Ugh! Need help with Singapore CWP-3 problem.

[Laurie in VA]

Good morning! I am embarrassed to say that I can not help DD figure out Topical Problems 1, #19 in CWP-3 (what will I ever do when she gets older)? I am sure it will be quite obvious once someone else points it out, but I am drawing a blank this morning. OK:

 

 

Jason and Bob have 193 marbles altogether. Bob has 47 marbles fewer than Jason. If Jason gives 15 marbles to Bob, how many more marbles will Jason have than Bob?

We figured out that Bob now has 32 less marbles, since Jason gave him 15. But after that we are both completely stumped. Can anyone explain this to me...without laughing ;)

[elegantlion]

Good morning! I am embarrassed to say that I can not help DD figure out Topical Problems 1, #19 in CWP-3 (what will I ever do when she gets older)? I am sure it will be quite obvious once someone else points it out, but I am drawing a blank this morning. OK:

 

Jason and Bob have 193 marbles altogether. Bob has 47 marbles fewer than Jason. If Jason gives 15 marbles to Bob, how many more marbles will Jason have than Bob?

We figured out that Bob now has 32 less marbles, since Jason gave him 15. But after that we are both completely stumped. Can anyone explain this to me...without laughing ;)

 

I took the challenge! I'm still on my first cup of coffee and we haven't gotten to that problem in the book so this was my morning brain teaser. Here's is how I derived the answer, there might be a simpler way, but not this early. I'm not even sure if I wrote out the math correctly, but it worked.

 

If the common amount of marbles is x, then

 

J (Jason) -47 = B (Bob)

 

 

To solve for x that 193 - 47= 146, then 146/2 = 73, then we have to add 47 + 73 to get J's . Clear as mud? x = 120

 

B = 73

J = 120

 

120-15 = 105

73 + 15 = 88

 

105- 88= 17

 

17 is the answer, I looked it up to be sure.

[TCoppock]

I know ow to get the answer but not sure if it is how Singapore teaches.

 

Given in Problem: J + B = 193 and J - 47 = B

 

So J + (J - 47) = 193

Add like terms 2J - 47 = 193

Add 47 from each side +47 +47

2J = 240

Divide each side by 2 2 2

J = 120

That give you originally J has 120 and B has 73

If J gives 15 to B then just subtract for new totals

 

J now has 105 and B has 88

 

Giving J 17 more than B

[TCoppock]

Paula your coffee must have kicked in before mine. :001_smile:

[Laurie in VA]

TY both! Technically we haven't worked on division in PM-3 - just addition/subtraction. So I was trying to do this without dividing, but after looking at your problems worked I see no way around it. We gave up and are on to penmanship now, but when we go back after lunch will re-work the problems we missed. Thanks again!

[Shasta Mom]

I was surprised to see these problems, too for grade 3! I can solve them using algebra, but my ds (4th gr) doesn't know real algebra yet. They are supposed to use bar diagrams, and for the more difficult ones, I wish there were some explanations. In each chapter, there are a couple that I have trouble with...

[nmoira]

 

Jason and Bob have 193 marbles altogether. Bob has 47 marbles fewer than Jason. If Jason gives 15 marbles to Bob, how many more marbles will Jason have than Bob?

 

 

You don't need to use the 193 to solve the problem.

 

Here's the diagram.

 

Jason:

|---------------------------------|

|---------------------------------|

 

Bob:

|---------------------| 47

|---------------------|<-------->

 

 

The difference is marked as 47.

 

Then indicate 15 fewer for Jason and 15 more for Bob.

 

Jason:

|-----------------------------|---|

|-----------------------------|---|

 

Bob:

|---------------------|---|

|---------------------|---|<->

The difference is now 47-15-15=17

[nmoira]

 

I was surprised to see these problems, too for grade 3! I can solve them using algebra, but my ds (4th gr) doesn't know real algebra yet. They are supposed to use bar diagrams, and for the more difficult ones, I wish there were some explanations. In each chapter, there are a couple that I have trouble with...

If you get stuck, Jenny on the Singpore Math Forums is a great help (this is the same Jenny who wrote the Home Instructor's Guides). Check first to see if the relevant problem has already been discussed.

[Shasta Mom]

Good point - I've communicated with Jenny before but never thought to do it for problems......

[dragon_horse_0002]

Both solutions from elegantlion and Tcoppod involve algebra, I think. They construct 2 equations and 2 unknowns. For 3rd grade level, I think nmoira's is the right way to solve the problem. I think the challenge of problems like this is how to solve it with the 3rd grade knowledge.

[at the beach]

I used a bar diagram as well. It's hard to draw here. But I'll try. I did:

 

J _____________________ ________

 

The second line would be the 47 more J has.

 

B ____________________

 

B only gets one line.

 

Then I determined that each of the identical lines equals 73 because if we take out 47 that J has more of that leaves us with 146 divided by two is 73 a piece. Now, we take 15 from J and give it to B, so I added the fifteen to the 73 to come up with B's total amount and subtracted the 15 from J's total to come up with his total amount. At that point, it's 105 minus 88, which equals 17.

 

Are you using CWP 3 with your dd age 6? That's a huge challenge IMHO for a 6 yo.

 

Anita

[arcara]

I'm learning to be a better math problem solver right along with my dd using CWP 3 this year! You have to use the bars, as in this example. They help me so much!!! I, too, was convinced when I first looked at this book when it arrived that these problems could only be solved with Algebra. My dd is doing very well now with this book, and loves doing the problems!! It helps that one day she got one right that I didn't and then she had to explain it to me :001_smile:

 

You don't need to use the 193 to solve the problem.

 

Here's the diagram.

 

Jason:

|---------------------------------|

|---------------------------------|

 

Bob:

|---------------------| 47

|---------------------|<-------->

 

 

The difference is marked as 47.

 

Then indicate 15 fewer for Jason and 15 more for Bob.

 

Jason:

|-----------------------------|---|

|-----------------------------|---|

 

Bob:

|---------------------|---|

|---------------------|---|<->

 

The difference is now 47-15-15=17

[elegantlion]

You don't need to use the 193 to solve the problem.

 

Here's the diagram.

 

Jason:

|---------------------------------|

|---------------------------------|

 

Bob:

|---------------------| 47

|---------------------|<-------->

 

 

The difference is marked as 47.

 

Then indicate 15 fewer for Jason and 15 more for Bob.

 

Jason:

|-----------------------------|---|

|-----------------------------|---|

 

Bob:

|---------------------|---|

|---------------------|---|<->

The difference is now 47-15-15=17

 

You are brilliant! As with most things in life I make them more complicated than they have to be. :lol:

[nmoira]

 

You are brilliant! As with most things in life I make them more complicated than they have to be. :lol:

Prior to using Singapore with DD the Elder, it would never have occurred to me to bring out anything other than the big guns. The thing I like best about the rod diagram approach is that it helps the child visualize what is going on the in the problem so they can literally see if the answer is a reasonable one.

[nmoira]

 

Are you using CWP 3 with your dd age 6? That's a huge challenge IMHO for a 6 yo.

It depends on the 6yo. Mine recently started 4A. OTOH, my younger child will be nowhere near that level at 6yo.

[TCoppock]

I understand the thought process behind Anita in Ohio's answer but nmoira why do you subtract 15 and 15 from the original difference to get the answer? Sorry I can't figure it out even with the diagram.

 

Sorry for hijacking. Just want to learn for the future.

[nmoira]

I understand the thought process behind Anita in Ohio's answer but nmoira why do you subtract 15 and 15 from the original difference to get the answer? Sorry I can't figure it out even with the diagram.

It's easier to see on paper.

 

Jason:

|-----------------------------|---|

|-----------------------------|---| Red portion is 15.

 

Old:

----------------------|-----------| (47)

New:

--------------------------|---|

 

Bob:

|---------------------|---|

|---------------------|---| Red portion is 15

 

The original difference, 47 is the purple line. The new difference is the green line. Remember that Bob gets 15 more marbles, but Jason also now has 15 fewer. Neither this way nor Anita's is better than the other. I prefer this way only because it shows what is going on, and the child can more easily understand why you don't just subtract 15 from the original difference of 47.

Edited October 31, 2008 by nmoira
clarity

[fractalgal]

Good morning! I am embarrassed to say that I can not help DD figure out Topical Problems 1, #19 in CWP-3 (what will I ever do when she gets older)? I am sure it will be quite obvious once someone else points it out, but I am drawing a blank this morning. OK:

Jason and Bob have 193 marbles altogether. Bob has 47 marbles fewer than Jason. If Jason gives 15 marbles to Bob, how many more marbles will Jason have than Bob?

We figured out that Bob now has 32 less marbles, since Jason gave him 15. But after that we are both completely stumped. Can anyone explain this to me...without laughing ;)

 

At the start Bob has 47 fewer marbles than Jason. If Jason gives Bob 15, the subtracted total will be 30 because now it will include the 15 more Bob has been given, and the 15 fewer that Jason will have.

 

47-30=17

 

The 193 total is there to throw you off - unnecessary information. That was my understanding.

 

I don't think the child is expected to do this with algebra yet.

Edited October 30, 2008 by fractalgal

[TCoppock]

I see it now! Thanks so much for showing me.