Daddy

Pls advise on workings/steps for 3 questions below.

Not sure how to explain.
Thanks in advance.

1. Mr Lim has the same number of cows and chickens. The total number of animal legs is 84. How many cows and chickens does Mr Lim have in all?

Ans: 28 cows and chickens

http://i58.tinypic.com/2e4zw2q.jpg\">

irbasg:

I need help to solve this problem


Angie, Brandon & Minah had $90 altogether. Angie gave Brandon $12, brandon gave Minah $13 and Minah gave Angie $5. The 3 children had the same amount of money in the end. How much money did Angie have at first?

This is an Internal Transfer problem. The initial sum will still be the same as the final sum.
In the end, all 3 children have $90. So each child has $30.

Angie has $30-$5+$12=$37 at first.

vaun:

This is the Question from NSW Math test for P3. Kindly help is showing the steps in getting the answer..


Amber planned a party. Each child was asked to bring a cupcake, juice and a present to the party. They were to bring one of each type of item.

Each child brought at least one item.
+ 16 children brought all 3 items.
+ 5 children brought the cupcake and the present only
+ 1 child brought the juice only
+ 1 child brought the cupcake only
+ 3 children brought the cupcake and the juice only

There were 25 cupcakes, 21 juicies and 26 presents to share altogether.

How many children were at the party ?
---------
Thank you

Rearranging to show all combinations:
+ 16 children brought all 3 items.

+ 5 children brought the cupcake and the present only
+ 3 children brought the cupcake and the juice only
+ ? children brought the juice and the present only

+ 1 child brought the juice only
+ 1 child brought the cupcake only
+ ? child brought the present only

http://i59.tinypic.com/13zblt3.jpg\">
1 Juice and 5 Presents are needed to meet the target total.
It means the only combination will be 1 child brought Juice and Present, and 4 children brought Present only.

There are 31 children in total.

muska:

The chairs in a hall are to be arranged in rows. If 6 chairs are put in a row, there will be 4 chairs left over.


If 7 chairs are put in a row, there will be 3 chairs short. How many chairs are there in total

Ans 46

Pls help to show the working for ques like this. Thanks lots

http://i58.tinypic.com/25hmzkj.jpg\">

Actually, for a young child, it's better to list them to give clarity to the rationale. Depending on formulas may lead to blind acceptance and less incentive to think.

Do note that \"different\" is not the same as \"no repetitions\". So it will be a matter of interpretation and defining the question properly.

If you ask \"How many different 2-digits numbers can you form using the digits 1,2?\"
Using the formula: We will get 2x1=2, which is wrong.
By listing them systematically, we will have 11,12,21,22 -> 4 different 2-digits numbers. To explain:
In the tens column, we have 2 choices of 1 & 2.
In the ones column, we have 2 choices of 1 & 2.
So there are 2x2=4 \"2-digits\" numbers.

If you ask \"How many 2-digits numbers can you form using the digits 1,2 without repetitions?\"
The answer will be 12 and 21 -> 2 \"2-digits\" numbers. You can't have 11 or 22 since they are repetitions. Then the formula will be 2x1, because in the tens columns we have 2 choices; and in the ones column, we are only left with the other remaining choice.

To the question: How many different 4-digit numbers can you form using the digits 1,3,4,5?
My answer is 4x4x4x4=256, because numbers like 1111, 3333, 4444, 5555, 1345, 1145 are included. You can repeat the digits and they will still end up different from each other!

\"different\" does not mean \"no repetitions\"

jolenekoh:

Appreciate for the model drawings. Thank you!


Alvin and Bobby had an equal number of marbles at first. Alvin then gave away 20 marbles and Bobby bought another 10 marbles. In the end, Bobby had thrice as many marbles as Alvin. How many marbles did Alvin have at first?

http://i61.tinypic.com/d61ed.jpg\">

Sruthi:

http://i61.tinypic.com/110lo93.jpg\"> . Please help with this one too . Thanks for all the help on advance [IMG]

http://i58.tinypic.com/2lsteu8.jpg\">

pinball:

Hi, is it possible to solve the following qn using model method?


There are 50 more carrot cakes than yam cakes on sale in a cafe. After selling three times as many carrot cakes as yam cakes, there were 14 more yam cakes than carrot cakes left. How many carrot cakes did the cafe sell?

http://i62.tinypic.com/taock7.jpg\">

scotia:

http://i59.tinypic.com/2rf5ydl.jpg\">


Liked this can?

Can't really see your model.
Anyway, here it is.
http://i62.tinypic.com/fvlqtk.jpg\">

ChiefKiasu:

JLxu70:

[quote=\"jolenekoh\"]Please help... How to draw model for this question or table method?


There is a total of 12 cows and ducks on a farm. They have 38 legs altogether. How many cows are there on the farm?

Appreciated

Another way that I read from my son's Math Olympia book, also called chicken and rabbit problem.

Step 1 - Assume they are same animals with the smallest number of legs. In this case, assume they are all ducks.
Step 2 - with all being ducks, total number of legs are: 12X2=24
Step 3 - there are 38-24=14 extra legs.
Step 4 - The difference of legs between cow and duck is: 4-2=2
Step 5 - use the result from step 3 and 4: 14/2 = 7, so there are 7 cows.
Step 6 - 12-7=5, there are 5 ducks.

To verify the result, 7X4+5X2=38 legs.

The above method may not make a lot of sense, but it is derived directly from the solving of simultaneous equations.
Eg. Let C=cows, D=ducks. So we have:
(1) : C + D = 12
(2) : 4C + 2D = 38
From (1) : 2C + 2D = 24
So 4C - 2C = 38-24
=> C = (38-24) / (4-2)

So there is a direct formula which you can use to get the answer immediately instead of guess and check.
Assuming that
C = Entity with the larger number of items
N = Number of \"extra\" items left over if the smaller entity makes up the entire population
D = Difference in items between the 2 entities
So C = N/D

Unfortunately, using this formula makes no sense since the intent of the exercise is to get the kid to do guess and check. Personally I feel such questions being asked at P2 or even P3 levels is ludicrous. Unless the kid is naturally gifted, encountering such a question for most kids will probably result in blank stares. I've asked many adults the same question and they have problems trying to solve the question too without using algebra. Unless the school has an explicit technique of training all children to be able to think creatively and solve this problem efficiently, such questions will only serve to differentiate kids who have tuition from those who do not.[/quote]
http://i57.tinypic.com/2d9d4jl.jpg\">

It is possible to solve by exhaustingly listing the entire range of possibilities, and identifying the correct combination (as in the image above). But this will take too long especially if the number of items is big (e.g. bigger than 20).

It may be pattern observation & efficiency are the key reasons for the non-intuitive method, so that an exhaustive table listing isn't required.

Let's start with the larger item.
- As you decrease the number of cows (decreases by 4 legs), and increase the number of ducks (increases by 2 legs), the total number of legs decreases by 2.
- The difference between the starting point of 48 legs and the target of 38 legs is 10 (i.e. the extra legs). So this means, row-wise, the number of cows must have decreased by 5 and the number of ducks increased by 5.
- You can get to this conclusion with just listing 2-3 rows.
- It's faster to reach the correct combination of 7 cows and 5 ducks. It's 'Guess & Check' method.
- There are other ways of 'Guessing' the starting point.

It won't be easy to understand this at P2. It's easier to just list the small number of combinations ('List & Check').