Grade 6 (or 7?) ratio problem (1 Viewer)

[oasfree]

Guys, I found this grade 6 or perhaps early grade 7 ratio problem. I'd like you guys to have a go and tell me how you feel. I am worried that mathematics teaching in NSW no longer has the punch to get through difficult problems. There seems to be too much rote-learning. I found that this little problem caused so much problems for many people I know from "smart" kids to adults. Can you guys solve this problem without resorting to using heavy tools like algebra simultaneous equations? Solving it is one thing. Solving it in an elegant and simple to understand way is another thing.

---
Fred and George had some savings in the ratio 3 : 4 respectively. They decided to buy a birthday cake for their little sister sharing the cost at ratio 2 : 3 respectively. After they have bought the present, Fred already spent half of his money. And George only has $21 left over. Find the price of the present.
---

 

[Vindictus]

Guys, I found this grade 6 or perhaps early grade 7 ratio problem. I'd like you guys to have a go and tell me how you feel. I am worried that mathematics teaching in NSW no longer has the punch to get through difficult problems. There seems to be too much rote-learning. I found that this little problem caused so much problems for many people I know from "smart" kids to adults. Can you guys solve this problem without resorting to using heavy tools like algebra simultaneous equations? Solving it is one thing. Solving it in an elegant and simple to understand way is another thing.

---
Fred and George had some savings in the ratio 3 : 4 respectively. They decided to buy a birthday cake for their little sister sharing the cost at ratio 2 : 3 respectively. After they have bought the present, Fred already spent half of his money. And George only has $21 left over. Find the price of the present.
---

Scratch that. I am an idiot.

60% of price is $21, Fred had $14 left and spent half, so $7.

Price of the present is $35? That's my guess.

 

[untouchablecuz]

price is $45

but i resorted to simultaneous equations tbh i don't think this is that heavy
-------------------------------------------------------------------------------------------------------------
let georges savings = G

let freds savings = F

George paid for 60% of the present

i.e. he paid 3P/5

Fred paid for 40% of the present

i.e. he paid 2P/5

we are told that half of freds savings is equal to what he paid

i.e. F/2 = 2P/5 => F=4P/5 [1]

we are told georges savings minus what he paid is 21

i.e. G - 3P/5 = 21 [2]

we are told the ratio of freds savings to georges savings is 3 to 4

i.e. F/G = 3/4 => F=3G/4 [3]

sub [3] in [1]

3G/4=4P/5 => G=16P/15 [4]

sub [4] in [2]

16P/15 - 3P/5 = 21 => P=45
-------------------------------------------------------------------------------------------------------------
this is quite easy to understand

 

[ninetypercent]

I resorted to simultaneous equations too, but it's only a one way substitution

total of savings = x
total of cost = y

Fred
3/7x - 2/5y = 1/2(3/7x)
3/7x - 2/5y = 3/14x
3/14x = 2/5y
x = 1.86677777 y [1]

George
4/7x - 3/5y = 21 [2]

substitute [1] in [2] for x
4/7(1.86677777y) -3/5y = 21
7/15y = 21
y = $45

 

[hermand]

did it in my head...

if fred has half his money left over, you want to change the ratios so that fred's side of the present ratio is half of his side of the savings ratio.
this changes the ratios into savings - 12:16 and present - 6:9.
if george has $21 left, and he's spent 9 parts of his 16, therefore has seven parts left over, and therefore each part is worth $3 [21/7].
therefore present parts - 6+9=15, therefore present was worth $45 [3x15]

 

[untouchablecuz]

did it in my head...

if fred has half his money left over, you want to change the ratios so that fred's side of the present ratio is half of his side of the savings ratio.
this changes the ratios into savings - 12:16 and present - 6:9.
if george has $21 left, and he's spent 9 parts of his 16, therefore has seven parts left over, and therefore each part is worth $3 [21/7].
therefore present parts - 6+9=15, therefore present was worth $45 [3x15]

haha, i got owned

i blame conics for my over complication

 

[ninetypercent]

that's awesome hermand

 

[hermand]

haha, i got owned

i blame conics for my over complication

haha, i blame four unit maths for all of my problems.

actually 3 unit is worse. stupid memorising formulas.

 

[oasfree]

did it in my head...

if fred has half his money left over, you want to change the ratios so that fred's side of the present ratio is half of his side of the savings ratio.
this changes the ratios into savings - 12:16 and present - 6:9.
if george has $21 left, and he's spent 9 parts of his 16, therefore has seven parts left over, and therefore each part is worth $3 [21/7].
therefore present parts - 6+9=15, therefore present was worth $45 [3x15]

Thanks for trying, guys. Hermand got it right in the shortest possible way. I came up with 3 solutions myself. Two of them have been offered here. The simultaneous algebraic equations offers the worst possible way to solve it. It's the way that many older students and adults tend to do (as programmed by the rote-teaching that teachers have been doing to students?). This problem was run acrossed smart kids, selective school kids, an adult and a University graduate. It stumped them all. The University graduate offered a correct solution like seen here with simultaneous algebraic equations.

The way that Hermand did was a very meaningful way as it highlights the true difficulty of fractions. Fractions split things into (often) different size sub-units making it impossible to work on them until you bring them to same-sized subunits using the rule of equivalent fractions. This is where rote-learning is so harmful in teaching fraction operations. Fraction multiplication and division are often taught in a rote manner. This knowledge of fraction translates directly to equivalent ratios that helps to solve this problem so elegantly.

There is a way to solve it that a smart grade 4-5 kids could use to solve this problem. Drawing a diagram will make the job easier and more suitable for kids at grade 5.

I hope some one here will offer this even simpler solution and post an illustration. If not, I will post my 3rd solution next week for every one to see how a young kid can solve this problem by drawing and reasoning along with the illustration.

I have a feeling that modern mathematics education put more emphasis on "flying over" mathematical knowledge accumulated for the past 2500 years rather than going deep and make sure students acquire strong ability to solve difficult problems. I suppose it makes sense to pose this question "Should a teacher be allowed to teach primary school mathematics if he/she cannot solve a problem for grade 6 like this in a proper way?". I think they proposed to test all teachers in Queensland on subjects that they teach but the teacher union was completely against the idea. They call this an insult to the teaching profession. In NSW there hasn't been talk about making sure that teachers know what they teach.

 

[oasfree]

price is $45

but i resorted to simultaneous equations tbh i don't think this is that heavy
-------------------------------------------------------------------------------------------------------------
let georges savings = G

let freds savings = F

George paid for 60% of the present

i.e. he paid 3P/5

Fred paid for 40% of the present

i.e. he paid 2P/5

we are told that half of freds savings is equal to what he paid

i.e. F/2 = 2P/5 => F=4P/5 [1]

we are told georges savings minus what he paid is 21

i.e. G - 3P/5 = 21 [2]

we are told the ratio of freds savings to georges savings is 3 to 4

i.e. F/G = 3/4 => F=3G/4 [3]

sub [3] in [1]

3G/4=4P/5 => G=16P/15 [4]

sub [4] in [2]

16P/15 - 3P/5 = 21 => P=45
-------------------------------------------------------------------------------------------------------------
this is quite easy to understand

My daughter at grade 5 had a look at this algebraic solution. As she had never seen this kind of algebra before, she said she would not attempt this way until she probably got to grade 10. But she came up with a solution that Hermand offered within 2 minutes after I told her that the answer was 45. She must have ran her technique through and verified that 45 was the answer before she confidently showed me how to do it.

She compared two ratios 3:4 and 2:3 and said that just by looking at 3: and 2: one must see that 12: and 6: were the respective multiples that fitted the idea that Fred had spent 1/2 of his money. From this she went on to work out 2 equivalent ratios 12:16 and 6:9 to get to the fact that 7 units of George's saving was left and that was $21. Therefore each unit was $3, and (6+9) units was $45.

 

[untouchablecuz]

Thanks for trying, guys. Hermand got it right in the shortest possible way. I came up with 3 solutions myself. Two of them have been offered here. The simultaneous algebraic equations offers the worst possible way to solve it. It's the way that many older students and adults tend to do (as programmed by the rote-teaching that teachers have been doing to students?). This problem was run acrossed smart kids, selective school kids, an adult and a University graduate. It stumped them all. The University graduate offered a correct solution like seen here with simultaneous algebraic equations.

The way that Hermand did was a very meaningful way as it highlights the true difficulty of fractions. Fractions split things into (often) different size sub-units making it impossible to work on them until you bring them to same-sized subunits using the rule of equivalent fractions. This is where rote-learning is so harmful in teaching fraction operations. Fraction multiplication and division are often taught in a rote manner. This knowledge of fraction translates directly to equivalent ratios that helps to solve this problem so elegantly.

There is a way to solve it that a smart grade 4-5 kids could use to solve this problem. Drawing a diagram will make the job easier and more suitable for kids at grade 5.

I hope some one here will offer this even simpler solution and post an illustration. If not, I will post my 3rd solution next week for every one to see how a young kid can solve this problem by drawing and reasoning along with the illustration.

I have a feeling that modern mathematics education put more emphasis on "flying over" mathematical knowledge accumulated for the past 2500 years rather than going deep and make sure students acquire strong ability to solve difficult problems. I suppose it makes sense to pose this question "Should a teacher be allowed to teach primary school mathematics if he/she cannot solve a problem for grade 6 like this in a proper way?". I think they proposed to test all teachers in Queensland on subjects that they teach but the teacher union was completely against the idea. They call this an insult to the teaching profession. In NSW there hasn't been talk about making sure that teachers know what they teach.

tbh, i don't think this is an issue with rote learning

rather, we (older students and uni kids) have developed the perspective algebra is a powerful tool that provides a systemic method for attacking a problem. hence, we seek to use it where possible. i personally prefer the algebra, regardless of its difficulty, because its much more ordered

 

[bell531]

I don't think needing simultaneous eqn's is too "messy". Its obviously very clever of you, dani, being able to do it in your head, but using algebra is simple enough, right?

 

[oasfree]

I don't think needing simultaneous eqn's is too "messy". Its obviously very clever of you, dani, being able to do it in your head, but using algebra is simple enough, right?

I agree with untouchablecuz about the point that older students and adults tend to fall for algebra. Perhaps this is because that's the last thing that we learn in secondary education. For a problem like this, the trouble is that the job of writing out the two relationships in algebraic way is worth more marks than what the question is designed for. That makes albegra looking like brute-force approach to a problem that is purely designed for ratio and fraction.

When I tried to find ways to solve this one. I had to keep in mind that the problem was for grade 6. The kids don't know much about algebra and they are supposed to do this kind of problem in at most 3 minutes (the question comes from a 2nd semester test in a Singapore school). I decided to use algebra as the last attempt. I found 3 solutions but there may be more. Only one of the 3 looks suitable for young kids to do.

 

[oasfree]

Here is my simplest solution. It does not involve algebra at all. It's a pure ratio and fraction method that kids at grade 5 should feel comfortable.

I drew an illustration of the ratios. From the attached image one can see
----
Fred used up 1 and a half of his money. This is 1.5 units. Then according to the ratio 2:3 (how much each contributed to the cake), we can see that 3:2 would be George's contribution against Fred's contribution. To work out George's contribution in terms of units, we do 3/2 x 1.5 = 2.25 (or 2 and 1/4). So I shaded George's contribution as well.

Now the picture shows clearly that the 7 small parts left over (from George's money) is also $21. So each small part is $3. We can see that all the shaded area together is 15 of the "small parts". Therefore the cake costs 15 x $3 = $45.
----

Refering back to the NSW mathematics syllabus for K-6, I could see that they mention about the ultimate outcome for each of the area is that students must be able to discover a way to solve a kind of problem that has not been taught by the teacher. If they can do this, they have completely achieve it. I am not sure if this is too ideal or realistic. I notice that some kids could discover their own ways to solve problems (that they had not been taught and does not have to be good ways). But this is only for the few top 1%-5% at most.

 

[rowdyroddy]

dude nothing near greade 6 or 7 mofo
ratios were done in those grades perhaps but not of this nature

 

[oasfree]

dude nothing near greade 6 or 7 mofo
ratios were done in those grades perhaps but not of this nature

What do you mean? Do you mean this is too hard for students in Australia at grade 6-7? Kids about grade 6 are expected to be able to deal effectively with this kind of problem in most Asian countries. As I said, I got this problem from grade 6 Singapore test.

 

[lolwth]

He means this is atleast hsc general maths level

 

[rowdyroddy]

i would say grade 10, in australia
but i have heard about asian mathematics, as a korean guy joined my school at about grade 9 in australia, and he had said he learnt all the stuff from grade 9 in grade 5-6 equivalent in korea
he was getting at least 95%, only errors were with english and understanding the question

 

[oasfree]

i would say grade 10, in australia
but i have heard about asian mathematics, as a korean guy joined my school at about grade 9 in australia, and he had said he learnt all the stuff from grade 9 in grade 5-6 equivalent in korea
he was getting at least 95%, only errors were with english and understanding the question

I find that top OC kids and selective school kids are learning mathematics about the level of normal kids in some places in Asia. While kids and their parents seem to have a false impression of how good they are, they just have no idea that in Japan, Taiwan, Korea and Singapore normal kids have to do in an average class.

In the last 20 years, there is a mentality among the new generation of primary school teachers who come from the lowest grade of HSC performers that worry me. They believe there is no need for them to know the subjects. They assume that all knowledge at primary school level is automatic (or built-in) to a mind that is older than 13 years old. So they learn the trade of teaching and go into the class assuming that God has given them all the knowledge they need because they are old enough.

It now looks funny because many of them cannot solve a difficult math problem for grade 5-6 students.

Anyway, this is changing. University of Sydney now only offers double-degree in teaching and something else to students with UAI above 85. It no longer stays with the low grade colleges which still offer teaching courses to students with UAI about 70. Inthe 80s, students with UAI under 50 could enroll in teaching degrees! God helps the current generation of primary school kids!

 

[ninetypercent]

I find that top OC kids and selective school kids are learning mathematics about the level of normal kids in some places in Asia. While kids and their parents seem to have a false impression of how good they are, they just have no idea that in Japan, Taiwan, Korea and Singapore normal kids have to do in an average class.

In the last 20 years, there is a mentality among the new generation of primary school teachers who come from the lowest grade of HSC performers that worry me. They believe there is no need for them to know the subjects. They assume that all knowledge at primary school level is automatic (or built-in) to a mind that is older than 13 years old. So they learn the trade of teaching and go into the class assuming that God has given them all the knowledge they need because they are old enough.

It now looks funny because many of them cannot solve a difficult math problem for grade 5-6 students.

Anyway, this is changing. University of Sydney now only offers double-degree in teaching and something else to students with UAI above 85. It no longer stays with the low grade colleges which still offer teaching courses to students with UAI about 70. Inthe 80s, students with UAI under 50 could enroll in teaching degrees! God helps the current generation of primary school kids!

yes I agree. I had a primary school teacher who couldn't do this question in the Year 3 Basic Skills test. My friends and I showed her how to do it (we were in Year 3!). Back then, I thought it was of no concern, but now that I think back to it, that teacher was seriously...retarded.