returning to the beginning, he ends up with 60+60=120 nickels and 60 dimes.

 

And somehow, again embarrassingly, I'm not comprehending. I think I'm having a really bad day.

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I don't think you should feel embarrassed at all thought Tammy, because the way math is taught these days can seem really confusing to anyone who didn't learn it that way. It took me at least 2 years of training in how to teach math before I really started to "get it" myself.

 

:grouphug:

Edited by jenbrdsly
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Hi Tammy,

 

I absolutely detest Singapore's bar method and I will **not** (ever ;) ) teach the method. It is not how I think and it is not how I want to teach. And, guess what, I have managed to raise incredibly strong math students in spite of the dire cries to the contrary heard on this forum.

 

I really like Hands-on Equations' approach and its methodology is how I think and SM problems are just as easily adaptable to their pawn/balance approach.

 

Here is how the problem would look solving it that way:

 

You are given WHITE pawns and BLUE pawns and an image of a balance scale. Children learn that both sides of the balance remain equal as long as everything is done to both sides of the balance. They also know that you have to solve in terms of only WH or BL, so they have to define in terms of a single one.

 

So, knowing that, the student is given that the number of nickels is equal to the number of dimes plus 43

N= D+43

 

They are also given that after he spends the 17 dimes that that number twice equals the number of nickels

N= D-17+D-17

 

The child would have WH=BL + 43 and WH=BL-17+ BL-17, so set up on the balance sheet, they would make the 2 sets of pawns equal each other

 

BL-17+BL-17 = BL + 43

 

They would remove the pawn from the left and therefore remove a pawn from the rt. Then they would add 17+ 17 to both sides and end up with

 

BL=43+17+17 , BL=77

 

So, WH= BL+43 or WH=77+43 or 120 nickels and since he spent 17 dimes, 77-17=60

 

This is a more algebraic way of teaching, but my kids and I think this way. ;)

 

HTH

Edited by 8FillTheHeart
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Hi Tammy,

 

I absolutely detest Singapore's bar method and I will **not** (ever ;) ) teach the method. It is not how I think and it is not how I want to teach. And, guess what, I have managed to raise incredibly strong math students in spite of the dire cries to the contrary heard on this forum.

 

I really like Hands-on Equations' approach and its methodology is how I think and SM problems are just as easily adaptable to their pawn/balance approach.

 

Here is how the problem would look solving it that way:

 

You are given WHITE pawns and BLUE pawns and an image of a balance scale. Children learn that both sides of the balance remain equal as long as everything is done to both sides of the balance. They also know that you have to solve in terms of only WH or BL, so they have to define in terms of a single one.

 

So, knowing that, the student is given that the number of nickels is equal to the number of dimes plus 43

N= D+43

 

They are also given that after he spends the 17 dimes that that number twice equals the number of nickels

N= D-17+D-17

 

The child would have WH=BL + 43 and WH=BL-17+ BL-17, so set up on the balance sheet, they would make the 2 sets of pawns equal each other

 

BL-17+BL-17 = BL + 43

 

They would remove the pawn from the left and therefore remove a pawn from the rt. Then they would add 17+ 17 to both sides and end up with

 

BL=43+17+17 , BL=77

 

So, WH= BL+43 or WH=77+43 or 120 nickels and since he spent 17 dimes, 77-17=60

 

This is a more algebraic way of teaching, but my kids and I think this way. ;)

 

HTH

 

 

Wow, you have no idea how helpful this was. My dd never learned bar models in ps (and neither did I, of course!) and when we have run across them in MM (luckily, not too often!) or in doing Singapore word problems, they have often mystified us both. Sometimes I am unable to figure out the bar models, but I know how to solve the problem using algebra, so I just do it and show her how to do it that way. Sometimes I have even said "in a couple of years you will learn a really easy way to solve this kind of problem, let's just not worry about it till then" but secretly, of course, I have worried that I am breaking her!! We just started using HOE and it makes *so much sense* to both of us!! Today when the bar models came up doing equations in MM, I told her to feel free to pull out her HOE materials to solve the problem, and she did! Perfectly, and with complete understanding! I think learning to do algebra is already rocking her world, and I am so grateful to hear that I might not be breaking her, after all, by not insisting she master bar models . . .

 

And Tammy, don't worry! :grouphug: I like to think I'm a smart little cookie, and those bar models kick my tuckus on many occaisions. Hang in there!

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Abel saved 43 more nickels than dimes. After he spent 17 dimes, he had twice as many nickels than dimes. How many coins does he have left.

 

 

I haven't read the responses, but here is how I solved it:

 

After Abel spent 17 dimes, he now had 60 more nickels than dimes (43 + 17). Since he has twice as many, then he must have 60 dimes and 120 nickels.

 

#nickels = #dimes + 60

#nickles = #dimes + #dimes

#dimes = 60, so

#nickles = 60 + 60 (120)

Abel had 180 coins.

 

 

 

I don't know how I would figure that out with a bar diagram.

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We are others where the bar method hasn't really clicked.

 

For the bar model to work well, there seems to need to be an intuitive sense that one needs to start the process at the point that there are twice as many nickels as dimes. How do people know to do that? If you do, then the bar model seems like it is easier. But unless you somehow know that is the right place to start, it seems to me that it is harder. ?????

Can someone who gets this shed some light?

 

But if you do not intuit the starting point to make the bar model work easily, then starting with any information given will get to a right answer using step by step algebra.

 

I would do, (starting in a less optimum way as the answer above this, but still getting to right place) where d is original dimes and n is nickels:

 

d+43=n

(d-17)2=n

 

therefore,

d+43=(d-17)2 or =2d-34

 

d-d+43+34=2d-d-34+34

 

43+34=d

77=d

 

77+43=n=120

 

77-17= final number of dimes =60

 

120 +60 = 180 coins left

Edited by Pen
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there seems to need to be an intuitive sense that one needs to start the process at the point that there are twice as many nickels as dimes.
I personally did not start with twice at all - I started with what I know as I read the problem - a pile of nickels and a pile of dimes (drawn simplistically as 2 rectangles :lol:)

and the first fact -- I have 43 less in one pile:

 

[Nickels--------------------------]

[Dimes --------------] <-- 43 --^ (difference of 43)

 

Then I see that he spent 17 dimes

[Nickels--------------------------]

[Dimes ------]<-17-><-- 43 --^

 

after which he had twice as many nickles as dimes

..................1/2...................... (dots only to try and keep 1/2 over the little line)

[Nickels------|-------------------]

[Dimes ------]<-17-><-- 43 --^

 

and then visually I see that the 17+43 is also = to 1/2 of the nickels

and also that 3*(17+43) = total

Edited by LaughingCat2
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I personally did not start with twice at all - I started with what I know as I read the problem - a pile of nickels and a pile of dimes (drawn simplistically as 2 rectangles :lol:)

and the first fact -- I have 43 less in one pile:

 

[Nickels--------------------------]

[Dimes --------------] <-- 43 --^ (difference of 43)

 

Then I see that he spent 17 dimes

[Nickels--------------------------]

[Dimes ------]<-17-><-- 43 --^

 

after which he had twice as many nickles as dimes

..................1/2...................... (dots only to try and keep 1/2 over the little line)

[Nickels------|-------------------]

[Dimes ------]<-17-><-- 43 --^

 

and then visually I see that the 17+43 is also = to 1/2 of the nickels

and also that 3*(17+43) = total

 

:iagree: That's how I thought of it as well. I tend to draw pictures and need some kind of visual scale to help me. I've heard a lot about the bar model in SM but the way some people talk about them has me very intimidated to even look into it. I was definitely not taught that way.

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I personally did not start with twice at all - I started with what I know as I read the problem - a pile of nickels and a pile of dimes (drawn simplistically as 2 rectangles :lol:)

and the first fact -- I have 43 less in one pile:

 

[Nickels--------------------------]

[Dimes --------------] <-- 43 --^ (difference of 43)

 

Then I see that he spent 17 dimes

[Nickels--------------------------]

[Dimes ------]<-17-><-- 43 --^

 

after which he had twice as many nickles as dimes

..................1/2...................... (dots only to try and keep 1/2 over the little line)

[Nickels------|-------------------]

[Dimes ------]<-17-><-- 43 --^

 

and then visually I see that the 17+43 is also = to 1/2 of the nickels

and also that 3*(17+43) = total

 

:iagree: I resisted the bar models for a while as well, but as we use them, I see that they are helping my math advanced child and my average math child.

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I'm so very thankful that I'm not the only one confused by the bar model. I'm going to come back to this later today, pencil and paper in hand and read through everything posted, all the links, and hopefully figure this stuff out thoroughly.

 

Thanks for all the posts so far!

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I tend to draw pictures and need some kind of visual scale to help me. I've heard a lot about the bar model in SM but the way some people talk about them has me very intimidated to even look into it. I was definitely not taught that way
To me, drawing a picture is the Singapore bar model method, except that the picture is always drawn as rectangles. As I put it in my PP, "a pile of nickels and a pile of dimes (drawn simplistically as 2 rectangles) ".
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I personally did not start with twice at all - I started with what I know as I read the problem - a pile of nickels and a pile of dimes (drawn simplistically as 2 rectangles :lol:)

and the first fact -- I have 43 less in one pile:

 

[Nickels--------------------------]

[Dimes --------------] <-- 43 --^ (difference of 43)

 

Then I see that he spent 17 dimes

[Nickels--------------------------]

[Dimes ------]<-17-><-- 43 --^

 

after which he had twice as many nickles as dimes

..................1/2...................... (dots only to try and keep 1/2 over the little line)

[Nickels------|-------------------]

[Dimes ------]<-17-><-- 43 --^

 

and then visually I see that the 17+43 is also = to 1/2 of the nickels

and also that 3*(17+43) = total

 

You had me to the very end.... Somehow, I get the beginning and then get lost putting in the equations. Can you break it down some more?

 

 

To Tammy---- you are not alone. Sometimes I am very confused about it and sometimes it feels straight forward.

 

I am glad to hear that you can find other ways to work these word problems without using the bar models. Thanks for this thread:)

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Ok, here's a slightly different method:

 

the first fact -- I have 43 less in one pile:

 

[Nickels--------------------------]

[Dimes --------------] <-- 43 --^ (difference of 43)

 

Then I see that he spent 17 dimes

[Nickels--------------------------]

[Dimes ------]<-17-><-- 43 --

 

after which he had twice as many nickles as dimes

[Nickels---------------------------] twice as many: nickels = 2x dimes

[Dimes----------][Dimes----------] <-means this line = the nickels line

[Dimes ---------]<-17-><-- 43 --^

and then visually I see one dimes section = (17+43) and I have the equivalent of 3 dimes sections so 3*(17+43) = total

 

 

Note: the difference between the two is, the first time I did the middle line in my head and realized "oh, 2x dimes = nickels is the same as dimes = 1/2 nickels").

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To me, drawing a picture is the Singapore bar model method, except that the picture is always drawn as rectangles. As I put it in my PP, "a pile of nickels and a pile of dimes (drawn simplistically as 2 rectangles) ".

 

Admittedly, I haven't looked into it at all...I was unaware that it was drawing rectangles in that way. For some reason, I had a different thing is mind when thinking of 'bar models'. Maybe it's something I'll have to introduce dd to as we tend to have the same perspective in how we approach math. Thanks for the clarification :001_smile:

Edited by waa510
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Hi Tammy,

 

I absolutely detest Singapore's bar method and I will **not** (ever ;) ) teach the method. It is not how I think and it is not how I want to teach. And, guess what, I have managed to raise incredibly strong math students in spite of the dire cries to the contrary heard on this forum.

 

I really like Hands-on Equations' approach and its methodology is how I think and SM problems are just as easily adaptable to their pawn/balance approach.

 

Here is how the problem would look solving it that way:

 

You are given WHITE pawns and BLUE pawns and an image of a balance scale. Children learn that both sides of the balance remain equal as long as everything is done to both sides of the balance. They also know that you have to solve in terms of only WH or BL, so they have to define in terms of a single one.

 

So, knowing that, the student is given that the number of nickels is equal to the number of dimes plus 43

N= D+43

 

They are also given that after he spends the 17 dimes that that number twice equals the number of nickels

N= D-17+D-17

 

The child would have WH=BL + 43 and WH=BL-17+ BL-17, so set up on the balance sheet, they would make the 2 sets of pawns equal each other

 

BL-17+BL-17 = BL + 43

 

They would remove the pawn from the left and therefore remove a pawn from the rt. Then they would add 17+ 17 to both sides and end up with

 

BL=43+17+17 , BL=77

 

So, WH= BL+43 or WH=77+43 or 120 nickels and since he spent 17 dimes, 77-17=60

 

This is a more algebraic way of teaching, but my kids and I think this way. ;)

 

HTH

 

I'm feeling very happy that we're doing HOE as well, because we have a much easier time solving things algebraically and I don't get the bar method either.

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To me, drawing a picture is the Singapore bar model method, except that the picture is always drawn as rectangles. As I put it in my PP, "a pile of nickels and a pile of dimes (drawn simplistically as 2 rectangles) ".

 

 

Me too. When we started CWP, I was so frustrated with my kids b/c I kept telling them to draw a picture of the problem and they would draw trees with apples, children in a classroom, etc. They couldn't figure out that the picture I wanted every single time was depicted as rectangles. Once we got on the same page, life was much smoother. :lol:

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Me too. When we started CWP, I was so frustrated with my kids b/c I kept telling them to draw a picture of the problem and they would draw trees with apples, children in a classroom, etc. They couldn't figure out that the picture I wanted every single time was depicted as rectangles. Once we got on the same page, life was much smoother. :lol:

 

OK, this can be our problem too sometimes, depending on the word problem. I guess I've been using the bar model without even knowing it! :tongue_smilie:

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after which he had twice as many nickles as dimes

[Nickels---------------------------] twice as many: nickels = 2x dimes

[Dimes----------][Dimes----------] <-means this line = the nickels line

[Dimes ---------]<-17-><-- 43 --^

and then visually I see one dimes section = (17+43) and I have the equivalent of 3 dimes sections so 3*(17+43) = total

 

 

Thank you Llolly ----- after staring at it for 15 mins, I think I got,:lol: except---> I can see 3 sections but I do not get the 3 sections because I thought it would be:

 

[Nickels---------------------------] twice as many: nickels = 2x

[Dimes ---------]<-17-><-- 43 --^

Then I see only 2 sections. Why do I need the middle section of dimes, is what I now stuck on.

 

I really appreciate your help!

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Let's put the two possible endings side by side:

 

[Nickels--------------------------] twice as many: nickels = 2x dimes

[Dimes----------][Dimes----------] <-means this line = the nickels line

[Dimes ---------]<-17-><-- 43 --^

 

 

[Nickels---------|----------------] <-- tick mark in middle = 1/2

[Dimes ---------]<-17-><-- 43 --^

 

If you look at them both together, maybe it will be more clear that the middle box is an intermediary way to think: nickels =2x dimes is the same as 1/2 nickels = dimes. Also if I was drawing it on paper with the middle step, I might have just put the middle box INSIDE the Nickels box - maybe that would be clearer.

 

Nickels: [[Dimes----------][Dimes---------]]

Dimes : [Dimes ----------]<-17-><-- 43 --^

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I thank everyone for the links and extra guidance. I sat down and really tried to think it through. I get it now, but I'm afraid I only get it because it's been hammered into my head a bunch of times here. I think I need a handful of similar problems for us to work through and make sure we really understand.

 

I think I was getting extra confused because of that 17 - I was somehow thinking I had to take that away from the dimes pile, rather than "add" it to the nickels. Heck, I'm still slightly confused.

 

Are there any other problems out there that I can grab that would help me with this concept? I do plan to review that pdf file also (I printed it) and go through all the Thinking Blocks videos, but I'd love more specific problems to solve :)

 

Thanks for all the support!

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I am slowly getting the bar method. Trial and error. However, both my kids hate CWP. So, I'm giving other books a try.

 

BA is challenging but engaging and funny for example.

 

I'm just not willing to die on the CWP hill with my kids. They hate it enough for me to let it go and find alternatives that will get them to the same place.....good at problem solving.

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Let's put the two possible endings side by side:

 

[Nickels--------------------------] twice as many: nickels = 2x dimes

[Dimes----------][Dimes----------] <-means this line = the nickels line

[Dimes ---------]<-17-><-- 43 --^

 

 

[Nickels---------|----------------] <-- tick mark in middle = 1/2

[Dimes ---------]<-17-><-- 43 --^

 

If you look at them both together, maybe it will be more clear that the middle box is an intermediary way to think: nickels =2x dimes is the same as 1/2 nickels = dimes. Also if I was drawing it on paper with the middle step, I might have just put the middle box INSIDE the Nickels box - maybe that would be clearer.

 

Nickels: [[Dimes----------][Dimes---------]]

Dimes : [Dimes ----------]<-17-><-- 43 --^

 

Thank you! That is very clear.

 

 

How would the bar models work for a non CWP problem such as this one:

 

In a traffic jam, you are able to drive only 1 3/4 miles in 2 1/2 hours. How many miles per hour on the average are you going?

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In a traffic jam, you are able to drive only 1 3/4 miles in 2 1/2 hours. How many miles per hour on the average are you going?

Ok, here's how I would do it. Started with the first fact, saying - 1 3/4 miles in whole increments = 7 increments of 1/4 miles and drew that :

[-|-|-|-|-|-|-] 1 3/4 miles in increments of 1/4 mile

 

and 2 1/2 hours into whole increments, 5 increments of 1/2 hour=>

[--|--|--|--|--] 2 1/2 hours in increments of 1/2 hour

 

Then I thought " hmm, miles per hour on average" for a little while before I realized that just translates in bar diagram speak to "how far in 1 hour" on the bar diagrams.

If you just look at those 2 diagrams you can see that 1 hour is ~= 3/4 miles(even with them not quite matching size wise - I could see it on my scribbled version and I wasn't being very careful at all).

[-|-|-|-|-|-|-] 1 3/4 miles in increments of 1/4 mile

[--|--|--|--|--] 2 1/2 hours in increments of 1/2 hour

 

However, I don't know a way do the exact calculation using those numbers without switching to decimal:

 

1.75 (total miles) / 5 (# of 1/2 hour increments)* 2 (to get 1 hour) = .7 miles/hour (I think this is the typical singapore bar model division - divide one number by the number of "parts")

 

Note: I did the decimal calculation by calculator, not by hand ;).

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Ok, here's how I would do it. Started with the first fact, saying - 1 3/4 miles in whole increments = 7 increments of 1/4 miles and drew that :

[-|-|-|-|-|-|-] 1 3/4 miles in increments of 1/4 mile

 

and 2 1/2 hours into whole increments, 5 increments of 1/2 hour=>

[--|--|--|--|--] 2 1/2 hours in increments of 1/2 hour

 

Then I thought " hmm, miles per hour on average" for a little while before I realized that just translates in bar diagram speak to "how far in 1 hour" on the bar diagrams.

If you just look at those 2 diagrams you can see that 1 hour is ~= 3/4 miles(even with them not quite matching size wise - I could see it on my scribbled version and I wasn't being very careful at all).

[-|-|-|-|-|-|-] 1 3/4 miles in increments of 1/4 mile

[--|--|--|--|--] 2 1/2 hours in increments of 1/2 hour

 

However, I don't know a way do the exact calculation using those numbers without switching to decimal:

 

1.75 (total miles) / 5 (# of 1/2 hour increments)* 2 (to get 1 hour) = .7 miles/hour (I think this is the typical singapore bar model division - divide one number by the number of "parts")

 

Note: I did the decimal calculation by calculator, not by hand ;).

 

 

Hmmmm. So, let me try to understand this. By drawing your model with at least some semblance of scale to it, you could get an estimate of the right answer, but not actually the right answer? To get the actual right answer, you actually had to use "standard" arithmetic, in your case decimals (I used fractions and got the same thing as 7/10 mph)?

 

I just tried a similar problem with the time reduced to 3/4 miles, still for 2 1/2 hours, and started with a bar model:

 

[----|----|----] 3/4 mi. in 1/4 increments

 

[--|--|--|--|--] 2 1/2 hours in half hour increments

 

Following what you did, it looks to me like the 1 hour mark is a bit past the 1/4 mile mark. And I do not see either how to actually solve the problem using the bar model beyond this estimate. Maybe someone else reading this will know.

 

How would you who use the bar models use that in reality, just as an estimate to make sure an answer otherwise obtained is not way off? Or perhaps as a real life estimate? Or maybe on a multiple choice type test it would help to choose a right answer-- if all but one were impossible?

 

 

The 'regular' arithmetic is just ...I have to give the reciprocal for the time figure since I don't seem to have a division symbol available on my keyboard:

 

3/4 x 2/5 = 6/20 = 3/10 or .3 mph

 

And, okay the bar model estimate appears to have been close since .3 is a bit more than .25

 

I see now more clearly how you would draw the model for this sort of problem...though not how you would use it to get something more than an estimate.

 

I guess I am wondering how much the bar models help in general. Have they much use where a problem is not specially designed to work with bar models? Do you find yourself and your children using bar models often to solve everyday arithmetic? I was using some Singapore word problems some last year, but ds didn't seem to like it, so I did not get any for this year. And I guess I am still wondering, is this really helpful to learn? Or not really. Or just one more tool out of many?

Edited by Pen
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Full disclosure, I've never done Singapore math or formally used the bar model, but I do have two degrees from MIT, so I'm pretty good at thinking about math.

 

To do this problem with a bar model:

In a traffic jam, you are able to drive only 1 3/4 miles in 2 1/2 hours. How many miles per hour on the average are you going?

 

I would start like llolly.

 

I made two equal length bars one over the other and labeled one 1 3/4 miles and the other 2 1/2 hours. I then split the 1 3/4 miles bar into 7 sections each 1/4 mile long. I split the 2 1/2 hour bar into 5 sections each 1/2 hour long.

 

At that point, I don't think you have to abandon the bar model or change to decimals. I visually decided I needed a common denominator. I further split the 1 3/4 miles bar. Each 1/4 hour got split into 5 pieces; (1/4)/5 = each piece = 1/20 mile. NOTE: I did not actually draw all those divisions, I just split one of the 1/4 mile pieces. Each 1/2 hour piece got split into 7; (1/2)/7 = 1/14 hour.

 

Now each full bar was divided into 35 pieces and I knew we drove 1/20 of a mile in 1/14 of an hour. How far did we drive in an hour? 1/20 * 14 = 14/20 = 7/10 miles/hour.

 

Wendy

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I didn't really go away from the bar model to do the math.

To compare to a more simplified example of bar division:

 

I want to give 4 kids with 24 pokemon cards, how many cards would I give each kid split evenly:

take the bar of 24 cards and split by 4

[--|--|--|--] 24/4 = 5 cards/child

 

is the same method as

 

I went 1.75 miles in 2 1/2 hours

take the bar of 1.75 miles and split by 2 1/2 hours - I just turned 2 1/2 into 5 equal increments to do that.

[--|--|--|--|--] 1.75/5*2 = .7 miles/hour

 

Most of these are easier with a real drawing imo, then these little typed ones. I'm more likely to put something showing relationship etc. Also the bars are just a symbolic visual way to think out the problem. So in this case I ended up with a bar I didn't really use* except to eyeball that my final answer is aprox correct - but no biggie because it was the thinking it out that was important.

 

*Although I also did use the other bar for a version more like Wendy did where I took my 1.75 split into 7 1/4 mile pieces and did the math with that as well - I got the same answer but it was too confusing when I tried to type it out.

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