Singapore 6A-6B, Specific Lessons

[Julie in MN]

Singapore Workbook 6A Help Needed!

I need help on a problem from the Singapore 6A WB, please.

• Review 3 Problem # 23: The ratio of the number of boys to the number of girls in a hall was 3:2 at first. After 30 boys left the hall, the ratio became 2:3. How many boys were there in the hall at first?

My 8th grader (not the one doing the Singapore 6A but who did Singapore up through 6th grade and is now doing Saxon) figured out how to do #23 using fractions. Very smart of him actually. Good to know that he is learning something! It is times like these that make all the hard work (and sometimes tears) worth it! Makes a mama proud!

If anyone has a different way of doing this problem (like using bar diagrams) and would like to share, please do. I can't help but think there's not a different "Singapore" way of doing it that we are just not seeing.

Gotta love Singapore. It really makes us think! Sometimes it takes the whole family!
Amy C.

Hi Amy,
It's fun for me to keep my math sharp, so I took a peek in my son's old 6A workbook.

• The ratio of the number of boys to the number of girls in a hall was 3:2 at first. After 30 boys left the hall, the ratio became 2:3. How many boys were there in the hall at first?

Okay, what we have is two ratios:
3:2
2:3

And all we know is that we lost 30 boys
And I guess it's good to notice again that the ratios are "boys:girls," so the 30 are lost from the left side

To compare ratios, it would help to have one side the same in both cases. Sort-of a common denominator. So that would probably be 6. We want to put the common "six" on the right, because the girls never change. If we make the left sides the same, then it will be strange and not helpful, because it will "appear" like the boys stayed the same.

So, 3:2 can be 9:6, because that's the same ratio as 3:2, right?
And 2:3 can be 4:6, because that's also the same as 2:3.

Now, we have these 2:
9:6
4:6

And what else do we know? The difference is 30 boys left the first group. So the "5" that's missing from the second ratio is equal to 30 boys.
If that's so, then "1" in the ratio would be equal to 6 boys, right? (We could draw this out with bars, if it helps.)
So the original amount was 9 (x6) and 6 (x6), does that make sense?
And before we do all that hard math (LOL), let's see "exactly" what we need...
(we only want to do the calculations we really need!)
We just need to know how many boys in the first group.
So the math is 9x6 (answer 54).

Does my thinking make any sense at all? All my son wrote in his workbook was "9:6 4:6 54" so he either did it a faster way or did it in his head (which he often does - he doesn't like pencils ).
Julie

[Amy C.]

Yes, Julie, that does make since. That is basically how my 8th grader worked it, but just placing the ratios in fraction form, but that was his thought process.

Thanks! That helps me know he was on the right track. I guess he did retain his Singapore math skills. So glad to know that!

Just the other day he helped us with a Singapore problem, and he said, "This is fun math! I wish my math was like this again." Um...OK...That was so not his attitude at the time when he was doing it. But I am thankful he sees it the way he does in retrospect and has fond memories of it AND that I can see the fruits from it.

Thanks again!

[Julie in MN]

Yay for boys who look back fondly

[kugoi]

You know, I always thought I was pretty good at math, until this year. There were several problems in 5A that took me a while to figure out how to do them. I eventually figured them all out, but I am not looking forward to what's coming up!

[Julie in MN]

Could someone that has done the 6B (us edition) please help me figure this problem out?
It is on page 82 of the workbook, number 12. Stating, The figure shows a quarter circle in a square. Find the area of the shaded part. (pi=3.14).

The little picture shows a 10cm square with a diagonal drawn as the radii of the quarter circle.

I hope this gives someone enough information to save my sanity on this one! I have a feeling that I am missing something obvious.

Laura

Hi Laura,
I have my son's workbook here but didn't keep the textbook or answer key. He's in calculus now LOL.

What threw me for a loop was thinking the quarter circle was shaded. But I decided that I don't think it is all shaded. So if the quarter circle is shaped like a normal quarter circle (where it comes to a point, like when you cut a pizza into 4 pieces), then I'm thinking you would:
1. Start with figuring the area of the circle (radius 10 cm., squared, times pi at 3.14).
2. Then divide the area of the circle by 4 to get the full quarter-circle area (shaded and unshaded).
3. Then figure the are of the triangle that is within the quarter circle but is NOT shaded (in this case I'd do the area of the full square, then cut in half, which is the same as 1/2 bh because it's a right triangle).
4. Then subtract the triangle area from the quarter circle area.

Would that be do-able? I didn't do the math because I don't have the answer key anyways. Or, I guess I do have ds's answer (28.5 cm. if I read his handwriting correctly), so I guess it's because I'm lazy LOL. Let me know if this makes sense and whether it works or not!
Julie

[Cyndi (AZ)]

We are not there, but close -- I have the book and the answer key.

Like Julie stated, the kicker is that the entire quarter circle is not shaded, only part of it.

The area of half of the triangle is 50cm squared, because it is 10x10x1/2=50.

The area of this quarter circle is pi times radius squared, so it is 3.14x10cm squared. If you mentally hide the unshaded piece of the square above the shaded part, you can see an actual quarter circle which has a radius of 10cm (along the bottom edge). So, the area of the circle is 314cm squared. Divide by 4 and the quarter circle is 78.5cm squared.

Then, take that 78.5cm squared quarter circle and subtract the triangle that is not shaded (50cm squared) and the answer is 28.5cm squared.

eta: Julie is never lazy. But I *am* totally avoiding housework at the moment.

[overholt]

Thank you both! I knew I was missing something obvious! The really sad part is that I worked it with my son a year ago and I didn't make proper notes as to how we figured it out before! I just kept assuming that the quarter circle was shaded and not that PART of the quarter was shaded. I would like to think that when I go through it with my next son this time next year I'll remember, but I made proper notes just in case! It's amazing how I can let an unsolved math "riddle" upset my day! Thanks again, Ladies!

[SarahP]

Is Singapore 6a/6b considered Pre-algebra?

No, Singapore Primary series doesn't include all of the traditional pre-algebra topics such as negative numbers, square roots, absolute value, and exponents. Those are in the next levels of Singapore math. MFW recommends Saxon 8/7 as a transition between Singapore Primary and Algebra.

[Julie in MN]

Can someone please check out Day 21, page 40, problem 25 in the workbook of Singapore 6a...... my son is getting 10cm squared for the area of the triangle but the book says 8.5 cm squared. He and I are probably making a silly mistake. But, could someone else please confirm or deny for me?

Thanks!
Sarah P

Hi Sarah,
How did he figure it out? Did he calculate the area of the three un-shaded triangles? That would be far easier because those are on straight lines. If he did that, what did he get for the area of each of those?

He just did the height of the whole rectangle by the width it and divided - the thinking being that the base and height must be perpendicular, therefore the rectangle represented those measurements....

He's right, the base and height need to be perpendicular. However, the base & height of the whole rectangle don't match the base & height of the gray triangle. He might trace and cut out the gray triangle, then turn it different ways to see if he could get the base & height to match what he had guessed (even cutting the gray one into pieces to try to fill half the rectangle). When my son got to this level, it was good for him to visually see that what looks "almost" right isn't going to work in math.

Have him notice, however, that all of the un-shaded areas (white triangles) have easy base & height measurements right along the edges of the rectangle. So, those are easier to figure out than the gray triangle. My son calculated the three white triangles, then subtracted those areas from the area of the whole rectangle (20 - 7.5 - 2 - 2), leaving the area of the gray triangle as the remainder (8.5).

Does that help? I'm sure there are several ways to do this.
Julie

[SarahP]

Thank you, we'll try this one again today with your suggestions and see how it goes!
Math is not a strong point for me, so I am trying to be extra diligent to help my boys have strong math skills!

Sarah

[SarahP]

Your tip worked, he successfully got the correct answer (after me prodding him to recount the side of 1 of the triangles half a dozen times! )

[Julie in MN]

Yay, so fun to hear the end of the story, thanks for posting! Yes, those adolescent brains need prodding sometimes. I always "tried" to remember I didn't answer every question in class when I was young, probably didn't pay attention wvery moment

But yay for moving forward in math skills.
Julie

[TriciaMR]

I have another Singapore math question! This problem is from day 33 of 6a - problem 3 in the workbook on page 65.

In a school choir, the number of boys was increased by 20% to 60 and the number of girls was decreased by 20% to 60. Find the overall increase or decrease in the membership of a choir.

My son got the answer: 0 increase or decrease and I understand his logic, though I get that it isn't correct. The answer key says that the correct answer is a decrease of 5. I can't wrap my own head around how to cipher this problem. I get that we need to figure out what the original number of each: boys and girls was. Which would be x + 20% = 60 (boys) and x-20% = 60 (girls). But we can't just subtract/add 20% to 60 because 20% of 60 isn't the same as 20% of whatever the original numbers were!

Help! (?) He got all the other problems for the day correct, but this one has us both stuck.
Thanks,

Sarah

Sarah,

I *think* for the boys its 120% times x = 60. The new 60 is 120% of the original amount of boys. And for the girls, the new 60 is 80% of the original number of girls...

So... 120/100 x Boys = 60

Multiply both sides by 100/120, solve for boys, and boys = 50.

Then, 80/100 x Girls = 60

Multiply both sides by 100/80, solve for girls, and girls = 75.

Old membership: 75 + 50 = 125

New membership: 120

-Trish

[SarahP]

Ooh, ooh, ooh! We figured out, while I awaited responses I decided we needed to take one more go at it! After I asked him to explain to me how he arrived at his other answers I figured out how we needed to work it.

for the girls 60 = 80% of the original number, so 60/80 = 1% of the original number .75 was 1% and 15 was 20% or a decrease of 15 members. The boys now have 60 members, and increased by 20% so 60 = 120% and 60/120 = .5 (or 1%) .5 x 20 = 10, so there was an increase of 10 boys and a net decrease of 5 choir members overall!

Whoo!

[SarahP]

singapore 6a problem #22 on page 88 of the text, #37 on page 90, and # 40 on page 91

anyone out there able to help us with singapore 6a problem #22 on page 88 of the text, #37 on page 90, and # 40 on page 91...we'll keep trying on this end and hope to hear from someone soon!

My son is in 6a right now and I just asked him to try to work through them to see if he could help you.

Kaleb (11) typed up his responses below. The answers are correct, and he ciphered them on his own, so the way he did it may not be the best, but it worked...

# 22 page 88
Ok, so to start, we figure out the amount of money he gets for every 30 magazines. We do this by multiplying 40 by 30, and then adding $3 to the answer. Now that we know that, we can expand things a bit. Change your 30 to 300, multiply that by 40, then multiply your answer by 3, finally, add $90. He sold 900 magazines total.


# 37 page 90

Boys = 130%
Girls = 100%
First, we divide 54(30%) by 3 so that we can find out what 10% is(10% = 18). Now that we know that, we can multiply 18 by 20, to find 200% Then, we add the remaining 10%(54) back, in order to get 414 kids total. Hope I helped.

[gorillamama]

you guys are just awesome! Thanks to Sarah (and son!) for numbers 22 and 37

Not sure how much work it would be to type out the problems, but there are a few of us who might be up to the challenge but don't have the textbook (for instance, I have the 6a workbook but didn't hang on to the textbook ).

julie - since those ones were already answered, i'll type 40 as best as I can as it is an area problem. there is a rectangle divided into 4 triangles and the goal is to find the area of the shaded triangle. honestly, i don't think i can type this out without confusing everyone reading it...18cm length with a 11cm width that is then divided into 6cm and 5cm. If anyone out there has the book and can offer a better description or a way to solve...we still need help with this one. in the future, i'll type out the questions as i am sure this will not be the last time we'll need help!!!

and thanks again!

[SarahP]

Yeah, that area problem is a tough one, Kaleb just completed day 43 today, so we aren't quite as far along in the book as are you, so we haven't been forced to solve this one yet. I can tell that we need to figure out the area of the whole rectangle and than each of the unshaded triangles and then add those up and subtract them from the area of the rectangle. I think the shaded triangle is an equilateral, but I am having a tougher time thinking about how to determine all the areas of the upper two of the 3 unshaded triangles..... I have been trudging through long division in 3a with my twin 8 year old boys...... it's 4:30 and we have been at today's math work since noon!

Hang in there!

Sarah

[Julie in MN]

I am so in love with having a student chime in here!

there is a rectangle divided into 4 triangles and the goal is to find the area of the shaded triangle. honestly, i don't think i can type this out without confusing everyone reading it...18cm length with a 11cm width that is then divided into 6cm and 5cm.

I can tell that we need to figure out the area of the whole rectangle and than each of the unshaded triangles and then add those up and subtract them from the area of the rectangle. I think the shaded triangle is an equilateral, but I am having a tougher time thinking about how to determine all the areas of the upper two of the 3 unshaded triangles.....

Of course it was hard to type, whew, thanks for trying!

I'd say with Singapore, just find "something" that you can measure. Generally, it's best not to over-think and try to figure out how to measure something that isn't clear, but instead to identify "what you do know" and go from there. So in other words, I would doubt you need to know the area of the rectangle and the area of every single triangle inside of it. Could happen, but it would rarely be so obvious or so time-consuming as that in Singapore. Instead, have your student inspect the information to find exactly what he can compute easily (what has an obvious base & height), and then how he can get from there to what he needs to know.

TRY THIS: A little birdie let me know that you can create one large triangle out of the shaded triangle and another unshaded triangle. After you calculate the area of that large triangle, then you can subtract the unshaded portion since that has a clear base & height, and the remainder will be the area of the elusive shaded triangle. See if your student discovers that, and if not maybe you can nudge him towards it.

HTH,
julie

[gorillamama]

Sarah- thanks for the encouragement and help (Kaleb, thank you as well!!!)

Julie- we just had time this afternoon to look at that area problem and with your insight (and that little birdie's!) we were able to get the answer! (all the while kicking ourselves for not seeing this before!) Thanks, again!

Crystal and family

[Julie in MN]

We are stuck on this one as I know another poster/user was a week or so ago.
It is an area problem, an unshaded rectangle with a shaded triangle within. The goal is to determine the area of the shaded triangle. The Rectangle is divided into 4 triangles (one of these being the shaded tri).
I cannot figure out how we are to work this problem, even with the tips given on the other post that included this question!

...Actually never mind. Apparently my kid is WAY better at math than am I.
I never gave him the suggestion given because it just didn't make any sense to me. However, I thought, why not let me see if it makes sense to him since others seem to get how making one larger triangle from the shaded and another unshaded would work. He got it immediately!

Now I will go hang the shame curtains. (and pat my son on the back)

Sarah

Hi Sarah,
I've so enjoyed reading about your son and math this past week or so. Maybe I'm more sentimental now that my youngest just graduated, but I can't imagine anything better than a student helping another student over the miles, or a student given the guidance by his teacher to surpass her own knowledge Thanks for sharing,
Julie

[TriciaMR]

That's exciting that your son understood it! yeah!

If you're wondering more, turns out the same little birdie from the other thread...would break it down even more, like this:

Erase the top horizontal line of the rectangle, and the left and right vertical lines of the rectangle.
When that is done, the only thing left is the shaded triangle and an unshaded triangle below it.
Put those 2 pieces together and you have big triangle.
The triangle is not a right triangle but the problem gives you the height as 6 + 5 cm.
The problem gives you the base as 18.
The entire big triangle area is 1/2 bh, which is 1/2 11*18 = 99.
The problem tells you the unshaded portion is triangle with height of 5 and base of 18. So, 1/2 bh = 45.
Take the whole thing (99) subtract the unshaded (45) and you are left with 54 cm.

And little birdie sends you a big hug.

(And I'm posting for the little birdie, not for me )

[TriciaMR]

Hello, friends,
Anyone able to help with this problem from Singapore Math 6B page 75, #27?????
The ratio of Dan's money to Jason's money was 3:2. After Dan gave Jason $15, Dan still had $10 more than Jason. How much money did Dan have at first?

Appreciate any and all input!. Thanks

So, if we draw the bar diagrams

| | | | (Dan has 3)
| | | (Jason has 2)

If Dan gave $15 to Jason, and yet still has $10 more dollars than Jason... Let's say Jason had no money (let's skip the 3:2 ratio for a second). If Dan gives Jason $15, and Dan still has $10 more than Jason, Dan must have had $40 to begin with, right? Because $15 + $15 + $10 is $40. So, each space in my bar diagram must be worth $40. So he had $120.

[Julie in MN]

Yes, and just in case visual is good, I thought I'd play with colors today

So, all the original bars are the same amount, and Trish is just looking at the one bar:

J /--/--/
D /--/--/ + /--/

J /--/--/ + /15 given to Jason/
D /--/--/ + /---same (15)-----/+10more/

J /--/--/
D /--/--/ + /---same (15)-----/+10more/ + /15 given to Jason/

J /--/--/
D /--/--/ + /-40-/

J /-40-/-40-/
D /-40-/-40-/-40-/

[gorillamama]

Julie - I Love You! Thank you (again!) - Crystal

[Julie in MN]

Singapore math 6b - Day 51
Textbook page 87 # 10

There were twice as many carnations as roses in a flower shop. After selling 50 carnations and 10 roses, there were 3 times as many roses as carnations left in the shop. How many roses were there in the shop at first?

This is what we have drawn out, but my son (nor I) is sure what to do next to figure out the original amounts!

/C/ /C/ - 50 = /c/
/r/ - 10 = /r/ /r/ /r/

Thanks for any help!

Sarah

Hi Sarah,
I really hope someone using 6B right now will chime in. Maybe Trish??? I have my son's level 6 workbooks here but not the textbooks. Sometimes the textbooks can help show me what skills the students are working on. I do remember equivalent fractions were big in level 6. So, I tried several bar diagrams and equivalent fractions and came up with nothing.

When that happened to us, we just tried good old fashioned plug-n-chug. There may be better methods, but the most I can say is that plug-n-chug requires a bit of logic. So here is what I ended up doing:

I need to subtract 50 carnations, so I know the number of carnations must be more than 50. I'll start with 60.
60 – 50 = 10
And the number of roses has to be half of 60, so 30 roses to start, then subtract 10.
30 – 10 = 20

No, the ration of 10:20 is not what I want. That is 1:2 and I'm looking for 1:3.
I will try starting with 70 carnations instead.
70 – 50 = 20
35 – 10 = 25

That went in the wrong direction. The ratio of 20:25 is farther from the goal (1:3) than 10:20 (in other words, 20 and 25 are closer to one another than 10 and 20 were, and I want them farther apart, like 10 and 30). So I'll try using "less" than 60 carnations. I like even numbers, so I'll go down to 58.
58 – 50 = 8
29 – 10 = 19

Okay, that's getting better, 8:19 is better than 1:2, but it's still not 1:3 (8:24 would be 1:3). I'll try going down 2 more:
56 – 50 = 6
28 – 10 = 18
Aha, 6:18 is 1:3, I won! (Woops, I mean I figured it out )

Again, hoping someone has a better way. Otherwise, you might tell your student that one mom used the guess-and-check method and see what happens. I wouldn't worry if a student started with 100 or even 1,000, as long as he or she can narrow it down fairly quickly. (And hopefully the student won't wear himself out starting with difficult numbers like 342.75 LOL.)
Julie