Q&A - P4 Math

[MathIzzzFun]

AgonyMum:

Please help to solve these 2 questions using both model and equation methods.P4 standard

Thanks very much.

1. Fiona had 3 times as many stickers as dolls. After she gave away 7 stickers and 10 dolls, the number of stickers was 4 times the number of dolls. How many stickers and dolls did she have at first?

2.There were 5 times as many candies as sweets in a store. After 16 candies were sold and 16 sweets were brought into the store, the number of candies was 3 times the number of sweets. If each item cost 80 cents, how much money would be collected from the sale of the sweets?

Q2.

http://i47.tinypic.com/zv3jpf.png\">

cheers.

[AgonyMum]

Thank you all so much for helping I am still trying to figure out the solutions.....lost right now.

Possible to solve using equation/ substitution/ algebra?
Thanks!

[BigDevil]

AgonyMum:

Thank you all so much for helping I am still trying to figure out the solutions.....lost right now.

Possible to solve using equation/ substitution/ algebra?
Thanks!

Fiona had 3 times as many stickers as dolls. S = 3D

After she gave away 7 stickers and 10 dolls, the number of stickers was 4 times the number of dolls.
S-7 = 4(D-10)

How many stickers and dolls did she have at first?
Substituting S,

3D - 7 = 4D - 40
D = 33
S = 33 x 3 = 99

[cimman]

AgonyMum:

Thank you all so much for helping I am still trying to figure out the solutions.....lost right now.

Possible to solve using equation/ substitution/ algebra?
Thanks!

hi AgonyMum,
have a look at this technique: http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=67&t=25121&start=1260

you can use the exact same technique for your problems.
First draw the table, then transfer all the values over to the table.

let me know if you need help in this area.

[optimistforum]

Hello friends


I have issues with DS1 (he will be 10 in summer 2013). The following two questions are symptomatic of his lack of problem-solving skills.

How do I communicate the following solutions to him. Are there other methods I can communicate and what would they be?

1) Mince costs £1.80 per 0.5Kg. Find the cost of mince weighing 600g.

Solution: £1.80 = 500g, so 100g = £1.80/5 = 36p
so 600 g = 6 X 36 = £2.16

2) A shopkeeper bought 6 balls for £1.32 and sold them to make a total profit of 48p. For how much did he sell each ball.

Solution: £1.32/6 = 22p intial cost for each ball.

profit of 48p for 6 balls = 48/6 = 8p profit per ball.

Therefore selling price of each ball is 22p + 8p = 30p

Regards
O

[dazzlego]

Hi optimistforum,


For Question 1, the way u solved it should be the best approach.
- Find the price for the nearest 100g ( common multiple)
I don't think it is necessary to mention the bold to the child
0.5 kg = 500g
500g --> £1.80 (Divide both sides by 5 to get 100g) (since 500g is 5x of 100g)
100g --> £1.80 / 5 = £0.36 (36p) (Multiply both sides by 6 to get 600g)
600g --> 36p x 6 = 216p (£2.16)


For Question 2, u may want to try this approach
- Find the total selling price for 6 balls
£1.32 + 48p = £1.80 (132p + 38p = 180p)

- Find the selling price for 1 ball
£1.80 / 6 = £0.30 (180p / 6 = 30p)

Hope it helps :celebrate:

optimistforum:

Hello friends

I have issues with DS1 (he will be 10 in summer 2013). The following two questions are symptomatic of his lack of problem-solving skills.

How do I communicate the following solutions to him. Are there other methods I can communicate and what would they be?

1) Mince costs £1.80 per 0.5Kg. Find the cost of mince weighing 600g.

Solution: £1.80 = 500g, so 100g = £1.80/5 = 36p
so 600 g = 6 X 36 = £2.16

2) A shopkeeper bought 6 balls for £1.32 and sold them to make a total profit of 48p. For how much did he sell each ball.

Solution: £1.32/6 = 22p intial cost for each ball.

profit of 48p for 6 balls = 48/6 = 8p profit per ball.

Therefore selling price of each ball is 22p + 8p = 30p

Regards
O

[cimman]

optimistforum:

Hello friends


I have issues with DS1 (he will be 10 in summer 2013). The following two questions are symptomatic of his lack of problem-solving skills.

How do I communicate the following solutions to him. Are there other methods I can communicate and what would they be?

1) Mince costs £1.80 per 0.5Kg. Find the cost of mince weighing 600g.

Solution: £1.80 = 500g, so 100g = £1.80/5 = 36p
so 600 g = 6 X 36 = £2.16

2) A shopkeeper bought 6 balls for £1.32 and sold them to make a total profit of 48p. For how much did he sell each ball.

Solution: £1.32/6 = 22p intial cost for each ball.

profit of 48p for 6 balls = 48/6 = 8p profit per ball.

Therefore selling price of each ball is 22p + 8p = 30p

Regards
O

you can give this approach a try. I developed this technique when I found that the conventional technique of explaining the solution did not work. While my child was able to understand the solution, he was unable to apply it to other problem sums of similar type. The conventional technique is what you have listed above. The conventional technique gives the child the fish, but does not teach the child how to fish.

This is a visual analysis technique that is similar to modelling in that it is visual but the approach is entirely different. It breaks down the analysis to 2 steps. The first step requires interpreting the individual values, ie. what does it mean ? how do I map the values in the problem sum to the table? The 2nd step involves logical reasoning purely from the relationships defined in the table. The student does not need to rely on the words from the Problem sum. It effectively isolates the mathematical relationships from the language comprehension aspect of the problem sum.

http://i47.tinypic.com/2u634nm.png\">
http://i48.tinypic.com/104oi1x.png\">
http://i47.tinypic.com/2ilodhs.png\">
http://i50.tinypic.com/10efv4j.png\">
After Step 4, the child can effectively stop reading the problem sum and proceed to anlayse the table to derive the required values to fill in the blank boxes.
http://i49.tinypic.com/veccc9.png\">http://i48.tinypic.com/2ibfabr.png\">

You will find that no words are used in the explanation. English language is not an effective medium to convey abstract mathematical relationships. The student draws the final table (sans the step by step explanation) [Step 6] and together with the arrows, provides a clear explanation of the solution. Black fonts denotes values transferred directly from the problem sum, while red fonts denotes calculated values.

http://i45.tinypic.com/1pj953.png\">
This is a more complex problem than the one above because it introduces a new concept: Profit. The child has to understand the meaning of Profit and how it relates to the original cost.
http://i46.tinypic.com/15f3q11.jpg\">
http://i46.tinypic.com/t00ye9.png\">
At Step 4, the child has to make inferences about the problem. What values can he copy from the 1st Case to the 2nd Case? typically, only one value is copied. The child has to understand from the context of the problem what that value is. This is actually quite simple. The table limits the choices to 3 values ie. Total, Quantity, Unit Value. The child just have to answer these questions: does the 1st and 2nd case have the same Unit Value ? does the 1st and 2nd Case have the same Quantity Value ? does the 1st and 2nd Case have the same Total Value ? or asked in another way : did the Unit Value change from 1st case to 2nd case ? did the Quantity Value change from 1st case to 2nd case ? did the Total Value change from 1st case to 2nd case ? if there is no change, then just copy the value over.

The table effectively becomes a scaffolding for the child to organize his thoughts. By transferring the values from the problem sum to the table, the child is forced to question what each value means and how it should map into the table. Once the values are in the table, the relationships are clearly shown. It becomes more of a puzzle approach, similar to a crossword puzzle where the child focuses on how to fill in the various blank boxes without the stumbling block of problem sum language comprehension, which is a major weakness in most children. The table remains the same for all such types of problem sums, as such, the analysis approach remains consistent from problem sum to problem sum.
The arrows are important, they show how values flow from one box to the next. The brain processes flows more easily than mathematical operations.

Parents who have attended my workshops (http://tinyurl.com/cnu9586) and who are conversant in basic algebra found this approach more intuitive. If your child likes puzzles, he might enjoy this approach.

[cimman]

cimman:

The child just have to answer these questions: does the 1st and 2nd case have the same Unit Value ? does the 1st and 2nd Case have the same Quantity Value ? does the 1st and 2nd Case have the same Total Value ? or asked in another way : did the Unit Value change from 1st case to 2nd case ? did the Quantity Value change from 1st case to 2nd case ? did the Total Value change from 1st case to 2nd case ? if there is no change, then just copy the value over.

here is another approach in Step 4, where you need to make some inferences about the problem sum, I call it the elimination approach, typically used in answering MCQ questions by eliminating the options.

In your first question, after drawing the table and transferring the values over, it is obvious that there is a change in the Quantity value. So that leaves out Quantity value that you can copy the value over. Next, you'll notice that the Total Value has a question mark in the 2nd case. It is unlikely that there is no change for this value since the it is something that you need to find. That just leaves the last option, Unit Value. This is the only option left, so just copy the value from 1st case to 2nd case.
The diagram below illustrates this process:
http://i45.tinypic.com/jub7uq.png\">

[snowball]

BigDevil:

AgonyMum:

Thank you all so much for helping I am still trying to figure out the solutions.....lost right now.

Possible to solve using equation/ substitution/ algebra?
Thanks!

Fiona had 3 times as many stickers as dolls. S = 3D

After she gave away 7 stickers and 10 dolls, the number of stickers was 4 times the number of dolls.
S-7 = 4(D-10)

How many stickers and dolls did she have at first?
Substituting S,

3D - 7 = 4D - 40
D = 33
S = 33 x 3 = 99

The \"substitution/ algebra\" method shown above is indeed very useful especially for parents. :xedfingers:
However, for Pr4, since they are not taught using this method, is it advisable for them to use it to solve their level questions?
pls advice. TIA :scratchhead: :idea:

[snowball]

MathIzzzFun:

AgonyMum:

2.There were 5 times as many candies as sweets in a store. After 16 candies were sold and 16 sweets were brought into the store, the number of candies was 3 times the number of sweets. If each item cost 80 cents, how much money would be collected from the sale of the sweets?

Q2.

http://i47.tinypic.com/zv3jpf.png\">

cheers.

Sorry, in the above diagram, why is it \"4units\" and not \"2units\" as the questions mentioned is (after: 3 times more candies than sweets ) ? :scratchhead:

Mmmm..... i think i understand but not so clear though could you help to explain in a \"before & after \" model? i think it will help slow learner like me to visualise :idea: