Math Question

[Rayleigh]

 

Bro, hope this helps.

 

Thank you Jp66. Appreciated.

[Angcheek]

 

Thank you Jp66. Appreciated.

 

 

waaa i catch no ball hahahaha pei seh

 

my way is this :

 

17 : 5 ... this tells you the answer must be a multi of this ratio

 

given that 20 x 10cent was changed to 4 50cent ...

 

so trial and error :

 

17: 5 ( minus 4 50cent and add back 20 10cent )

 

become 13 : 25 (dont match equally)

 

 

so 17:5 X 2 on both side = 34 : 10

 

now again try ( minus 4 50cent and add back 20 10cent )

 

30 : 30 ........... jackpot !

[Rayleigh]

 

 

waaa i catch no ball hahahaha pei seh

 

my way is this :

 

17 : 5 ... this tells you the answer must be a multi of this ratio

 

given that 20 x 10cent was changed to 4 50cent ...

 

so trial and error :

 

17: 5 ( minus 4 50cent and add back 20 10cent )

 

become 13 : 25 (dont match equally)

 

 

so 17:5 X 2 on both side = 34 : 10

 

now again try ( minus 4 50cent and add back 20 10cent )

 

30 : 30 ........... jackpot !

I was on this solution too and was trying hard to move away from the trial and error method. Thanks anyway

[Angcheek]

I was on this solution too and was trying hard to move away from the trial and error method. Thanks anyway

 

 

another way is algebra ... but that will be even harder to explain ....

[Ender]

Hi MCF Bro, need your help with this math working. While i am able to get the answer, i cant explain it to a P5 boy.

 

A box contain the same number of ten-cent coins and fifty-cent coins. A boy took 20 ten-cent coins and changed into fifty-cent coins of the same value and put them back into the box. Hence the ratio of the number of fifty-cent : ten-cent is 17:5. How much money was in the box.

 

Answer: While I know there are 30 ten-cent coins and 30 fifty-cent coins, i have difficulty in presenting the solution to a P5 level,

My son is doing PSLE this year, I'm kinda familiar with the sylabus, and the method I'm showing is somewhat familiar to most of us from the old school, i.e simultaneous equation, but they call it the bag and box, or the u and p. This method can solve most of the ratio related problems.

 

 

10C : 50c

1 : 1 <----Designate this row as the "u"

-20 +4

5 17 <----Designate this row as the "p"

 

From the above ratio table, we can derived two equations.

 

5p + 20 =1u -----(1)

17p - 4 = 1u ------(2)

 

You can now use simultaneous equatuon to solve for p and u.

 

Since they are equal,

5p + 20 = 17p - 4

24 = 12p

p=2

 

from eqn(1),

5(2) +20 =u

u=30

 

Initially there were 30X 10c coins and 50c coins at first.

 

 

 

 

Edited April 27, 2015 by Ender

[1fast1]

another way is algebra ... but that will be even harder to explain ....

I'm sorry, but I *strongly* disagree with this. Algebra is the easiest method to grasp, and once basic algebraic manipulation is learnt (bringing over terms, opening brackets, negative numbers, etc.) it becomes the most generalisable one-size-fits-all solution. Furthermore, it is the *only* way that will stand the test of time and be carried over into higher level classes and even adult life. Not this piecemeal "model-drawing" crap that will be discarded as soon as they become teenagers.

 

Model drawing is like a complex set of cutlery you have to choose from before you start attacking your food. No one tool can help you tackle all foods, and often you may have to try different pieces before you find the best one. In contrast, algebra is like a bloody swiss-army knife. You see a problem, you pull it out and just solve it.

 

The emphasis on this backwards method here boggles the mind. It's sad, really. I believe they are handicapping our future generations by not equipping them with basic algebraic concepts early on (say at Pri 3) and "forcing" them into a mindset that only this unwieldy, un-generalisable crap can solve complex word problems.

Edited April 28, 2015 by Turboflat4

[1fast1]

My son is doing PSLE this year, I'm kinda familiar with the sylabus, and the method I'm showing is somewhat familiar to most of us from the old school, i.e simultaneous equation, but they call it the bag and box, or the u and p. This method can solve most of the ratio related problems.

 

 

10C : 50c

1 : 1 <----Designate this row as the "u"

-20 +4

5 17 <----Designate this row as the "p"

 

From the above ratio table, we can derived two equations.

 

5p + 20 =1u -----(1)

17p - 4 = 1u ------(2)

 

You can now use simultaneous equatuon to solve for p and u.

 

Since they are equal,

5p + 20 = 17p - 4

24 = 12p

p=2

 

from eqn(1),

5(2) +20 =u

u=30

 

Initially there were 30X 10c coins and 50c coins at first.

No need simultaneous equations. You can reduce the transfer step to a non-algebraic one by simple division, i.e. 20 ten-cent coins is equivalent to 4 fifty-cent coins. Then if the original number of coins of each denomination was x, (x+4)/(x-20) = 17/5, cross-multiply and solve and Bob's your uncle. Or Ah Bob.

Edited April 28, 2015 by Turboflat4

[Ender]

No need simultaneous equations. You can reduce the transfer step to a non-algebraic one by simple division, i.e. 20 ten-cent coins is equivalent to 4 fifty-cent coins. Then if the original number of coins of each denomination was x, (x+4)/(x-20) = 17/5, cross-multiply and solve and Bob's your uncle. Or Ah Bob.

You're right, and as in this case the ratio of 10c and 50c started as 1:1 which your method applies.

 

But the method I'm showing useful if the question started with a different ratio, like suppose the ratio of 10c : 50c started out to be 5:8 . Just hope it useful for the poster to tackle a variety of ratio questions.

Edited April 28, 2015 by Ender

[1fast1]

You're right, and as in this case the ratio of 10c and 50c started as 1:1 which your method applies.

 

But the method I'm showing useful if the question started with a different ratio, like suppose the ratio of 10c : 50c started out to be 5:8 . Just hope it useful for the poster to tackle a variety of ratio questions.

Well, that's easily tackled by letting one be x and the other be 8x/5, but good to show a variety of methods.

[Muddy]

Hi MCF Bro, need your help with this math working. While i am able to get the answer, i cant explain it to a P5 boy.

 

A box contain the same number of ten-cent coins and fifty-cent coins. A boy took 20 ten-cent coins and changed into fifty-cent coins of the same value and put them back into the box. Hence the ratio of the number of fifty-cent : ten-cent is 17:5. How much money was in the box.

 

Answer: While I know there are 30 ten-cent coins and 30 fifty-cent coins, i have difficulty in presenting the solution to a P5 level,

 

 

[Porker]

I'm sorry, but I *strongly* disagree with this. Algebra is the easiest method to grasp, and once basic algebraic manipulation is learnt (bringing over terms, opening brackets, negative numbers, etc.) it becomes the most generalisable one-size-fits-all solution. Furthermore, it is the *only* way that will stand the test of time and be carried over into higher level classes and even adult life. Not this piecemeal "model-drawing" crap that will be discarded as soon as they become teenagers.

 

Model drawing is like a complex set of cutlery you have to choose from before you start attacking your food. No one tool can help you tackle all foods, and often you may have to try different pieces before you find the best one. In contrast, algebra is like a bloody swiss-army knife. You see a problem, you pull it out and just solve it.

 

The emphasis on this backwards method here boggles the mind. It's sad, really. I believe they are handicapping our future generations by not equipping them with basic algebraic concepts early on (say at Pri 3) and "forcing" them into a mindset that only this unwieldy, un-generalisable crap can solve complex word problems.

 

A business associate commented that Singaporeans are too inflexible. And I say this because our education system is a huge influence

Edited April 28, 2015 by Porker

[1fast1]

A business associate commented that Singaporeans are too inflexible. And I say this because our education system is a huge influence

But you very flexible hor. Very easily bend over for me.

[Porker]

But you very flexible hor. Very easily bend over for me.

 

I love fat men with short d i c k s

Edited April 28, 2015 by Porker

[Jp66]

I learnt the modelling method from my kids few year back. Realized it is very useful in solving ratio problem and easy to understand.

 

Initially, I also taught them trial & error and simultaneous equation method but they catch no ball

I was on this solution too and was trying hard to move away from the trial and error method. Thanks anyway

 

[Vid]

 

I love fat men with short d i c k s

 

 

[Darryn]

calling the math pros

@turboflat4,

I need some help with the below problem -

 

I know that the angles of the equilateral triangle are 60 each - and think that you should be able to work it out with either congruency or reciprocity - but I'm a little bit stuck -

Any clues?

 

[Ender]

ED = DC, Therefore △CDE is an isosceles triangle.

Angle EDC = 90 + 60 =150

Angle CED = (180 - 150)/2 = 15

? = 60 - 15 = 45

Edited October 3, 2015 by Ender

[Darryn]

 

ED = DC, Therefore △CDE is an isoscelese triangle.

Angle EDC = 90 + 60 =150

Angle CED = (180 - 150)/2 = 15

? = 60 - 15 = 45

 

Crap, I knew I was missing something about one of the properties of the shape -

 

Didn't think it would be that easy though!

 

Thanks