Q&A - P5 Math

[buds]

Gemini11:

Hope it helps. Gemini11 :evil:

Hi Gemini! Thanks! :please:
Yes, it was very helpful.

While i usually very much look forward to tianzhu's beautiful model drawings and workings, yours were pleasantly simple to comprehend and tongue in cheek some more. I loike too!

DD says thank you very much and had used your workings in her worksheets.

Gemini11:

PS: Sorry tianzhu hope you dont mind i hijacked your questions.

The more the merrier. So we can all choose from a variety of solutions from the many knowledgable parents here in the forum. There's always more than one way to reach the correct answer. Our kiddies are spoilt for choice. They can use the one solution they can understand the best. :celebrate:

[buds]

MathIzzzFun:

http://i42.tinypic.com/35mo390.jpg\">

MathIzzFun, thank you so much for the alternative solutions! :please:

Appreciate it a lot.

[buds]

cimman:

buds:

I am thinking of a fraction.
The sum of its numerator and denominator is 20.
When i add 22 to its denominator, the fraction becomes 1/6.
What is the fraction i am thinking of?

I've recently developed a way to solve problem sums with algebra and simultaneous equations. It offers a simple way of forming the necessary equations. For the above problem, it is an equation with one unknown.
http://i40.tinypic.com/3347uxs.png\">
The fonts in black are values transferred directly from the problem to the table. The fonts in red are calculated values.
The calculation is based on this basic formula: Before + Transfer = After
how did I get (6 units - 22) for Denominator (Before) ?
it is from Before = After - Transfer
= 6 units - (+22)
= 6 units - 22
Since there was no mention of any changes to the Numerator, we put a 0 in the Transfer box for numerator.

The red circle simply circles the relevant equation.
All equations are dervied directly from the table.

If you're open to algebra and simultaneous equations, this method can solve complex problems easily, ie. the equations can be formed easily through the table alone. From the table, the relationships can be clearly seen without a deep comprehension of the problem sum and the relationships of the data values.
The relationships are derived from the table instead of from the problem sum.

Wow! You developed this transfer solution yourself? :salute:
I loike it! :love: Nice to see different ways to come to the same answer.
Looking at the different perspectives from everyone who has shared just shows
that Math isn't that impossible to comprehend.

[YumYum]

tianzhu:

YumYum:

Hi, Can someone pls help with this question:


Farmer Tan has ducks and cows on his farm. The total number of ducks and cows is between 20 and 40. The ratio of the total number of ducks' legs to the total number of cows' legs is 5:9. What is the total number of ducks and cows on the farm?

Thanks

Hi

One needs to differentiate the ratios representing the number of legs and the number of animals. The ratios for the number of legs and the number of animals are of different measures.

The total number of ducks' legs to the total number of cows' legs is 5:9 ------ 20:36 (equivalent ratios)

A duck ----- 2 legs, a cow ----- 4 legs

This means the total number of ducks to the total number of cows ----- 10:9 -----20:18

Ducks ----- 20
Cows ----- 18

There are 38 ducks and cows.

Best wishes

thank you

[metz]

buds:

Helloooo again, tianzhu! :peekaboo:


I have a few more for you here.

I am thinking of a fraction.
The sum of its numerator and denominator is 20.
When i add 22 to its denominator, the fraction becomes 1/6.
What is the fraction i am thinking of?

This question carries 4 marks! :yikes:

Once your dd learns Ratio, it might be easier for her to solve such questions.

ratio of numerator to denominator after adding 22 to denominator = 1:6
New total = 20+22 =42
(Total units = 1 + 6 = 7)
7 units -> 42
1 unit -> 6
nominator = 6
Denominator = 36 - 22 = 14
fraction = 6/14

[cimman]

buds:

Wow! You developed this transfer solution yourself? :salute:
I loike it! :love: Nice to see different ways to come to the same answer.
Looking at the different perspectives from everyone who has shared just shows
that Math isn't that impossible to comprehend.

if you're interested in this technique, I'm conducting a workshop to teach parents on how this technique works.

The same technique works on problems like these:
1.\tJohn and Sherry shared some sweets. John has 180 sweets less than Sherry. If Sherry gives 1/7 of her share to John, John will have 20 more sweets than Sherry How many sweets did Sherry have at first? Ans: 700

2.\tDoris and Annie have each planned a list of words to learn. If Doris learnt 40 words per day while Annie learnt 20 words per day, Doris would have 70 words left to learn while Annie had completed learned all the words. If Doris learnt 20 words per day while Annie learnt 40 words per day, Doris would have 130 words left to learn while Annie had completed learned all the words. How many words do Doris and Annie have to learn respectively? Ans: Doris:150, Annie: 40

This technique works, because all Transfer type problems are bounded by this very simple equation:
Before + Transfer = After
...... and thus, we can make use of that very fact to solve all Transfer Type problems. A Transfer Type problem is defined by a problem in which some value is decreased or increased or both. This covers a very wide range of problems in PSLE Maths.

There's no need to use different techniques for different problem patterns. It makes it a lot easier for a child to understand and drill on just one method.

My child would find it difficult to identify unchanged total or unchanged difference or unchange quantity in a problem. This method was created to remove that layer of linguistic inference of problem sums.

I created a thread in the Happenings forum.
http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=43&t=32278
You can register for the next workshop. Just drop me a PM and I'll alert you on the next workshop date.

[cimman]

since you're comfortable with equations, here is an alternative solution:


Jason and Kevin were given some money each. If Jason and Kevin spent $100 and $50 each day respectively. Jason would have $1300 when Kevin had spent all his money. If Jason and Kevin spent $50 and $100 each day respectively, Jason would have $3700 when Kevin had spent all his money. How much money was given to Jason at the beginning.
Model Answer: $4500
http://i40.tinypic.com/35hif0o.png\">

the key thing to note is:
Before + Transfer = After
thus:
Before = After - Transfer

Also, since this is a Double If question (there are 2 Ifs in the question),
the 1st IF Before = 2nd IF Before.

Note that the concept of negative numbers is very important here.

u and p represents the number of days for the 2 Ifs respectively.

Note that the fonts in Black represents numbers transferred directly from the problem sum. Fonts in Red represents numbers that are derived or calculated from values in the table.

[cimman]

for those who are interested in how this technique can be applied to Double If questions (ie. questions which literally have 2 If statements), you can look at my post in this thread: http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=68&t=33165&p=730810#p730810


the technique is scalable, meaning that it can handle simple equations with one unknown, as in Bud's question and it can scale to handle problems with 2 unknowns, as in the Double If problem. It can also handle more than 2 unknown variables as well. Once it is in that territory, > 2 unknowns, you will have 3 or more equations. However, I find that, fortunately, in Primary school Maths, the multiple equations are quickly reduced down to 2 equations, 2 unknowns, so that it is manageable for our children.

[tianzhu]

ariasnow:

Another question.


Tom was reading a book. He read 1/8 of it on the first day, 1/5 of it on the 2nd day and 1/4 of it on the 3rd day. After that, he still had 68 pages to read. How many pages does the book have?

Hi

Good Morning.

A common multiple of 4, 5 and 8 is 40.

First day ------- 5 units (1/5 of 40)
Second day ------- 8 units
Third day ------- 10 units

40 – 23 ------ 17

17 units ------ 68
1 unit ------ 4

Number of pages ------- 40*4 ------- 160

Best wishes

[tianzhu]

ariasnow:

Need help on this. thanks.


Ali, Bala and Charles shared a sum of $. Ali recieved 1/3 of the amt that Bala and Charles recieved altogether. The amt Bala recieved is 1/2 of the total amt that Ali and Charles recieved. If Charles recieved $110, find the total amt that Ali and Bala recieved altogether.

Hi

Ali : (Bala + Charles) ------- 1:3 (Total 4 unit)
(x3)
Ali : (Bala + Charles) ------- 3:9 (Total 12 unit)

Bala : (Ali + Charles) ------ 1:2 (Total 3 units)
(x4)
Bala : (Ali + Charles) ------ 4:8 (Total 12 units)

Make the total units the same.

Ali ------ 3 units
Bala ------- 4 units
Charles ------- 5 units

5 units ------ 110
1 unit ------- 22

(Ali + Bala) ------- 7 units ------- 7*22 ------- 154

Best wishes