Math Question

[Fastnfaster]

  On 10/15/2013 at 7:45 AM, Raymondism said:

Let the increase factor be I.

 

barry salary x 130% = 2600

Barry Salary = 2000

 

2600I - 2000I =780

600I=780

I=780/600

I=780 ÷ 600 =1.3

 

their salary increased by 1.3 times

increase percentage = 30%

thanks for helping!

[1fast1]

  On 10/10/2013 at 5:43 PM, Manmaster said:

Are you sure this is P5 math? It's too easy for P5. Let me show u something which is suitable for P5.

 

There are 764 marbles to be shared among a,b,c and d. If A has 20 marles lesser, B has 30 marbles more, C has to halve his marbles & D has to double his marbles, all the boys will have equal share of the marbles at the end. How many marbles does each boy have initially?

 

I can only give you a partial answer which is 1 unit = 86 marbles. I leave it for you to work it out. That's the math I had to solve for my friend's daughter.

 

Since noone seems to have answered this one yet,

 

Consider only the addition/subtraction operations on A and B. Net result = +10 (+30 - 20). That gives a total of 774 marbles.

 

Consider the situation of C and D. Halving C would give the same no. as doubling D. Which means that C started with 4 times as many marbles as D.

 

Let D represent the basic unit size of marbles.

 

So initial C = 4 units, initial D = 1 unit (heh - Initial D! )

 

Halving C/ doubling D would leave them with 2 units each. So this is what final A and final B would be as well.

 

So:

 

A (after -20) = 2 units

 

B (after +30) = 2 units

 

C (initial) = 4 units

 

D (initial) = 1 unit

 

Total = 9 units.

 

And this total is 774.

 

So 1 unit = 774/9 = 86.

 

Now reverse the add/subtract operations to get initial A and B.

 

Hence,

 

A initial = 2*86 + 20 = 192

 

B initial = 2*86 -30 = 142

 

C initial = 4 *86 = 344

 

D initial = 86

Edited October 15, 2013 by Turboflat4

[Raymondism]

  On 10/15/2013 at 8:28 AM, Fastnfaster said:

thanks for helping!

 

but i am not sure what was taught in school

[Rayleigh]

Thank you all for your help.

[Mcf777]

Bro/sis,

 

anyone can help me with this P5 question. TIA.

 

7 blouses and 4 skirts cost S$121.

The cost of 3 skirts and 2 blouses is S$68.

What is the cost of each item?

 

Answer

Blouses : ?

Skirt : ?

[Sosaria]

  On 2/8/2014 at 4:34 AM, Mcf777 said:

Bro/sis,

 

anyone can help me with this P5 question. TIA.

 

7 blouses and 4 skirts cost S$121.

The cost of 3 skirts and 2 blouses is S$68.

What is the cost of each item?

 

Answer

Blouses : ?

Skirt : ?

 

(B)louse and (S)kirt...

 

7B + 4S = 121

2B + 3S = 68

Of course we can solve it via simultaneous equations, but for primary level, they can actually do by model as well, as follows:

We know that 13B (7B+2B+2B+2B) + 13S (4S+3S+3S+3S) = $325

So 1B + 1S = 325 / 13 = 25

It is then very easy to find that 2B + 2S = 50,

so 1S = 68 - 50 = $18, and 1B = $7

At least, my P5 kid solved it this way. I would've just do by simultaneous.

Edited February 8, 2014 by Sosaria

[Freestylers09]

  On 2/8/2014 at 4:34 AM, Mcf777 said:

Bro/sis,

 

anyone can help me with this P5 question. TIA.

 

7 blouses and 4 skirts cost S$121.

The cost of 3 skirts and 2 blouses is S$68.

What is the cost of each item?

 

Answer

Blouses : ?

Skirt : ?

 

 

no algebra for p5 rite? reverse the method by draw diagram lor

[Mcf777]

Bro Sosaria,

Kam sia.......good job.

 

Freestyler

P5 still havnt learn algebra

[Wind30]

  On 2/8/2014 at 5:26 AM, Sosaria said:

 

 

(B)louse and (S)kirt...

 

7B + 4S = 121

2B + 3S = 68

Of course we can solve it via simultaneous equations, but for primary level, they can actually do by model as well, as follows:

We know that 13B (7B+2B+2B+2B) + 13S (4S+3S+3S+3S) = $325

So 1B + 1S = 325 / 13 = 25

It is then very easy to find that 2B + 2S = 50,

so 1S = 68 - 50 = $18, and 1B = $7

At least, my P5 kid solved it this way. I would've just do by simultaneous.

 

 

I think your method is long... a lot of calculations.

 

I think the method to teach is to try to make either the blouse or skirt the same number so we can cancel one of them

 

Looking at the problem you can get 1S+5B=121-68=53 very fast

From there is easy to cancel out the skirts by 3S+15B=53*3=159

13B=159-68=91

 

1B=91/13=7

[Little_prince]

  On 10/15/2013 at 2:59 PM, Rayleigh said:

Thank you all for your help.

 

knn. i look at the question my heart think that i'll probably end up in em3 if i were to retake my psle this yr

[1fast1]

  On 2/8/2014 at 9:23 AM, Wind30 said:

 

I think your method is long... a lot of calculations.

 

I think the method to teach is to try to make either the blouse or skirt the same number so we can cancel one of them

 

Looking at the problem you can get 1S+5B=121-68=53 very fast

From there is easy to cancel out the skirts by 3S+15B=53*3=159

13B=159-68=91

 

1B=91/13=7

 

This is essentially elimination, a well-known method of solving simultaneous equations. When the manipulations are carried out formally via matrix operations, it is known as Gaussian elimination or Gauss-Jordan elimination.

 

Whatever the name, it *is* algebra. In fact, even though people somehow seem to have developed the notion that Primary school students are not supposed to use algebra, a lot of the actual methods are indeed algebraic. It doesn't matter whether one uses blocks, symbols or the letter 'x' : as long as an unknown quantity is symbolically represented and manipulated to arrive at a solution that quantifies the unknown, it is algebra. I find that end run around "algebra" to be hypocritical and completely baffling, given that, often, highly inefficient and convoluted methods of thought are required to solve problems that would be much more directly solved by just casting them as algebraic equations from the start. It is completely ridiculous to teach students to use a trial and error method to solve a problem that is directly solvable with algebra.

 

In short, I think the way in which Math is being taught in our primary schools is *very* misguided and unsystematic. I just wanted to get that little rant out of my system.

Edited February 8, 2014 by Turboflat4

[Spiderman302]

  On 8/8/2013 at 5:12 PM, Watodo said:

2(A - 2/3A) = B - 1/4B

2A - 4/3A = B - 1/4B

2/3A = 3/4B

A = 9/8B

 

If A + B = 425

 

9/8B + B = 425

17/8B = 425

B = 200

 

To find out how much B has left = 3/4B = 200 * 3/4 = 150.

Pri sch are not supposed to use this method. Your method is for sec sch.

[Manmaster]

  On 2/8/2014 at 4:34 AM, Mcf777 said:

Bro/sis,

 

anyone can help me with this P5 question. TIA.

 

7 blouses and 4 skirts cost S$121.

The cost of 3 skirts and 2 blouses is S$68.

What is the cost of each item?

 

Answer

Blouses : ?

Skirt : ?

This math is easy.

 

Info A : 7B + 4S = 121

Info B : 2B + 3S = 68

 

The difference between Info A & B = Info C

Info C: 5B + 1S = 53

 

Use Info C to compare with Info B or A. I will compare Info C with Info B. So Info C has to multiply by 3 so that it has 3S.

 

For comparison with info B, Info C multiply by 3 becomes 15B + 3S = 159.

 

The difference between Info C & B : 13B = 91, so B = 7

 

If B = 7, that will mean Info C will be 1S = 53 - (7x5) = 18

 

If you use this method, you don't have to make both B & S equal before you can resolve this problem.

[Sosaria]

What turboflat4 said is true. And i also agree.

 

But unfortunately 'singapore math' that emphasizes heuristic methods is now being popularised in several other countries as well.

 

I think using models to visualize algebra is beneficial for preschool and lower primary, but moving on to P4 and on, should just introduce 'real' algebra.

 

I was recently looking at how they solve ratio problems using a very convoluted (to me) way of 'units' and 'parts' - until i also blur...

Edited February 8, 2014 by Sosaria

[Mcf777]

Wa lao.....all kao shou in math.

 

Respect!

[1fast1]

  On 2/8/2014 at 2:57 PM, Sosaria said:

What turboflat4 said is true. And i also agree.

 

But unfortunately 'singapore math' that emphasizes heuristic methods is now being popularised in several other countries as well.

 

Yeah, the problem is this: our students have always been excellent in technical subjects like Math and Science. This was true even when our math was a bit more traditional (rather than emphasising this heuristic nonsense). So other countries got the notion that our educational methods were fantastic, which was probably true back then. Keep in mind that at the same time, the "developed" countries were going through weird educational experiments of their own. So they all decided to adopt Singapore's schooling methods hook, line and sinker. Selling "Singapore Math" became a profitable enterprise to some. Unfortunately, that trend has continued while Singapore's math education itself has gone into mini-crisis mode. Clearly, there are just as many uncritical educators in other countries as there are in Singapore.

Edited February 8, 2014 by Turboflat4

[TJX]

  On 2/8/2014 at 4:34 AM, Mcf777 said:

Bro/sis,

 

anyone can help me with this P5 question. TIA.

 

7 blouses and 4 skirts cost S$121.

The cost of 3 skirts and 2 blouses is S$68.

What is the cost of each item?

 

Answer

Blouses : ?

Skirt : ?

For P5, i think u should use this method

 

u have to find the common multiple for the blouse or skirt.

I will use the common multiple for the blouse which is 14.

So i will make both 7 and 2 into 14.

eg.

 

Step 1

7 x 2 blouse + 4 x 2 skirts = 121 x 2

14 blouse + 8 skirts = 242

 

 

Step 2

 

2 x 7 blouse + 3 x 7 skirts = 68 x 7

14 blouse + 21 skirts = 476

 

Since the number of blouses are the same, the difference in cost must be the difference in the skirts.

Step 3

 

21 skirts - 8 skirts = 476 - 242

13 skirts = 234

1 skirt = 18.

 

Since 1 skirt = 18,

3 skirt + 2 Blouse = 68

2 blouse = 68-54

2 blouse = 14

1 blouse = 7.

 

hope this helps.

Edited February 9, 2014 by TJX

[Rayleigh]

Hi guys, need some help with a P6 math.

 

The ratio of number of male to number of female in the hall was 5:3. The number of male to number of female in auditorium was 7:1. There were twice as many people in the hall as in the auditorium. There determine the followings:

 

a. What was the ratio of the number of female in the hall to the number of male in the auditorium?

 

b. When 18 male left the auditorium and 16 female went into the auditorium, the ratio of the number of male to the number of female in the auditorium became 2:1. How many male were there in the auditorium in the end?

 

I only manage to solve a) and i hope it is correct. I'm stuck at b).

 

For a)

 

Hall => Male : Female => 5: 3 ==> 10:6 (double)

Auditorium => Male : Female => 7:1

 

Therefore the ratio of female in hall to male in auditorium is 6: 7 (correct?)

 

For b) please help

Edited April 24, 2014 by Rayleigh