Singapore 5A, Specific Lessons

[cbollin]

Problem #3

Again, this will become an equivalent fractions problem.

1/3 of something
2/3 of the rest of it.

So if you start off with 3 bars for the 1/3
__ ___ ___


well, that leaves 2 full bars. Hmmm… I want those “2 bars into 3 pieces each so that I can see each thing in thirds.”
Start over with 9 bars
____ ___ ___ (for the 1/3) 3 bars
___ __ ___ ___ ___ ___ (for 2/3) 6 bars

2/3 of the six bars is 4 bars. That means that 2 bars is the money left.

We are told that the money left = 20 dollars.
20 divided by 2 = 10 dollars per bar.

So 90 total
70 spent
20 left.


Problem #4… I can’t remember what the problem asked because the book went missing.

The key on it is that you have to think of the Calculator as “3 pens”

Let me see if I remember the problem enough to fake it.

Pen + Calculator = 2/3 of total.
Rest is 1/3.

So draw 3 bars to start
___ ____| ____

now we are told that the calculator = 3 pens (it costs 3 times as much) and equals 24 dollars.
So that means a Pen is 8.

8+ 24 =32 for the left side of the original bars
take that 32 and divide it over those 2 bars
that means 16 in each bar

so the bar left over is 16.
Total amount is 48 dollars.



I do enjoy from time to time to mention it..... I had to learn bar diagrams when we first started Singapore. I found Julie's description of them helpful and got on fast for solving these kinds of problems. Much better than the plug and chug method I learned in "advanced math" in high school.

The "trick" in 5A/5B is to remember it's ok to draw the bars a second time with "equivalent" fractions.

[Julie in MN]

We are doing Singapore 5A & 5B this year. I ran into this problem last year already and had my college graduate son who has always been good in math figure out one of the word problems we couldn't get. It took him about 20 minutes to do it. So I don't feel too bad when we don't get one of the problems. This year I ordered the Home Instuctor's Guides for 5A & 5B. It has made my life as a teacher a little easier.

Kimberly,
My son who's an engineer doesn't do math the Singapore way, either. Makes me kinda proud of my youngest
But I think you'll do better with that HIG or asking over here, or on other forums with folks who've done Singapore before, rather than asking someone who hasn't learned the Singapore methods. Singapore teaches you to lay out what you know and get to what you need (often using "bar diagrams"), not to put the numbers through all the computations that regular math students (like our older boys) might expect.

Julie

[Julie in MN]

The problem says: Larry spent 1/2 of his money on a camera and another 1/8 on a radio. The camera cost $120 more than the radio. How much money did he have at first?

I am once again stuck on my 6th graders math problem! We finally did look at the answer, and could work backwards from there, or use the process of elimination. But I just want to know the right way of doing it. I did figure it out by using some algebra, with the equation process.....but how about figuring it out by using a graph?

Thanks for the help!!

Hi Jamie,
I posted my method of thinking over here on this thread, and so did a couple of others. So you aren't the first to ask!
http://board.mfwbooks.com/viewtopic.php ... 838#p71890

If you have any more, ask away -- I like to challenge my brain
Julie

[cbollin]

This is one of the "let's pull it all together with everything we've learned"... so agreeing with Julie, you're not the first to ask

I'm guessing I could find my link somewhere.... it might be on the link Julie gave, but I didn't see the one that I drew some bars..

Basically, the big thing (in my opinion) is to remember Equivalent Fractions and "more than".

so 1/2 can mean 4/8

let's draw those 4 bars
camera ___ ___ ___ __
and 1 bar for radio
radio ____

we are told the camera is $120 more than radio. that means 3 bars (which is 4 minus the one) equals that 120. each bar is 120 divided over 3. 120/3=40
each bar is 40.
so there are 8 bars, right? 40*8 = 320


-crystal

[cbollin]

Okay, I know this is pitiful. I have already had one ds who has done this problem, but when I look back in his notebook, he did not complete all the steps and just has the right answer written down and circled. I am assuming he figured it out without having to work it through. However, he is not able to help us and neither is my husband. It is probably obvious, but none of our brains are working in the "Singapore" way today.

So here it is:

The total weight of Peter, David and Henry is 123 kg. Peter is 15 kg heavier than David. David is 3 kg lighter than Henry. Find Henry's weight.

Thanks!
Amy C.

The total weight of Peter, David and Henry is 123 kg. Peter is 15 kg heavier than David. David is 3 kg lighter than Henry. Find Henry's weight.
start with something and adjust...

let's say David is the bar ________
that means Henry is ____ + 3
Peter is _____ + 15.

so.
3 bars + 18 = 123
3 bars = 123-18 = 105
one bar is 35.

one bar is David, so David is 35 kg. So Henry is 3 kg heavier than that... Henry is 38 KG?

Let's try again...
what if we did Henry as the bar
David is ________ - 3
Peter is ________ minus 3 plus 15 more... which means ____ +12

____ + ___ -3 + ____ +12 = 123

3 bars + 9 =123
3 bars = 114
1 bar is 38.
this time the bar was Henry.

but that way is kinda complicated to me for this age with the bar minus and bar plus..... so I'd make the bar be David.

here's some help on 1d prob 10 if you need it.
http://board.mfwbooks.com/viewtopic.php ... 838#p74568

-crystal

[Amy C.]

Thanks, Crystal! I knew I could count on you. I was working it with the bars but doing something goofy with the 3. I can't even remember now how I was trying to work it out. Very little sleep last night. Glad to have other "brains" out there to help me out!

Thanks, again!
Amy C.

[Amy C.]

First let me say, I hate word problems!

In Singapore 5A Textbook, page 25, Practice 1D #6, how do you work that problem? I tried. My son (in Saxon Algebra 1) tried. I cannot figure out the problem to get the answer in the book.

Hi, CarIa! First of all (((hugs)))! Singapore math can really make my brain hurt sometimes.

I looked back in my ds's work. This is how we worked it. And I say we because I had to help my ds with this one. I think I may have even had to ask on here. I can't remember if it was this exact one or not, but I know I have had to ask for help more than once on here.

So this is how we worked it....


Peter has twice has much as Joe ____ ____
Joe has 40 more than Emily ____ + 40
Emily ____
They have 300 altogether.


Peter's amount + Joe's amount + Emily's amount = 300

____ + ____ + ____ + 40 + ____ = 300

Subtract 40 from each side

____ + ____ + ____ + ____ = 260

260 divided by 4 = 65

So one ____ = 65

Double because Peter has twice as much as Joe ____ + ____ = 130

Then add the 40 to it because we have to include that in Peter's amount as well.
130 + 40 = 170

Hope this helps!
Amy C.

[Julie in MN]

I don't see this particular problem on the thread, but some of the upcoming problems are there so you might want to keep this bookmarked for Singapore 5A: http://board.mfwbooks.com/viewtopic.php?f=23&t=6838

The word problems will tax your brain, but they are where Singapore shines, in my opinion. They are where the "real math" is. My son's algebra 2 video teacher said something last year that I thought was important - even if you plan to use a computer to solve your real-world problems, you have to know what to tell the computer to do, you have to know how to formulate the equation and where the computer is supposed to plug in which numbers. That's sort-of what you are figuring out with Singapore, I think. In the end, Singapore students often have to do less computation, because they've figured out what piece they really need to find out.

Just trying to encourage you as you sweat through your mental workout
Julie

P.S. Just because sometimes it helps to see things from more than one perspective:

Peter has twice as many stickers as Joe. Joe has 40 more stickers than Emily. They have 300 stickers altogether. How many stickers does Peter have?

Okay, I'll try to draw the first sentence:
Peter /-----/-----/
Joe /-----/

The second sentence:
Joe /--E--/ + 40
Emily /--E--/

And the third sentence:
Peter + Joe + Emily = 300

Okay, the closest thing to a real number so far is the 2nd sentence,
so I'm going to plug that back into the 1st sentence...
Emily /--E--/
Joe /--E--/+40
Peter /--E--/+40 /--E--/+40

Okay, if the total is 300,
I do know where 120 of that is, so I'll take that out...
300-120=180
Now I see that what is still in question is 4 bars, and that has to equal the 180,
so each bar is 45

Now, what was the question? Oh, how much does Peter have?
Peter is 2 bars plus 2 40's, so
45 + 40 + 45 + 40, which is fairly easy to add in my head... 85 + 85, or 80 + 90, which equals 170.

Let's see if the rest works,
Emily = 45
Joe = 45 + 40 = 85
170 + 45 + 85 = trying to do it in my head, 170 + 130 = 300, yes!

[asheslawson]

I just want to double check this problem....

Singapore, Primary Math 5A, WB, exercise 5, pg 15, #4

It reads: "Give a quick estimate of the area of a rectable with length 114 in. and width 92 in."

My son answered 9,000 because he rounded 114 to 100 & 92 to 90 resulting in 100 X 90 = 9,000

I believe that to be correct - HOWEVER - using the Sonlight HIG the author said he sould have said 100 X 92 = 9200 ?

I have two concerns...first in the problem itself the author rounds the 114 to the nearest 100, not the nearest 10 or it would have been 110 X 90... and then the 90 was used without rounding at all. Makes no sense to suddenly not round one specific number...which brings me to my secong problem...

Throughout the questions in the TB & WB for this exercise - all numbers have been rounded uniformly...such as the one above was 229 X 28 & the author said the student should have done 200 X 30 = 6,000 (rounded to nearest 100 & nearest 10). The one before that was 4165 X 53 & the author said it s/b 4000 X 50 = 200,000 (rounded to nearest 1,000 & nearest 10).

So - I believe my son was correct with rounding 114 X 92 to 100 X 90 = 9,000

BUT - I want to make sure I am not missing something. Math is sometimes a struggle for me.

Thanks - ash

[Julie in MN]

I just want to double check this problem....
Singapore, Primary Math 5A, WB, exercise 5, pg 15, #4

It reads: "Give a quick estimate of the area of a rectangle with length 114 in. and width 92 in."
My son answered 9,000 because he rounded 114 to 100 & 92 to 90 resulting in 100 X 90 = 9,000
I believe that to be correct - HOWEVER - using the HIG the author said he should have said 100 X 92 = 9200 ?

I have two concerns...first in the problem itself the author rounds the 114 to the nearest 100, not the nearest 10 or it would have been 110 X 90... and then the 90 was used without rounding at all. Makes no sense to suddenly not round one specific number...which brings me to my secong problem...

Throughout the questions in the TB & WB for this exercise - all numbers have been rounded uniformly...such as the one above was 229 X 28 & the author said the student should have done 200 X 30 = 6,000 (rounded to nearest 100 & nearest 10). The one before that was 4165 X 53 & the author said it s/b 4000 X 50 = 200,000 (rounded to nearest 1,000 & nearest 10).

So - I believe my son was correct with rounding 114 X 92 to 100 X 90 = 9,000
BUT - I want to make sure I am not missing something. Math is sometimes a struggle for me.
Thanks - ash

I hope you get answers from those doing 5A right now. For us, it's kind of ancient history, but in general I have some thoughts on "estimating" in math.

Main thought: It's always a headache. LOL. I think every math book we've used has had some inconsistencies on this topic, including high school, including public schools. My best counsel to my kids is to figure out why the author may have chosen that route.

In the case of an equation that includes 100, my guess is that it is very easy to multiply anything by 100, so you can afford to be slightly more precise in the *other* number than you might be if you weren't working with 100. With 100, there is virtually no other multiplying involved, besides counting zeros. Whereas if it were 200, then I might stick to multiplying just one digit in each number when estimating.

The best reason for learning to estimate seems to me to give the student the ability to glance at an answer and notice whether it is logically way off the mark - whether in a future math class or in the supermarket. The details don't seem so important, other than an ability to watch what a teacher expects and follow their lead. And it looks like you have those covered very well! I'd count the answer correct!

Julie

[Poohbee]

Hi! We're using 5A right now. My dd also gave 100 x 90 = 9000 for her answer. In the Singapore Primary Mathematics Answer Keys 4A-6B book, the answer is given as 9000. So, I think your son did just fine with his estimation on this problem.

[asheslawson]

Thank you Poohbee. I don't have Singapore's official answer guide or MFW's this year as I found a good deal on HS classifieds & was able to get several of the Home instructors guides by buying several. Most of the lower levels I also have the MFW answers or I have Singapore's 1a-3B answers...but in 5A & 5B all I have is the HIG. Since I had no second source to refer too, which has helped me in past when answer guides had errors, I needed clarification. I really felt like his was more correct than the 9,200 the HIG offered.

Thanks again, everyone is so helpful on this board when I get stumped.

[Julie in MN]

IP 5A, pg 14, #10

This one, from Intensive Practice 5A, is stumping us! I am also doing 5A textbook along with 5A Intensive Practice. I also have the HIG from Sonlight - which is helpful if I'm stumped on a problem in the TB or WB, but there is not help for IP!

How do I help my son solve: Tim & Shelly have a total of 3450 coins in their piggy banks. Shelley & Nina have a total of 5130 coins. Nina has 5 times as many coins as Tim. How many coins does Shelley have?

I posted to Singapore's forum for this problem & #9 as both left me clueless. The responder was VERY helpful with #9. However - this one does not make sense - I'm trying to get it un-muddled in my head - but I'm clueless! Here is what she sent me....

"Have you done the Primary Mathematics? The Standards edition HIG has a lot of pointers. Or you could do Process Skills at a lower level until they become more familiar. This one is like one in the 5A textbook, are you doing that?

Nina has 5 times as much as Tim
|----| Tim
|----|----|----|----| Nina
Now add Shelley to both. I am going to add her to the beginning. The point is that the difference stays the same if you add the same to both.
|---------|----| Shelley and Tim = 3450
|---------|----|----|----|----| Shelley and Nina = 5130
Can you see that the difference is 3 units and solve from there? "

I have 3 issues...
First: I still don't get it. I have the answer in my IP answer section - but I don't see how to find it.
Second: based on the other problem when a skirt that cost 4 times the blouse in her response was represented by 4 units - THEN in this problem, with Nina having 5 times as much as Tim - shouldn't Nina be represented by 5 units?
Third: How does she know to show Shelley & Tim (with 3450 coins) as 2 units & Shelley & Nina (with 5130 coins) with 5 units?

I know this is a LOT - but I am feeling so much like my head is full of concrete & it isn't letting this in!

Aw, hugs for your spinning brain.

I like to just do stream-of-consciousness typing, the way I would think thru a problem on the marker board with my son. Then maybe I'll end up with diagrams similar to hers... or not.

So, I start with the beginning,
Tim & Shelly have a total of 3450 coins in their piggy banks.
/---Tim---/---Shelly---/ = 3450 coins
Nothing very helpful there, we don't even know if the bars are equal.

So moving on to the next piece.
Shelley & Nina have a total of 5130 coins.
/---Shelley---/---Nina---/ = 5130 coins
Again, not too helpful, we don't know if they're equal, but I've gotta start somewhere, so I draw everything and keep those drawings in case they help later.

Next,
Nina has 5 times as many coins as Tim.
/---Nina---/---Nina---/---Nina---/---Nina---/---Nina---/
/---Tim---/
Okay, this is the first time we know the bars are actually equal to one another.
(By the way, I think the gal above made a typo here - you're right about the 5.)

Now, I don't really have a direction here, but I do know we could plug those bars into the other diagrams and see what we get.
So here are the first 2 diagrams we had.
/---Tim---/---Shelly---/ = 3450 coins
/---Shelley---/---Nina---/ = 5130 coins
I'm wondering where I could plug in those "truly equal" Tim/Nina bars from the 3rd diagram?

Okay, I'll try subbing in Nina for Tim:
/---(1/5 of Nina)---/---Shelly---/ = 3450 coins
/---Shelley---/---Nina---/ = 5130 coins
Naw, I don't like that, the fraction gives me a headache to think past that.

Let me try the other way, subbing in Tim for Nina:
/---Tim---/---Shelly---/ = 3450 coins
/---Shelley---/---Tim---/---Tim---/---Tim---/---Tim---/---Tim---/ = 5130 coins
Do you see how I could exchange "Nina" for 5 Tims, because what she has is 5x as much as Tim? I hope that makes sense to you?!

Okay, now we have Shelly & Tim equal to 3450.
Then we have the difference between 5130 and 3450 equals what? Let me see if I can do it in my head...
3450 (+ 50) = 3500, then (+500) is 4000, then (+1130) is 5130,
so that means the difference is 1680 (50 + 500 + 1130).
Now, 1680 has to be equal to 4 of the Tims, can you spot that?
So, doing it in my head again,
1600 is easily divisible by 4, so that's (400), and 80 divided by 4 is (20), so 420 (400 + 20) must be one Tim.

Okay, it's late and I'd better stop typing until you confirm that 420 is the correct answer in the back of your book? We never did the IP books. I could, of course, check my answer by plugging that back into the original word problem, if I were a good math student

And hopefully others will chime in coming from different angles, and one way of thinking will be something you relate to.
Julie

[asheslawson]

Hallelujah!!!!!! Julie - I was doing the chart process too - that is how I usually figure these things out - but I just wasn't getting this one. UGH!

The answer in the book is actually 3,030! However - your chart helped me figure out how many coins Nina had (or 5 Tims as you showed it - and YES that made perfect sense!!!)

When I multiply 420 X 5 = 2,100
Then subtract 5,130 - 2,100 = 3,030

This is how many coins Shelley has - which is actually what the question asked. But, of course, we had to discover the 'unit' value of Tim (& therefore Nina) in order to get that.

I do appreciate that at least I caught that her bars showed 4 times & should have been 5 times - but her difference picture wasn't working to get the answer even when I corrected it to 5. I just could not figure it out & I've worked on it off and on all day! SO thankful for your help!

We are adding the IP to the TB & WB, because he struggles a little and seems to do better with a little more practice. This is the 1st time I've tried the IP (intensive practice) books - and they are ok - except that there is no help for them in the HIG if you do get stumped.

Thanks again so much! I can't wait to help my ds see the light tomorrow!!!

[Julie in MN]

So glad I made sense And glad you were on-the-ball and fixed my answer - how many times have I told my son to look at what the original question is asking LOL?!

I also thought of another way of looking at this:

Let me try the other way, subbing in Tim for Nina:
/---Tim---/---Shelly---/ = 3450 coins
/---Shelley---/---Tim---/---Tim---/---Tim---/---Tim---/---Tim---/ = 5130 coins
Do you see how I could exchange "Nina" for 5 Tims, because what she has is 5x as much as Tim? I hope that makes sense to you?!

Okay, now we have Shelly & Tim equal to 3450.

I could have followed that with this:

Okay, now we have Shelly & Tim equal to 3450, so I'll sub 3450 in for each Shelley/Tim pair:
/---3450 (Shelley/Tim)---/ = 3450 coins
/---3450 (Shelley/Tim)---/---Tim---/---Tim---/---Tim---/---Tim---/ = 5130 coins

Oh, well, you got it already, see how addicting this can be? Have fun in math today,
Julie

[asheslawson]

Oh, well, you got it already, see how addicting this can be? Have fun in math today,

It is addicting!! I actually couldn't wait to teach math today - despite the fact that he tries everything to get out of it. I actually think he caught on to a little of my excitement. I had written a few hints on the dry erase board by the table so he saw them when he woke. He laughed at me! I told him I couldn't sleep till I'd solved it & that I was so excited to have it actually make sense! (Math never did for me as a child - so you can imagine how much I've been terrified to teach it when it got to the level I started struggling, which was about now - his age!)

However - today's problems in the IP were equally tough - or even tougher. And we solved them completely on our own and got the correct answers! Woohoo! Then - I checked back in with the administrator @ Singapore Math forum & in one of her replies - she said the IP problems tend to be a little harder than the TB & WB. That makes me feel better about adding them. We just do one page a day in the IP to accompany his regular math - because when I tried more it was too much & he got bogged down. But now - I feel like we're giving BOTH of our brains a good, healthy challenge! I love that - I think my brain got a little softer & less rigid today!! Maybe I'll finally get math down yet!

[Julie in MN]

I will appreciate help in teaching txt book 5a. Page 69. # 4b
Thank you.

Area of triangle. I think my husband and I over thought this one. All is well now. Thanks!

Yes, as long as you have a base and a height, you are good to go -- even if the triangle seems to be "leaning over"
Julie

[Julie in MN]

My ds just finished a review in 5A. It was a really long one...he counted 117 problems.

PROBLEM: He missed over half the problems.

EXCUSES: 1) He was overwhelmed by the sheer size of the review.
2) We had just finished fractions (which was tough) and then had taken a week off for Thanksgiving. The first day back to school after the break he started the review.
3) We had a lot going on the days he was working on the review (we spread it out over a couple of days) so I feel he may have rushed through it.

What do I do? Should I have him redo the review? Do we just work on each problem he missed together? Has anyone else had this happen? Just looking for some seasoned advice....thanks,
Kathy

Hi Kathy,
My first curiosity was what exactly he had trouble with. I know "half the problems" might lead to the conclusion that he had trouble with "everything," but I wonder if you see a trend. Did he miss all the word problems? Did he have trouble with basic math computation? Were they silly oversites?

To me, silly mistakes are normal for a growing boy. I started to build the discipline, or "math endurance," in 7th grade at my house. On the other hand, not understanding concepts means going back and reteaching would probably be helpful, and problems with computation may mean more drill on the math facts would be useful.

I'm looking at my son's 5A workbook, the first review (there are 2 total reviews in his 5A workbook), and here are the things I notice:

+ I let my son just do one each in the first two sections (writing out long numbers in words and vice versa)- though I still had to have him re-do one of them for "comma and spelling" (fourty LOL).
+ In #7, he didn't write out those numbers in order; instead, he took the lazy man's way and just wrote "3,2,1,4" on top.
+ In #8, I circled the word "twelve" so I suspect he didn't write all 12 multiples the first time thru.
+ In #11, I wrote in the margin, "1 m = 100 cm, 1 kg = 1,000 g," so apparently he needed that info to answer those problems correctly.
+ In #15, I wrote "square?" Apparently he had originally put cm instead of cm2.
+ In #16, I circled "2-cm" because apparently he originally thought they were 1 cm. squares, even though it is very clear in the diagram. Even worse, the corrected answers are in my handwriting, so apparently we did them on the marker board or orally and I scribed for the poor young man.
+ Then in #18, I wrote "of what" because apparently he wrote "15" instead of "15 in."
+ It looks like he did the first 3 word problems in his head, but the last one has a detailed drawing that is mostly in my handwriting

Does that help you see the kinds of mistakes a typical kid might make? My son was taking his math through a local Christian college by 11th grade, so he turned out okay. Just thought that might ease some worries, in case the problem isn't really math skills but simply young man syndrome.

Julie