Q&A - P4 Math

[chloecube]

tianzhu:

chloecube:

so can i conclude that model drawing is better than using multification?

Hi

Whether you go the MD or MM way, the answer should be the same.

Just like porridge, some like it hot, some like it cold.

There are many ways to solve such questions and students should choose the method they are comfortable with.

Best wishes

hi tianzhu,
thks for all the help.
i am just a little confused of when to use MM and MD
for example this question below, it is quite impossible to solve by MD rite?
i use MM to solve and i got the answer as 32, but if i use MD, i cant.

Lester had some toy animals. If he put them into groups of 5, he would have 2 toy animals left over. . If he put them into groups of 7, he would be short of 3 toy animals. What is the smallest number of toy animals he could have?

[MathIzzzFun]

chloecube:

tianzhu:

[quote=\"chloecube\"]
so can i conclude that model drawing is better than using multification?

Hi

Whether you go the MD or MM way, the answer should be the same.

Just like porridge, some like it hot, some like it cold.

There are many ways to solve such questions and students should choose the method they are comfortable with.

Best wishes

hi tianzhu,
thks for all the help.
i am just a little confused of when to use MM and MD
for example this question below, it is quite impossible to solve by MD rite?
i use MM to solve and i got the answer as 32, but if i use MD, i cant.

Lester had some toy animals. If he put them into groups of 5, he would have 2 toy animals left over. . If he put them into groups of 7, he would be short of 3 toy animals. What is the smallest number of toy animals he could have?[/quote]Hi

for this question, there is a shortcut.

groups of 5 --> 2 left = 3 short
groups of 7 --> 3 short

in both cases, there is a shortage of 3. So, if we add 3 to the total number, the new total will be a multiple of both 5 & 7.

Lowest common multiple of 5 & 7 = 35, 35-3 = 32
answer is 32

cheers.

[MathIzzzFun]

AgonyMum:

Hi, need guidance and enlightenment with these 2 questions.Pls help.Thanks in advance.


I can solve them using Guess and Check but is there a better and more efficient method?

Qn1. Every 10 children shared a big plate of fried rice and every 8 children shared a big plate of chicken wings. If 9 plates were used altogether, how many children were there?

Qn2. Ahmad has 20 toy wheels. He wants to make some model toy cars and motorcycles.How many of each type of toys can he make using all the 20 wheels? How many different answers can you find?

Hi
Q1.
lowest common multiple of 10 & 8 = 40
40 children --> 4 plates of fried rice
40 children --> 5 plates of chicken wings
total 40 children & 9 plates of food

Q2
Max number of toy car --> 20/4 = 5
max number of toy motorcycles --> 20/2 = 10

so, you can start from either one and then list. If we start with toy cars and reduce to 1, every reduction of 1 toy car is replaced by 2 toy motorcycles.
I suppose there must be at least one of each,

4 toy cars 2 toy motorcycles
3 toy cars 4 toy motorcycles
2 toy cars 6 toy motorcycles
1 toy cars 8 toy motorcycles
total 4 possible ways.

cheers.

[Thein]

Hi chloecube


I prefer MM method. I don't know whether this is hot or cold

Regarding the problem:

James has some marbles. If he packed them into box of 6, he will have 2 left. If he pack them in box of 7, he will be short of 11. How many marbles does he have?

There is a chance of placing 7 marbles as one group if we pack them 7 as 1 group. So the remaing marbles should be 4 not 11(As nebbermind said)...
Hence the answer can be 38 too....

Regarding this problem :
Lester had some toy animals. If he put them into groups of 5, he would have 2 toy animals left over. . If he put them into groups of 7, he would be short of 3 toy animals. What is the smallest number of toy animals he could have?

Using MM method the answer shud be 32.

If you go for MD.... u get 2u = 5.... quite confusing....

[CoffeeCat]

Thein:

Hi chloecube


I prefer MM method. I don't know whether this is hot or cold

Regarding the problem:

James has some marbles. If he packed them into box of 6, he will have 2 left. If he pack them in box of 7, he will be short of 11. How many marbles does he have?

There is a chance of placing 7 marbles as one group if we pack them 7 as 1 group. So the remaing marbles should be 4 not 11(As nebbermind said)...
Hence the answer can be 38 too....

Regarding this problem :
Lester had some toy animals. If he put them into groups of 5, he would have 2 toy animals left over. . If he put them into groups of 7, he would be short of 3 toy animals. What is the smallest number of toy animals he could have?

Using MM method the answer shud be 32.

If you go for MD.... u get 2u = 5.... quite confusing....

Hi, I believe there is some confusion here.
These 2 problems are distinct, although there is a bit of similarity, which leads to the confusion. As noticed in such questions, there is the idea of groups and remainder(excess/shortage). However for the first question, it is suggested (by the 7 / 11 thingy) that the number of groups must be equal, whereas for the second question, the number of groups need not be equal.
I saw the model you drawed, it is suitable for the first type of question (called excess-shortage problem by terry chew) where the number of groups are the same. The MM method ( which i assumed is just the listing multiples method) is suitable for the second type, where the number of groups need not be equal. You can tweak MM method for excess-shortage, you just need to check the number of groups everytime you found the common number, which is plainly inefficient.

[chloecube]

thks for the short-cut method

meaning for this question, MD is not possible rite?
i am trying to understand this and how to tell my DS when to use MM or MD.

[chloecube]

Thein:

Hi chloecube


I prefer MM method. I don't know whether this is hot or cold

Regarding the problem:

James has some marbles. If he packed them into box of 6, he will have 2 left. If he pack them in box of 7, he will be short of 11. How many marbles does he have?

There is a chance of placing 7 marbles as one group if we pack them 7 as 1 group. So the remaing marbles should be 4 not 11(As nebbermind said)...
Hence the answer can be 38 too....

Regarding this problem :
Lester had some toy animals. If he put them into groups of 5, he would have 2 toy animals left over. . If he put them into groups of 7, he would be short of 3 toy animals. What is the smallest number of toy animals he could have?

Using MM method the answer shud be 32.

If you go for MD.... u get 2u = 5.... quite confusing....

hi,
ya, that is what i am trying to say. sometime MD dun work, so have to try MM
ok, i will tell DS to try one method first, if cannot, then try another.
hope is not too confusing for him

[CoffeeCat]

chloecube:

thks for the short-cut method

meaning for this question, MD is not possible rite?
i am trying to understand this and how to tell my DS when to use MM or MD.

Ya.
When to use a method depends on the question.
So first your son must figure out whether the groups must be equal or need not be equal.
Equal use MD, need not equal use MM... Hope this is simple

Cannot anyhow try methods recklessly in random order, because wrong method can get an answer. Most students will just submit that first answer.

[chloecube]

CoffeeCat:

chloecube:

thks for the short-cut method

meaning for this question, MD is not possible rite?
i am trying to understand this and how to tell my DS when to use MM or MD.

Ya.
When to use a method depends on the question.
So first your son must figure out whether the groups must be equal or need not be equal.
Equal use MD, need not equal use MM... Hope this is simple

Cannot anyhow try methods recklessly in random order, because wrong method can get an answer. Most students will just submit that first answer.

see some light now
:thankyou: everyone

[AgonyMum]

Thanks so much, Thein

And Thanks so much too MathIzzzFun