Visiting B&M mom asking Singapore MIF (5th grade) Q

[SKL]

Hi, my kids are in 5th grade at a school that uses Math in Focus.  They just finished the "order of operations" stuff and now they are doing "real-life" word problems that involve multiple steps.  (Book 5a.)

 

Thing is, some of these problems are algebra problems, where the only way I can think to answer them is to set up equations such as 7+b = 2(2+b) and then solve for b.  (This was an actual problem on yesterday's homework.)  Problem is that I don't think the kids have ever seen this kind of process, so how are they supposed to figure this out?

 

The above question stumped both of my kids and my slower kid's [experienced adult] math tutor.  In addition, there was a question that the teacher had solved for them (d + 2d + (2d+250) = 1,800), but the teacher had done it wrong!  For both of these questions, I walked my kids through the solution a few times, but I don't think they get it, because they don't have the foundation.

 

My question:  are we missing a piece of the puzzle that MIF provides, but their school apparently didn't cover?  Or are they really just trying to find out if the kids can figure out algebra on their own?

 

Or is there some other trick besides algebra that I just don't know because I don't know the MIF method?

 

Any info would be appreciated.  :)

[letsplaymath]

Singapore math uses bar model diagrams (also known as strip models or tape models, or in Singapore they just call them "models"). This is a type of visual algebra that lets the student solve problems by manipulating virtual blocks (like Legos-of-the-mind).

 

I have a series of posts on my blog that explain bar modeling from simple to challenging. They might give you the foundation you need to understand what the math book has in mind:

• Elementary Problem Solving: Review

[Janeway]

You showed us what the how you would set up the problem but not the actual original problem. What was the actual original problem?

[Janeway]

Hi, my kids are in 5th grade at a school that uses Math in Focus.  They just finished the "order of operations" stuff and now they are doing "real-life" word problems that involve multiple steps.  (Book 5a.)

 

Thing is, some of these problems are algebra problems, where the only way I can think to answer them is to set up equations such as 7+b = 2(2+b) and then solve for b.  (This was an actual problem on yesterday's homework.)  Problem is that I don't think the kids have ever seen this kind of process, so how are they supposed to figure this out?

 

The above question stumped both of my kids and my slower kid's [experienced adult] math tutor.  In addition, there was a question that the teacher had solved for them (d + 2d + (2d+250) = 1,800), but the teacher had done it wrong!  For both of these questions, I walked my kids through the solution a few times, but I don't think they get it, because they don't have the foundation.

 

My question:  are we missing a piece of the puzzle that MIF provides, but their school apparently didn't cover?  Or are they really just trying to find out if the kids can figure out algebra on their own?

 

Or is there some other trick besides algebra that I just don't know because I don't know the MIF method?

 

Any info would be appreciated.  :)

If the teacher could not solve the problem you showed, then I would question her ability to teach. The problem with D is something my 12 yr old could solve easily and he has done Singapore Math and just now has finished 6B. I suspect you have a situation of a teacher who has failed to teach. If she did not teach the skills needed before now, they would not get what they are doing now.

[SKL]

Janeway (and anyone else interested):  the problems were as follows:

 

1) [the one my kids' tutor couldn't do]:  Jane had $7, her sister had $2.  Their parents gave them each the same amount of money.  Then Jane had twice as much money as her sister.  How much money did the parents give them?

 

[ETA I shouldn't say the tutor *couldn't* do this, rather that the tutor couldn't help my kid solve it, which could be because it was not explainable without higher math than what they've learned so far.]

 

2) [the one my kids' teacher got wrong]:  (This is simpler in my opinion.)  Joe bought a washer, dryer, and fridge for $1,800.  The fridge was $250 more than the washer.  The washer was twice as much as the dryer.  How much was the washer?

 

My kid did set up #1 with the bar diagram, but that wasn't enough to solve the problem.  You still have (2+g)+(2+g) = (7+g).  What then?

Edited October 22, 2016 by SKL

[SKL]

If the teacher could not solve the problem you showed, then I would question her ability to teach. The problem with D is something my 12 yr old could solve easily and he has done Singapore Math and just now has finished 6B. I suspect you have a situation of a teacher who has failed to teach. If she did not teach the skills needed before now, they would not get what they are doing now.

 

Yeah, I'm not sure if he just made a mistake or was trying to simplify the problem for the kids.  Either way I'm not comfortable with it, but he seems competent in general.

 

I understand the "d" problem should be easy for someone in 6B, but what are we missing back in 5A or before?  Especially for the "g" problem.  I could figure out the "d" problem without resorting to algebraic equations if I needed to.

 

Edited October 22, 2016 by SKL

[TheAttachedMama]

IF I were to help a child with this problem (who had not had algebra), I would have talked out various ways to solve this problem.    Singapore math often puts emphasis in the process rather than just "getting the right" answer.   In other words, math can be a creative subject because there are often more than one ways to solve a problem.   The fun is in talking out various ways to solve it and even trying to come up wtih different alternatives in solving it.   Sometimes, given our US math education, we become so close minded because we have practiced a process or a procedure so often we can't think of other ways to solve it.   One acceptable way to solve this, for example, is through trial and error.   That may not be the most efficient method, but it would certainly get you to the correct answer without using algebra.   AND...I thnk problems like this prepare children for thinking about problem solving in algebraic terms.

 

This is how I would explain the problem to a child with no knowledge of algebra:

 

1)   I would start out reading the first sentence of the problem and drawing a bar model showing how much the sisters had to start with:

"Jane had $7, her sister had $2."

Jane's bar  {__________$7________}

Sister's bar {__$2__} 

 

2)  Next, I would continue reading aloud the problem, "Their parents gave them each the same amount of money."  And I would say to the kids, "OK, so lets add that information to our bar model.  We don't know how much money the parents gave them, but we know it was the same amount.   So lets draw our bars roughly the same size to represent that."

Jane's bar  {__________$7________} +  {___?___}

Sister's bar {__$2__} +  {___?___}

 

3)  Then I would say, "Now, lets look at the other information we are given.  "Then Jane had twice as much money as her sister.  How much money did the parents give them?"

So, what are some ways we could sove this?  Does anyone have any ideas?    (If someone has a suggestion, we explore it.   If no one offers up any suggestions, I would make my own." )

 

So we know that 2 X (2 + ?) = 7 + ?, right.    (Writing this, even though they haven't' been introduced to algebra leads them to start exploring it.)

 

Money given by parents      Sister's $   Janes $

$1                                         $3              $8     (NOT twice as much)

$2                                         $4              $9     (NOT twice as much)

$3                                         $5              $10   (Bingo!)

 

 

 

[Clemsondana]

It's been a few years since my older child did Singapore 5, and my younger one isn't there yet.  I do remember a few problems that I stared at because I could solve them quickly with algebra and I had to stop and figure out how to turn them into something that my child knew.  But, Attachedmama's drawing for the first problem is right. 

 

 You draw the sister as

_____I_____ l 4 l       because you need 2 parent gifts plus 2x2 to equal Jane.  Then under it draw Jane so that it is 

_____l    7          l

 

so that the bar (what the parents give) is 3 - one bar is 7-4, because both sisters have the first bar in common, and then the sister would need another bar +4 to equal Jane's 7.

 

For the bottom problem, you would first draw a box + 250 to equal the fridge, with a box the same as the first box to equal the washer.  Then reading that the washer is twice the dryer, you would draw a 1/2 sized box for the dryer, and divide the big boxes for the washer and fridge in half.  Then you could start solving - 1800 - 250 equals the 5 boxes.  Divide by 5 to figure out how much is in each box. 

[wendyroo]

As others have said, bar models.

 

The easier second problem first:

"Joe bought a washer, dryer, and fridge for $1,800."

When we have a big bar broken into smaller bars, and we don't initially know their relative sizes, then I have my kids sub-divide the large bar with dotted lines to remind us that the divisions will probably need to move...I could not really show that in this model.

[--------------------1800-------------]

[-----W-----][-----D-----][-----F-----]

 

"The fridge was $250 more than the washer."

Now we re-label as possible to show all the amounts in relationship to one variable.  It isn't technically necessary to move the sub-divisions to reflect more accurate ratios, but it does help to visualize the problem.

[--------------------1800-------------]
[---W---][-----D-----][-------F-------]

[---W---][-----D-----][---W+250----]

 

"The washer was twice as much as the dryer."

Again, try to re-label so that everything is shown as multiples of your smallest variable.

[--------------------1800-------------]

[----W-----][--D-][-----W+250------]

[--D-][--D-][--D-][--D-][--D-][-250-]

 

D+D+D+D+D+250 = 1800

D+D+D+D+D = 1550

D = 310

 

"How much was the washer?"

W = 2 * D = $620

 

Wendy

[wendyroo]

The first problem:

 

"Jane had $7, her sister had $2."

J -> [-------7-------]

S -> [-2-]

 

"Their parents gave them each the same amount of money."

J -> [-------7-------][---?---]
S -> [-2-][---?---]

 

"Then Jane had twice as much money as her sister."

J * 1 -> [-------7---------][--?---]
S * 2 -> [-2-][--?---][-2-][--?---]

 

Now it becomes obvious that the ?'s at the end of the two bars line up...they have to since they are the same amount, so we can rearrange:

[-------7--------][--?---]
[-2-][-2-][--?---][--?---]

 

"How much money did the parents give them?"

2+2+?=7

4+?=7

?=$3

 

Plug in to check:

Jane starts with 7 and ends up with 10.

Sister starts with 2 and ends up with 5.

5 * 2 = 10

Check.

 

Wendy

[ondreeuh]

Janeway (and anyone else interested):  the problems were as follows:

 

1) [the one my kids' tutor couldn't do]:  Jane had $7, her sister had $2.  Their parents gave them each the same amount of money.  Then Jane had twice as much money as her sister.  How much money did the parents give them?

 

[ETA I shouldn't say the tutor *couldn't* do this, rather that the tutor couldn't help my kid solve it, which could be because it was not explainable without higher math than what they've learned so far.]

 

2) [the one my kids' teacher got wrong]:  (This is simpler in my opinion.)  Joe bought a washer, dryer, and fridge for $1,800.  The fridge was $250 more than the washer.  The washer was twice as much as the dryer.  How much was the washer?

 

My kid did set up #1 with the bar diagram, but that wasn't enough to solve the problem.  You still have (2+g)+(2+g) = (7+g).  What then?

 

We just did this earlier this year. It is tricky! This has been the hardest part of 5A so far, and we have just finished chapter 4.

 

For the first problem: you set up a bar diagram with the $7 and $2 and show that there is a $5 difference. There will always be a $5 difference, because they are being given equal amounts of money. Then you show that for Jane to have twice as much as her sister, twice $5 is $10, so she has to have $10 and her sister has to have $5, which means you have to add $3 to each kid's total.

 

For the second problem: you set up three bars for the washer, dryer and fridge.

Fridge has a nice long bar.

The washer has a shorter bar that is $250 less than the fridge.

The dryer has a bar half as long as the washer.

Now show that the washer's price is twice the dryer's price (divide the washer bar into two dryer-size chunks)

Now divide up the fridge's bar into two chunks that line up with the washer, and the extra $250

 

-------------------- ----------------- + $250     (fridge)

-------------------- -----------------                 (washer)       = $1800

--------------------                                       (dryer)

 

At this point you should have five equal-size chunks and one that is worth $250. Subtract $250 from each side and you see that $1550 is the cost of five dryers (and the amount of each dryer-sized chunk). Now divide the $1550 into five equal pieces - and label the bar diagram. Each chunk is worth $310. So the dryer cost $310, the washer cost $620, and the fridge cost $870.

[SKL]

We just did this earlier this year. It is tricky! This has been the hardest part of 5A so far, and we have just finished chapter 4.

 

For the first problem: you set up a bar diagram with the $7 and $2 and show that there is a $5 difference. There will always be a $5 difference, because they are being given equal amounts of money. Then you show that for Jane to have twice as much as her sister, twice $5 is $10, so she has to have $10 and her sister has to have $5, which means you have to add $3 to each kid's total.

 

For the second problem: you set up three bars for the washer, dryer and fridge.

Fridge has a nice long bar.

The washer has a shorter bar that is $250 less than the fridge.

The dryer has a bar half as long as the washer.

Now show that the washer's price is twice the dryer's price (divide the washer bar into two dryer-size chunks)

Now divide up the fridge's bar into two chunks that line up with the washer, and the extra $250

 

-------------------- ----------------- + $250     (fridge)

-------------------- -----------------                 (washer)       = $1800

--------------------                                       (dryer)

 

At this point you should have five equal-size chunks and one that is worth $250. Subtract $250 from each side and you see that $1550 is the cost of five dryers (and the amount of each dryer-sized chunk). Now divide the $1550 into five equal pieces - and label the bar diagram. Each chunk is worth $310. So the dryer cost $310, the washer cost $620, and the fridge cost $870.

 

For the second one, that's how my kids had it on their papers, but the teacher had told them to divide the $1,800 by 5 and the answer was 360*2=720.  Hopefully just a one-time clerical error by the teacher.

 

I appreciate the different suggestions for the 1st one.  I think I will go over this again with my kids today, using one or more of these methods.  I still think it's a pretty big leap to expect all kids in 5A to come up with those types of answers themselves.  Maybe the authors intended these to be done in class with the teacher guiding them.

Edited October 23, 2016 by SKL

[ondreeuh]

For the second one, that's how my kids had it on their papers, but the teacher had told them to divide the $1,800 by 5 and the answer was 360*2=720.  Hopefully just a one-time clerical error by the teacher.

 

I appreciate the different suggestions for the 1st one.  I think I will go over this again with my kids today, using one or more of these methods.  I still think it's a pretty big leap to expect all kids in 5A to come up with those types of answers themselves.  Maybe the authors intended these to be done in class with the teacher guiding them.

 

I definitely had to walk my kid through those problems, and bar models are really necessary at that point. I do think they are advanced - the word problems at the end of chapter 4 (multiplying & dividing fractions) have been easy in comparison. But the problems where *something magical happens* and then your result is given in terms of proportions - those are hard!

I would chalk that teacher's error to a one-time thing.

[MeganW]

Look at the books FAN Math Process Skills in Problem Solving.  Those are the easiest way to understand bar models.